# Find E[XZ] if X,Z are not independent and we do not know their covariance

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)

Find Cov(X,Z)

The proposed solution is 16.

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

• Are $Y$ and $Z$ the same? Are they independent or not? Jan 20, 2016 at 14:35
• Use dollar signs in the beginning and end of your equations. Jan 20, 2016 at 14:36
• 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/…. Jan 20, 2016 at 14:45
• 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!
– SW7
Jan 20, 2016 at 15:44

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