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I am very unsure on how to answer the following question. X and Y are independent normally distributed random variables with the properties listed below. How do I find E[XY] to find the covariance? I tried multiplying the two PDFs and the integrating over X and Z, which only gave me 1 as an answer.

So far I have found: $\mu_z = -4$ and $\sigma_z^2 = 13$

$\mu_x=0$
$\mu_y = 2 $ $\sigma_x = 2 $ $\sigma_y = 3 $

$Z=4X-3Y+2$

Assume: E(XY)=E(X)*E(Y)

Task:

Find Cov(X,Z)

The proposed solution is 16.

I am very lost on this, thank you very much in advance for any help!

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  • $\begingroup$ Are $Y$ and $Z$ the same? Are they independent or not? $\endgroup$ Jan 20, 2016 at 14:35
  • $\begingroup$ Use dollar signs in the beginning and end of your equations. $\endgroup$
    – Arbuja
    Jan 20, 2016 at 14:36
  • $\begingroup$ Sorry I am trying to edit this but I cannot understand your equations. Don't you mean "sum" instead of "sigma" and why would you have $o\sigma_{z^2}$? Just in case you should check meta.math.stackexchange.com/questions/5020/…. $\endgroup$
    – Arbuja
    Jan 20, 2016 at 14:45
  • $\begingroup$ Sorry about the notation, I had it as LaTex code and wanted to just attach an image at first but I did not have enough rep for that. Thanks for the pointer, I hope it is more readable now! $\endgroup$
    – SW7
    Jan 20, 2016 at 15:44

1 Answer 1

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If $X$ and $Y$ are independent, it holds that $Cov(X,Y)=0$. Also, for any constant $a \in \mathbb{R}$, $Cov(X,a)=0$. Moreover, $Cov(X, aY+bR) = aCov(X,Y) + bCov(X,R)$. Putting this together gives us the following:

$Cov(X,Z) = Cov(X, 4X-3Y+2) = 4Cov(X,X) - 3Cov(X,Y) + Cov(X,2) = 4Var(X) - 3\cdot0 + 0 = 4\cdot 4 = 16$

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