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Let $\theta_1$ be an estimator for $\theta$. We call $\theta_1$ inadmissible if there exists another estimator $\theta_2$ such that $MSE(\theta_2) < MSE(\theta_1)$ for all possible values of $\theta$.

(a) Let $X_1,\cdots, X_n \sim U(0, \theta)$, and define two estimators for $\theta$: Let $\theta_1 = 2\bar{X}$, and let $\theta_2 = \max (X_1,\cdots,X_n)$. Does either estimator prove the other is inadmissible?

(b) Let $Y \sim Expo(\theta)$, and define two estimators for $\theta$: Let $\theta_1 = \frac{1}{X}$, and let $\theta_2 = 4$. Does either estimator prove the other is inadmissible?

What I have so far: For part a, $2\bar{X}$ is an unbiased estimator since the sample mean is an unbiased estimator of the population mean. So the $MSE$ of $2\bar{X}$ would just be the variance of $2\bar{X}$ which is just 2*(variance of sample mean). I am not sure about the $MSE$ of $\theta_2$. How would I figure out its variance and its bias?

For part b, I assume $\theta_2$ is in admissible because it is a constant. How do I compute its mean squared error? I assume it has no variance. What about $\theta_1$? How do I figure out its variance and bias?

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  • $\begingroup$ I think the definition of inadmissible is:Let θ1 be an estimator for θ. We call θ1 inadmissible if there exists another estimator θ2 such that $MSE(θ2) \leq MSE(θ1)$ for all possible values of θ and for some $\theta$ , $MSE(θ2) < MSE(θ1)$ $\endgroup$
    – Masoud
    Commented Jun 20, 2020 at 4:10
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    $\begingroup$ The first one is also discussed here: math.stackexchange.com/q/3237797/321264. In Exp$(\theta)$, is $\theta$ the rate or mean? $\endgroup$ Commented Jun 20, 2020 at 7:13

1 Answer 1

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Part a)

Let $X_1,\cdots ,X_n\sim U(0,\theta)$ and consider

$$T_1=2\bar{X} ,\quad T_2=X_{(n)}=\max(X_1,\cdots , X_n)$$

Since $T_1$ is unbiased

$$MSE_{\theta}(T_1)=Var(T_1)=Var(2\bar{X})=4\frac{\frac{\theta^2}{12}}{n} =\frac{\theta^2}{3n}.$$

From the fact that $\frac{X_i}{\theta} \sim Uniform(0,1)$ so $\frac{T_2}{\theta}\sim Beta(n,1)$ so $E(T_2)=\frac{n}{n+1} \theta$ and $Var(T_2)=\frac{n\theta^2}{(n+2)(n+1)^2}.$ So

$$MSE_{\theta}(T_2)=Var(T_2)+(E(T_2)-\theta)^2=\frac{n\theta^2}{(n+2)(n+1)^2}+\frac{\theta^2}{(n+1)^2} =\frac{\theta^2}{(n+1)^2}(\frac{2n+2}{n+2})=\frac{2\theta^2}{(n+1)(n+2)}$$

So for every $\theta$ $$\frac{MSE_{\theta}(T_2)}{MSE_{\theta}(T_1)} =\frac{6n}{(n+1)(n+2)}\leq 1$$ So $T_2$ dominate $T_1$ and hence $T_1$ is inadmissible.

Part b)

$S_1=\frac{1}{X}$ and $S_2=4$.

$S_1$ can not dominate $S_2$ for some $\theta$ near $4$ like $\theta=4$. Since for $\theta=4$ $MSE(S_2)=0$.

Let $\theta \in \Theta$, every Constant estimator $S=cte$ is admissible if $cte \in \Theta$. Since when $\theta=cte$ ,$MSE(S)=0$ and no estimator can dominate it.

Finally $S_1$ can not Dominate $S_2$ too.

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