# Find the posterior distribution of θ

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 $$\theta$$?

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 $\{0,4,3\}$.
• @ Did: What is the "loss function"? Why are the left side of the bars in $(|\theta)$ and $q(|x)$ left out? – Parseval Dec 3 '18 at 13:01
• @Parseval "Loss function" is referring to a previous version of the question, which I just reinstated. // The notation $q(\ \mid x)$ stands for the function $\theta\mapsto q(\theta\mid x)$, as opposed to a value $q(\theta\mid x)$ of this function at a specific argument $\theta$. – Did Dec 3 '18 at 14:03