# Bayesian Statistics: what is it about?

First of all, if you are unfamiliar with Bayes’ Law, here is a very nice video that explains both the formula AND the concept. Yes, the end of the video goes into social commentary (and makes some interesting points) but the math before it is very good.

If you’ve already familiar with those ideas, you can start here.

Let’s start with an example from sports: basketball free throws. At a certain times in a game, a player is awarded a free throw, where the player stands 15 feet away from the basket and is allowed to shoot to make a basket, which is worth 1 point. In the NBA, a player will take 2 or 3 shots; the rules are slightly different for college basketball.

Each player will have a “free throw percentage” which is the number of made shots divided by the number of attempts. For NBA players, the league average is .672 with a variance of .0074.

Now suppose you want to determine how well a player will do, given, say, a sample of the player’s data? Under classical (aka “frequentist” ) statistics, one looks at how well the player has done, calculates the percentage ( $p$) and then determines a confidence interval for said $p$: using the normal approximation to the binomial distribution, this works out to $\hat{p} \pm z_{\frac{\alpha}{2}} \sqrt{n}\sqrt{p(1-p)}$

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Yes, I know..for someone who has played a long time, one has career statistics ..so imagine one is trying to extrapolate for a new player with limited data.

That seems straightforward enough. But what if one samples the player’s shooting during an unusually good or unusually bad streak? Example: former NBA star Larry Bird once made 71 straight free throws…if that were the sample, $\hat{p} = 1$ with variance zero! Needless to say that trend is highly unlikely to continue.

Classical frequentist statistics doesn’t offer a way out but Bayesian Statistics does.

This is a good introduction:

But here is a simple, “rough and ready” introduction. Bayesian statistics uses not only the observed sample, but a proposed distribution for the parameter of interest (in this case, p, the probability of making a free throw). The proposed distribution is called a prior distribution or just prior. That is often labeled $g(p)$

Since we are dealing with what amounts to 71 Bernoulli trials where p = .672 so the distribution of each random variable describing the outcome of each individual shot has probability mass fuction $p^{y_i}(1-p)^{1-y_i}$ where $y_i = 1$ for a make and $y_i = 0$ for a miss.

Our goal is to calculate what is known as a posterior distribution (or just posterior) which describes $g$ after updating with the data; we’ll call that $g^*(p)$.

How we go about it: use the principles of joint distributions, likelihood functions and marginal distributions to calculate $g^*(p|y_1, y_2...,y_n) = \frac{L(y_1, y_2, ..y_n|p)g(p)}{\int^{\infty}_{-\infty}L(y_1, y_2, ..y_n|p)g(p)dp}$

The denominator “integrates out” p to turn that into a marginal; remember that the $y_i$ are set to the observed values. In our case, all are 1 with $n = 71$.

What works well is to use the beta distribution for the prior. Note: the pdf is $\frac{\Gamma (a+b)}{\Gamma(a) \Gamma(b)} x^{a-1}(1-x)^{b-1}$ and if one uses $p = x$, this works very well. Now because the mean will be $\mu = \frac{a}{a+b}$ and $\sigma^2 = \frac{ab}{(a+b)^2(a+b+1)}$ given the required mean and variance, one can work out $a, b$ algebraically.

Now look at the numerator which consists of the product of a likelihood function and a density function: up to constant $k$, if we set $\sum^n_{i=1} y_i = y$ we get $k p^{y+a-1}(1-p)^{n-y+b-1}$
The denominator: same thing, but $p$ gets integrated out and the constant $k$ cancels; basically the denominator is what makes the fraction into a density function.

So, in effect, we have $kp^{y+a-1}(1-p)^{n-y+b-1}$ which is just a beta distribution with new $a^* =y+a, b^* =n-y + b$.

So, I will spare you the calculation except to say that that the NBA prior with $\mu = .672, \sigma^2 =.0074$ leads to $a = 19.355, b= 9.447$

Now the update: $a^* = 71+19.355 = 90.355, b^* = 9.447$.

What does this look like? (I used this calculator) That is the prior. Now for the posterior: Yes, shifted to the right..very narrow as well. The information has changed..but we avoid the absurd contention that $p = 1$ with a confidence interval of zero width.

We can now calculate a “credible interval” of, say, 90 percent, to see where $p$ most likely lies: use the cumulative density function to find this out: And note that $P(p < .85) = .042, P(p < .95) = .958 \rightarrow P(.85 < p < .95) = .916$. In fact, Bird’s lifetime free throw shooting percentage is .882, which is well within this 91.6 percent credible interval, based on sampling from this one freakish streak. ## Author: oldgote

I enjoy politics, reading, science, running, walking, (racewalking and ultrawalking) hiking, swimming, yoga, weight lifting, cycling and reading. I also follow football (college and pro), basketball (men and women) and baseball (minor league and college)