# Thread: Finding the correlation coefficient

1. ## Finding the correlation coefficient

Hello

I've been working on this problem all day but, honestly I don't know what else to try.

Let Z~N(0,1) and $Y= a+bZ+cZ^2$

find $\rho(Y,Z)$ (Correlation coefficient

I think I need to find the the joint PDF of X and Y then calculate the Cov(XY), maybe this will let me cancel what's left of the denominator and obtain a number. I'm quite lost.

*****EDIT******

I should probably mentioned that i tried the linear properties of expectation and arrived to the conclucion that it was not correlated but i think my application is wrong.

2. You will need the covariance divided by the st. deviations.

The covariance is $E(YZ)-E(Y)E(Z)=E(YZ)$ since E(Z)=0.

Now $E(YZ)= E((a+bZ+cZ^2)Z) =aE(Z)+bE(Z^2)+cE(Z^3)=0+b+0=b$
since Z is symmetric and the odd moments should be zero.

Next $\sigma_Z=1$ and $\sigma_Y=\sqrt{Var(a+bZ+cZ^2)}$

Then $Var(a+bZ+cZ^2)= Var(bZ+cZ^2)= b^2Var(Z)+c^2Var(Z^2)+2bcCov(Z,Z^2)$

Where you will need $Cov(Z,Z^2)=E(Z^3)-E(Z)E(Z^2)=0-(0)(1)=0$.

That's most of this problem. You should not even consider integrating it, whatsoever.

Fine, no one's going to finish this... $Var(Z^2)=E(Z^4)-(E(Z^2))^2$.

I was too lazy to compute that fourth moment, so I looked it up at http://en.wikipedia.org/wiki/Normal_distribution
It's 3, so $Var(Z^2)=E(Z^4)-(E(Z^2))^2=3-1=2$.

This seems to be (or not 2b, that is your question) ${b\over \sqrt{b^2+2c^2}}$.

3. Thanks very much for your answer. I see it clearly now. This should help me inn next week's final

4. that was fun, any more of these?