Exercise: Suppose $X_1,\ldots,X_n|\Theta = \theta \stackrel{iid}{\sim} \operatorname{Pois}(\theta)$ and $\Omega\sim\operatorname{Ga}(\alpha, \beta)$. Derive the posterior distribution and compute the posterior mean.

What I've tried: I know that the posterior distribution is equal to $$f_{\Theta|X}(\theta|x) = \dfrac{f_{X|\Theta}(x|\theta)\,f_\Theta(\theta)}{\int f_{X|\Theta}(x|\theta)\,f_\Theta(\theta)d\theta}$$ so with $f_{X|\Theta}(x|\theta)\,f_{\Theta}(\theta) = \dfrac{\theta^{\sum x_i}e^{-n\theta}}{\prod x_i !}\dfrac{\beta^\alpha}{\Gamma(\alpha)}\theta^{\alpha-1}e^{-\beta\theta}$ this means that $$f_{\Theta|X}(\theta|x)=\dfrac{\dfrac{\theta^{\sum x_i}e^{-n\theta}}{\prod x_i !}\dfrac{\beta^\alpha}{\Gamma(\alpha)}\theta^{\alpha-1}e^{-\beta\theta}}{\displaystyle\int\dfrac{\theta^{\sum x_i}e^{-n\theta}}{\prod x_i !}\dfrac{\beta^\alpha}{\Gamma(\alpha)}\theta^{\alpha-1}e^{-\beta\theta}d\theta}.$$ I think this is some sort of Beta distribution. However, I'm not sure which one and how to get there.

Question: How do I solve this exercise?

Thanks in advance!

  • $\begingroup$ Brief summary of Answ by @Ceph (+1): Gamma prior on rate with Poisson data gives gamma posterior. Gamma prior and Poisson likelihood are 'conjugate' (mathematically compatible) so it is sufficient to look only at the kernel of posterior (pdf without const of integ) to recognize exact posterior distribution. (Hence not necessary to evaluate integral in denom of posterior, which is a const.) $\endgroup$
    – BruceET
    Dec 28, 2017 at 19:08

1 Answer 1


You've mostly got this; the key to the problem is to ignore irrelevant information. Right now, you're trying to keep track of too much information you don't need -- which makes the problem much harder than it needs to be.

The denominator of the RHS of your first equation is some constant that doesn't depend on $\theta$ -- it doesn't matter what constant it is, since ultimately you know it will be whatever constant is required to turn the numerator into a proper pdf. So instead of working with: $$ f_{\Theta|X}(\theta|x) = \frac{f_{X|\Theta}(x|\theta) f_\Theta(\theta) } {\int f_{X|\Theta}(x|\theta)f_\Theta(\theta)d\theta}$$ you should be working with $$ f_{\Theta|X}(\theta|x) \propto f_{X|\Theta}(x|\theta) f_\Theta(\theta) $$ that is, all you need to know is that the LHS is proportional to the RHS, i.e. they are equal up to some normalizing constant. As you have found: $$f_{\Theta|X}(\theta|x) \propto \theta^{\sum x_i}e^{-n\theta}\theta^{\alpha-1}e^{-\beta\theta}$$ Notice that I left off not only the denominator from the equation you found in your question, I left off a bunch of stuff from the numerator, too. Why? Because all the stuff I left out is constant with respect to $\theta$, and all we care about is the distribution of $\theta$. So all that other stuff is again just part of the normalizing constant for the pdf we care about.

Simplifying the above, we get: $$f_{\Theta|X}(\theta|x) \propto \theta^{\alpha+\sum x_i-1}e^{-(n+\beta)\theta}$$

Now, what kind of distribution has $f(x) = c \cdot x^{a} e^{-bx}$ for some constant $c$? (Hint: it's not Beta, as you suggested.) Once you know that, you can easily figure out what $c$ should be in this case -- but really, who cares what the normalizing constant is, since you will already know the exact distribution (including parameters) of $\Theta|X$. And once you have the distribution, it will be trivial to recover the posterior mean.

  • $\begingroup$ Thanks for your answer! However, I can't find/see what distribution $f(x) = c\cdot x^a e^{-b}$! Could you tell me which one it is? $\endgroup$
    – titusAdam
    Dec 30, 2017 at 10:29
  • $\begingroup$ You definitely can find this distribution, since you use two distributions in your question, and this is one of those two! :) $\endgroup$
    – Ceph
    Dec 31, 2017 at 0:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .