0
$\begingroup$

If X and Y are jointly normal random variables with:

$E[X]=E[Y]=0$, $E[X^2]=9$, $E[Y^2]=1$, $\rho_{XY}=0.5$. Find the joint distribution of $W=X+Y$ and $Z=X-Y$.

1- I first tried to find the covariance(X,Y) but I am getting zero since E[X]=0, so how can there be a correlation coefficient?

First time I have seen a question like this so I don't know how to proceed.

  • 0
    Covariance, like variance, is something that occurs _around_ the mean. It doesn't matter whether the mean is $E[X]=0$ or $E[X]=100$. Try looking up the relevant formulas. – 2017-01-19
  • 0
    David K: The equation for Pxy in my book is Cov(X,Y)/E[X]*E[Y], since E[X] is zero then I have a problem. what am I missing? – 2017-01-19
  • 0
    @user3064033 The denominator should be $\sqrt{Var(X) * Var(Y)}$. – 2017-01-19
  • 0
    I edited the question to reflect your comment "The question says find the joint distribution of variables W=X+Y and Z=X-Y." Please check that it is what you wanted to ask. – 2017-01-19
  • 0
    @user3064033 You are missing that your book is wrong (or you've read it so). $\rho_{X,Y} =\dfrac{\mathsf{Cov}(X,Y)}{\sigma_X~\sigma_Y} = \dfrac{\mathsf{Cov}(X,Y)}{\sqrt{\mathsf{Var}(X)\mathsf{Var}(Y)}}$ Did you misread $\sigma_\Box$ as the mean, rather than the deviantion? – 2017-01-19

1 Answers 1

0

Claim: $(W,Z)$ is jointly normal.

To prove this, it is sufficient to show that any linear combination $aW+bZ$ is also Gaussian. This comes from $aW+bZ = (a+b)X + (a-b) Y$ which is Gaussian since $(X,Y)$ is jointly normal.

Now that you know it is jointly normal, all you have left to specify are the means, variances, and covariance.

The means are easy to compute.

The variances are not too bad either.

$$Var(W)=E[(X+Y)^2] - E[X+Y]^2 = E[X^2] + 2E[XY] + E[Y^2].$$ The only term that is not given explicitly is $E[XY]$. But you can show this is equal to $Cov(X,Y)$, which can be computed using the formula $Cov(X,Y) = \rho_{XY} / \sqrt{Var(X) Var(Y)}$. Computing $Var(Z)$ is similar.

Finally, you need to compute $Cov(W,Z)$. \begin{align} Cov(W,Z) &= E[WZ]-E[W]E[Z]\\ &=E[(X+Y)(X-Y)] - E[X+Y]E[X-Y]\\ &=E[X^2-Y^2] - 0\\ &= E[X^2] - E[Y^2] \end{align}

  • 0
    Could you please explain the last line? I don't see how you got that end result. – 2017-01-19
  • 0
    thanks, that really helped. – 2017-01-19