Let $X$ have the distribution $N(\theta,1)$ where $\theta \ge 0$.
Is $T=X$ an admissible estimator with respect to the mean squared error?
Construct an estimator that respects the assumption $\theta \ge 0$.
Compare the maximum risk of the estimator in 2. and $T$.
I'm rather puzzled about 2 and hence 3: It asks about some estimator, but I guess that I shouldn't choose any arbitrary estimator, but one that is suggested by 1?
What estimator could be meant? I guess I don't know because I can't properly solved 1. So, about that:
The Risk is given by $$R(\theta,T)=E[(\theta-T)^2].$$ I want to find out if there is an estimator $T'$ such that we have $$R(\theta,T') \le R(\theta,T)$$ for all $\theta \in \Theta$ and $$R(\theta,T') < R(\theta,T)$$ for at least one $\theta \in \Theta$.
In this case, $T$ would not be admissible (I hope this is the correct English term). Sadly, I can't find such $T'$, nor prove that it doesn't exist. I tried to compute the risk of $T$ like this:
We have $f_X(x)=\frac{1}{2\pi}\exp(-\frac{1}{2}(x-\theta)^2).$ So $$E_\theta[(\theta-X)^2]=E_\theta[\theta^2]-2E_\theta[\theta X]+E_\theta[X^2]=\theta^2-2\theta\theta+\theta^2+1=1.$$
I can treat $\theta$ as a constant, right? It seems to make sense intuitively, too, given that $\sigma^2=1$. But how do I proceed from here?
The idea is that since you know that $\theta \geq 0$, your estimator should take this in consideration. Here, $T=X$ is not admissible because $T'= \max \{X, 0 \}$ has a smaller or equal MSE for all $\theta \geq 0$.