# help computing the beta likelihood when we only observe the number of successes and failures (not the latent probability of success)

Imagine we have a biased coin that generates heads with unknown probability $$\theta$$ where $$\theta$$ is drawn from a beta distribution with known parameters $$(\alpha, \beta)$$. Next imagine that we flip the coin some number of times and we get $$m$$ heads and $$n$$ tails. I'd like to compute the quantity $$p(m,n|\alpha,\beta)$$.

What I have so far is \begin{align} p(m,n|\theta) &= \begin{pmatrix}m+n\\n\end{pmatrix} \theta^m (1-\theta)^n \\ &\\ p(\theta|\alpha,\beta) &= \frac{\theta^{\alpha-1}(1-\theta)^{\beta-1}}{B(\alpha,\beta)} \\ &\\ p(m,n|\alpha,\beta) &= \int_0^1 p(m,n|\theta) \, p(\theta|\alpha,\beta)\,d\theta \\ &= \int_0^1 \begin{pmatrix}m+n\\n\end{pmatrix} \theta^m (1-\theta)^n \frac{\theta^{\alpha-1}(1-\theta)^{\beta-1}}{B(\alpha,\beta)} \, d\theta\\ &= \begin{pmatrix}m+n\\n\end{pmatrix}\bigl(B(\alpha,\beta)\bigr)^{-1} \int_0^1 \theta^{m+\alpha-1}(1-\theta)^{n+\beta-1}\,d\theta \\ &= \begin{pmatrix}m+n\\n\end{pmatrix} \frac{B(m+\alpha, n+\beta)}{B(\alpha,\beta)}. \end{align}

The final expression is pretty nice, but I'm wondering if it can be simplified even further. For instance, is it equivalent to a binomial distribution with some parameter $$\theta'$$?

• You'd have to show that the CDF, MGF or pmfs of the two candidate distributions are equal. These expressions are fussy, but it should be apparent that they are not equal. Intuitively, the beta-binomial has variance at least as large as the binomial, so they cannot be the same distribution.
– Sycorax
Commented Jan 15, 2022 at 4:15

The beta binomial and binomial distributions are distinct. The binomial distribution has mean $$kp$$ and variance $$kp(1-p)$$ where $$k$$ is the number of trials and $$p$$ is the probability of success.

The pmf that you have written is a beta-binomial distribution (as you know). The beta binomial distribution has mean \begin{align} \mu &=(n+m)\frac{\alpha}{\alpha+\beta} \\ &=q(n+m) \end{align} This is suggestive! Perhaps we can use the substitution $$q = \frac{\alpha}{\alpha+\beta}$$ and show that the beta-binomial distribution is secretly a binomial distribution.

Unfortunately, this plan falls apart when we look at the beta-binomial variance \begin{align} \sigma^2 &= (n+m)\frac{\alpha}{\alpha+\beta}\cdot\frac{\beta}{\alpha+\beta}\cdot\frac{\alpha+\beta+n+m}{\alpha+\beta+1} \\ &= q(n+m)(1-q)\frac{\alpha+\beta+n+m}{\alpha+\beta+1} \end{align} and the first three factors are exactly what we want to find in a binomial distribution... but the fourth factor is not equal to 1 for $$n + m > 1$$.