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## Introduction

Statistics is one of those subjects that has the potential to be the most obscure thing you will ever see. Yet without it, you can't really do an experiment or make an observation and come up with a meaningful conclusion.

The core concept of classical statistics has to do with the idea of an ensemble of experiments, or measurements. Statistics concerns how to use the results of the ensemble to make conclusions, and to understand aspects of probability. For instance, you might want to know the probability that someone in your class is a Republican. You can count the number of Republicans, and divide by the total number of students, but that won't give you a probability - it will only give you the fraction of students in this class that are Republican. What we mean by probability is something else: if we had an ensemble of classes, and all classes are equal, then we can count the number of Republicans in each class, divide by the total to get a percentage, then make a histogram (frequency plot) of how many times we see 1%, 2%, etc. The histogram might look like the one below:

If we divide the number in each bin by the total number of classes sampled in the ensemble, we can then extract the probability distribution. If we used the fraction that had the greatest probability (the mean, or the value at the peak), when we could say that that is the probability for a class to have Republicans. Be careful here - the probability is only defined for an ensemble, and it leaves out information because it tells you nothing about how the number of Republicans are distributed. In other words you need to know now only the value at the peak, but also the shape of the distribution (a distribution that has only values at the peak will have the same mean as the one above but will have a very different distribution, and tells you a lot!).

So here are two experiments, but the results might have very different accuracies, as well as different precision. You will find that most people in science tend to think that this added information (precision and accuracy) are important when reporting the result. Of course, in our world, knowing the time to less than a minute or two is usually not important. But in the world of science it is amazingly important, for a very simple reason: we often do experiments to find out something that we do not already know! In which case we would get a result, and compare it to what was already believed. For instance, if we drop two balls of different weight off the Tower of Pisa, and we want to see if they do hit the ground at the same time, we would make 2 measurements, $t_1$ and $t_2$, take the difference $\Delta t = t_1 - t_2$, and compare to what we expect, which is that $\Delta t=0$, also known as the "null" result ("null" meaning "nothing new"). What we want to be able to know, and to report in a systematic way, is the uncertainty in $t_1$ and $t_2$: $\delta t_1$ and $\delta t_2$. If we know these uncertainties, we can form the uncertainty in the difference, and report $\Delta t \pm \delta\Delta t$ where $\delta\Delta t$ is the uncertainty in $\Delta t$, defined in some reasonable manner (but surely a function of $\delta t_1$ and $\delta t_2$). And given these quantities, we want to understand whether the measurements yield any significant new information. And typically, we use the concepts of probability, and looking at an ensemble of measurements, to understand how to do this. Anyway the goal here is to give you some understanding and tools so that you can do just this, for this simple situation but also in general.

This is what "statistics", in the sciences, is really about: how do we characterize and understand probability, and when we make a measurement, what are the uncertainties, how do you calculate and report them, and how do you use them to understand the significance of a measurements, especially relative to the "null" hypothesis.

## Useful Definitions

We start with the definitions that are most useful to know in order to understand statistics:

### Mean

The "mean" of a set of $N$ numbers $x_i$ is the same as the "average", defined as: $$\bar x = \frac{\sum_{i=0}^N x_i}{N}\label{emean}$$ Image we have 5 children all with different ages: $1, 3, 6, 2, 10$, and we want to calculate the average (mean) age. That will be given by by $(1+3+6+2+10)/5=22/5=4.4$.

The mean is also sometimes called the "first moment" due to its analog with mechanics. To see this, imagine you have a collection of identical masses, $m$, you place them all at different distances from a pivot, and calculate the torque about the pivot. The picture looks like this:

The torque $\tau$ about the pivot would be given by:

 $\tau$ $=$ $x_1\cdot mg + x_2\cdot mg + x_3\cdot mg + x_4\cdot mg$ $=$ $mg\cdot (x_1+x_2+x_3+x_4)$

You could also place a mass that is the sum of all the other masses at some distance $\bar{x}$ from the pivot, and find $\bar{x}$ such that the torque was the same, as in the following picture:

The calculation for that torque would be: $$\tau=\bar{x}\cdot (4m)g\nonumber$$ Equating the two equivalent torques gives $4\bar{x}=x_1+x_2+x_3+x_4$, or $\bar{x}=(x_1+x_2+x_3+x_4)/4$, and $\bar x$ is the "first moment of inertia" (the "second moment" relates to the variance and standard deviation, see below). This equation is the same equation for how to calculate the mean, hence the name "1st moment".

In considering the mean, one can also consider the probability for any particular $x_i$ to occur. In the case of the average age of a group of people, when you add the ages together, you are assuming that each age is equally likely to occur, or equivalently, that one age is no more likely to occur than any other. We can calculate the probability for any age to occur in the sample easily by requiring that the sum of all probabilities should add up to 1. So in our case above, we can write $P_i$ as a shorthand for $P(x_i)$, and impose: $$\sum_{i=0}^N P_i=1\label{eprob}$$ The sum of all probabilities adds to 1 because given a set of probabilities, if they describe all possibilities, then since something has to happen, the sum of all probabilities have to add to 1. This is a key formula not only in statistics, but also in quantum mechanics.

So, if we assume that all of the $P_i$ are equal (all ages equally probable, so $P_i$ is a constant), then this sum evaluates to: $$\sum_{i=0}^N P_i=P_i\sum_{i=0}^N =P_iN=1\nonumber$$ Note that $\sum_{i=0}^N$ can be written as $\sum_{i=0}^N 1$, which is clearly just $N$.

So combining the two equations above gives us: $$P_i = \frac{1}{N}\nonumber$$ Now, we can go back to equation $\ref{emean}$ and write it a little differently: $$\bar x = \frac{1}{N}\sum_{i=0}^N x_i\nonumber=\sum_{i=0}^N \frac{1}{N}x_i =\sum_{i=0}^N P_ix_i\nonumber$$ Now we have an equation which is even more general than equation $\ref{emean}$, that tells us how to calculate the mean even if all of the measurements are not equally likely: $$\bar x = \sum_{i=0}^N P_ix_i\label{emean2}$$ This formula says something more than what's in equation \ref{emean}: it says that the mean is given by the sum of each $x_i$ "weighted" by some probability $P_i$ that tells you how likely it is that that particular $x_i$ is seen. This is very different than equation $\ref{emean}$, which tell us how to take all of the data and form the mean. In equation $\ref{emean2}$, if each of the $P_i$ are equal, then the sum is over all of the data. However, if instead we sum over all of the data that are distinctly unique, then it's an entirely different (but equally valid) equation.

To illustrate, let's calculate the $P_i$ for the two sets ${1,3,6,2,10}$ and $4,4,5,5,4$. For the first set, $P_1$ would be the probability that we see the number $1$ in the set, $P_3$ would be the probability we see the number $3$, and so on. We only have 5 numbers, and each is distinct (not repeated), so we would have $P_i \equiv P = 1/5 = 0.2$. The mean is then given by

$\bar{x}=P_1\cdot 1+P_3\cdot 3+P_6\cdot 6+P_2\cdot 2+P_{10}\cdot 10 = P\cdot (1+3+6+2+10)=0.2\cdot 22=4.4$

as before.

For the second set, we see the number $4$ occur 3 times, so $P_4=3/5=0.6$, and the number $5$ occurs twice, so $P_5=2/5=0.4$. Notice that $P_4+P_5=1$ as required. Then the mean is given by

$\bar{x}=P_4\cdot 4+P_5\cdot 5=0.6\cdot 4+0.4\cdot 5=2.4+2.0=4.4$.

Voila. This way of looking at means and variances will come in handy later.

To finish the analogy, let's go back to our example where we find the torque about a pivot point with a set of 4 masses at different locations. In that example, all of the masses were equal. Now let's imagine that they are not equal, and we have masses $m_1, m_2, m_3, m_4$ at positions $x_1, x_2, x_3, and x_4$. The torque will be given by:

$\tau=x_1\cdot m_1g + x_2\cdot m_2g + x_3\cdot m_3g + x_4\cdot m_4g$

We want to place a mass $M=m_1+m_2+m_3+m_4$ at some location $\bar x$, with resultant torque given by

$\tau=\bar xMg$

Equating the two torques give:

 $\bar x$ $=$ $\frac{x_1\cdot m_1 + x_2\cdot m_2 + x_3\cdot m_3 + x_4\cdot m_4} {m_1+m_2+m_3+m_4}$ $=$ $\frac{m_1}{m_1+m_2+m_3+m_4}x_1 + \frac{m_2}{m_1+m_2+m_3+m_4}x_2 + \frac{m_3}{m_1+m_2+m_3+m_4}x_3 + \frac{m_4}{m_1+m_2+m_3+m_4}x_4$ $=$ $P_1x_1 + P_2x_2 + P_3x_3 + P_4x_4$ $=$ $\sum P_ix_i$

where $P_i = m_i/(m_1+m_2+m_3+m_4)$. Here, the weights are truly weights!

### Binning, and Histograms

Equation $\ref{emean2}$ is extremely important, especially when it comes to histograms, and what is known as "binned data". To illustrate, imagine we have a population of $N$ students, and we tabulate the age $Y_i$ of each, for example $Y_0=15.32, Y_1=12.42$, etc. Note that we use real numbers, with units in years. We can measure the mean $\bar Y$ using equation $\ref{emean}$: $$\bar{Y}=\sum_{i=0}^NY_i/N\nonumber$$ However, we also might want to look at the population distribution, to visualize the average, and how the various values are distributed about the average. To do this, we would make a histogram, which would tell us how many students have the "same" age. Histograms are all about counting, and is nothing more than a frequency distribution: how many times you see the value of some data come up. However, when we say "counting", what we mean is that we want to count the number of times we see a particular value within some range.

To make the concept of "same age" well defined, we first have to decide on the "binning", which in our example means that we need to decide on the interval of ages. For instance, we could count the number of students who are between 9 and 10, 10 and 11, 11 and 12, and so on. In this example, the bin size $\delta b$ would be 1 year. Or we could instead decide to count the number between 9 and 9.5, 9.5 and 10, and so on: bin size $\delta b=0.5$ years. This bin size is a crucial parameter in a histogram.

Then, we will want to decide on the number of bins, and the range: 0-20, or 5-15, etc. And lastly, we want to know where the bin boundaries are. For instance, if we have bins of size $\delta b=1$, then we would need to decide whether someone who is 9 years old appear in the 8-9 bin, or the 9-10 bin, or even the 9.5-10.5 bin. The standard thing to do is to decide on the bin range, and include the low edge of the bin, but not the high edge. So if the bin size is 1, and there are 10 bins, then that means the histogram will go from 0 to 10, and any age that is greater than or equal to 0, but less than 1 will fall within the 1st bin. If you have an age that is greater or equal to 2 but less than 3, it falls within the 3rd bin (2-3). And so on.

So now that we know how to specify the histogram, we can set aside some memory (an array), start the counting using the value of each each age to determine which bin it will fall in, and incrementing the corresponding array address (the bin). There is an easy way to do this: we know the bin size (call it $\delta b$), so we first take each age and divide by $\delta b$. For instance, if $\delta b=1$, and we have age $Y_i=10.42$, then $Y_i/\delta b=10.42$. If we then lop off the decimal, we have an integer, here it's 10. That's the bin number to increment! So the program would look something like this in javascript:

	//
// assume all ages are in an array, called "year[i]".
// and the histogram will be in the array hist[i], initialized to 0.
//
nbins = 20;
db = 1.0;
npts = 1000;   // assume all ages are in an array, called "year[i]"
for (var i = 0; i < npts; i++) {
thisbin = int(year[i]/db);
hist[thisbin] = hist[thisbin] + 1;
}


where the function "int" just truncates to an integer. (In JavaScript, you would use Math.floor().)

To illustrate, in the figure below we have 1000 ages, distributed in a classic bell curve shape with a true mean age of exactly 10 years. Above the histogram, the mean $\bar{Y}$ is calculated using equation $\ref{emean}$. The "Generate" button will simply generate a new list to histogram, and recalculate the mean. You can see how it fluctuates around the true value 10.

Check to show probability distribution:
$\bar Y=$
$\bar b=$

We would like to know how to how to calculate $\bar Y$ using the histogram bin values (and also how to characterize the fluctuation in $\bar Y$). This is where equation $\ref{emean2}$ can be used.

Back to equation $\ref{emean2}$: to find the mean using the histogram (let's call this $\bar b$, the mean from the bin values, to differentiate it from $\bar Y$, the mean as calculated using the raw data), we use equation $\ref{emean2}$ as a guide and form the following: $$\bar b = \sum_{i=1}^{N_b} P(b_i)b_i\nonumber$$ where $N_b$ is the number of bins, not the number of data points (here $N_b=20$ and $N=1000$), $b_i$ is the value of the $i$th bin (remember, the bin is the variable that has units of years), and $P(b_i)$ is the probability weight for bin $b_i$. Histograms make $P(b_i)$ easy: using the constraint described in equation $\ref{eprob}$ that all of the probabilities sum to 1, and given that the number of counts in any bin give the frequency for seeing the data in that bin range, all we have to do is divide each bin by the total number of counts (which is total number of data points generated): $$P(b_i) = \frac{N_i}{N}\nonumber$$ where $N$ is the total number, and $\sum_{i=1}^N N_i=N$.

One thing left to determine: what do we use for $b_i$? The index $i$ goes from $1$ to $N_b$, and each bin $i$ has a width $\delta_b$. We could use the left edge of the bin, or the right edge, or the middle. It is probably better to use the middle: $b_i = \delta_b(i-\half)$, and we use the $-$ sign because we are summing bins from $1$ to $N$ (if we were summing from $0$ to $N-1$ then we would use $+$). So the average $\bar Y$ will be given by: $$\bar b = \frac{\sum_{i=1}^{N_b} N_i\delta_b(i-\half)}{\sum_{i=1}^{N_b} N_i} = \frac{\delta_b}{N}\sum_{i=1}^N N_i(i-\half)\label{emeanH}$$ As a check, imagine we have a "flat" distribution, where all bins are equally likely, and so have the same number of entries. Summing the value in all bins should lead to the total number $N$, which tells us what $N_i$ is: $$N = \sum_{i=1}^{N_b} N_i = N_i\cdot N_b\nonumber$$ which gives us $N_i=N/N_b$. (Note the sum is over the number of bins, $N_b$). In this case, the probability per bin $P(b_i)=N_i/N = 1/N_b$. For instance, if we have 1000 points, and 20 bins, then each bin should hold 50 if all bins are equally likely (50x20=1000). In this case (flat distribution, constant $N_i$), our average is given by: $$\bar b = \delta_b\sum_{i=1}^{N_b} P_i(i-\half) = \frac{\delta_b}{N_b}\sum_{i=1}^{N_b} (i-\half)\nonumber$$ This is an easy thing to sum: $\sum_{i=1}^m i = m(m+1)/2$ and $\sum_{i=1}^m = m$. So our average bin will be $$\bar b = \frac{\delta_b}{N_b}\big[ \frac{N_b(N_b+1)}{2} - \half N_b \big] = \frac{N_b\delta_b}{2}\nonumber$$ In our example, if we have 20 bins each with a width of $\delta_b=1$, then the average from the perfect histogram will be $10$, which is what we would expect.

In the figure above, below the mean $\bar Y$ is shown the mean as calculated using the histogram bin contents, called $\bar b$. That the two are different is due to the fact that in making a histogram, we are throwing away some a small amount of information by binning.

### Variance and Standard Deviation

The mean of a set of numbers sometimes does not always tell the whole story. For example, in the case of age distributions as in the figure above, imagine that you go to 2 different schools, each with around 200 4th graders, collect the list of ages, and make a frequency plot for each school as in the two plots below:

The two have the same average over the same sample size. But the distributions are very different: the probability of finding a 10 year old (age between 10 and 11) in the first distribution is around 0.2 (40/200), whereas it's around 0.1 in the second (20/200), and you are much more likely to find a 19 year old in the 2nd distribution than the first. The lesson here is that you can't just use the mean $\bar x$ to describe things completely.

This motivates some way to characterize the "spread" around the mean. The usual thing to do is to use the "variance", defined as the average of the square of the deviations from the mean: $$VAR\equiv\sigma_V^2 = \frac{\sum_{i=0}^N (x_i-\bar{x})^2}{N}\label{evar}$$ where $x_i$ is the $i$th age. If you define $d_i \equiv x_i-\bar{x}$, then the equation for the variance becomes: $$\sigma_V^2 = \frac{\sum_{i=0}^N d_i^2}{N}$$ This looks like you are taking the mean of the square of the quantity $d_i$. Since the variance $\sigma_V^2$ will have different units than $x$ (if $x$ is in meters, the variance $\sigma_V^2$ will be in meters squared), we often take the square root of $\sigma_V^2$ so that we can quote the mean and the "root mean squared" (RMS) in the same units: $$\sigma_{RMS}=\sqrt{\sigma^2}\label{erms}$$ The RMS (square root of the mean of the quantity squared) is also called the "standard deviation", and tells you something about how the values are spread about the mean.

We can take equation $\ref{evar}$ and rewrite it in the following way:

 $\sigma_V^2$ $=$ $\frac{1}{N}\sum_{i=0}^N (x_i-\bar{x})^2$ $=$ $\frac{1}{N}\sum_{i=0}^N (x_i^2-2\bar{x}x_i+\bar{x}^2)$ $=$ $\frac{1}{N}\sum_{i=0}^N x_i^2 -2\frac{1}{N}\bar{x}\sum_{i=0}^N x_i + \frac{1}{N}\bar{x}^2\sum_{i=0}^N \bar{x}^2$
The first term in the last line can be written as $\bar x^2$, the average value of the quantity $x_i^2$. The second term is $-2\bar{x}^2$, and the 3rd term is $+\bar{x}^2$ (remember that $\sum_{i=0}^N = N$). So we can write the variance as: $$\sigma_V^2 = \overline{x^2}-\bar{x}^2\label{evardiff}$$ or in words, it's the difference between the average $x^2$ and the (average $x$) squared. The quantity $\overline{x^2}$ is called the "2nd moment", analogous to the quantity $\bar x$ being the "1st moment".

This equation is useful in computations, allowing you to run through the list of numbers once to calculate the mean, variance, and standard deviation, as shown below:

	//
// assume all ages are in an array, called "year[i]".
//
//
sumy = 0;
sumx = 0;
sumx2 = 0;
for (var i = 1; i < nbins+1; i++) {
bin_i = db*(i-0.5);				// we sum from 1 to nbins, db is the bin width
sumy = sumy + year[i];
sumx = sumx + bin_i
sumx2 = sumx2 + bin_i*bin_i;
}
mean = sumx/sumy;
sumx2 = sumx2/sumy;
variance = sumx2 - mean*mean;
stand_dev = sqrt(variance);

The above code assumes that the low edge of the first bin is 0. If it's not, then you have to add that value to $bin_i$.

It is worth noting that value for $\overline{x^2}$ will always be bigger than the value $\bar{x}^2$ ($x^2$ is always positive, whereas $x$ does not have to be). So you don't have to worry about the situation where you are taking the square root of a negative number.

### Variance in Binned Histograms

Equation $\ref{evardiff}$ can be used to calculate the variance in histograms using the bin contents. To accomplish this, we already know how to calculate $\bar{b}$ using equation $\ref{emeanH}$, s owe need to construct a similar equation involving the 2nd moment. Using equation $\ref{emean2}$ as a guide, we can form the average over any quantity by weighting the bin value by the bin probability, which is just the number of events in that bin. So in general, we can form the variance by: $$\sigma^2_V = \sum_{i=1}^m P(x_i)(x_i-\bar{x})^2\label{evar2}$$ For histograms, we can form the 2nd moment by taking $x_i\to b_i$: $$\overline{b^2} = \sum_{i=1}^m P(b_i)b_i^2\nonumber$$ where $m$ is the number of histogram bins, $N$ is the total number of entries, $P(b_i)=N_i/N$ and $b_i=\delta_b(i-\half)$, same as above. This gives: $$\overline{b^2} = \frac{\delta_b^2}{N}\sum_{i=1}^m N_i(i-\half)^2\nonumber$$ As above, we can check how this works when we have a histogram where all bins are the same, given by $N_i=N/N_b$: $$\overline{b^2} = \frac{\delta_b^2}{N_b}\sum_{i=1}^{N_b} (i-\half)^2\nonumber$$ When we expand the $(i-\half)^2$ term, and use the series sum for the square of the index $\sum_{i=1}^{N_b} i^2=N_b(N_b+1)(2N_b+1)/6$, we get: $$\overline{b^2} = \frac{\delta_b^2}{12}(4N_b^2-1)\nonumber$$ The variance then will be: $$\sigma_V^2 = \frac{\delta_b^2}{12}(4N_b^2-1) - N_b^2\big(\frac{\delta_b}{2}\big)^2 = \frac{\delta_b^2}{12}(N_b^2-1)$$ To make this formula a bit more clear, note that $N_b\delta_b$, the number bins times the bin width, is equal to the range of the histogram which we can call $R$: $$R=N_b\delta_b\nonumber$$ We can then rewrite the above equation as: $$\sigma_V^2 = \frac{R^2}{12}(1-\frac{1}{N_b^2})\nonumber$$ That means that the standard deviation of a flat distribution will be given by $$\sigma = \frac{R}{\sqrt{12}}\sqrt{1-\frac{1}{N_b^2}}\nonumber$$ and as the number of bins gets large, $\sigma \to R/\sqrt{12}$. If we have $R=1$, then the average and standard deviation of a flat signal will be given by $1/2$ and $1/\sqrt{12}$, respectively, a well known result. We will make use of this below.

The following figure sums up the situation for a flat distribution with $N_b=20$ bins: the bin width $\delta_b=1/20=0.05$, the average $\bar x$ is 0.5 (halfway), and the standard deviation $\sigma_x = 1/\sqrt{12} = 0.289$. The bin boundaries to the left of the value, so for example "0.2" is the bin for $0.2\le x\lt 0.25$.

## Probability Distributions

Sometimes, events (numbers) are produced from some kind of systematic process with constraints. For instance, if you randomly pulled cards out of a deck, the process limits the cards to being Ace, 2-10, J, Q, and K only. Another example, an even simpler one, might be an experiment where the answer is either yes, or no, 0 or 1. In that case, we way that the experiment is binary: 1 of 2 possible choices only. What kind of questions can you ask about the results of such an experiment? Not much, since the answer is either yes, or no. If you did the experiment $N$ times, then the number of yes results ($N_y$) plus the number of no results ($N_n$) should add to $N$: $N_n+N_y=N$. And given some number of yes and no results, you might want to ask the important question of how this compares to what we expect. Which brings you to the question of how to calculate what to expect.

To understand the answer here, let's go to a simple case where the probability of a yes ($P_y$) is equal to the probability of a no ($P_n$): $P_y=P_n$. Since probabilities sum to 1, we would have $P_y+P_n=2P_y=1$, or $P_y=P_n=\half$. This would be the case for example if you were flipping a coin.

For coin slips, we can define $P_h$ = "heads" and $P_t$ = tails. If you flipped the coin 2 times, you might want to calculate the probability that you get 2 heads in a row (call it $P_{hh}$). The way to calculate that probability is to go through all the possible outcomes, and calculate the fraction of times that that outcome exists. Each fraction will be the probability (it will add to 1!). So in an experiment with 2 coin tosses, with tosses $T_1$ and $T_2$, we could get the following combinations:

HH: $T_1=h, T_2=h$
HT: $T_1=h, T_2=t$
TH: $T_1=t, T_2=h$
TT: $T_1=t, T_2=t$

Of the 4 possibilities, HH occurs once, so the fraction of times you get 2 heads is 1/4. The fraction of times you get 2 tails is also 1/4. The fraction of times you saw one heads and one tails, without caring about which one came first, would be 2/4 (TH and HT). Another way to see this: of the 3 different combinations, the HH and TT probabilities add to 1/2, so the remaining probability (TH+HT) has to be 1/2.

This gets to a very important point: if order doesn't matter, and all you care about is the probability of getting a result, you have to keep track of the "combinatorics". More on this below.

As an interesting aside, imagine you did the same 2 flip experiment, and on the 1st flip you got a tails. What's the probability that on the 2nd flip you got a tails? You might think that it's 1/4, since after 2 flips the probability of seeing 2 tails is 1/4. Do the experiment, see what you get, but the answer will be that the probability of seeing a tails on the 2nd flip is 1/2, not 1/4. Why is this? Because the two flips are uncorrelated. And the probability of seeing a tails after a tails is not the same as the probability of seeing 2 tails in a row. The former is very specific: flip a tails, then ask for the probability of seeing another tails. The latter says: flip the coin twice, what's the probability of seeing 2 tails. So be careful with statistics. It can really cause headaches!

### Let's Make A Deal

An interesting aside: in the 1960s, there was a game show on TV called "Let's Make a Deal", starring Monty Hall as the host. What would happen during the show is that contestants who won some kind of prize would, at the end of the show, be shown 3 doors on the stage. They would be told that 2 of the doors contained something worthless (called a "zonk"), and 1 of the doors would be a grand prize like a vacation, or a new color TV (this was the 60s - color TVs were very novel!). The contestant was told that they could choose any of the doors, or stick with what they had. Of course, the initial probability of choosing the door with the grand prize was 1/3. After they would make their choice, Monty would then say that to help them out, he was going to show them what's behind one of the doors they did not pick. Since there was only 1 grand prize and 2 zonk prizes, he could always find a zonk to show them whether they had guessed right in the initial pick or not.

Then comes the interesting part: he would say that the contestant could switch their choice, or stick with what they had picked initially. The question for us is this: what should you do based on maximizing the probability of getting the grand prize?

The way to answer this is to do the experiment many times, and see what happens. You will need a partner! But you can also apply what we've learned and calculate whether it's better, on average, to switch or to stay. Here's how to think about it: before you chose the door, you had a 1/3 chance of getting the grand prize. Then you choose, and then the host opens another door. There are 2 possibilities here: 1) you chose correctly (1/3 chance), and so the host could have opened either of the other two zonk doors; or 2) you did not choose correctly the first time (2/3 chance), so of the 2 unopened doors, the host showed you the one that had the zonk, the other unchosen unopened door containing the grand prize. Since the initial probability of guessing right initially was 1/3, and that probability doesn't change (you've already guessed!), logic dictates that you should always switch, and that if you do your chances of getting the grand prize goes from 1/3 to 2/3.

### Binomial Distribution

If we have $N$ measurements each of which can come out $m$ ways, we will want to be able to calculate the probability of seeing one of those possibilites $n$ times. For instance, if we flip $N$ coins, then $m=2$ (2 possible outcomes), and so we want to calculate the probability that we will get $n$ heads. This probability is denoted as $P(n,N)$, and is called the binomial probability. The graph of this probability as a function of $n$ is called the binomial distribution. It's one of the most important probability distributions in all of science, and it turns out that many other probability distributions can be derived from it.

To start, let's say that we have 1 coin, that heads and tails are equally probable, and that we flip them 4 times ($N=4$). Here are the 16 different possibilities for what you should see, where H means heads and T means tails:

HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT

The binomial probability $P(n,N)$ is the probability of seeing $n$ heads after $N$ flips, which we can get by looking at the number of times we see $n$ heads, and dividing by the number of flips. So the first thing we need to know is the number of times we see $n$ flips, and to get that all we have to do is count: there are 6 combinations of seeing 2 heads (HHTT, HTHT, HTTH, THHT, THTH, and TTHH). This turns out to be easy to put into a formula, called the combinatoric formula: $$C(n,N)=\frac{N!}{n!(N-n)!}\nonumber$$ $C(n,N)$ is just the combinatoric function for the number of ways you can arrange $N$ things so that you have $n$ of them showing a particular value (here it's heads). The $N!$ notation means "factorial", so that for example $4!=4\times 3\times 2\times 1=24$. There are many ways to motivate this particular formula, but they all have to do with the idea of keeping track of permutations. For our example, $N=4$, $n=2$, we would have $$C(2,4)=\frac{4!}{2!(4-2)!} = \frac{4\cdot 3\cdot 2}{2\cdot 2} =6\nonumber$$ This agrees with what we get from just counting.

$C(n,N)$ tells you the number of ways you can see $n$ out of $N$, but it doesn't tell you $P(n,N)$, the probability of seeing $n$ out of $N$. For that, you have to multiply by the probability $P_n$ of seeing $n$, so that we would have $$P(n,N) = C(n,N)\cdot P_n\nonumber$$ Calculating $P_n$ is a bit more subtle than calculating $C(n,N)$, which is just combinatorics. To start, we first remember that probabilities are multiplicative. So imagine you have a variable that can have the value $R$, with probability $p_R$, or $B$, with probability $p_B$. Imagine that you have 2 of these variables. Then the probability for one of them to be $R$ and the other to be $B$ would be given by $P(RB)=p_Rp_B$ - the probabilities are multiplicative, providing that they are uncorrelated (which means that the probability of one of them being $R$ in no way effects the probability of the other being $B$). If you have 2 coins that can be either heads or tails with probability $p_H$ and $p_T$ respectively, then the probability of 1 heads and 1 tails will be given by $p_Hp_T$. If you have 4 coins, then the probability of seeing 2 heads and 2 tails will be given by $p_H^2\times p_T^2$.

Now we can simplify it even more: let's define $p\equiv p_H$, and remember that $p_T=1-p_H=1-p$. Also, if $n$ is the number of heads, then $m=N-n$ is the number of tails. So the probability of seeing any combination of $n$ heads and $m$ tails will be given by $P_n=p^n(1-p)^m$, and since $n+m=N$, we can write $$P_n = p^n(1-p)^{N-n}\nonumber$$ For our case of 2 heads and 2 tails, $n=2$ (number of heads) and $m=N-n=2$ (number of tails), so $P_2 = (\half)^2(1-\half)^2 = 1/16$.

Now that we also know $P_n$ we can write $P(n,N)$: $$P(n,N) = C(n,N)\cdot P_n = \frac{N!}{n!(N-n)!}p^n(1-p)^{N-n}\label{ebinomial}$$ For $N=4$, $n=2$, $p=(1-p)=\half$, we would have $P(2,4)=6\cdot 1/16=0.375$.

The graph below shows the binomial distribution (normalized to 1) for the situation where $N=10$ (10 coins) and $p=\half$ (equal probability for heads and tails). Changing either $N$ or $p$ using the clider will redraw it, allowing you to see how the distribution chnages. Notice what happens when you make $p$ very small (or very large), and also when you let $N$ become large. Note that you can click on any of the points and read off the values for $n$ and $P(N,n)$, and if you click on the title in white, it will display all of the points (this is so you can cut and paste, if desired).

Log scale?

Overlay curve?

N=

p=0.50

$X_{max}$=

The mean and variance of the binomial distribution are not all that difficult to calculate. To get the mean, we start with the definition that the mean of anything is given by the sum of that thing times the probability for that thing: $$\bar{n} = \sum_{n=0}^\infty n\cdot P(N,n) = \sum_{n=0}^\infty n\cdot \frac{N!}{n!(N-n)!}p^n(1-p)^{N-n}\nonumber$$ The way to calculate this is to note that $n/n!=n/[n(n-1)(n-2)...]=1/(n-1)!$. Since dealing with $(n-1)!$ seems complicated, we can use the following trick: if we factor $N$ and $p$ out of equation \ref{ebinomial}, and use $N-n=N-1-(n-1)$, we get the following: $$\bar{n} = Np\sum_{n=1}^\infty \frac{(N-1)!}{(n-1)!(N-1-[n-1])!}p^{n-1}(1-p)^{N-1-[n-1]}\nonumber$$ This is a nice little trick. Notice first that in the original sum of $n=0$ to $\infty$, since we've multipled by $n$ to get $\bar{n}$, the part for $n=0$ vanishes. So the sum now really goes from $n-1$ to $\infty$. Now all we have to do is make the substitution $m=n-1$ and $M=N-1$, and note that now $m$ goes from $0$ to $\infty$. This gives us: $$\bar{n} = Np\sum_{m=0}^\infty \frac{M!}{m!(M-m)!}p^m(1-p)^{M-m}\nonumber$$ Since the sum is over the binomial distribution, which is normalized to 1, then the sum sums to 1. So the final answer is simple: $$\bar{n} = Np\label{emeanbinomial}$$ To get the variance, we would do something analogous to what we just did to calculate the mean, except as in equation $\ref{evar}$, we replace $n$ by the square of the difference between $n$ and $\bar{n}$: $(n-\bar{n})^2$. Then we have to evaluate the following: $$\sigma^2 = \sum_{n=0}^\infty (n-\bar{n})^2\cdot P(N,n) = \sum_{n=0}^\infty (n-\bar{n})^2\cdot \frac{N!}{n!(N-n)!}p^n(1-p)^{N-n}\nonumber$$ We first multiply out the square of the difference: $(n-\bar{n})^2=n^2-2n\bar{n}+\bar{n}^2$, giving 3 terms: $$\sigma^2 = \sum_{n=0}^\infty (n^2-2n\bar{n}+\bar{n}^2)\cdot \frac{N!}{n!(N-n)!}p^n(1-p)^{N-n}\nonumber$$ Let's label them $S_1$, $S_2$, and $S_3$, where the $S_1$ term has the $n^2$, $S_2$ has the $-2n\bar{n}$ part, and $S_3$ has the $\bar{n}^2$.

For $S_3$, we can pull the $\bar{n}^2$ out of the sum, leaving just the binomial distribution, which sums to 1. So $S_3=\bar{n}^2$. For $S_2$, we can pull the factor $-2\bar{n}$ out of the sum, leaving the $n$ along with the binomial distribution. However, that is the same sum we evaluated above, yielding the same $\bar{n}$, which gives us $S_2=-2\bar{n}^2$. This gives us $S_2+S_3=-\bar{n}^2$.

The first term with the $n^2$ part is slightly more complicated, and the solution goes like this:

 $S_1$ = $\sum_{n=0}^\infty n^2\frac{N!}{n!(N-n)!}p^n(1-p)^{N-n}$ = $pN\sum_{n=1}^\infty n\frac{(N-1)!}{(n-1)!(N-1-[n-1])!}p^{n-1}(1-p)^{N-1-[n-1]}$ = $pN\sum_{m=0}^\infty (m+1)\frac{L!}{m!(L-m)!}p^L(1-p)^{L-m}$ ($L\equiv N-1$, $m\equiv n-1$) = $pN[(\sum_{m=0}^\infty m\frac{L!}{m!(L-m)!}p^L(1-p)^{L-m})+1]$ = $pN(pL+1)$ = $pN(p(N-1)+1)$ = $pN(pN-p+1)$ = $pN(1-p)+(pN)^2$
Adding all 3 terms and using $\bar{n}=pN$ gives:

$S_1+S_2+S_3=pN(1-p)+(pN)^2 -(pN)^2=pN(1-p)$

or $$\sigma^2=pN(1-p)=\bar{n}(1-p)\label{evarbinomial}$$ Equations $\ref{emeanbinomial}$ and $\ref{evarbinomial}$ are very useful, and tells you that the mean of a binomial distribution is just the total number of possible combinations, $N$, times the probability $p$ of any one event. In the coin experiment, $p$ is the probability of getting a heads, and $N$ is the number of coin tosses. $P(n,N)$ tells you the probability that you will see $n$ heads (and therefore $N-n$ tails), $\bar{n}=pN$ tells you the average number of heads (if $p=\half$, then $\bar{n}=N/2$), and $\sigma^2=\bar{n}(1-p)=N/4$ is the variance on the average number of heads.

It is interesting that the variance $\sigma^2$ will be equal to the mean $\bar{n}$ when $p\lt\lt 1$. This is going to be important when we next consider the Poisson distribution.

### Combinatorics

Imagine doing an experiment where you ask some number of strangers for their birthday, and make a histogram of what how many people had birthdays on any particular day (where day means day and month as a sequential number from 1 to 365). The histogram will have 365 bins, and the sum of all the entries in the bins adds up to the total number of people you queried ($N$). One would expect that all days should have the same probability $p=1/365$ of being a birthday. Of course, there could be biases that favor some days over others, for instance one could imagine that maybe more people were conceived on holidays than non-holidays, therefore the birth dates should cluster around holidays plus 9 months. But let's assume the probability $p(n)$ of any random birthday to be on day $n$ is given by $p(n)=p=1/365$.

For instance, let $N=1000$ people. If we make the plot of the number of birthdays $N$ on any given day $n$, then if all days are equally likely, $N(n)$ will look something like this:

Figure 1. Histogram of birthdays of $N=1000$ sample.

The horizontal axis is the day of the year (1-365) and the vertical tells you how many times we found a person with a birthday at that day. We can see right away that it's much more likely that we see something between 1 and 3 counts than that we see something between 8 and 10 counts.

You might then want to make a histogram of the distribution of counts (how many days had 0 counts, how many had 1, etc), called the "frequency". In the above plot, it looks like no day had more than 10 counts, so our frequency histogram will be from 0 to 10. It will look like this:

To turn our frequency histogram into a probability distribution, all we have to do is divide each bin by the total number in the sample, which in this case is 1000 (if you sum over all the bin contents in the frequency histogram, you will get 1000 since the x axis tells you the number of days that had 0, 1, 2, etc birthdays, and you are summing over all possible days). The result is shown next:

To understand the shape of this probability distribution, let's formulate the problem slowly. We are asking each person from a sample of $N$ people for a date corresponding to their birthday. Each answer has a probability $p$, and since all dates are equally likely if there is no bias, then the probability for any given date coming up will be given by $p=1/365$. This "stream" of 1000 is no different than the "stream" of 1000 coin tosses, each being either heads or tails. However, the difference is that for the coin tosses, there are 2 possible outcomes, while for the birthday experiment, there are 365.

If we ask $N$ people for their birthday, and want to predict the number of people who have the same birthday as Mozart (January 27), then after asking all $N$ people ($N$ samples), the the answer will be that on average, there will be $N/365$ with that particular birthday. If $N=365$, then on average there will be only 1 person out of a sample of $365$ that have that particular birthday.

Now, let's consider how to measure the probability that in a sample of $N$ people, that at least 2 of them have the same birthday as Mozart. That's equivalent to looking at figure 1, finding the right bin (the one that corresponds to Jan 27), and running countless experiments with $N$ people to see how many times you get at least 2 entries. The probability will be given by that number divided by the number of times you've tried.

To calculate what you would expect for that probability, let's start with some small number of people, say $N=5$, and calculate the probability that we have 0, 1, 2, 3, 4, or 5 with that particular birthday (call these probabilities $P(0)$, $P(1)$, $P(2)$, $P(3)$, $P(4)$, and $P(5)$):

• $P(0)$. Here we want to know the probability that no one has a Jan 27 birthday. We start with the probability $p$ that the 1st person has that birthday, which will be $p=1/365$. So the probability that the 1st person does not have that birthday (call it $\bar p$) will be given by $\bar p=1-p=364/365$. For 5 people, then the probability that none of them will have that birthday is given by $\bar p^5=(364/365)^5=98.63\%$.
• $P(1)$. If the probability for the 1st person to have that birthday is $p$, and the probability that the next 4 people do not have that birthday is $\bar p^4$, then the joint probability will be given by $p\cdot \bar p^4$. Since there are 5 ways we can have only 1 person with that birthday, the total probability $P(1)=5p\cdot\bar p^4=1.35\%$.
• $P(2)$. We want 2 people with that birthday ($p^2$), and 3 without ($\bar p^3$), and there are $C(2,5)=5!/(2!3!)=10$ ways to arrange these 5 people in pairs. So the joint probability will be given by $P(2)=C(2,5) p^2 \bar p^3=C(2,5) p^2(1-p)^3=2.33\%\cdot$, and now we see that we are dealing with the binomial distribution!
• $P(3)$ is obviously given by $P(3)=C(3,5)p^3(1-p)^2$, $P(4)=C(4,5)p^4(1-p)$, and $P(5)=p^5$.
So we can see how the binomial distribution can be used to solve problems that involved probabilities and combinatorics. Now we can calculate the probability (call this $M(5,2)$) that in a room of $N=5$ people, at least 2 of them have the same birthday as Mozart. Since "at least 2" means 2, 3, 4, or 5, then we add those probabilities together: $M(5,2)= P(2)+P(3)+P(4)+P(5)$. Or, since the sum of all probabilities is 1, we can instead use $M(5,2)= 1 - P(0) - P(1)$. Since $P(0)=98.63\%$ and $P(1)=1.35\%$, then $M(5,2)=1-99.993\%=0.0075\%$.

### The Birthday Problem

To finish this example, let's say you have a room of $N$ people, you make a histogram of birthdays, normalize it so that it's a probability distribution, and ask for the probability that at least 2 people have the same birthday. Or better yet, you ask for the number of people (what is $N$) such that the probability that any 2 have the same birthday be at least 50%.

The easiest way to solve this is to use the property that probability distributions are normalized to 1:

P(at least 2 people having the same birthday) = 1 - P(no two people have the same birthday), because the latter probability, that no two people have the same birthday, is easier to calculate!

To start, let's take 2 people in the room. The first person will have some birthday (it doesn't matter what the date is), so the probability that they both have the same birthday is just the probability that the 2nd person have a birthday on that particular date, which is $1/365$, if of courseall birthdays are equally likely. Then the probability that they do not share the same birthday will be $p_2=1-1/365-364/365$. Now, introduce the 3rd person. If these first two do not have the same birthday, then there are 363 days left for the 3rd person, so the probability that that person does not have either of the 1st 2 birthdays will be given by the number of possible dates left, $363$, divided by all possible dates, $365$, or $p_3=363/365$. Then the joint probability is given by $p_2\cdot p_3$, since we want birthday 1 not equal to birthday 2 AND birthday 3 not equal to both 1 and 2. If we have 4 people, then then $p_4=362/365$ (3 dates are eliminated due to 1, 2, and 3), and the joint probability will be given by $p_2\cdot p_3\cdot p_4$. We can now extend this to an unknown number of people, $N$: $$P(N) \equiv p_2\cdot p_3\cdot ... \cdot p_N = \frac{364}{365}\cdot\frac{363}{365}\cdot\frac{362}{365} ... \cdot\frac{365-(N-1)}{365}\nonumber$$. We can rewrite this as $$P(N) = \frac{365\cdot 364\cdot 363 \cdot 362 \cdot ... \cdot 365-(N-1)}{365^N}\nonumber$$ This is the probability that for $N$ people, no two share the same birthday. The probability of having 2 people share the same bithrday will be $1-P(N)$.

The figure below shows a plot of $P(N)$ vs $N$. You can see that for $N=23$, the probability just barely exceeds $50\%$. This is perhaps a surprisingly small number, however it's all due to the combinatorics: as you add people, you increase the number of ways (the combinatorics) for $2$ out of $N$ to share the same birthday.

### Combinatorics and Throwing Dice

Unfortunately there are characters in the world who cheat, and to cheat in dice all you have to do is "load" the dice so that you get the right face showing more often than not. Imagine that your cousin who owns a gambling establishment offers to pay you to test some dice to see if they are loaded, you will need to know what you are doing! So now let's attack the problem of calculating the expected probability distribution for situations with combinatorics using the dice.

We can start with a single die, and roll it $N$ times. If each face is equally likely to show up, then the probability $p(n)$ for seeing face $n$ appear will be independent of which face: $p(n)=1/6$. What we want to do is to calculate the following quantities after $N$ rolls:

• $\bar{f}(n)$: the average of number of times we see face $n$ appear
• $\delta^2(n)$: the variance on $\bar{f}(n)$
Then we will do the experiment, roll the die $N$ times, and measure the frequency $f(n)$ for seeing face $n$ appear, and compare it to $\bar{f}(n)\pm\delta(n)$. This will give us an idea of whether the die is loaded or not.

When you roll the die $N$ times, you will be looking at a stream of numbers, each with $M=6$ different possible values, and you want to know the probability that we see face $n$ appear $m$ times. For example, say you roll the dice 5 times ($N=5$) and ask for the probability that you should see face 1 appear $m=2$ times. As usual, you can calculate these probabilities by adding up the number of different ways something can be seen, and multiplying by the probability of seeing just that thing. So for $m=2$, we could see face 1 on trial 1, and then on trial 2,3,4,5, or 6. Or, we could see some other face on trial 1, see face 1 on trial 2, and then on 3,4,5, or 6. And so on. If you add up all possible combinations, you will get 5 (1 shows up the first time) plus 4 (2nd), plus 3 + 2 + 1 (face 1 shows up on trial 5 and 6). So there are 5+4+3+2+1=15 ways to see face 1 show up twice. This is exactly what the combinatorics part of the binomial distribution calculates for you, the number of ways you can take $M=6$ things and arrange it so you see $m=2$ of them: $$\frac{M!}{m!(M-m)!}=\frac{6!}{2!(6-2)!}= \frac{6\cdot 5\cdot 4\cdot 3\cdot 2}{2(4\cdot 3\cdot 2)}=15\nonumber$$ To calculate the probability of seeing face $n=1$ show up $m=2$ times, we would need to multiply by the probability of seeing face 1 showing up twice, and any other face showing up 4 times, which would be given by $p^2(1-p)^4$. Putting this together gives you the binomial probability distribution $P_n(M,m)$, where $M$ is the number of different die in the set ($M=6$), and $m$ is the number of times you would see face $n=1$: $$P_1(m,M)=\frac{M!}{m!(M-m)!}p_1^m(1-p_1)^{M-m}\nonumber$$ Since all faces are equally likely, you can drop the subscript (here it's $n=1$), and just write the probability of seeing any of the faces snow up $m$ times is: $$P(m,M)=\frac{M!}{m!(M-m)!}p^m(1-p)^{M-m}\nonumber$$ where $p=1/6$.

Since we know that the mean and variance of a binomial distribution will be given by equations $\ref{emeanbinomial}$ and $\ref{evarbinomial}$, we can write that

• $\bar{f}(n) = \bar{m} = Np=N/6$
• $\delta(n) = \sqrt{\sigma^2} = \sqrt{Np(1-p)}=\sqrt{5N}/6$.
All you have to do now is roll the die $N$ times, make a histogram of how many times you see face $n$ apear, and compare that to what the binomial distribution predicts, which is that each face will have an equal mean and variance after $N$ rolls. If you find one of the faces that is very different from any of the others, then that die is loaded. Of course, we have to define what we mean by "very different", and this turns out to be where science meets art (more below)!

### Poisson Distribution

The binomial distribution is a general case, and somewhat complex. Sometimes, however, the situation you are in will have a small probability $p\lt\lt 1$, a large number $N$, and you want to know the probability $P(n,N)$ where $n\lt\lt N$. For instance, imagine we were to ask $N$ people for their birthday, and tablulate the date as the ith day of the year (from 1 to 365). Then we would want to calculate the expected frequency probability distribution for any particular day to have 2, 3, 4, 5, etc birthdays. For example, you ask 365 people for their birthday, and want to know the probability of there being 2 birthdays in any given day. Or 3 birthdays. And so forth. This will again involve the binomial distribution, because you are dealing with the combinatorics (the combinatorics and associated probabilities for taking $N$ things $n$ ways). But here, the probability of any given day will be quite small, and if all days are equal, we should have $p=1/365$. If we took the data ($N$), made the histogram of how many days had 2, 3, etc ($n$) people, and compared to what you would expect, you could maybe see if there was some reason why some days would be more preferred or not (for insance, perhaps more birthdays 9 months from holidays). Here, the number of possibilities in the stream is 365 ($p=1/365$), which is quite a small number. If you took $N$ samples, then the average for each day will be given by the mean of the binomial distribution, or $\bar{n} = Np$. So if you sampled 365 people, you would expect $\bar{n}=1$, but you might actually find that some days had 2, even maybe 3. But you would not expect that some days would have anywhere near 365!

So in our case here, we have the following conditions:

• $p \lt\lt 1$
• $n \lt\lt N$
We can investigate how the binomial distribution behaves by doing the relevant expansions, and there are several equally complex ways to do this. One very straightforward way is to do make the following approximations:
1. For $n \lt\lt N$, the factor $\frac{N!}{(N-n)!}$ will have $n$ terms, and they will all be large compared to $n$. For instance, if $N=100$ and $n=2$, then $\frac{100!}{98!}=100\cdot 99$ which is almost equal to $100^2$. So a reasonable approximation here is $\frac{N!}{(N-n)!}\to N^n$
2. That leaves $P(N,n)=\frac{N^n}{n!}p^n(1-p)^{N-n}$
3. The factor $(1-p)^{N-n}\to (1-p)^N$ for $n\lt\lt N$.
4. For $p\lt\lt 1$, we can expand $(1-p)^N$ using the expansion for the natural log $ln$. We take $ln(1-p)^N = Nln(1-p)$, and expand $ln (1-x) \sim -x - \half x^2$. So the expansion for $(1-p)^N\sim e^{-Np}e^{-Np^2/2} \sim e^{-Np}$ as long as $Np^2\lt\lt 1$, or $Np\lt\lt 1/p$.
5. So replacing $(1-p)^{N-n}\to e^{-Np}$, we have $P(N,n)=\frac{N^n}{n!}p^ne^{-Np}=\frac{(Np)^n}{n!}e^{-NP}$
6. Now replace $\mu = NP$
A brief word on the approximation made above, that $Np = \mu \lt\lt 1/p$. The probability distribution will max out in the region where $n\sim \mu$, and will fall to zero way outside that region. So as long as we keep away from distributions where $\mu$ is close to $N$, we are ok. For the birthday example, this approximation will hold as long as the average number in any day is less than $1/p=365$. This also tells us that $N\lt\lt 1/p^2\sim 10^5$, so it sets a limit on $N$

With these approximations, the binomial distribution then goes to: $$P(n,N)\to \frac{\mu^ne^{-\mu}}{n!}\nonumber$$ as $n/N\to 0$, $N\lt\lt 1/p^2$, and $p\lt\lt 1$, and the definition $\mu\equiv Np$ is the mean. This is called the Poisson distribution, and is one of the more important distributions in science (up there with the binomial, and gaussian).

Why the poisson has only 1 parameter $\mu$ instead of the 2 parameters $p$ and $N$ for the binomial is because of the limiting condition: $p$ is very small, $N$ is very large, but the product $\mu=pN$ is finite. Also, if the variance of the binomial distribution is given by equation $\ref{evarbinomial}$, then in the limit of $p\to 0$, then $Np(1-p)\to Np = \mu$ so the variance of the Poisson will be given by $\sigma^2 = \mu$, and there really is only 1 free parameter ($\mu$).

So the poisson distribution is: $$P(n,\mu)= \frac{\mu^ne^{-\mu}}{n!}\label{epoisson}$$ and the mean and RMS values are: $$\mu = Np\nonumber\label{mupoisson}$$ $$\sigma_{RMS}=\sqrt{\mu}\label{vpoisson}$$ For this to be a real distribution, we would want $\sum P(n,\mu)=1$, and this is the case (in the sum, the factor $e^{-\mu}$ factors out, leaving a sum over $\mu^n/n!$, which is the expansion for $e^{+\mu}$).

The physical significance of these limits is interesting. For an experiment like the birthday experiment, it describes a situation with a very small probability $p$ but a large number of trials compared to the combinations ($n\lt\lt N$). For something like a radioactivity experiment, where you count the number of disintegrations in some time window, the poisson also applies, since the limiting conditions means that the probaiblity for any given particle of the sample to decay is small, and number of counts in any particular time window will be small as well.

In the following plot, you can change the mean value $\mu$ and see the shape of the poisson distribution $P(n,\mu)$. Notice how the distribution becomes more symmetric as $\mu$ gets larger.

 Log scale? $\mu =$10 # pts to plot =25

The poisson distribution is over discrete values for $n$. However, as $\mu$ gets larger, the term $e^{-\mu}$ sends the probability to 0 except in the region near $n\sim\mu$ (play with the poisson distribution above, driving $\mu$ up). To see this more explicitly, we can use Stirling's formula for the factorial: $$n!\to x! \to \sqrt{2\pi x}e^{-x}x^x\label{estirling}$$ Substituting, we have $$P(n,\mu)= \frac{\mu^ne^{-\mu}}{n!}\to\frac{\mu^ne^{-\mu}}{\sqrt{2\pi n}e^{-n}n^n} = \frac {e^{-(\mu-n)}} {\sqrt{2\pi n}} \Big(\frac {\mu}{n}\Big)^n \nonumber$$ Now that we have an equation that substitutes for the integer $n$, we can consider the poisson as a probability over a continuous variable $x$ instead of $n$. This will be true when $\mu \gt\gt 1$. This allows us to write the poisson distribution as $$P(x,\mu)= \frac{\mu^xe^{-\mu}}{x!}\label{epoissonx}$$ Note that even though we derived the poisson starting with the binomial, the poisson can also be derived from first principles for counting experiments where the numbers counted are independent of each other (for instance, the number of photons from radioactive decay, or the number of people in line at the grocery store), and the average rate is constant (and so you are just seeing fluctuations about the average).

### Gaussian Distribution

As we just saw, the poisson distribution is the binomial distribution in the limit of $p\lt\lt 1$ and $n\lt\lt N$, but $pN=\mu$ is finite. Both are distributions over discrete values, and usually have to do with predicting the results from counting experiments. Using Stirling's rule, we can turn the poisson into a distribution over continuous variables, or we can calculate it from first principles. One can then look at poisson distributions when the mean $\mu$ is quite large, and when $n$ is large, and after taking the limits one can get the following distribution: $$P(x,\mu)=\frac{e^{-(x-\mu)^2/\mu}}{\sqrt{2\pi\mu}}\nonumber$$ If we use the fact that both the mean $\bar{x}$ and the variance $\sigma^2$ are both equal to $\mu$, we can substitute in the above formula to get: $$P(x,\mu)=\frac{e^{-(x-\bar{x})^2/\sigma^2}}{\sqrt{2\pi\sigma^2}}\label{egaussian}$$ This is called the gaussian distribution, and is probably the most important and ubiquitous distribution in all of science.

The Gaussian and the Poisson distributions are very similar. In the plot below, the Poisson is drawn in red and the Gaussian in yellow. The mean is the same for both, and the width of the gaussian will be given by $\sigma = \sqrt{\mu}$. As you can see, they are similar, however the poisson variance for the gaussian gives you a slightly wider function, showing that $\sigma = \sqrt{\mu}$ is not true for the gaussian.

Gaussian and Poisson $\mu$ =

Instead of deriving the gaussian starting with the binomial, or poisson, there's another way to do it, and this will help you to appreciate the importance of the gaussian. We start with a rather simple and interesting problem: consider a variable $x$, distributed in some random way. If you made a histogram of $x$ and normalize to 1, you will see the probability distribution function $P(x)$. In this example, we do not have to specify $P(x)$: it could be totally random, in which case $P(x)=$constant, or it could have some totally weird shape. The key thing is that the shape is irrelevant. Next, we generate a large number of $x_i$ according to $P(x)$. Let's start with a probability distribution in which $x_i$ is a random number, and so $P(x)$ is constant, or "flat". In the world of computer programming, there are always built-in functions that generate random numbers between 0 and 1, so each $x_i$ would be some random number in that interval. Next, we take all of these $x_i$ and form an average over $N$ of them, and call this average $X_N$: $$X_N = \frac{1}{N}\sum_{i=1}^N x_i\nonumber$$ For instance, if $N=10$, and the $x_i$ are random numbers "flat" over the interval between 0 and 1, then the probability of any particular $x_i$ value seen is the same for all $x_i$, the average $X_N$ would also be a number between 0 and 1. If we take numerous numbers $x_i$ and form lots of different $X_N$ averages, and then make a histogram of $X_N$, we could see the probability distribution $P(X_N)$. We might expect the average $X_N$ to be near 0.5, however what would you expect for the distribution of $X_N$? The answer is surprising: $P(X_N)$ turns out to be a gaussian!

In the histogram below, you can see a distribution of random numbers. It is "flat", but of course there are fluctuations, so the number in each histogram bin will fluctuate somewhat. However the more numbers you generate, the "flatter" the distribution becomes, and you can try this by changing the box below with the "# to generate# (call this $N_{tot}$) and hitting the "Regenerate" button. The "# of bins in histogram:" (call this $N_b$) is used by the code to make the histogram, so you can change that as well and see how the "flatness" changes. The number of entries in each bin will fluctuate around the mean for each bin, given by $\mu = N_{tot}/N_b$.

Generate distribution of random numbers flat between 0 and 1
 # to generate: # bins in histogram: # to sum: 1

Average number per bin $\bar {N}:$
$\bar{x}=$  $\sqrt{\sigma^2}=$

### Central Limit Theorem

Next, we run the same random number generator, but instead of making a histogram of $x_i$, we will take an average over some number of points ($N$), and histogram that average. For instance, we could generate 10000 points, calculate $X$ as the average over every 10 points, (call this $X_{10}$) and histogram $X_{10}$.

In the above, if you change the slider called "# to sum:" to anything beyond 1, you will see what happens when you change from histograming $x_i$ to histogramming $X_N$, with $N\gt 1$. As you can see, the distribution over the averages becomes highly peaked, more and more as you increase the number of points in the average, $N$.

This is straight forward to understand: when you calculate the average over say 10 points, then much of the time you will get something near the true average, which will be at 0.5 if $0\le x_i\lt 1$ But not always. Consider how you would get an average that is far away from 0.5, say at 0.1. In that case, you will need to get pretty lucky to generate 10 random values that are all clustered around 0.1. There are not many ways to do this, as opposed to the many ways you could generate numbers that have an average near 0.5. So the probability of getting something far from the true average falls fast as you go away from that average. And so combinatorics plays a big role here: there are many more ways to generate 10 points that form an average near 0.5 than there are ways to generate 10 points to form an average near 0.1.

This effect is the main subject of what is known as the "Central Limit Theorem" (CLT). There are two amazing things about CLT:

• The probability distribution of averages approaches the gaussian probability distribution as described above in equation $\ref{egaussian}$, and the more you increase $N$ (the number of points in the average) the more gaussian the distribution becomes
• The gaussian is independent of the shape of the distribution for $x_i$
It can be shown that the distribution for the average approaches the gaussian as the number of points in the average ($N$) approaches $N_{tot}$, with a variance given by: $$\sigma^2_{Gaussian} = \frac{\sigma^2_{random}}{N}\nonumber$$ where $N$ is the number of random values to include in each mean that is histogrammed. In the above, if you select "Averages", then the histogram will be of the averages $X$, and the line will be a gaussian that uses the mean $\bar x$ and RMS $\sqrt{\sigma^2}$ as the 2 parameters. You can see that the difference is pretty small! It is educational to play with the parameters to get a feeling for this: try increasing the number of points to generate, and the number of points for averaging, and notice how the variance changes: you should see it get smaller and smaller as the number of points in the average gets larger and larger. This reflects the falling probability of seeing combinations of $x_i$ that are all far away from $\bar{x}$.

So we see here that the gaussian distribution is not merely something derived by taking the limits for binomial distributions, but is in fact much more basic and fundamental, and is all tied up with the notion of randomness. This is what makes the gaussian so important: if you are doing an experiment and want to see if what you are measuring is according to some known process, or something new, you want to understand the random fluctuations in the known process and compare to the experimental observations. Those fluctuations will be (under most circumstances) described by gaussian (random) fluctuations. If the comparison is good, then what are measuring is just a fluctuation. The question is, what does "good" mean?

The more you get to work with and know statistics, the more you will use the gaussian distribution. And you will get a feeling for how the gaussian quantifies fluctuations in measurements. For instance, if you have a quantity $x$ with a mean $\bar x$, and RMS $\sigma_x$, then the gaussian describes the probability of any particular measurement yielding the value $x_i$: $$P(x_i) = \frac{1}{\sqrt{2\pi\sigma_x}}e^{-(x_i-\bar{x})^2/2\sigma_x^2}\nonumber$$ The implicit assumption here is that the true value of $x$ will be $\bar x$, so that when you make a measurement, what you want to know is the probability that the measured value $x_i$ will be within some finite distance from the mean $\bar x$. That means you have to integrate the probability distribution from $\bar x-x_i$ to $\bar x+x_i$.

Log scale:
between $\pm$

## $\chi^2$

When we make measurements, we almost always want to and compare to theory. Our theory could be based on some real effect, or it could be that we just want to compare to the null result. This chapter introduces techniques of comparison, and the idea of "significance".

To start, imagine you take some data and want to compare to a theory. For instance, you might make a series of experiments ($N$ measurements, each one denoted by the subscript $i$) and for each measurement "i" you set an acceleration $a_i$ on some object of unknown mass, and measure the force $F_i$ with some uncertainty (aka "error bar") $\delta F_i$. Each data point should obey Newton's law, to within the uncertainties from fluctuations: $$F_i=ma_i\nonumber$$ How can we extract the mass from all this data? One possibility is to take the ratio of each measurement of $F_i$ and $a_i$, and extract the average mass $\bar{m}$ by averaging over all ratios: $$\bar{m} = \frac{1}{N}\sum_{i=1}^N F_i/a_i\label{embar}$$ This is a mathematically well defined thing to do, however it suffers from one big drawback: we are assuming that the relationship between $F_i$ and $a_i$ is linear, described by a line with a zero y-intercept. It is also susceptible to ratio's where the acceleration fluctuates low, approaching a "divide by zero" problem and skewing the average ratio.

Another possibility is to average over all $F_i$, and over all $a_i$, and take the ratio of the averages: $$\bar{m} = \frac{\sum_{i=1}^N F_i}{N}/\frac{\sum_{i=1}^N a_i}{N}= \frac{\sum_{i=1}^N F_i}{\sum_{i=1}^N a_i}\nonumber$$ This is a bit better behaved (and we will see has some validity), but still suffers from the fact that we are assuming here that there is no y-intercept in the linear relationship between $F_i$ and $a_i$.

At this point, let's think through what we are trying to accomplish, and see if we can come up with a better method. To start, the plot below shows the set of data for $F_i$ and $a_i$ measurements. The $a_i$ are assumed to be "dialed up", so they have a negligible uncertainty, and the $F_i$ are the measured forces, each with an uncertainty $\delta F_i$. Each time you hit the "Reset" button, it will generate an entirely new set of data, randomly:

Generate straight line to: $F=m\cdot a+b$
 Slope $m$: Y intercept $b$: $\chi^2=$ DOF=

The task at hand is to determine the "best" straight line that describes a linear relationship between $F$ and $a$. Your intuition here is useful: if you were to lay a pencil down on top of the distribution of $F$ vs $a$, and eyeball the line that would best fit the data, you would probably try for the line that goes through the most points. And you would be pretty close. But to get the very best fit, what you would want to do is draw a line that minimizes the distances between the y-value of the line ($y=mx_i + b$), and the y-value of the data point ($y_i$). So if you pick a slope $m$ and intercept $b$, then you would perhaps form the following quantity: $$d_i = y_i - y = y_i - (mx_i + b)\nonumber$$ Summing over all points (all $N$ $d_i$'s) would give a measure of how well the theory (the straight line) fits the data, so you might form the sum: $$D = \sum_{i=1}^N d_i = \sum_{i=1}^N [y_i - (mx_i+b)]\nonumber$$ If you then varied the slope slope and y-intercept, and recalculated $D$ for each new choice of $m$ and $b$, you could find the $m$ and $b$ that minimize $D$. The problem with this particular choice of measuring how well the theory and data match (using $d_i$) is that in the sum, some of the time the line will be below the point ($d_i\gt 0$), and some of the time above ($d_i\lt 0$). So you could have the situation where the individual $d_i$'s are large, but by (bad) luck cancel each other out, giving a small value for $D$. We need another measure!

An alternative would be to just take the absolute value of the various $d_i$ values. However it turns out that the technique that has the most statistical validity uses the square of the $\delta_i$'s, something intrinsically positive: $$\sum_{i=1}^N d_i^2 = \sum_{i=1}^N \Big(y_i - (mx_i+b)\Big)^2\nonumber$$ This is close to being the best thing to do, however how will you judge whether the value of the sum is "good" enough? And this particular technique, using $d_i^2$, does not take into account the uncertainty $\delta_i$ in the measurement of each $y_i$.

A sensible (and as it turns out, statistically valid) way to solve both problems is to first divide $d_i$ by the uncertainty $\delta_i$, and then square forming the "residual" $r_i$: $$r_i\equiv \frac{d_i}{\delta_i}\nonumber$$ By doing that, $r_i$ is measuring the vertical distance between the measured $y_i$ point at a given $x_i$, and the $y(x_i)$ value calculated from that $x_i$ and the line parameters $m$ and $b$, in units of the uncertainty in $y_i$: $r_i=d_i/\delta_i$ tells you "how many uncertainties" separate $y_i$ from $y(x_i)$. Then we form the sum over all $r_i^2$ (squared, to avoid the cancellation problem above), and call this value $\chi^2$: $$\chi^2 = \sum_{i=1}^N r_i^2 = \sum_{i=1}^N \Big(\frac{d_i}{\delta_i}\Big)^2 = \sum_{i=1}^N \Big(\frac{y_i - (mx_i+b)}{\delta_i}\Big)^2\label{echi2}$$ The procedure would then be to vary $m$ and $b$ independently, forming a new $\chi^2$ each time, and choose our final values for $m$ and $b$ that gives the smallest possible $\chi^2$. This is what the "Solver" does for you in Excel. Notice that in the above plot of "Force vs Acceleration", the $\chi^2$ is given (once you input a non-zero slope and y-intercept). The assumption made in the code is that all of the $\delta_i$'s are the same (just to make things simple), and equal to around $2$, just as a guess. But in reality, you have to come up with a reasonable value for the $\delta_i$'s, and this is one of the most important parts of doing experiments: knowing your uncertainties. (Note: scientists often say "know your errors", a cute expression that is actually the same thing as "know your uncertainties".)

In the plot above, you can type in a non-zero value for the slope and y-intercept, the line will be displayed, and the code will tell you the $\chi^2$ and number of degrees of freedom. You can change the slope and y-intercept and see where the $\chi^2$ is minimized for the best fit.

### Linear Fit, Analytically

For a linear fit, you can actually find the best slope analytically. To keep it simple and illustrate the concept, let's choose a constant value for all of the $\delta_i$'s (call it $\delta$). There are 2 parameters here, $m$ and $b$, so to calculate values of $m$ and $b$ that minimize the $\chi^2$, we set the 1st derivative $\partial\chi^2/dm$ and $\partial\chi^2/db$ to 0, and solve for the best value of $m$ and $b$. The calculation goes like this:

 $\chi^2$ $=$ $\sum_{i=1}^N \Big(\frac{y_i - (mx_i+b)}{\delta_i}\Big)^2$ $=$ $(1/\delta^2)\sum (y_i - mx_i-b)^2$
Now we take the 2 derivatives and set them equal to 0: $\partial\chi^2/dm=0$ and $\partial\chi^2/db=0$: $$\frac{\partial\chi^2}{dm}=-(2/\delta^2)\sum x_i(y_i - mx_i-b)=0\nonumber$$ $$\frac{\partial\chi^2}{db}=-(2/\delta^2)\sum (y_i - mx_i-b)=0\nonumber$$ We can drop the $-2/\delta^2$, and get 2 equations with 2 unknowns after some rearranging: $$m\sum x_i^2 + b\sum x_i= \sum x_iy_i\nonumber$$ $$m\sum x_i + b\sum = \sum y_i\nonumber$$ The $\sum$ term in the 2nd equation takes some getting used to, but if you were to write it as $\sum_{i=1}^N 1$, you can see that it is equal to $N$. So our two equations are: $$m\sum x_i^2 + b\sum x_i= \sum x_iy_i\nonumber$$ $$m\sum x_i + bN = \sum y_i\nonumber$$ With some linear algebra, you can then solve for $m$ and $b$ to get: $$m = \frac{N\sum x_iy_i - \sum x_i\sum y_i}{N\sum x_i^2 - (\sum x_i)^2}\nonumber$$ $$b = \frac{\sum x_i^2\sum y_i - \sum x_iy_i\sum x_i} {N\sum x_i^2 - (\sum x_i)^2}\nonumber$$ These are pretty busy formulae, but they can be simplified a bit, starting with noticing that the denominator can be written as: $$N\sum x_i^2 - (\sum x_i)^2 = N^2\big(\frac{\sum x_i^2}{N}-\frac{(\sum x_i)^2}{N^2} \big) = N^2(\overline{x^2}-\overline{x}^2)=N^2\sigma_x^2\nonumber$$ where $\sigma_x^2$ is the variance in $x$, as defined above in equation $\ref{evardiff}$. We can then write the best fit for the slope and y-intercept as: $$m = \frac{ \overline{xy}-\overline{x}\cdot\overline{y}}{\sigma_x^2}\nonumber$$ $$b = \frac{ \overline{x^2}\cdot\overline{y}-\overline{x}\cdot\overline{xy}}{\sigma_x^2}\nonumber$$ Notice that if you set $b=0$ by construction, after some algebra you would get the following for the slope $m$: $$m = \overline{y}/\overline{x}\nonumber$$ which is identical to equation $\ref{embar}$ above! Note that if each of the $\delta_i$ are not different, then it doesn't matter, you can carry out the calculate just like above, it gets a bit more complicated but you get the same answer.

These formula for $m$ and $b$ are interesting, and deserve a closer look to see if we can understand them better. For instance, let's first define a quantity that calculates the 2nd moment like this: $$M2(x,y) \equiv \overline{x}\cdot\overline{y} - \overline{xy}\nonumber$$ Then we can show that $M2(x,x)=\sigma_x^2$, and we can rewrite the above formula for $m$: $$m = \frac{ \overline{xy}-\overline{x}\cdot\overline{y}}{\sigma_x^2}=\frac{M2(x,y)}{M2(x,x)}\nonumber$$ This tells you something about the slope, that it is related to the ratio of the 2nd moments of $xy$ to $xx$. But the slope doesn't tell the whole story. For the y-intercept $b$, we first add and subtract $\overline{x}^2\cdot\overline{y}$ from the numerator and collect terms, and after some algebra, we get $$b = \overline{y}-m\overline{x}\nonumber$$ or in other words, $\overline{y} = m\overline{x} + b$, which tells you that the best fit is such that the average $x$ and average $y$ fit the straight line!

The above formulae are very useful if you have an array of $x$ and $y$ values and want to calculate the best fit for the slope and y-intercept hypothesis: just loop over all values, calculate the averages $\overline{xy}$ etc, and out pops the result without having to do double loops:

	//
// "x" and "y" are in arrays x[i] and y[i], and we sum from 0 to nbins-1
//
sumy = 0;
sumx = 0;
sumx2 = 0;
sumxy = 0;
for (var i = 0; i < nbins; i++) {
sumx = sumx + x[i];
sumy = sumy + y[i];
sumx2 = sumx2 + x[i]*x[i];
sumxy = sumxy + x[i]*y[i];
}
avx = sumx/nbins;
avy = sumy/nbins;
avx2 = sumx2/nbins;
avxy = sumxy/nbins;
variance2 = avx2 - avx*avx;
slope = (avxy - avx*avy)/variance2;
y_intercept = (avx2*avy - avxy*avx)/variance2;


Once you have a value for the parameters that minimize the $\chi^2$, you still need to ask whether that minimum $\chi^2$ is describing a "good" fit, or not. Note that you have taken the sum of $(d_i/\delta_i)^2$. If the fit is "good", and your uncertainties are accurate, then you would expect that the fit curve will go through most of the points within $\pm\delta_i$, more or less, so that each of the $r_i$ will be "close" to 1, on average. Of course sometimes you will have points where the fit is more than 1 $\delta$ away, and sometimes you will have points that are less than 1 $\delta$ away, but on average, a "good fit" should give you a $\chi^2$ where each of the $r_i$ are of order 1. So we can expect a good fit to have the sum $\sum_{i=1}^N$ will be close to $N$.

Actually, you have to take care here, because if you are fitting for 2 parameters (as in a straight line, $m$ and $b$), then you need at least 2 points. If you have only 2 points, then you have no freedom to move the slope $m$ and y-intercept $b$ around (2 points determine a straight line), and so the $\chi^2$ for 2 points ($N=2$) will be identically 0.0. So the number of "degrees of freedom" will be, for a straight line fit, 2 less than the total number of points, and for a good fit you would expect that the $\chi^2$ will add up to something very close to $N_{dof}=N-2$. And so you can look at the quantity $\chi^2/N_{dof}$, and that should be close to 1. If it is, then it's a good fit. If it's not, then it's not a good fit! If that ratio is much larger than 1, it means that either the points do not describe the fit function, or it could mean that your uncertainties $\delta_i$ are too small and in fact you have bigger uncertainties than you thought. If on the other hand the ratio is close to 0, it could mean that your uncertainties are too large (or that your points are too good, which should make any scientist nervous!).

## Uncertainties and Error Propagation

Sometimes we make measurements on more than 1 quantity at a time, and combine them to form something that is a function of the measured quantities. For instance, we measure the width $W$ and length $L$ of a parking lot, with corresponding uncertainties $\delta W$ and $\delta L$. Then we want to combine them to form the perimeter $P=2(W+L)$. To keep it simple, let's define $P_h=W+L$, and so $P=2P_h$. Knowing the uncertainties in $W$ ($\delta W$) and $L$ ($\delta L$, we want to know the uncertainty in $P_h$, and deal with the uncertainty in $P$ later.

One sensible thing to do is to say that the extreme values of $W$ are given by $W_+=W+\delta W$, and $W_-=W-\delta W$. And similarly for $L_+$ and $L_-$. Then the extreme values of $P_h$ would be given by: $$P_{h+}=P_h+\delta P_h = W_++L_+ = W+\delta W + L + \delta L = W+L+(\delta W+\delta L)\nonumber$$ and $$P_{h-}=P_h-\delta P_h = W_-+L_- = W-\delta W + L - \delta L = W+L - (\delta W+\delta L)\nonumber$$ This suggests we use $\delta P_h=\delta W + \delta L$.

However, there is more to the story, going back to what $\delta W$ and $\delta L$ mean. These quantities are usually given as the "$1\sigma$" uncertainties of the quantities $W$ and $L$, meaning that the measured values of $W$ and $L$ will fluctuate within the uncertainties such that $68\%$ of the time, $W$ and $L$ will be between $W\pm\delta W$ and $L\pm\delta L$. (See the above discussion on gaussian intervals.) To be consistent, we would want to have a $\delta P_h$ that also describes the $1\sigma$ in $P_h$, however $\delta P_h=\delta W + \delta L$ clearly does not accomplish that.

The following shows this explicitly. We generate random numbers that are normally distributed around $8$ for the width $W$, and around $16$ for the length $L$, and we use $\sigma=1$ for both distributions. We calculate $P_h=L+W$ and plot this as well. The mean and standard deviation are calculated from the histogram and reported in the table above it. The fact that the standard deviations are not exactly at $1$ for the $W$ and $L$ histograms reflect the statistical fluctuations in making measurements based on the bin contents of the histogramming.

Mean$\sigma$
$W$
$L$
$P$

As you can clearly see, the standard deviation of $P=W+L$ does not equal the sum of the standard deviations of $W$ and $L$: $\sigma_{W+L} \ne \sigma_W+\sigma_L$. But this should be expected: it is not very probable that when $L$ fluctuates by the maximum, that $W$ will do the same, because $W$ and $L$ are uncorrelated. That is, the fluctuation in the measurement of $L$ does not have any effect on the fluctuation of the measurement of $W$, and so the two fluctuations are uncorrelated and independent. This is the key reason why simply adding the uncertainties does not work, and so we need a better way to form uncertainties (so called "errors").

To develop a way to add uncertainties, we can define the problem in the following way:

• We start with 2 sets of variables, $x_i$ and $y_j$
• We want to form a new variable that is a function of $x$ and $y$: $z = f(x,y)$.
• We know the uncertainties $\sigma_x$ and $\sigma_y$
• We know the probabilities $P(x_i)$ and $P(y_i)$
How do we form the uncertainty in $z$, called $\sigma_z$?

To keep it simple, let's start with a function that is linear, like $z=x+y$. Going back to the quation for the variance ($\ref{evar2}$), we have the following 3 equations: $$\sigma_x^2 = \sum_{i=1}^n P(x_i)\delta x_i^2\nonumber$$ $$\sigma_y^2 = \sum_{i=1}^n P(y_i)\delta y_i^2\nonumber$$ $$\sigma_z^2 = \sum_{i=1}^n P(z_i)\delta z_i^2\nonumber$$ where $\delta x_i=x_i-\bar{x}$, $\delta y_i=y_i-\bar{y}$, and $\delta z_i=z_i-\bar{z}$. We can easily calculate $\bar x$ and $\bar y$, and then it's not difficult to see that $\bar z = \bar x + \bar y$. Since we know that $z=x+y$, then $\delta z = \delta x + \delta y$, and we have the equation: $$\sigma_z^2 = \sum_{i,j=1}^n P(z_{ij})(\delta x_i + \delta y_j)^2\nonumber$$ Note that the sum over $ij$ means we are summing over the two distributions, $x_i$ and $x_j$, hence the peculiar notation. Let's drop this notation and just sum, keeping in mind that there is a dual sum here.

The quantity $P(z_{ij})$ is related to the probability of seeing any particular value of $z_i$, which involves the probability of seeing any particular value of $x_i$ and $y_i$. In the world of probability, probabilities multiply. For instance, if you have a particular $x,y$ pair that has probabilities of $\half$ each, then the probability of both will be given by $P_xP_y=1/4$. So we can write $$P(z_{ij})=P(x_i)\cdot P(y_j)\nonumber$$ The above equation for $\sigma_z^2$ becomes $$\sigma_z^2 = \sum P(x_i)P(y_j)(\delta x_i + \delta y_j)^2\nonumber$$ Expanding the square gives: $$\sigma_z^2 = \sum P(x_i)P(y_j)(\delta x_i^2 + 2\delta y_j\delta x_i + \delta y_j^2)\label{esum3}$$ There are 3 quantities here. The first is $$\sum P(x_i)P(y_j)\delta x_i^2\nonumber$$ which can be written as $$\sum P(x_i)\delta x_i^2\sum P(y_j)=\sum P(x_i)\delta x_i^2=\sigma_x^2\nonumber$$ Here we made explicit use of the fact that $\sum_{j=1}^n P(y_j)=1$ by definition. Similarly, the 3rd equation in our troika equation $\ref{esum3}$ above is: $$\sum P(y_j)\delta y_j^2\sum P(x_i)=\sum P(y_j)\delta y_j^2=\sigma_y^2\nonumber$$ The middle equation is actually pretty easy to understand: $$2\sum P(x_i)P(y_j)\delta y_j\delta x_i\ =2\sum P(x_i)\delta x_i\sum P(y_j)\delta y_j=0\nonumber$$ Each term in the sum is zero by definition of the mean $\bar x$ and $\bar y$. So, equation $\ref{esum3}$ becomes: $$\sigma_z^2 = \sigma_x^2 + \sigma_y^2\label{equad}$$ If we look at the above histograms of $W$, $L$, and $P=W+L$, we can see that indeed $\sigma_P^2 = \sigma_W^2 + \sigma_L^2$, and now we know that uncertainties for sums of variables add in quandrature. And what makes this happen is that the middle term of equation $\ref{esum3}$ is equal to 0, which comes from the fact that the individual $x_i$ and $y_j$ are all independent of each other, or in other words uncorrelated. So we should amend our rule and say that uncertainties in sums of uncorrelated variables add in quadrature.

We can now try to generalize equation $\ref{equad}$ to arbitrary functions $f(x,y)$. The equation to start with is $$\sigma_z^2 = \sum P(z)\delta f^2\nonumber$$ where $\delta f^2 = (f-\bar{f})^2$. We can expand $f$ around the mean $\bar{f}$ using the Taylor expansion: $$f(x,y) = f(\bar x,\bar y) + \frac{\partial f}{\partial x}\delta x + \frac{\partial f}{\partial y}\delta y + ...\nonumber$$ where we have stopped after 1st order in $\delta x$ and $\delta y$, and we keep in mind that $\partial f/\partial x$ is evaluated at $x=\bar{x}$ and the same for $\partial f/\partial y$. Note that $\bar f$ really is $f(\bar x,\bar y)$ to first order. This means we can then write $$\delta f^2 = (f-\bar f)^2 = \big(\frac{\partial f}{\partial x}\delta x + \frac{\partial f}{\partial y}\delta y\big)^2\nonumber$$ When we expand this squared quantity, we get: $$\sigma_f^2 = \sum P(x)P(y)\big[ \big(\frac{\partial f}{\partial x}\delta x\big)^2 + \big(\frac{\partial f}{\partial y}\delta y\big)^2 + \big(\frac{\partial f}{\partial x}\delta x\frac{\partial f}{\partial y}\delta y\big) \big]\nonumber$$ As we noted above, the quantities that are linear in $\delta_x\delta_y$ will vanish if $x$ and $y$ are uncorrelated. Expanding what's left (the first 2 parts of the above equations) and summing $\sum P(y)=\sum P(x)=1$, we are left with: $$\sigma_f^2 = \sum P(x)\big(\frac{\partial f}{\partial x}\delta x\big)^2 + \sum P(y)\big(\frac{\partial f}{\partial y}\delta y\big)^2\nonumber$$ One last thing to do: the quantity $\partial f/\partial x$ (and $y$) is evaluated at $\bar x$ (and $\bar y$), which means they are not a function of $x_i$ or $y_i$, which means you can bring them out of the sum. What's left is just $\sum P(x)\delta_x^2=\sigma_x^2$ (and same for $y$), which gives us: $$\sigma_f^2 = \big(\frac{\partial f}{\partial x}\delta x\big)^2 + \big(\frac{\partial f}{\partial y}\delta y\big)^2\label{eerrprop}$$ and again, the main thing here to remember is that this is correct if $x$ and $y$ are uncorrelated, and of course is only good to 1st order in the uncertainties. That means that if the uncertainties are of order $10\%$ of the values, then the correction to this equation would be of order $(10\%)^2 = 1\%$.

As an example, say that $f(x,y)=x/y$. Then $\partial f/\partial x = 1/y$ and $\partial f/\partial y=-x/y^2$, which gives us the relation $$\sigma_f^2 = \delta x^2/y^2 + x^2\delta y^2/y^4\nonumber$$ If we form $\sigma_f/f$, we get $$\frac{\sigma_f^2}{f^2} = \frac{\delta x^2}{x^2} + \frac{\delta y^2}{y^2}\nonumber$$

## The $p$-value

What we want to do in physics (and any other science) is to try to understand nature by coming up with theories, doing experiments, and seeing if the theories and experimental results agree. For instance, if we apply a force $F$ on a mass $m$ and measure the acceleration $a$, we could use the ratio $F/a$ and compare to the mass $m$. If they agree, then our theory $F=ma$ is valid. If they don't agree, it's back to the drawing board.

But the world is complicated, because any measurement will come with a corresponding uncertainty, and any particular value measured will come from some true value that has fluctuated. So even if our theory is correct, we will often find that $F/a$ is not equal to $m$ for any particular measurement! To be concrete, when we make a measurement, say $M_i$, we are thinking that there exists a true value, $T$, and that the difference between the two comes from fluctuations: $$M_i = T + \delta_i\label{xfluctuation}$$ We measure $M_i$, and we have an understanding of the uncertainty $\sigma$, but we don't know $T$ (we are trying to find it!) and so we don't know the size of the fluctuation $\delta_i$ that caused the true value to come out as $M_i$. Note that $\sigma$ is not the same as $\delta_i$: $\sigma$ is an educated guess about the average fluctuation, and how precise is our measurement $M$, whereas $\delta_i$ is the actual fluctuation from $T$ for measurement $i$.

Back to our example of trying to see if $F=ma$ holds, we can make a series of measurements where we place a force $F_i$ on a mass $m$, and measure the acceleration $a_i$. Even if each individual measurement fluctuates, it is probably the case that sometimes the fluctuations will be high, and sometimes low, but that on average the true physics will come out. So after some number of measurements, your data might look like the following plot.

The standard theory from Newton's laws of gravity tells us that $a=F/m$, so that a plot of $a_i$ vs $F_i$ should show a straight line with slope $1/m$. However, experimental uncertainty means that each measurement will fluctuate, so even though any given data point might deviate from the true linear relationship with slope $1/m$, the data should show a linear trend on average.

The plot above does look like there's a linear relationship, but you can imagine that there are many different slopes that can go through all the values, each with a different $\chi^2$ (see above). What we want to do is find the slope that minimizes the $\chi^2$, and compare that $\chi^2_m$ per DOF to 1 to get an idea of whether the theory is consistent with the data - the further away from 1, the worse the fit. So ultimately, to quantify the extent to which this minimum $\chi^2_m$ is consistent with our theory given our uncertainties, we need to come up with a probability that a fluctuation could give us this particular $\chi^2_m$, or larger. If that probability is small, then it means it's not so likely a fluctuation, which could indicate that the theory is not a good fit to the data. Of course "too small" is subjective, and a subject of great almost religious debate. This probability (that the data fit the theory) is called the $p$-value. What we want to do in this chapter is to figure out how to calculate the $p$-value, and related concepts (the "confidence level", and "significance"). Note that it is up to the experimenter, and the community of scientists, to agree on a cutoff value for $p$ to indicate agreement with the theory. In the world of biology, a $p$-value less than $0.05$ (5\%) is often the metric for whether the experiment is "significant" or not. But be careful here, because a $p$-value of $0.05$ means that if you do the experiment 20 times, then you should see this kind of fluctuation once, on average! That doesn't mean $0.05$ is not a good cutoff. Since a small $p$-value means that the $\chi^2_m$ (per DOF) is somewhat large, a $p$-value of $0.05$ is an indication that either your theory is looking like it might be incorrect (large numerator in the $\chi^2$) or your uncertainties are too small (small demoninator), and this could serve as a good guide for checking things as you go along.

To investigate how we can calculate a $p$-value, we go back to the discussion of the $\chi^2$ and $N_{dof}$, and consider an experiment with 2 measurements $x_1$ and $x_2$, with uncertainties $\sigma_1$ and $\sigma_2$. If we form the mean $\bar x=\half (x_1+x_2)$, we can then form the $\chi^2$ with two terms: $$\chi^2 = \frac{(x_1-\bar x)^2}{\sigma_1^2} + \frac{(x_2-\bar x)^2}{\sigma_2^2}\nonumber$$ The quantities $x_1$ and $x_2$ are both distributed normally about the mean with widths $\sigma_1$ and $\sigma_2$, and since we've already used $x_1$ and $x_2$ to form $\bar x$, our $\chi^2$ has 1 degree of freedom, which we can see explicitly by substituting for $x_2 = 2\bar{x}-x_1$ to get $$\chi^2 = \frac{(x_1-\bar{x})^2}{\sigma_1^2}\big(1-\frac{\sigma_1^2}{\sigma_2^2}\big)\nonumber$$

If we define $y \equiv (x_1-\bar{x})/\sigma_1$, then $y_1$ will be distributed normally about $y_1=0$ with width $1$, and we can write the probability distribution of $y$ as: $$P(y_1) = \frac{1}{\sqrt{2\pi}}e^{-y_1^2/2}\label{pofy}$$ and we can write the $\chi^2$ as: $$\chi^2 = y_1^2\times k\nonumber$$ where $k = 1-(\sigma_1/\sigma_2)^2, which is just a number$. So if we know $P(y)$, then we should be able to calculate $P(y^2)$ and that will give us $P(\chi^2)$ for 1 DOF. There are many ways to do the math here, so let's try the simplest: since $P(y)$ is a probability, and since probabilities all have to integrate to 1, we can write: $$\int_{-\infty}^{+\infty} P(y)dy = 1\nonumber$$ where $P(y)$ is given by equation $\ref{pofy}$ above. If we substitute $z\equiv y^2$, use $dz = 2ydy$, and note that the integral is symmetric over the range between $-\infty$ and $+\infty$, we have $$2\int_{0}^{+\infty} \frac{1}{\sqrt{2\pi}}e^{-z/2}\frac{dz}{2\sqrt{z}} = 1\nonumber$$ which means we can read off the probability: $$P(z) = \frac{1}{\sqrt{2\pi z}}e^{-z/2}\nonumber$$ This derivation works well for our simple case with 2 measurements $x_1$ and $x_2$, where we have 1 degree of freedom, but for more degrees of freedom, the complications due to the nature of probability set in, and the $\chi^2$ per degree of freedom probability distribution gets much more complex. Now we can calculate the $p$-value easily: just integrate $P(\chi^2)$ from your $\chi^2_m$ to infinity to get the probability that our $\chi^2$ came out at $\chi^2_m$ or larger. For completeness, here is how you calculate the $p$-value for various degrees of freedom ($p(N)$ where $N$ = DOF): $$p(\chi^2_m,1) = \int_{\chi^2_m}^\infty \frac{1}{\sqrt{2\pi z}}e^{-z/2}dz\label{p1}$$ $$p(\chi^2_m,N) = \int_{\chi^2_m}^\infty \frac{z^{(N/2-1)}e^{-z/2}}{2^{(N/2)}\Gamma(N/2)}\label{pN}$$ where $\Gamma(N)=(N-1)!$ for integers, and for non-positive integers is defined as: $$\Gamma(x) = \int_0^\infty y^{x-1}e^{-y}dy\nonumber$$ Below we see the distribution of $p$-value vs $\chi^2_m$ for the situation where you have 1 degree of freedom (see equation $\ref{p1}$ above).

You can see that you get a $p$-value of $0.05$ if your $\chi^2_m$ is around $4$ ($3.841$ to be exact) or greater. This is easy to understand: 1 degree of freedom means you have one more data point than parameters you are fitting for. Imagine you have 2 data points, and you are fitting for a single parameter, like $\mu$ as above. If each data point was about as far away from $\mu$ as the uncertainty $\delta_i$, then each would contribute $1.0$ to the $\chi^2$, and we would have $\chi^2_m~\sim 2$ ($2$ terms, each is $1.0$). For a $p$-value of $0.05$, however, the $\chi^2_m$ is around $4$, and so each of the 2 terms is approximately equal to $2$. Since each term is the square of the distance between the data point and $\mu$, then each term is $\sqrt{2}$ times the uncertainty $\delta_i$ from $\mu$. The gaussian probability that a quantity with $\mu=0$ and $\sigma=1$ would fluctuate to $1.4$ is given by $$P(\sqrt{2}) = \frac{1}{\sqrt{2\pi}}e^{-(\sqrt{2})^2/2} \sim 14.7\%$$ For 2 terms to both fluctuate by that amount, we would expect the probability to be $P(\sqrt{2})^2 \sim 2\%$. But the $p$-value tells you the probability that the $\chi^2$ for that number of degrees of freedom was $\chi^2_m$ or greater, which means each term fluctuated by on average $\sqrt{2}$ or more. There are so many ways to get a $\chi^2_m$ of $4$, so it should be no surprise that a $p=0.05$ is consistent with fluctuations on each term that gives a $\chi^2_m$ of around $3.84$.

In Excel, you can use the CHIDIST function to give you the $p$-value. You give it the $\chi^2_m$ and $N_{dof}$, and it integrates the probability between your value and infinity.

For the case where there are 3 data points and 1 parameter, the number of degrees of freedom is 2, and the 3 data points are all allowed to fluctuate independent of each other. If each data point fluctuated $\sqrt{2}$ relative to the uncertainty $\delta_i$, then each term would contribute 2, and the $\chi^2_m$ would be at around 6. In fact, the $\chi^2$ that gives a $p$-value of $p=0.05$ for 2 degrees of freedom is 5.99, and the reduced $\chi^2$ ($\chi^2/N_{dof}$) is around $3.0$ If we go to 4 data points, or 3 degrees of freedom, then we would expect $p=0.05$ would give a $\chi^2$ of around $8$, and in fact the real value is $7.81$ with a reduced $\chi^2$ of $2.6$. That the $\chi^2$ for $p=0.05$ is less than $8$ is because for so many points, you could have the situation where one fluctuations low and another high, and the low fluctuation pulls the $\chi^2$ down a bit. So as we go to a larger and larger number of degrees of freedom, we might expect that the reduced $\chi^2$ approaches $1$. This is exactly what we see in the following plot of the reduced $\chi^2$ vs the number of degrees of freedom that gives $p=0.05$ - it does tend towards $1.0$ as $N_{dof}$ increases to $100$ and beyond.

There's nothing all that special about $p=0.05$. If we used $p=0.01$ for our cutoff, then that means we would require a bigger fluctuation to refute the null hypothesis, but the reduced $\chi^2$ would still approach $1.0$ in the limit of a large number of data points, for the same reason that it did for $p=0.05$, namely that for so many points there are many ways to fluctuate. For instance, for $N_{dof}=50$, the $\chi^2$ for $p=0.05$ is $67.5$, and for $p=0.01$ it's $76.2$.

## Significance and Confidence Interval

Ultimately, what we want to do is to make a measurement, and find some way to quantify whether this measurement is "significant", that is, whether it adds up to anything, or not. And of course, this means that we have to factor into it what we expect. So we will need to define what significant means, relative to what we expect, and this leads us into the idea of confidence intervals, and the concept of "how many sigmas".

To be more specific, say we have a theory (let's call it the "standard theory") that predicts that an ensemble of experiments will yield a value $\mu$ for some quantity. Each measurement we make will yield a different $x_i$ with an uncertainty ("error") of $\delta_i$. We know how to make a $\chi^2$ that tells us how close each $x_i$ is to $\mu$ in terms of the uncertainty $\delta_i$:

$$\chi^2 = \sum_{i=1}^N \frac{(x_i-\mu)^2}{\delta_i^2}\nonumber$$

Conceptually, what we are doing here is using classical statistics and probability to quantify how well our theory ($\mu$) fits our data.

When we do such experiments comparing theory to data, if we find that the comparison is good, then this tells us that the theory holds some water. If on the other hand we find that the comparison is not so good, then this tells us 1 of 2 possible things, either:

1. our standard theory is not such a good representation of nature, and that some other theory might be better (that is, based on our standard theory, something new is happening) or
2. our theory is ok but we got unlucky and had a pretty big (and unlikely from a probability sense) fluctuation.

Note that in this case where the data and the standard theory are not so consistent, you can't know in principle whether item 1. or 2. is correct! A common thing to do is to assume the standard theory is ok, use the $\chi^2$ approach to find the minimum $\chi^2_m$, and calculate the probability that we should get such a value or larger for $\chi^2_m$ for the number of degrees of freedom we have, which is how we calculate the $p$-value: $$p = P(\chi^2_m,N_{dof})\nonumber$$ If the $p$-value comes out to be $0.05$ or less, then in some cultures people say that the data and theory are consistent. If the $p$-value is greater than $0.05$, then it means the standard theory is either "probably" not correct, or it's actually ok but we got unlucky and saw a large fluctuation. Sometimes people will call the standard theory the "null hypothesis", where the use of the word "null" means "not new". So a cut of $0.05$ is used such that for $p > 0.05$, we have consistency with the null hypothesis, and below we do not. However, we have to be careful: a $p$-value of $0.05$ means that there was only a 1 in 20 chance that the data fluctuated from what the theory says it should be (or bigger), and what we found. But $0.05$ is not so unlikely: if we did 20 different experiments comparing a correct theory to data, then on average 1 out of every 20 will show a fluctuation with a $p$-value of $0.05$. Of course all you have to do is decrease the $p$-value cutoff from $0.05$ to something smaller to designate that the null hypothesis is so unlikely that we are sure we are seeing something new.

Copywrite Drew Baden, Feb 13, 2017