Good morning,
I have a problem regarding the MSE of the estimator of the population mean, $$\hat\theta=\frac{1}{n}\sum_{i=1}^n X_i=\bar X$$ for which I want to compute the MSE. I know that it is an unbiased estimator, so that the MSE equals the variance. For the $MSE(\bar X)$ I receive $$MSE(\bar X)=E[(\bar X-\mu)^2]=\frac{\sigma^2}{n}$$ which corresponds whith the solution in my textbook. Now I tried to figure out $$MSE(\frac{1}{n}\sum_{i=1}^n X_i)$$ which of course should have the same variance. Unfortunately it seems that I am making a mistake, which I cannot figure out. I proceeded as follows $$MSE(\frac{1}{n}\sum_{i=1}^n X_i)=E[(\frac{1}{n}\sum_{i=1}^n X_i-\mu)^2]$$ $$=E[\frac{1}{n^2}\sum_{i=1}^n (X_i)^2-2\mu\frac{1}{n}\sum_{i=1}^n X_i+\mu^2]$$ $$=\frac{1}{n^2}\sum_{i=1}^n E(X_i^2)-\mu^2$$ $$=\frac{1}{n^2}\sum_{i=1}^n(\sigma^2+\mu^2)-\mu^2$$ $$=\frac{1}{n}(\sigma^2+\mu^2)-\mu^2\neq\frac{\sigma^2}{n}$$ Thanks!
Assume that $X_1,...,X_n$ are i.i.d sample with $\mathbb{E}X=\mu$ and $var(X)=\sigma^2_X$. Next, observe that $\bar{X}_n$ is an unbiased estimator of $\mu$, hence $$ MSE(\bar{X}_n)=var(\bar{X}_n)=\frac{1}{n^2}\sum var(X)=\frac{n}{n^2}\sigma^2_X=\frac{\sigma^2_X}{n}. $$