Could anyone help with this?
If X1, X2, and X3 constitute a random sample of size n=3 from a normal population with the mean m and the variance s2, find the efficiency of (X1+2X2+X4)/4 relative to (X1+X2+X3)/3.
$\displaystyle T_1=(X_1+2X_2+X_3)/4$ and $\displaystyle T_2=(X_1+X_2+X_3)/3$ are unbiased estimators for \mu. (As can be seen since $\displaystyle E(T_1)=\mu$ and $\displaystyle E(T_2)=\mu$ )
Then the relative efficiency $\displaystyle e(T_1,T_2)$ of $\displaystyle T_1$ and $\displaystyle T_2$ is:
$\displaystyle e(T_1,T_2)=\frac{E((T_2-\mu)^2)}{E((T_1-\mu)^2)}$
CB