Let $X_1, \dots, X_n$ a sample from the density distribution: $$ f(x) = 1/\theta \quad \theta < x < 2\theta, \quad \theta>0 $$ Found $ \hat\theta = \frac{2}{3}\bar X$ with method of moments, I calculate MSE of $\hat \theta$.
The proposed solution is $\frac{\theta^2}{27n}$ but I get $\frac{\theta^2}{27}$
What am I doing wrong? Where does the $n$ cames from?
Here is some help:
First note that: $MSE = E[(\hat{\theta} - \theta)^2] = Var(\hat{\theta}) - \left(E[\hat{\theta}] - \theta\right)^2$.
Now, compute $Var(\hat{\theta}) = Var\left(\frac{2}{3} \bar{X}\right) = \frac{4}{9}Var(\sum \frac{X_i}{n}) = \frac{4}{9n^2}\left( n \cdot Var(X_1) \right)$
where $Var(X_1)$ is the variance of $X_1 \sim uniform(\theta,2\theta)$
Compute also $E[\hat{\theta}] = E[\frac{2}{3}\bar{X}]$.
The remaining of the solution should be easy.
Observation: Note that $Var(X_1)$ is not the same of $Var(\bar{X})$ (it is probably from here that you are missing a $1/n$ in your solution).