variance of estimator $\hat{\mu}_1=\bar{X_1}$

Question

From a population with mean $\mu$ and variance $\sigma^2$, an independent random sample of size $n_1$ is extracted. The mean sample is $\bar{X_1}$. The next estimator is proposed: $$\hat{\mu}_1=\bar{X_1}$$

a)Find the expected value of the estimator.

b)Find the variance of the estimator.

For a):

$$E[\hat{\mu}_1]=E[\bar{X_1}]=E[\frac{1}{n_1}\sum_{i=1}^{n_1} X_i] =\frac{1}{n_1}\sum_{i=1}^{n_1}E[X_i]=\frac{n_1}{n_1}\mu=\mu$$

For b):

$$Var(\hat{\mu}_1)=E[\hat{\mu}_1^2]-E[\hat{\mu}_1]^2=E[\hat{\mu}_1^2]-\mu^2$$

When I develop $E[\hat{\mu}_1^2]$ I get to $$\frac{1}{n_1^2}\sum_{i=1}^{n_1}E[X_i^2]$$ and from there I don't know how to proceed. I know that $Var(\bar{X_1})=\frac{\sigma^2}{n_1}$, but I wonder if I can get to the same result using the expected value.

Answer 1

Seems this thread got abandoned without a resolution but for those who come here seeking answers.

$Var(\overset{\hat{}}{\mu})= Var(\frac{1}{n}\sum_{i=1}^{n}X_{i})$ $=\frac{1}{n^2}\sum_{i=1}^{n}Var(X_{i})=\frac{1}{n^2} n \sigma^2=\frac{\sigma^2}{n}$.

This of course assumes an independent random sample of $X_i$

Answer 2

I just stumbled upon this post. Here is an answer using matrix notation. Let $$\boldsymbol X = \begin{pmatrix} X_1 \\ X_2 \\ \vdots \\ X_n \end{pmatrix}\qquad\text{and}\qquad \boldsymbol 1 = \begin{pmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{pmatrix}.$$ Then $$\bar X = n^{-1} \boldsymbol 1'\boldsymbol X.$$ The mean is given by $$\mathbb E[\bar X] = n^{-1} \boldsymbol 1'\mathbb E[\boldsymbol X] = \bar X = n^{-1} \boldsymbol 1'(\mu\boldsymbol 1) = \mu$$ and the variance by $$\mathbb V[\bar X] = n^{-2} \boldsymbol 1'\mathbb V[\boldsymbol X]\boldsymbol 1 = n^{-2}\boldsymbol 1'\boldsymbol\Sigma\boldsymbol 1.$$ In the case of uncorrelated $X_i$s, $\boldsymbol\Sigma$ becomes a diagonal matrix, and the main diagonal elements are all equal in case identically distributed $X_i$s.