Find $\mathbb{E}[\lambda \mid X=n],~\text{where}~ X \sim \operatorname{Poisson}(\lambda), \lambda \sim \Gamma(a,b).$

Question

Let $X \sim \operatorname{Poisson}(\lambda)$ where $\lambda \sim \Gamma(a,b)$. Find $$\mathbb{E}[\lambda \mid X=n].$$

Thanks to StubbornAtom hint I have solved this problem.

Solution:

By definition: $$\mathbb{E}[\lambda \mid X=n] = \frac{\mathbb E\left[\lambda\mathbf1_{X=n}\right]}{\mathbb P(X=n)}.$$

So we have to find distribution of random variable $X$ and $\mathbb E\left[\lambda\mathbf1_{X=n}\right].$ Let's start by finding distribution of $X$. We have

\begin{align} P(X=n) &= \mathbb{E}[\mathbb{E}[\mathbf1_{X=n} \mid \lambda]] = \int_{0}^{\infty} P(X=n \mid \lambda = x) \mu_{\lambda}(dx) \\ &= \int_{0}^{\infty} e^{-x}\frac{x^n}{n!}\frac{b^a}{\Gamma{(a)}}x^{a-1}e^{-bx}dx \\ &= \frac{b^a}{\Gamma{(a)}n!}\int_{0}^{\infty} e^{-x(1+b)}x^{n+a-1}dx \end{align} substitut $t=x(1+b)$ \begin{align} &\frac{b^a}{\Gamma{(a)}n!}\int_{0}^{\infty} e^{-t}\big(\frac{t}{1+b}\big)^{n+a-1}\frac{1}{1+b}dt = \frac{b^a}{\Gamma{(a)}n!} \frac{\Gamma{(n+a)}}{(1+b)^{n+a}} \\ &= \frac{b^a}{a!n!} \frac{a(a+n-1)!}{(1+b)^{n+a}} = \frac{(a+n-1)!}{(a-1)!n!}\big(\frac{b}{b+1}\big)^{a}\big(1-\frac{b}{b+1}\big)^{n} \\ &= {{n+a-1}\choose{n}}\big(\frac{b}{b+1}\big)^{a}\big(1-\frac{b}{b+1}\big)^{n}. \end{align}

Now

\begin{align} \mathbb{E}[\lambda \mathbf1_{X=n}]&=\mathbb{E}[\mathbb{E}[\lambda \mathbf1_{X=n}\mid \lambda]] \\&=\mathbb{E}[\lambda\mathbb{E}[\mathbf1_{X=n}\mid \lambda]] \\&= \int_{0}^{\infty} xe^{-x}\frac{x^n}{n!}\frac{b^a}{\Gamma{(a)}}x^{a-1}e^{-bx}dx \\&= \frac{b^a}{\Gamma{(a)n!}}\frac{\Gamma{(n+a+1)}}{(b+1)^{n+a+1}} \\& = \frac{(a+n)!}{(a-1)!n!}\frac{b^a}{(b+1)^{n+a+1}}. \end{align} Hence

\begin{align} \mathbb{E}[\lambda \mid X=n] &= \frac{\mathbb E\left[\lambda\mathbf1_{X=n}\right]}{\mathbb P(X=n)} \\ &= \frac{(a+n)!}{(a-1)!n!}\frac{b^a}{(b+1)^{n+a+1}}\frac{(a-1)!n!}{(a+n-1)!}\frac{(b+1)^{n+a}}{b^a} \\ &= \frac{a+n}{b+1}. \end{align}

Answer 1

You are given that the random variable $X$ conditioned on the random variable $\Lambda$ has a Poisson distribution where $\Lambda$ itself has a Gamma distribution.

As mentioned in comments, you have by definition $$\mathbb E\left[\Lambda\mid X=n\right]=\frac{\mathbb E\left[\Lambda\mathbf1_{X=n}\right]}{\mathbb P(X=n)}$$

Now you can use the law of total expectation:

\begin{align} \mathbb E\left[\Lambda \mathbf1_{X=n}\right]&=\mathbb E\left[\mathbb E\left[\Lambda \mathbf1_{X=n}\mid \Lambda\right]\right] \\&=\mathbb E\left[\Lambda\mathbb E\left[ \mathbf1_{X=n}\mid \Lambda\right]\right] \\&=\cdots \end{align}