Given $f(x|\theta)= \frac{\mathrm{e}^{-\theta}\theta^x}{x!} , x = 0,1,2,…,\theta$ and $\mathcal{L}(a-\theta)^2$
a) Calculate $\mathcal{R}(\theta,d)$ for $d(x)=x$
b) Given the gamma density $\pi(\theta) = \frac{\lambda^\alpha\theta^{\alpha-1}\mathrm{e}^{-\lambda\theta}}{\Gamma(\alpha)}$. $\mathcal{L}>0, \lambda>0, \theta>0$.
Calculate $\mathcal{r}(\pi,d)$ for $d(x)=x$.
From what I know I guess:
$\mathcal{R}$ is Risk Function and $\mathcal{L}$ is Loss function.
This is an exercise problem in Introduction to Statistical Theory by Charles J. Stone, Paul Gerhard Hoel, and Sidney Charles Port. (Chapter 1 Exercise 7). Someone please help me understand and solve this problem.
Best Answer
a)
$\mathcal{R}(\theta, d) = \sum_{x=0}^{\infty} \mathcal{L}(\theta, a)f(\theta|x)$ - This is the formula to calculate Risk Function for discrete distributions.
Given, $\mathcal{L}(a-\theta)^2$ , we can substitute this in the formula
$ = \sum_{x=0}^{\infty} (x-\theta)^2 f(\theta|x)$
$= \sum_{x=0}^{\infty} (x^2+\theta^2-2x\theta) f(\theta|x)$
$= \sum_{x=0}^{\infty} x^2f(\theta|x) + \sum_{x=0}^{\infty} \theta^2f(\theta|x) - \sum_{x=0}^{\infty} 2x\theta(\theta|x) $
Actually if we look closely we can tell that $f(\theta|x)$ is Poisson Distribution. Therefore,
$= E[X^2] + \theta^2(1) - 2\theta E[X] $
$= \theta + \theta^2 + \theta^2 - 2\theta^2$
$= \theta$
$\mathcal{R}(\theta, d) = \theta$
b)
$r(\pi, d) = E[\mathcal{R}|\theta, d]$ - This is the formula to calculate mean risk
We calculated $\mathcal{R}(\theta, d) = \theta$, in section (a). Using that
$= \int_{0}^{\infty} \theta \pi(\theta) \, d\theta $
$= \int_{0}^{\infty} \frac{\lambda^{\alpha}\theta^{(\alpha +1)-1}\mathrm{e}^{-\lambda\theta}}{\Gamma(\alpha)} \, d\theta $
$= \frac{\alpha}{\lambda}\int_{0}^{\infty} \frac{\lambda^{\alpha + 1}\theta^{(\alpha +1)-1}\mathrm{e}^{-\lambda\theta}}{\Gamma(\alpha + 1)} \, d\theta $
$= \frac{\alpha}{\lambda} (1)$ -- (Because of Gamma Distribution - pdf)
$= \frac{\alpha}{\lambda}$
$r(\pi, d) = \frac{\alpha}{\lambda}$