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$\newcommand{\Var}{\operatorname{Var}} \newcommand{\Cov}{\operatorname{Cov}}$

If I'm given $\Var(\hat {\beta_1}-3\hat {\beta_2})$ how do I find the variance of this? I have gotten this far I'm just not sure about the covariance part. I know that $$\Var(\hat {\beta_1}-\hat {\beta_2})=\Var(\hat {\beta_1})+\Var(\hat {\beta_2})-2\Cov(\hat {\beta_1},\hat {\beta_2})$$ so I came up with the following: $$\Var(\hat {\beta_1}-3\hat {\beta_2})=\Var(\hat {\beta_1})+9\Var(\hat {\beta_2})-{}???$$ I thought it would be $6\Cov(\hat {\beta_1},\hat {\beta_2})$ but I'm just really not sure what I would do for the covariance part.

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2 Answers 2

up vote 3 down vote accepted

The map $(X,Y)\mapsto \operatorname{Cov}(X,Y)=E[(X-EX)(Y-EY)]$ is bilinear (that is, linear once we fixed one argument). So it's indeed $6\operatorname{Cov}(\widehat {\beta_1},\widehat{\beta_2})$.

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Given two random variables $X$ and $Y$ and $a,b \in \mathbb R$ we have that $Var(aX+bY)=a^2\cdot VarX+ ab\cdot cov(X,Y)+b^2 \cdot VarY$. So you are right. If you don't know the $cov(X, Y)$ you cannot estimate $Var(X+Y)$. But if $X$ and $Y$ are known to be independent, then the covariace is zero

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