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Suppose that $X_1, X_2, \ldots, X_9$ denote a random sample from from a population having mean $\mu$ and standard deviation $\sigma$. Consider the following four estimators of $\mu$:

$$\hat{\mu}_1=X_2,\,\,\hat{\mu}_2=\frac{X_1+X_2}{2},\,\,\hat{\mu}_3=\frac{2X_1-X_2+3X_6}{3},\,\,\hat{\mu}_4=\frac{\sum_{i=1}^9X_i}{9} $$

1) Which of these estimators are unbiased?

2) Among the unbiased estimators found in the preceding part, which one has the smallest variance?

Attempt:

1) I found that $\hat{\mu}_1$, $\hat{\mu}_2$, and $\hat{\mu}_4$ are the unbiased estimators.

2) I'm not sure how to figure out the variance of these estimators?

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Just treat the estimators as random variables and calculate their variance.

Your first estimator is $X_2$. But by assumption $\text{Var}(X_2)=\sigma^2$.

Your second estimator is $(X_1+X_2)/2$. But

$$\text{Var}\left(\frac{X_1+X_2}{2}\right)=\frac{1}{4}\text{Var}(X_1+X_2)=\frac{1}{4}\big(\text{Var}(X_1)+\text{Var}(X_2)+2\,\text{Cov}(X_1,X_2)\big)$$

But by assumption $X_1$ and $X_2$ are independent, so their covariance is zero, and again by assumption, their variances are both $\sigma^2$. Hence the variance of your second estimator is $\frac{1}{4}(\sigma^2+\sigma^2)=\frac{\sigma^2}{2}$.

Continue this way for your other two estimators.

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