MSE of uniform distribution I would like to know how to calculate the MSE for a Uniform Distribution on $(θ,2θ)$
I know that MSE is the variance of the Method of Moments Estimator (MME). I have found this to be $3θ/2$.
Then I think I should calculate the $Var(2/3X\bar)$. Assuming that the bias of the MME is $0$. I think the answer is $3(θ^2) / 48n$. However I am unsure if this is the correct computation.
 A: First of all, the mean squared error (MSE) does not apply to a distribution, but to an estimator.  Usually, we are interested in the estimator of a parameter of a distribution, but not always:  I could, for example, ask for the MSE of the maximum likelihood estimator of the variance of a parametric distribution.
Thus, your question is improperly posed.  The "MSE of a distribution" makes no sense.  You have to specify the estimator of interest, and it cannot be assumed that this estimator is a method of moments (MoM) estimator; moreover, it cannot be assumed that the estimator estimates a parameter.
If I make the assumption that you are interested in the MSE of the MoM estimator of the parameter $\theta$ from a continuous uniform distribution on $(\theta,2\theta)$, then we first have to determine what the MoM estimator is:  for a sample $\boldsymbol x = (x_1, \ldots, x_n)$ of IID observations from $$X \sim \operatorname{Uniform}(\theta,2\theta),$$ the sample mean is set equal to the expectation of $X$; i.e., $$\bar x = \frac{1}{n} \sum_{i=1}^n x_i = \operatorname{E}[X] = \frac{3\theta}{2}.$$  The formal calculation of $\operatorname{E}[X]$ is simple; we write $$\operatorname{E}[X] = \int_{x=\theta}^{2\theta} x \frac{1}{2\theta - \theta} \, dx = \left[\frac{x^2}{2\theta} \right]_{x=\theta}^{2\theta} = \frac{4\theta^2 - \theta^2}{2\theta} = \frac{3\theta}{2}.$$  Consequently, the MoM estimator of $\theta$ is $$\tilde \theta = \frac{2}{3}\bar x.$$  The MSE is given by $$\operatorname{MSE}[\tilde \theta] = \operatorname{Var}[\tilde\theta] + \operatorname{E}[\tilde\theta - \theta]^2;$$ that is, it is the sum of the variance and squared bias.  We have $$\operatorname{E}[\tilde \theta] = \frac{2}{3}\operatorname{E}[\bar x] = \frac{2}{3n} \sum_{i=1}^n \operatorname{E}[x_i] = \frac{2}{3n} \cdot n \frac{3\theta}{2} = \theta,$$ so the MoM estimator is unbiased, and the MSE equals the variance of $\tilde\theta$:  $$\operatorname{Var}[\tilde\theta] = \left(\frac{2}{3}\right)^2 \operatorname{Var}[\bar x].$$  Since the observations are independent, the variance of their sum is equal to the sum of their variances; i.e., $$\operatorname{Var}[\bar x] = \frac{1}{n^2} \sum_{i=1}^n \operatorname{Var}[x_i] = \frac{1}{n} \operatorname{Var}[X] = \frac{\theta^2}{12n}.$$  The calculation of this variance is simple:  $$\operatorname{Var}[X] = \operatorname{E}[X^2] - \operatorname{E}[X]^2 = \int_{x=\theta}^{2\theta} \frac{x^2}{\theta} \, dx - \left(\frac{3\theta}{2}\right)^2 = \left[\frac{x^3}{3\theta}\right]_{x=\theta}^{2\theta} - \frac{9\theta^2}{4} = \frac{\theta^2}{12},$$ as claimed.  We therefore conclude $$\operatorname{MSE}[\tilde\theta] = \frac{4}{9} \operatorname{Var}[\bar x] = \frac{4}{9} \cdot \frac{\theta^2}{12n} = \frac{\theta^2}{27n}.$$
