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?

• 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 $\theta$ , $MSE(θ2) < MSE(θ1)$ Jun 20, 2020 at 4:10
• The first one is also discussed here: math.stackexchange.com/q/3237797/321264. In Exp$(\theta)$, is $\theta$ the rate or mean? Jun 20, 2020 at 7:13

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.