$ E(\hat \theta_1 ) = E(\hat \theta_2) = \theta_1 $ , $ Var (\hat \theta_1) = \sigma_1 , Var(\hat \theta_2) = \sigma_2, Cov(\hat \theta_1, \hat \theta_2) = \sigma_{12}$

$\hat \theta_3 = a \hat \theta_1 + (1-a) \hat \theta_2 $

The question asks me to find the value of a that minimizes the variance. I have the solutions to the answer but I don't understand how the covariance is used in calculating $V(\hat \theta_3) $

$V(\hat \theta_3) $ = $ V(a \hat \theta_1 + (1-a) \hat \theta_2 ) = $ $a^2 V( \hat \theta_1) + (1-a)^2 V(\hat \theta_2) + 2(1-a)Cov(\hat \theta_1, \hat \theta_2)$

My issue isn't with determining the minimum value of alpha, that's just found by setting the derivative of $V(\hat \theta_3)$ to $0$ and checking 2nd deriviative for if it is a minimum, my issue is with the $(1-a)$

I do know that since both $\hat \theta_1, \hat \theta_2$ are not independent, there exists a covariance $> 0$, hence obviously in the calculations of adding 2 variances, you must take into account covariance and hence the theorem;

$Var( X + Y ) = $Var(X) + Var (Y) + 2 Cov(X,Y)

but where is the (1-a) coming from in $2(1-a)Cov(\hat \theta_1, \hat \theta_2)$ ?


\begin{align} V(a\hat{\theta}_1+(1-\theta_1)\hat{\theta_2})&=a^2V(\hat{\theta_1})+(1-a)^2V(\hat{\theta}_2)+2Cov(a\hat{\theta}_1, (1-a)\hat{\theta}_2)\\ &=a^2V(\hat{\theta_1})+(1-a)^2V(\hat{\theta}_2)+2a(1-a)Cov(\hat{\theta}_1, \hat{\theta}_2)\\ \end{align}

I believe it's a mistake, it should be $a(1-a)$.

  • $\begingroup$ yea that would make more sense. I think it is a mistake because I noticed some other errors in these solutions I was given. Thank you! $\endgroup$ – ajdawg Feb 8 at 2:10

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