# 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.

## 1 Answer

In your argument for the first problem you write

[...] Then $$R(c,\eta)=E_{\theta}^{X}(c-\eta)^2=0$$ and $\eta=c$ with probability 1.

I don't see how this follows. You should be worried about the fact that you haven't used any properties of the Normal distribution. Given that your argument as written does in fact apply just as well to the second problem (where the result it would prove is false) as this problem, your argument is obviously flawed.

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