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.



Poisson distribution and conditional expected value
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 $\mathbb{E}a = 24/8 = 3$.


