Baysian Inference with beta-binomial model with different number of trials

Question

Let's imagine that we have created and estimated a posterior with a beta prior and binomial likelihood function. The posterior is given by $p(\mu |D) = Beta(2,99)$. Now, I want to do inference with that model. But the number of trials has changed and I want to know the probability of having 3 success from 40 trials. Can I still use that estimated model, i.e., will just plugging in $\mu$ from $Beta(2,99)$ into $Binom(3|\mu, 40)$ work?

I was looking at a Bernoulli example and there it is clear. We have Bernoulli distribution parameterized as $Bern(x| \mu)$ so just one parameter $\mu$. In the Binomial case it is $Binom(x|\mu, N)$ so N is also playing a role.

Answer 1

You observed some Binomial data and ended up with a posterior of $\mathrm{Beta}(2,99)$ for your parameter $\mu$, which I assume is the probability of success.

If you wanted to give a point estimate for $\mu$, you could choose to use the posterior mean, which is $\hat{\mu}=2/(2+99)=0.0198$ (or, alternatively, the posterior median or mode).

You are now going to observe a further 40 trials; let $X$ be the number of successes. You are interested in the probability of observing exactly 3 successes, which is equal to $$ \mathrm{P}(X=3|\mu)={40 \choose 3} \mu^3 (1-\mu)^{40-3} $$ While you could plug your point estimate into the formula, this is not quite right, as it does not take into account the uncertainty around $\mu$.

What you should do is compute $$ \int_0^1 \mathrm{P}(X=3|\mu) f(\mu) d\mu $$ where $f(\mu)$ is your posterior (Beta) density for $\mu$. In other words, you want to integrate $\mu$ out of the joint mass/density of $X$ and $\mu$.

Answer 2

Not sure if I follow you, but as I understand your question you observed $k$ successes in $n$ trials and used this data in a beta-binomial model. If your prior was $\mathsf{Beta}(\alpha, \beta)$, then the posterior is $\mathsf{Beta}(\alpha + k, \beta + n-k)$. Now, let's go one step back, your model is

$$\begin{align} \mu &\sim \mathsf{Beta}(\alpha, \beta) \\ X_n &\sim \mathsf{Bin}(\mu, n) \end{align}$$

where $X_n = \sum_{i=1}^n Y_i$ for $Y_i \underset{i.i.d.}{\sim} \mathsf{Bern}(\mu)$. In plain English, you assume that $\mu$ is the probability of "success" for $n$ independent Bernoulli trials. The sum of the ones and zeros from the Bernoulli trials $X_n$ makes a binomial distribution. If now you want to make a guess on the number of successes in $m \ne n$ independent Bernoulli trials, you can just plug in $\mu$ into the binomial distribution $\mathsf{Bin}(\mu, m)$. The trials are identical and independent, so it is like tossing a coin $m$ times instead of $n$, each individual toss behaves the same regardless of how many times you toss.

You can also look at it through the lens of linearity of expectation. The expected value of a single Bernoulli trial is $E[Y_i] = \mu$, for $n$ trials it is $E[nY_i] = n E[Y_i] = n\mu$ and for $m$ trials $m\mu$. Those correspond to the means of the binomial distributions.

So yes, you just plug in the estimated $\mu$.