# Poisson-Gamma conjunction - calculating posterior [duplicate]

How to calculate posterior distribution step-by-step while given:

• some observed numbers of customers from the last days
• that number of clients is distributed by Poisson($$\lambda$$) ($$\lambda$$ is not given)
• that prior distribution of $$\lambda$$ is Gamma(a, b)

And after that, how to calculate ptobabilities of getting x number of clients the next day?

• Next time, please add the "self-study" flag and show us what you've done and where you're stuck. This is not a site for answering homework questions, and, if it's self-study, we prefer to help you work through the problem rather than just give you the answer. Nov 18, 2019 at 14:40

Let $$X_1,\dots,X_n \mid \lambda \sim \mathcal{P}(\lambda)$$

The likelihood of this model is then \begin{align*} L(\lambda ; X) &= \prod_{i=1}^{n} e^{-\lambda} \frac{\lambda^{X_i}}{X_i!} \\ &= e^{-n \lambda} \frac{\lambda^{ \sum X_i}}{ \prod X_i!} \end{align*}

The prior on $$\lambda$$ being a $$\mathcal{G}(a,b)$$, we have $$p(\lambda ; a,b) = \frac{b^a \lambda^{a-1} e^{-b \lambda }}{\Gamma(a)}$$

From Bayes formula the posterior is \begin{align*} p( \lambda \mid X ;a,b) &\propto p(\lambda ; a,b) L(\lambda ; X)\\ &\propto \lambda^{\sum X_i + a-1} e^{-\lambda(n+b)} \end{align*} The posterior distribution of $$\lambda$$ is then $$\mathcal{G}(\sum X_i + a,n+b)$$

One interesting property of the Gamma-Poisson mixture is that the marginal distribution is Negative Binomial. That is, given a particular value of $$\lambda$$, $$X \mid \lambda$$ will follow a Poisson distribution. But if we average over the distribution of $$\lambda$$, then the marginal distribution of $$X$$ is Negative-Binomial.

From the section 4 this pdf , if $$\lambda \sim \mathcal{G}(a,b)$$ the marginal distribution of $$X$$ is $$\mathcal{NB}(a, \frac{1}{b+1})$$.

Thus since in our case we have $$\lambda \sim \mathcal{G}(\sum X_i + a,n+b)$$, the posterior predictive distribution of $$X$$ knowing $$(X_1,\dots,X_n)$$ is $$\mathcal{NB}(\sum X_i + a, \frac{1}{1+n+b})$$

And you can use that to compute the probability of getting $$k$$ customers :

$$\mathbb{P}\big (X_{n+1} = k \mid (X_1,\dots,X_n) \big) = \binom{k+\sum X_i + a-1}{k}\Big(1-\frac{1}{1+n+b}\Big)^{\sum X_i + a}\frac{1}{(1+n+b)^k}$$ which is the probability mass function of a Negative Binomial distribution.