# Variance and Covariance

• CescGoal
In summary, this conversation discusses the calculation of Cov(X,Z) and the correlation coefficient of X and Z, given the variances of independent random variables X and Y. The value of Cov(X,Z) is determined to be 0, as Cov(X,Y) is 0 due to independence. The correlation coefficient is calculated to be Cov(X,Z)/12. The conversation concludes with a suggestion to write out the definition of Cov to understand why Cov(X,Z) = Cov(X,X-Y).

#### CescGoal

Very sorry that I've double posted but I realized i placed the original post in Precalculus.

1. Homework Statement
Question

Let X and Y be independent random variables with variances 9 and 7 respectively and let
Z = X - Y

a) What is the value of Cov(X,Z)
b) What is the value of the correlation coefficient of X and Z?

I've been stuck on this one question for 2-3 hours; its ridiculous, I know. Here's my terrible try.

3. The Attempt at a Solution
a)

Var(X) = 9
Var(Y) = 7

Var(X-Y) = Var(X) + Var(Y) = Var(Z)
Therefore, Var(Z) = 7 + 9 =16
Cov(X,Z) = E[XZ] - E[X]E[Z]

and b) $$\rho$$XZ = $$\frac{Cov(X,Z)}{\sqrt{Var(X)*Var(Z)}}$$

= $$\frac{Cov(X,Z)}{\sqrt{9}*\sqrt{16}}$$
= $$\frac{Cov(X,Z)}{12}$$

Since last topic, I've realized that Cov(X,Y) = 0 due to independency. But I don't know how to use it.

Cov(X,Z)=Cov(X,X-Y). Cov(X,X-Y)=Cov(X,X)-Cov(X,Y), right?

I didn't know about that law; thankyou very much.

CescGoal said:
I didn't know about that law; thankyou very much.

It's pretty obvious if you write out the definition of Cov. You should try and do that so you can see why it's true.