Let $X_1$, $X_2$, $X_3$, ... , $X_n$ be a random sample from a distribution having mean µ and variance $σ^2$. Also, Let $Y_1$, $Y_2$, $Y_3$, ... , $Y_m$ be a second random sample from the same distribution that is independent from the first sample.
A. Under what condition(s) on a and b is $W_1$ = $a\bar{X} + b\bar{Y}$ an unbiased estimator for µ?
B. Assuming your condition(s) in part a are met, what values of a and b make Var($W_1$) a minimum? (Hint: do this directly using the tools of calculus.)
C. Let n = 6 and m = 8. If $W_1$ = $a\bar{X} + b\bar{Y}$ is the unbiased estimate from part b which has the minimum variance, find the relative efficiency of $W_1$ with respect to $W_2$ = $\frac{\bar{X} + \bar{Y}}{2}$
So far, for A. I solved for the expected value of $W_1$ and got E($W_1$) = a$\mu$ + b$\mu$. The conditon that needs to be met is that $ a+b=1$ or $ b = 1 - a$. Assuming that is right, for B. my problems start when trying to take the Var($W_1$). My issue is that even though the I can substitute $1-a$ for $b$ the $b$ portion is still over m an not n. Therefore when I try to take the derivatives in terms of $a$ the m is still limiting what I can cancel out and I can't get it to work out so that the second derivative is greater. Am I even headed down the right path? The use your calculus tools hint isn't very specific so I'm not sure how far to take it since I imagine that I'm suppose use the C.R.L.B at some point.
This is my first post so forgive me if my formatting is off, or didn't put enough details. I really am lost though so thanks in advance for the help.