1
$\begingroup$

Assume you have $x_i \sim \operatorname{Bernoulli}(p_i)$ with $p_i \sim \operatorname{Beta}(\alpha,\beta)$.

I am exploring $Z=X_1+ \dots +X_n$

According to this page, it is $Z \sim \operatorname{BetaBinomial}(n,\alpha,\beta)$

but according to this page, with simulation, it is $Z \sim \operatorname{Binomial}(n,\frac{\alpha}{\alpha+\beta})$

Two different distributions. Which one is correct?

Also equally important, what is the difference in assumptions for each? So I know when to use them correctly and can simulate either one in the right context. Thanks!

$\endgroup$
0
6
$\begingroup$

Short summary: if the $p_i$s are independent, it's the binomial. If the $p_i$s are all equal, it's the beta-binomial.

By $X_i \sim \textrm{Bernoulli}(p_i)$, you must mean the conditional distribution $X_i\mid p_i \sim \textrm{Bernoulli}(p_i)$. The marginal distribution of $X_i$ (that is, the distribution obtained by averaging over different values of $p_i$s) is obtained, e.g., by noting that $X_i$ is Bernoulli, and computing the expectation using the tower law: \begin{equation} \mathbb{E}(X_i) = \mathbb{E}(\mathbb{E}(X_i \mid p_i)) = \mathbb{E}(p_i) = \frac{\alpha}{\alpha+\beta}. \end{equation} So, $X_i \sim \textrm{Bernoulli}(\frac{\alpha}{\alpha+\beta})$, for all $i$. However, this does not yet determine the distribution of $Z$ as you have not specified enough information to deduce the joint distribution of $X_i$s. Two additional things are needed:

  1. The conditional distribution of $X_i$s given the $p_i$s, $p(X_1,\ldots,X_n \mid p_1,\ldots,p_n)$.
  2. The joint distribution of the $p_i$s: $p(p_1,\ldots,p_n)$.

For the first part, I guess&assume you intend the $X$s to be conditionally independent given the $p$s. The difference of the two distributions mentioned in your question stems from the second point.

If we assume the $p$s to be mutually independent, then the $X$s will be mutually independent, too, as each $X$ depends on only one $p$ and the $X$s are conditionally independent given the $p$s. Then, $Z$ is just the sum of iid. Bernoulli random variables. But this is the definition of the binomial distribution, and thus indeed \begin{equation} Z \sim \textrm{Binomial}\left(n,\frac{\alpha}{\alpha+\beta}\right). \end{equation} Note that in this case it did not make much sense to define the $p$s in the first place: if each $p_i$ influences only the corresponding $X_i$, nothing is learned about the distribution of $p_i$ except via the value of $X_i$. Thus, the $p_i$s are useless in the sense that the exactly same model would have been easier to specify by just stating that $X_i$s are independent Bernoulli random variables with parameter $\alpha/(\alpha+\beta)$.

The beta-binomial distribution is actually defined so that there is only one $p$ parameter drawn from the beta distribution, and then all $X_i$s are Bernoulli with this $p$. This is obtained as a special case of your question by stating in my "step 2" that the joint distribution of the $p_i$s are all the same.

By defining some other dependence structures for the $p_i$s (not independent, but not constrained to be equal either), other distributions for $Z$ would be obtained, but I don't know if any of these have special names.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.