Suppose we have two independent binomial distribution given p, i.e. $X_1|p \sim Bin(n_1, p)$, $X_2|p \sim Bin(n_2, p)$. We also know the prior distribution for p is $p \sim U(0,1)$. Now I would like to find the posterior distribution P|(X1, X2). Here is my calculation.

Our prior density for p is \begin{align*} \pi(p) = I(1 > p > 0) \end{align*}

The likelihood for $(X_1, X_2)$ given p is \begin{align*} f(x_1, x_2|p) &= P(X_1=x_1, X_2=x_2|p) \\&= P(X_1=x_1|p)P(X_2=x_2|p) \\&= {{n_1}\choose{x_1}}p^{x_1}(1-p)^{n_1-x_1} {{n_2}\choose{x_2}}p^{x_2}(1-p)^{n_2-x_2} \end{align*}

Thus, the posterior distribution is \begin{align*} \pi(p|x_1,x_2) &\propto f(x_1, x_2|p)\pi(p) \\&\propto p^{x_1+x_2}(1-p)^{n_1+n_2-x_1-x_2} \end{align*}

Thus, $P|X_1,X_2 \sim Beta(X_1+X_2+1,n_1+n_2-X_1-X_2+1)$

Finally, the posterior mean, i.e. the Bayes estimator for p is the mean of $Beta(X_1+X_2+1,n_1+n_2-X_1-X_2+1)$, \begin{align*} E(P|X_1,X_2) = \frac{X_1+X_2+1}{n_1+n_2+2} \end{align*}

I am just wondering if my calculation is correct. I saw examples from the textbook, there is the example for n iid Bernoulli(p) trial with $p \sim U(0,1)$ prior, and there is also the example for $X|p \sim Bin(n, p)$ with $p \sim U(0,1)$ prior. But I haven't seen an example of my derivation.

Suppose now $n_1=400$, $n_2=600$, $X_1=10$, $X_2=200$, then my Bayes estimator for p is $\frac{211}{1002}$. While if we only consider $X_1$ and $n_1$, the Bayes estimator for p is $\frac{X_1+1}{n_1+2} = \frac{11}{402}$, if we only consider $X_2$ and $n_2$, the Bayes estimator for p is $\frac{X_2+1}{n_2+2} = \frac{201}{602}$. The three Bayes estimators are quite different. That's why I doubt if my calculation makes sense.

For my case, $X_1|p \sim Bin(n_1, p)$, $X_2|p \sim Bin(n_2, p)$, the p in $Bin(n_1, p)$ is the same number as the p in $Bin(n_2, p)$, they not only share the same $U(0,1)$ distribution, but also are identical number, is my interpretation correct? Or they only share the same $U(0,1)$ distribution, but are not identical number.

  • $\begingroup$ Your posterior $Beta(X_1+X_2+1,n_1+n_2-X_1-X_2+1)$ looks correct and its mean is indeed $\frac{X_1+X_2+1}{n_1+n_2+2}$. Meanwhile your $n_1=400$, $n_2=600$, $X_1=10$, $X_2=200$ example is extreme in the sense that it is unlikely to occur if $p$ is the same in both cases, so you should not be at all surprised that using one of $X_1$ or $X_2$ would give a very different result compared to using both together $\endgroup$
    – Henry
    Commented Oct 4, 2022 at 1:14
  • $\begingroup$ Indeed, and the "extremity" observed here also is unrelated to the Bayesian route taken. The MLEs (ie, the sample proportions computed in the answer by jbowman below) are also vastly different. $\endgroup$ Commented Oct 4, 2022 at 9:17

1 Answer 1


To answer your last question first - the way you have written it, the $p$ is the same in the two distributions, so they share the same prior.

Your calculation is indeed correct; you've found the posterior distribution and posterior mean.

As for why the three estimators give different values - look at the sample proportions of $X_1$ and $X_2$ in your example. For $1$, you have 2.5% of the observations for which $X_1 = 1$; for $2$, you have 33% of the observations for which $X_2 = 1$. This is extremely unlikely to happen with $n_1 = 400$, $n_2 = 600$, and the same probability, to put it mildly. Given the widely different results of your two samples, your sample sizes in the hundreds, and the uninformative prior on $p$, it's natural that the estimators will be very different as well.

  • $\begingroup$ For the last question, I mean if p are also the identical number. For example, suppose we generate a random number 0.3 from U(0, 1), then p for $X_1$ and $X_2$ are all 0.3. Or we generate two random number 0.3, 0.5 from U(0,1), p for $X_1$ is 0.3, p for $X_2$ is 0.5. Thanks for your help! $\endgroup$
    – Mizzle
    Commented Oct 4, 2022 at 1:21
  • $\begingroup$ I would interpret it as being that $p$ is the identical number in the two distributions; the way the formulae are written, that would be the correct interpretation at any rate. $\endgroup$
    – jbowman
    Commented Oct 4, 2022 at 1:45

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