# A constant as an admissible estimator

This is a homework question so I would appreciate hints. I believe I have the first part correct, but I fail to see how the second part is different.

Assume square error loss, $L(\theta ,a)=(\theta - a)^2$.

1. When $X|\theta \sim N(\theta , \sigma^2)$, show $\delta (X) = c$, where $c$ is a constant, is an admissible estimator.

Since $\delta$ is not random, the risk is

$$R(\theta,\delta)=E_{\theta}^{X}(\theta-\delta)^2=(\theta-c)^2$$

which is zero for $\theta=c$. Suppose there exists an estimator $\eta$ such that $R(\theta,\eta)\leq R(\theta,\delta)$ for all $\theta$, and $R(\theta,\eta)<R(\theta,\delta)$ for some $\theta$. Then

$$R(c,\eta)=E_{\theta}^{X}(c-\eta)^2=0$$

and $\eta=c$ with probability 1. Then any estimator that dominates $\delta$ must be a constant a.s. and is then R-equivalent to $\delta$.

1. When $X|\theta \sim U(0,\theta)$, show $\delta (X) = c$, where $c$ is a constant, is an inadmissible estimator.

I feel like the same proof as before holds. The only difference in the questions is that the support of a $U(0,\theta)$ distribution depends on $\theta$, but I don't see how that is important here.

[...] Then $$R(c,\eta)=E_{\theta}^{X}(c-\eta)^2=0$$ and $\eta=c$ with probability 1.
For the second problem, consider the estimator $\eta = \max\{c, X_{(n)}\}$, where $X_{(n)}$ is the largest value in the sample.
• For your first comment, if $f(x)$ is the $N(\theta,\sigma^2)$ density, then $E_{\theta}^{X}(c-\eta_{X})^2=\int(c-\eta_{X})^{2}f(x)dx$. Since $f(x)>0$ for all $x\in \mathbb{R}$, and the integral is zero, then the term $(c-\eta{X})^2$ must be zero except on a set of lebesgue measure zero. Well done on the choice of $\eta$ for the second case. I am going to write up that part right now. – caburke Jan 23 '13 at 0:33