Poisson distribution and conditional expected value [closed]

Question

I have a task:

Lat's take independent variables $X_{i}$ with Poisson distribution $Poiss(a)$. Distribution of $a$ has density $p(a)=\frac{8}{3}a^3e^{-2a}, a\ge0$. Calculate: $E(a|X_1=3, X_2=2, X_3=5, X_4=5, X_5=1, X_6=4)$.

I tried to calculate this in this way: $E(a|X_1=3, X_2=2, X_3=5, X_4=5, X_5=1, X_6=4)=\int{af(a|X_1=3, X_2=2, X_3=5, X_4=5, X_5=1, X_6=4)da}= \int{a \frac{f(X_1=3, X_2=2, X_3=5, X_4=5, X_5=1, X_6=4|a) \cdot f(a)}{P(X_1=3, X_2=2, X_3=5, X_4=5, X_5=1, X_6=4)}}da$, but i do not know how to calculate $P(X_1=3, X_2=2, X_3=5, X_4=5, X_5=1, X_6=4)$. Could you help me in this way or show other solution?

Thanks in advance.

Answer 1

Your problem seems to be a special case of https://web.stanford.edu/class/stats200/Lecture20.pdf about prior and posterior distributions in Bayesian analysis. By assumption $X_1,\ldots,X_6 \sim Poisson(a)$ and $a \sim \Gamma(\alpha,\beta) = \Gamma(4,2)$, the a priori distribution. Then with $s = x_1+\ldots+x_n$ where $x_1 = 3, \ldots , x_6 = 4$, hence $n = 6$ we get that the a posteriori distribution of $a$ is $\Gamma(s+\alpha,n+\beta) = \Gamma(s+4,n+2) = \Gamma(20+4,6+2) = \Gamma(24,8)$. In particular $\mathbb{E}a = 24/8 = 3$.