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I want to solve the following variance of an estimator but after many trials I still didn't succeed...

$V(a\hat{\theta}_1 + (1-a)\hat{\theta}_2)$

We have that:

  • $V(\hat θ_1) = σ_1^2$
  • $V(\hat θ_2) = σ_2^2$
  • $\operatorname{Cov}(\hat θ_1, \hat θ_2) = c ≠ 0$

Any idea?

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$\newcommand{\v}{\operatorname{var}} \newcommand{\c}{\operatorname{cov}}$ \begin{align} & \v(a\hat{\theta}_1 + (1-a)\hat{\theta}_2) \\[6pt] = {} & \v(a\hat\theta_1) + \v((1-a)\hat\theta_2) + 2\c(a\hat\theta_1,(1-a)\hat\theta_2) \\[6pt] = {} & a^2 \v(\hat\theta_1) + (1-a)^2\v(\hat\theta_2) + 2a(1-a)\c(\hat\theta_1,\hat\theta_2) \\[4pt] = {} & a^2\sigma_1^2 + (1-a)^2\sigma_2^2 + 2a(1-a)c. \end{align} (A standard exercise asks you to find the value of $a$ that minimizes this.)

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  • $\begingroup$ When you extract from the covariance $a$ and $(1-a)$, is that a property of covariance? I don't get that step... Thanks for you answer by the way! :) $\endgroup$ – JaviOverflow Aug 9 '17 at 18:14
  • $\begingroup$ @JaviOverflow : Another standard exercise is to use the definition of covariance to show that $$\operatorname{cov}(aX, bY) = ab\operatorname{cov}(X,Y). $$ That's when $a$ and $b$ are real numbers. If they are matrices, then one gets $a \operatorname{cov}(X,Y) b^T. \qquad$ $\endgroup$ – Michael Hardy Aug 9 '17 at 18:17

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