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Let $X$ and $Y$ be independent random variables with means $\mu_X$ and $\mu_Y$ and variances $\sigma_X^2$ and $\sigma_Y^2$. Find an expression of correlation of $XY$ and $Y$ in terms of these means and variances. When I follow the definition for correlation, I get a zero in the denominator, which does not make sense. Please help.

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Could you type up where it is you get a zero in the denominator? –  Stefan Hansen Nov 24 '12 at 19:41
    
Which term(s) amongst the covariance of XY and Y, the variance of XY and the variance of Y do you have problems with? –  Did Nov 24 '12 at 19:48
    
@did By definition, correlation of XY and Y=COV(XY,Y)/σ(xy)*σ(y) So, σ(xy) =zero –  JouJ FarrouJ Nov 24 '12 at 19:56
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If you by $\sigma(xy)$ mean the standard deviation of $XY$, then this is not true. See André's answer below. –  Stefan Hansen Nov 24 '12 at 20:27
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2 Answers 2

The denominator is $\sigma_{XY}\sigma_Y$.

Let us compute $\sigma_{XY}$, since that is undoubtedly the source of the problem. We get $$\sigma_{XY}^2=E((XY)^2)-(E(XY))^2.$$ By independence, $$E((XY)^2)=E(X^2)E(Y^2)=(\sigma_X^2+\mu_X^2)(\sigma_Y^2+\mu_Y^2).$$ And $(E(XY))^2=\mu_X^2\mu_Y^2$. Subtract. We get $$\sigma_{XY}^2=\sigma_X^2\sigma_Y^2+\mu_X^2\sigma_Y^2+\mu_Y^2 \sigma_X^2.$$

The calculation for the numerator is quite a bit easier.

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That helps a lot. Thanks guys! –  JouJ FarrouJ Nov 26 '12 at 7:32
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The correlation is given by: $$ \mathrm{Corr}(XY,Y)=\frac{\mathrm{Cov}(XY,Y)}{\sqrt{\mathrm{Var}(XY)\mathrm{Var}(Y)}}. $$ Now you just have to use that for any random variable $Z_1$ and $Z_2$ we have $$ \mathrm{Var}(Z_1)=E[Z_1^2]-E[Z_1]^2,\qquad \mathrm{Cov}(Z_1,Z_2)=E[Z_1Z_2]-E[Z_1]E[Z_2] $$ along with the fact that $X$ and $Y$ are independent (in particular this implies that $X^2$ and $Y^2$ are independent and also $X$ and $Y^2$ are independent, which you will have to use).

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