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I am trying to prove this equation $$\mathrm{Var}(aX+bY) = a^2\mathrm{Var}(X)+b^2\mathrm{Var}(Y)+2ab\mathrm{Cov}(X,Y),$$ where $\mathrm{Cov}(X,Y):= E(XY)-EX\cdot EX$ and $E[\cdot]$ denotes the mean of a random variable. Also $\mathrm{Var}(X)$ is defined like this: $$\mathrm{Var}(X)=E(X^2)-[E(X)]^2$$ which is the variance of the random variable $X$.

I started this way:

$$\mathrm{Var}(aX+bY) = E[((aX+bY)-E(aX+bY))^2] = E[(aX-bY-aEX-bEY)^2] = $$

Please help me showing that this equals $a^2\mathrm{Var}(X)+b^2\mathrm{Var}(Y)+2ab\mathrm{Cov}(X,Y)$.

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The first equality in your proof is wrong - you're not using your definition of $\mathrm{Var}(X)$ correctly. Instead it should go something along the lines of: $$ \begin{align} \mathrm{Var}(aX+bY)&={\rm {\rm E}}[(aX+bY)^2]-\left({\rm {\rm E}}[aX+bY]\right)^2\\ &={\rm E}[a^2X^2+b^2Y^2+2abXY]-\left(a{\rm E}[X]+b{\rm E}[Y]\right)^2\\ &=a^2{\rm E}[X^2]+b^2{\rm E}[Y^2]+2ab{\rm E}[XY]-a^2{\rm E}[X]^2-b^2{\rm E}[Y]^2-2ab{\rm E}[X]{\rm E}[Y]. \end{align} $$ I'm sure you can take it from here.

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  • $\begingroup$ perfect, thanks, now i got it $\endgroup$ – doniyor May 30 '13 at 9:00

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