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Problem:

If $\hat{θ}_1$ and $\hat{θ}_2$ are unbiased estimators of $θ$, and $\hat{θ}_1$ and $\hat{θ}_2$ are antithetic, we derived that $c^∗ = 1/2$ is the optimal constant that minimizes the variance of $\hat{θ}_c = c\hat{θ}_1 + (1 − c)\hat{θ}_2$. Derive $c^∗$ for the general case. That is, if $\hat{θ}_1$ and $\hat{θ}_2$ are any two unbiased estimators of $θ$, find the value $c^∗$ that minimizes the variance of the estimator $\hat{θ}_c = c\hat{θ}_1 + (1 − c)\hat{θ}_2$ in equation $var(c\hat{θ}_1 + (1 − c)\hat{θ}_2)=Var(\hat{θ}_2) + c^2Var(\hat{θ}_1 − \hat{θ}_2) + 2c Cov(\hat{θ}_2,\hat{θ}_1 − \hat{θ}_2)$ (c∗ will be a function of the variances and the covariance of the estimators.)

My understanding:

In the special case of antithetic variates, $\hat{θ}_1$ and $\hat{θ}_2$ are iid and $cor(\hat{θ}_1,\hat{θ}_2) = −1$. Thus, $cov(\hat{θ}_1,\hat{θ}_2) =−var(\hat{θ}_1)$, and so the variance of $\hat{θ}_c$ equals $(4c2 − 4c + 1) ∗ var(\hat{θ}_1)$. I am not sure how to relate this special case with the above general case. I appreciate your suggestions. Thanks!

ForestGump
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    The appropriate calculations are explained at https://stats.stackexchange.com/a/185000/919. – whuber Nov 15 '21 at 14:23
  • @whuber The link u gave is just showing the computation of variance-covariance relationships, but it doesn't show how to derive general case to minimize the variance of the estimator. – ForestGump Nov 15 '21 at 15:08
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    Once you have done that calculation--which appears to completely answer your stated question of "relate this special case [to] the general case," finding the minimum can be done in several ways, amounting to finding the vertex of the parabola. – whuber Nov 15 '21 at 15:16
  • @whuber So if I can compute $Var(\hat{θ}_2) + c^2Var(\hat{θ}_1 − \hat{θ}_2) + 2c Cov(\hat{θ}_2,\hat{θ}_1 − \hat{θ}_2)$, then i will have the answer for the general case. Right? – ForestGump Nov 15 '21 at 15:29
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    I don't understand exactly what you mean by "compute" here: the values of both variances and the covariance are determined by the two estimators, so it only remains to find the value(s) of $c$ that minimize that expression. – whuber Nov 15 '21 at 16:46
  • @whuber $(4c^2 − 4c + 1)$ is minimum for $c^{*}=\frac{1}{2} $ since $4c^2 − 4c + 1$ is a quadratic function. How can we derive this for general case? – ForestGump Nov 15 '21 at 16:53
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    I don't understand what you mean by "general case:" isn't that exactly what you wrote in your second comment in this thread? The coefficients of the quadratic involve the variances and covariance, but nothing really changes: use your knowledge of quadratics or even Calculus to find the minimum. – whuber Nov 15 '21 at 16:59
  • @whuber Also, I don't see how they got the variance. It doesn't match with formula in the link u gave earlier. In the link the formula is: $Var(aX + bY) = a^2 \Var(X) + b^2 \Var(Y) + 2ab \Cov(X,Y)$ – ForestGump Nov 15 '21 at 17:09
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    It's a *perfect* match with that formula, applying it to $Y=\hat\theta_1-\hat\theta_2$ and $X=\hat\theta_2.$ – whuber Nov 15 '21 at 17:32
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    @whuber Got it! Thank u so much! – ForestGump Nov 15 '21 at 17:52

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