# Limit of beta-binomial distribution is binomial

I am trying to understand the relationship between the beta-binomial and the binomial distribution. More specifically, I am trying to show that the limit of the beta-binomial distribution, with $p=a/(a+b)$ is binomial as $a+b$ goes to infinity. I am having trouble showing it. Any useful hints would be very helpful.

For this, I believe I should take the limit of the $\text{beta}(a,b)$ function as $a+b$ go to infinity. Does this exist? According to an answer below this does not exist. I am also hesitant to use the MGF as it seems nasty.

• The MGF (or better, the CF) is not as nasty as it might seem: it's a hypergeometric function, which means its power series has a nice form. However, you could just apply Stirling's approximation to the Gamma function to show directly that the probability mass function converges to the Binomial distribution. – whuber Jan 10 '18 at 16:59
• I am trying through sterling's formula. For B(a,b) I seem to get something sensible. However for B(k+a,n-k+b) I don't see how this helps. – DanRoDuq Jan 10 '18 at 18:05

There are at least two ways of seeing this.

The urn interpretation of the distribution can be shown to be

The beta-binomial distribution can also be motivated via an urn model for positive integer values of $\alpha$ and $\beta$, known as the Polya urn model. Specifically, imagine an urn containing $\alpha$ red balls and $\beta$ black balls, where random draws are made. If a red ball is observed, then two red balls are returned to the urn. Likewise, if a black ball is drawn, then two black balls are returned to the urn. If this is repeated $n$ times, then the probability of observing k red balls follows a beta-binomial distribution with parameters $n$,$\alpha$ and $\beta$.

However if $n$ is negligible compared to the number of balls in the urn, adding a few balls back to the urns makes negligible difference for the next draw. It follows that the distribution is simply that of drawing with replacement, which is binomial.

From an algebraic viewpoint, the distribution is $${n \choose k} \frac{B(k + \alpha, n - k + \beta)}{B(\alpha, \beta)} .$$

$$B(x + 1, y) = B(x, y) \frac{x}{x + y}, \\ B(x, y + 1) = B(x, y) \frac{y}{x + y}$$

Specifically,

$$B(i + \alpha, n - k + \beta) = B(i - 1 + \alpha, n - k + \beta) \frac{i - 1 + \alpha}{i - 1 + n - k + \alpha + \beta},$$

and for large $\alpha, \beta$, taking into account the Taylor series of $\frac{1}{1 + x}$:

$$\frac{i - 1 + \alpha}{i - 1 + n - k + \alpha + \beta} = \frac{i - 1 + \alpha}{\alpha} \frac{\alpha}{(\alpha + \beta) \left(1 + \frac{i - 1 + n - k}{\alpha + \beta}\right)} \sim \frac{i - 1 + \alpha}{\alpha} \frac{\alpha}{(\alpha + \beta)} \left(1 - \frac{i - 1 + n - k}{\alpha + \beta}\right) \sim \frac{\alpha}{(\alpha + \beta)} .$$

Continuing this,

$$\frac{B(k + \alpha, n - k + \beta)}{B(\alpha, \beta)} \sim \frac{B(\alpha, \beta)}{B(\alpha, \beta)} \left( \frac{\alpha}{\alpha + \beta}\right)^k \left( \frac{\beta}{\alpha + \beta} \right)^{n - k} ,$$ and the distribution is approximately binomial.

• I guess I am having one issue with this. Although for large he large $\alpha$ $\beta$ the approximation above makes intuitive sense, really for the limit to converge we should have the condition that $p=\alpha/(\alpha+\beta)$. With this in mind I am having trouble formally justifying the last approximation. For example, how can I show that $(\alpha+k)/(\alpha+k+\beta)$ tends to p. – DanRoDuq Jan 10 '18 at 20:09
• @DanRoDuq I expanded this a bit in the answer. – Ami Tavory Jan 10 '18 at 21:33