Tuesday, January 15, 2008

Are you a Bayesian or a Frequentist? (Or Bayesian Statistics 101)

OK, the previous post was actually a brain teaser given to me by Roy Radner back in 2004, when I joined Stern, in order to teach me the difference between Bayesian and Frequentist statistics. It actually illustrates nicely how the two techniques lead to different conclusions. For completeness, let me list the question again:

You have a coin that when flipped ends up head with probability $p$ and ends up tail with probability $1-p$. (The value of $p$ is unknown.)

Trying to estimate $p$, you flip the coin 14 times. It ends up head 10 times.

Then you have to decide on the following event: "In the next two tosses we will get two heads in a row."

Would you bet that the event will happen or that it will not happen?

One of the basic differences of Bayesian and Frequentists is how they treat the parameters. Using frequentist statistics (the statistics that I learned in school :-) we would say that the best (maximum likelihood) estimate for $p$ is $p=\frac{10}{14}$, i.e., $p \approx 0.714$.

In this case, the probability of two heads is $0.714^2 \approx 0.51$ and it makes sense to bet for the event.

Bayesian approach: $p$ is not a value, it's a distribution

A Bayesian approach, instead of considering only the ML estimate for $p$, it would treat $p$ as a random variable with its own distribution of possible values. The distribution can be defined by the existing evidence. The logic goes as follows. What is the probability of a given value of $p$, given the data?

$Pr\{ p | data \} = \frac{ Pr\{ data | p \} \cdot Pr\{ p \}} { Pr\{ data \} }$

Since the term $Pr\{ data \}$ is going to be the same for all values of p, we can for now ignore it. We will see that we can infer it later, as the probability distribution when integrated over all values of $p$ will need to be equal to 1.

The value $Pr\{ data | p \}$ is easy to compute using simple statistics. The result of the experiment is a binomial distribution and we have

$Pr\{ data | p \} = \binom{14}{10} \cdot p^{10} \cdot (1-p)^{4}$

Again, ignoring the binomial coefficient $\binom{14}{10}$, we have:

$Pr\{ p | data \} \propto p^{10} \cdot (1-p)^{4} \cdot Pr\{ p \}$

where the $\propto$ symbol means proportional.

Now, the factor $Pr\{ p \}$, the prior distribution, comes into play. A very convenient prior distribution for this scenario (also known as the conjugate prior of the binomial) is the Beta distribution, $Beta(p;a,b)$ defined as:

$Beta(p;a,b) = \frac{ \Gamma(a+b) } {\Gamma(a) \cdot \Gamma(b)} \cdot p^{a-1} \cdot (1-p)^{b-1}$

In this case, we have:

$Pr\{ p \} = \frac{ \Gamma(a+b) } {\Gamma(a) \cdot \Gamma(b)} \cdot p^{a-1} \cdot (1-p)^{b-1}$

Dropping once again the constant:

$Pr\{ p | data \} \propto p^{10} \cdot (1-p)^{4} \cdot p^{a-1} \cdot (1-p)^{b-1}$

$Pr\{ p | data \} \propto p^{(10+a-1)} \cdot (1-p)^{(4+b-1)}$

Now, what distribution is that? We need the normalizing factor to get the proper distribution. Looking at the form of the Beta distribution above, we can see that this is again a Beta distribution, and the normalizing factor is

$\frac{ \Gamma((10+a)+ (4+b)) } { \Gamma(10+a) \cdot \Gamma(4+b) } = \frac{1}{B(10+a, 4+b)}$

where B(x,y) is the Beta function (not to be confused with the Beta distribution!)

Therefore the distribution for the parameter $p$ is $Beta(p;a+10, b+4)$. If we assume that we know nothing about $p$, we can assume that the prior is a uniform distribution, and the uniform distribution is a special case of Beta, with $a=b=1$. In this case the distribution for $p$ is $Beta(p; 1+10, 1+4) = Beta(p; 11, 5)$, which has the following form:

Probability of "Two Heads in a Row"

Now, we need to compute the probability of the event "two heads in a row", given the data that we have, which is $Pr\{HH|data\}$. To simplify:
  • We do not update $p$ between the two tosses.
  • We assume conditional independence that, given $p$, the two tosses are independent.
Given that we will not update the value of $p$, we can break the process into two steps. First, we compute the possible values of $p$ as described above; then, for each value of $p$ we compute the probability of two heads in a row. These steps are encoded into the following integral computation:

$Pr\{HH|data\} = \int_0^1 Pr\{HH|p\} \cdot Pr\{p|data\} dp$

Given the conditional independence of the two tosses:

$Pr\{HH|p\} = [Pr\{H|p\}]^2 = p^2$

So, we have:

$Pr\{HH|data\} = \int_0^1 p^2 \cdot Pr\{p|data\} dp$

From above, we know that:

$Pr\{ p | data \}= \frac{ 1 } { B(10+a, 4+b) } \cdot p^{10+a-1} \cdot (1-p)^{4+b-1}$

By replacing this value, we get:

$Pr\{HH|data\} = \frac{ 1 } { B(10+a, 4+b) } \int_0^1 p^{(10+a-1)+2} \cdot (1-p)^{4+b-1} dp$

Solving the integral, we have:

$Pr\{HH|data\} = \frac{ B(10+a+2, 4+b) } { B(10+a, 4+b) }$

If we assume that we know nothing about $p$, we can assume that the prior is a uniform distribution, and the uniform distribution is a special case of Beta, with $a=b=1$. In this case, the probability of the event "two heads in a row" is $\frac{ B(13, 5) } { B(11, 5) } = 0.485$ and it makes sense to bet against the event. So, the Frequentist approach gives probability 51% and the Bayesian approach with uniform prior gives 48.5%. Kudos to Roy for coming up with example, and shame on me for screwing up the initial posting!

(Update based on Foster's comment below: instead of using the uniform distribution as a prior, we can be even more agnostic. In this case, we can use the Beta(0,0) distribution as a prior. Such a distribution corresponds to the case where any mean of the distribution is equally likely. In this case, the two approaches, Bayesian and frequentist give the same results.)

(Update 2 based on multiple comments below: I changed the example to be more "fully Bayesian". Initially, I was estimating the full distribution for $p$, but then, in order to compute the "two heads in a row" probability, I was collapsing $p$ into a point estimate, using its mean value. Now, with the help of the commenters, the example follows closer the Bayesian paradigm. I had to change the number of heads and tails for the example to still be valid. WolframAlpha was surprisingly easy to use for identifying numbers that will work for this example. The values (3 heads, 1 tails),  (5 heads, 2 tails), (8 heads, 3 tails), (try your own value of tails)... also work.)