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IF $X$ and $Y$ are independent random variables such that $E(X)=\lambda_1$,$E(Y)=\lambda_2$ and the variances of $X$ and $Y$ are $\sigma_1^2$ and $\sigma_2^2$ respectively and $\sigma_{12}$ is the variance of $XY$ ,then how can we prove that $$\sigma_{12}^2=\sigma_1^2\sigma_2^2+\lambda_1^2\sigma_2^2+\lambda_2^2\sigma_1^2$$?

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  • $\begingroup$ Is $\sigma_{12}$ the variance of the product of $XY$? $E(XY)=E(X)E(Y)$ $\endgroup$ – HipsterMathematician Nov 10 '12 at 17:14
  • $\begingroup$ @Charlie, you guessed it right. $\endgroup$ – user48012 Nov 10 '12 at 17:18
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$Var(XY)=E(X^2Y^2)-E^2(XY)$

Since they are independent:

$E(X^2)E(Y^2)-E^2(X)E^2(Y^2)$ (*), but

$Var(X)=E(X^2)-E^2(X)=\sigma _1 ^2$,then we have:

$E(X^2)=\sigma _1 ^2 +\lambda _1 ^2$

The same way with $Y$:

$Var(Y)=E(Y^2)-E^2(Y)=\sigma _2 ^2$, same way :

$E(Y^2)=\sigma _2 ^2+\lambda _2 ^2$

So we have using (*)

$(\sigma _1 ^2 +\lambda _1 ^2)(\sigma _2 ^2+\lambda _2 ^2)-\lambda _ 1^2 \lambda _2^2= \sigma _1^2\sigma_2^2+\lambda_1^2\sigma_2^2+\sigma_1^2\lambda_2^2$

As desired.

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