I've got this question that says both $X$ and $Y$ are random variables with mean $\mu$ and variance $\sigma^2$, and $\mu \ne 0$, the correlation is $p, -1<p<1$
We have $$Q=aX+bY$$ and it is unbiased estimator of $\mu$.
we want to determine the values of $a$ and $b$ so that $Var(Q)$ is minimised.
I'll show what I did, which doesn't feel right near the end,
$E(Q)=\mu$ , since $Q$ is unbiased.
$Var(Q)=Var(aX+bY)$
$Var(Q)=a^2Var(X)+b^2Var(Y)+2abCov(X,Y)$
$Var(Q)=a^2\sigma^2+b^2\sigma^2+2abCov(X,Y)$
$Cov(X,Y)=Corr(X,Y)\sqrt{Var(X)Var(Y)}$, which gives us $Cov(X,Y)=p\sigma^2$
now $Var(Q)=a^2\sigma^2+b^2\sigma^2+2ab\sigma^2p$
I think to get $Var(Q)$ at minimum, the first derivative needs to equal to zero, which doesn't feel right for me for some reason.
$d\over dp$ $(a^2\sigma^2+b^2\sigma^2+2ab\sigma^2p)$
$=2ab\sigma^2=0$, which gives me either $a$ or $b$ or $\sigma=0$
to confirm my findings, the second derivative needs to be larger than zero, which doesn't work in this case.
what am I doing wrong?
Thanks a lot for help :)
You forgot that $a$ and $b$ are functionally connected by a relation that $Q$ is unbiased esimator: $$ \mathbb E[Q] = a\mathbb E[X]+b\mathbb E[Y]=(a+b)\mu = \mu. $$ so $a+b=1$ or $b=1-a$.
Therefore $$ \text{Var}(Q)=a^2\sigma^2+(1-a)^2\sigma^2+2a(1-a)\sigma^2p. $$
Equate first derivative to zero and get $a=\frac12$, what could be predicted in advance by virtue of symmetry.