Minimum mean squared error of uniform distribution Let $X_1,X_2,\ldots,X_n$ be i.i.d $\operatorname{Uniform}(-\theta,0)$.
Now consider all of the estimates of the form $S_\rho=\rho \hat{\theta}_\text{MLE}$.
I have to find which of these estimates has the minimum mean squared error.
I found that the MLE of $\theta$ is $\hat{\theta}_\text{MLE}=-x_{(1)}$.
I know that the mean squared error is $\operatorname{MSE}(\hat{\theta})=\operatorname E_{\hat{\theta}}[(\hat{\theta}-\theta)^2]$ and that the minimum (best) mean squared error is $\operatorname{MMSE} = \operatorname E[X\mid Y]$. However I wouldn't know how to apply these to $S_\rho=\rho \hat{\theta}_\text{MLE}$. Is there a better (more intuitive) way to see which estimate is the best without using these?
Is the best estimator simply the MLE?
This is from a past exam of 5 years ago and seems very foreign to me, so it is possible that this was never covered in class. Any hint is appreciated.
 A: You showed already that $\rho \hat{\theta}_{MLE} = - \rho X_{(1)}$, so the MSE is
$$E[(-\rho X_{(1)} - \theta)^2].$$
It suffices to compute $E [X_{(1)}^2]$ and $E[X_{(1)}]$.

You can compute the density of $X_{(1)}$ by
$$f(t) = \frac{d}{dt} P(X_{(1)} \le t) = -\frac{d}{dt} P(X_{(1)} > t) = - \frac{d}{dt} (-t/\theta)^n = (-1)^{n+1} n t^{n-1}/\theta^n,$$
so we have
\begin{align}
E[X_{(1)}] = (-1)^{n+1} \frac{n}{\theta^n} \int_{-\theta}^0 t^n \mathop{dt} = -\frac{n}{n+1} \theta.
\\
E[X_{(1)}^2] = (-1)^{n+1} \frac{n}{\theta^n} \int_{-\theta}^0 t^{n+1} 
\mathop{dt}
= \frac{n}{n+2} \theta^2.
\end{align}

Thus the MSE is
$$E[(-\rho X_{(1)}-\theta)^2] = \theta^2 \left(\frac{n}{n+2} \rho^2 - 2  \frac{n}{n+1} \rho + 1\right).$$
Now find $\rho$ that minimizes the quadratic on the right-hand side.
I believe it should be $\rho=\frac{n+2}{n+1}$.

Interpreting the answer / sanity check: The MLE $-X_{(1)}$ underestimates $\theta$ since $X_{(1)} \ge - \theta$ almost surely. Thus it makes sense that introducing a little inflation by multiplying by a factor $\rho > 1$ might improve the MSE. Of course finding the specific value of the optimal $\rho$ requires computation.
A: HINT
Do the mirror image problem of $U(0,\theta)$ and find what $\rho$ makes the estimator $\rho x_{(n)}$ have lowest expected mean squared error. The answer to this question will be the same $\rho$ by symmetry.
The PDF of the max is $\frac{nx^{n-1}}{\theta^n}$ for $0<x<\theta$ so $E(X_{(n)} )$ and $ E(X_{(n)}^2)$ can be readily calculated. 
Then we have $$E((\rho X_{(n)}-\theta)^2) = \rho^2E(X_{(n)}^2) -2\rho\theta E(X_{(n)})+\theta^2$$
So you just need to minimize this wrt $\rho.$
