# Covariance and correlation

• Nov 11th 2009, 12:33 PM
chella182
Covariance and correlation
I know how to calculate covariance and correlation, just not in this question... I'll explain why after I write out.

Suppose that $\displaystyle X_1$, $\displaystyle X_2$, $\displaystyle X_3$, $\displaystyle X_4$, $\displaystyle X_5$ are independent random variables with variance $\displaystyle \sigma^2$. Determine the covariance of $\displaystyle X_1+3X_2+2X_3$ and $\displaystyle 2X_1-X_2+3X_4+X_5$? Determine the correlation between these two variables.

Now, as far as I'm aware, you need to know the means to calculate the covariance (Worried) am I being an idiot and missing something completely obvious?
• Nov 11th 2009, 12:39 PM
Laurent
Quote:

Originally Posted by chella182
I know how to calculate covariance and correlation, just not in this question... I'll explain why after I write out.

Suppose that $\displaystyle X_1$, $\displaystyle X_2$, $\displaystyle X_3$, $\displaystyle X_4$, $\displaystyle X_5$ are independent random variables with variance $\displaystyle \sigma^2$. Determine the covariance of $\displaystyle X_1+3X_2+2X_3$ and $\displaystyle 2X_1-X_2+3X_4+X_5$? Determine the correlation between these two variables.

Now, as far as I'm aware, you need to know the means to calculate the covariance (Worried) am I being an idiot and missing something completely obvious?

You don't need the mean here, simply use the fact that the covariance is bilinear: $\displaystyle {\rm Cov}(aX+bY,cZ+dT)=ac{\rm Cov}(X,Z)+ad{\rm Cov}(X,T)+...\mbox{(two terms with$Y$)}$, and $\displaystyle {\rm Cov}(X,Y)=0$ if $\displaystyle X,Y$ are independent, and $\displaystyle {\rm Cov}(X,X)={\rm Var}(X)$.
• Nov 11th 2009, 01:03 PM
chella182
Quote:

Originally Posted by Laurent
You don't need the mean here, simply use the fact that the covariance is bilinear: $\displaystyle {\rm Cov}(aX+bY,cZ+dT)=ac{\rm Cov}(X,Z)+ad{\rm Cov}(X,T)+...\mbox{(two terms with$Y$)}$, and $\displaystyle {\rm Cov}(X,Y)=0$ if $\displaystyle X,Y$ are independent, and $\displaystyle {\rm Cov}(X,X)={\rm Var}(X)$.

Okay, so am I doing $\displaystyle 2\times Cov(X_1,X_1)+(-1)\times Cov(X_1,X_2)+3\times Cov(X_1,X_4)+Cov(X_1,X_5)...$ then with terms starting with $\displaystyle 3X_2$ and $\displaystyle 2X_3$ et cetera?
• Nov 12th 2009, 03:06 AM
Laurent
Quote:

Originally Posted by chella182
Okay, so am I doing $\displaystyle 2\times Cov(X_1,X_1)+(-1)\times Cov(X_1,X_2)+3\times Cov(X_1,X_4)+Cov(X_1,X_5)...$ then with terms starting with $\displaystyle 3X_2$ and $\displaystyle 2X_3$ et cetera?

Yes, but since the covariance of independent terms is 0, you can write the terms like $\displaystyle {\rm Cov}(X_1,X_1)$ only, and there are just two of them.
• Nov 12th 2009, 10:30 AM
sirellwood
Is the correlation -1/$\displaystyle \sqrt{210}$?