I have this problem
Given a loss function
\begin{align}l(a,\theta) = |a-\theta|\end{align} for a ∈ A = {4,5,6} and $\theta$ ∈ $\theta$ = {4,5,6}
Assume the prior distribution is
\begin{align}Pr(\theta=i)=\pi_i=\begin{cases}
0.5, & \text{for i=4}.\\
0.3, & \text{for i=5}.\\
0.2, & \text{for i=6}.\\
\end{cases}\end{align}
How would I find the posterior distribution of $\theta$ given a sample of 3 observations on a RV X: $X_1=x_1=4,X_2=x_2=0,X_3=x_3=3$. Note that $X_1,X_2,X_3$ are IID Poisson so
\begin{align}
f(x_i|\theta=\frac{e^{-\theta}\theta^{x_i}}{x_i!}) for x_i=0,1,2,..
\end{align}
How would I find the posterior distribution of θ?
Thoughts so far:
I have done other examples where the prior is a continuous distribution and applied it in the formula
$f(\theta|x)=\frac{\pi(\theta)f(x|\theta)}{f(x)}$, but I am not sure regarding a discrete prior and what to do with the 3 sample observations
On a side note, would the prior distribution be the natural conjugate for $\theta$?
Best Answer
If the prior distribution $\pi$ of $\theta$ and the conditional distributions $p(\ \mid\theta)$ of the observations are discrete, the posterior distribution $q(\ \mid x)$ of $\theta$ given some observations $x=(x_i)$ is $$ q(\theta\mid x)\propto p(x\mid\theta)\pi(\theta)=\pi(\theta)\prod_ip(x_i\mid\theta), $$ that is $$ q(\theta\mid x)=\frac{\pi(\theta)}{z(x)}\prod_ip(x_i\mid\theta),\quad z(x)=\sum_\alpha\pi(\alpha)\prod_ip(x_i\mid\alpha). $$ In your case, the sum over $\alpha$ has three terms for $\alpha$ in $\{4,5,6\}$ and the product over $i$ has three terms for $x_i$ in $\{0,4,3\}$.
Can you carry on from here? (Note that the loss function is irrelevant.)