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Let $V$ and $W$ be independent standard normal random variables where $X=V+W$ and $Y=3W$

This is what I did:

$$M_{x,y}(s,t)=E(e^{sx+ty})=E(e^{s(v+w)+t(3w)})=E(e^{sv+sw+3tw})=E(e^{vs+w(s+3t)})$$

and

$$M_{v,w}(s, s+3t)=M_{v}(s)M_{w}(s+3t)=e^{\frac{1}{2}s^{2}}e^{\frac{1}{2}(s+3t)^{2}}=e^{\frac{1}{2}s^{2}+\frac{1}{2}(s^{2}+6st+9t^{2})}$$

$e^{\frac{1}{2}(2s^{2}+6st+9t^{2})}$

So then the joint distribution would be $(X,Y)\sim BVN(0,0,2,9,3)$ but that can't be right since $\mathrm{corr}(x,y)>1$. I can't really figure out what I'm doing wrong.

Daniel W. Farlow
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1 Answers1

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$$\mathrm{var}(X)=2,\ \mathrm{var}(Y)=9,\ \mathrm{cov}(X,Y)=3 \implies\mathrm{corr}(X,Y)=\frac3{\sqrt{2\cdot9}}=0.707...$$

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