# Independence of Combination of Normal Random Variables

$\newcommand{\Cov}{\operatorname{Cov}}$

I have a practice question I'm trying to answer in studying for an upcoming exam:

$X\sim N(0,1)$ and $Y\sim N(0,1)$ and I have $\rho(X,Y)=0.4$. Define a random variable $Z = X + wY$ that also has a normal distribution. What value of $w$ makes $Z$ independent of $X$?

My thoughts is solve for when $\Cov(X,Y) = 0$, so I was going to set it up as: $\rho(X,Y) = \frac{\Cov(X,Y)}{\sigma_X \sigma_Y}$ and solve $0.4 = \frac{\Cov(X,Y)}{1}$ so $\Cov(X,Y) = 0.4$ and then I'd have $\Cov(X,Y) = \Cov(X,wY) = w\Cov(X,Y) = 0.4w$ and solve $0.4w = 0$ so $w = 0$ is when $Z$ is independent of $X$? But I have a feeling I'm not approaching this correctly and am doing something wrong. Can someone please help guide me in the right direction?

• $w=0$ doesn't make sense at all, since this would mean $Z=X$ and $X$ is clearly not independent of $X$. – saz Mar 18 '13 at 17:38

## 1 Answer

$\newcommand{\Cov}{\operatorname{Cov}}$

You need $\Cov(Z,X)=0$, not $\Cov(X,Y)=0$.

Then $$0=\Cov(Z,X)= \Cov(X+wY,X)=\Cov(X,X)+w\Cov(Y,X),$$ and recall that $\Cov(X,X)=1$, etc.

Also, you omitted to add that the pair $(X,Y)$ is jointly normally distributed. Without that, your conclusion that $X+wY$ is necessarily normal is not justified.

• Thank you so much for the help! – Abi Mar 18 '13 at 18:43