Correlation coefficient of normal random variables

Anyone knows about the following property?

If $X$, $Y$ jointly are normal with zero means and correlation coefficient $\rho$, then the correlation coefficient between $X^2$ and $Y^2$ is $\rho^2$.

Thank you

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Assume without loss of generality that $X$ and $Y$ are standard normal, then $$Y=\varrho X+\sqrt{1-\varrho^2} Z,$$ where $(X,Z)$ is i.i.d. standard normal and $\sigma^2=1-\varrho^2$. Recall furthermore that, for any standard normal random variable $U$, $\mathbb E(U)=\mathbb E(U^3)=0$, $\mathbb E(U^2)=1$ and $\mathbb E(U^4)=3$.

Thus, the expansion $X^2Y^2=\varrho^2X^4+2\varrho\sqrt{1-\varrho^2} X^3Z+(1-\varrho^2)X^2Z^2$ implies readily that $\mathbb E(X^2Y^2)=1+2\varrho^2$.

Likewise, $\mathbb E(X^2)=\mathbb E(Y^2)=1$ and $\mathbb E(X^4)=\mathbb E(Y^4)=3$ hence $\mathrm{var}(X^2)=\mathrm{var}(Y^2)=2$.

Finally, $$\mathrm{Corr}(X^2,Y^2)=\frac{\mathbb E(X^2Y^2)-\mathbb E(X^2)\mathbb E(Y^2)}{\sqrt{\mathrm{var}(X^2)\mathrm{var}(Y^2)}}=\frac{1+2\varrho^2-1}{\sqrt{2\cdot2}}=\varrho^2.$$

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Since $X$ and $Y$ are jointly normal, you can write $X = aM + bN$ and $Y = cM + dN$ where $a, b, c, d \in R$ and $M$ and $N$ are independent normals with mean $0$ and variance $1$. Next,

$$\mathrm{Cov}[X,Y] = ac + bd$$

$$\mathrm{Var}[X] = a^{2}+b^{2}$$

$$\mathrm{Var}[Y] = c^{2}+d^{2}$$

$$\rho_{X,Y} = \frac{ac+bd}{\sqrt{a^{2}+b^{2}}\sqrt{c^{2}+d^{2}}}$$

Next, the $X^{2}$ and $Y^2$ part...

$$\mathrm{Cov}(X^{2},Y^{2}) = \mathrm{Cov}(a^{2}M^{2}+b^{2}N^{2}+2abMN,c^{2}M^{2}+d^{2}N^{2}+2cdMN) =$$

$$= a^{2}c^{2}\mathrm{Var}[M^{2}] + (2a^{2}cd+2b^{2}cd+2abc^{2}+2abd^{2}) \mathrm{Cov}(M^{2},MN) +b^{2}d^{2}\mathrm{Var}[N^{2}] + 4abcd\mathrm{Cov}(MN,MN)$$

Observe that $M^{2}$ and $N^{2}$ are a chi-squares with $1$ degree of freedom. Hence, $\mathrm{Var}[M^{2}]=\mathrm{Var}[N^{2}]=2$.

Also observe that:

$$\mathrm{Cov}(M^{2},MN) = E[M^{3}N] - E[M^{2}]E[MN] =$$

$$= E[M^{3}]E[N] - E[M^{2}]E[M]E[N] = 0$$

and,

$$\mathrm{Cov}(MN,MN) = E[M^{2}N^{2}] - E[MN]E[MN] =$$

$$= E[M^{2}]E[N^{2}] - E[M]^{2}E[N]^{2} = 1$$

Hence,

$$\mathrm{Cov}(X^{2},Y^{2}) = 2a^{2}c^{2}+2b^{2}d^{2} +4abcd$$

Finally,

$$\mathrm{Var}[X^{2}] = \mathrm{Var}[a^{2}M^{2}+ 2abMN + b^{2}N^{2}] = 2a^{4} + 4a^{2}b^{2} + 4b^{4}$$

$$\mathrm{Var}[Y^{2}] = 2c^{4} + 4c^{2}d^{2} + 2d^{4}$$

Thus,

$$\rho_{X^{2},Y^{2}} = \frac{2a^{2}c^{2}+2b^{2}d^{2} +4abcd}{\sqrt{2a^{4} + 4a^{2}b^{2} + 2b^{4}} \sqrt{2c^{4} + 4c^{2}d^{2} + 2d^{4}}} =$$

$$= \frac{(ac+bd)^2}{(a^{2}+b^{2})(c^{2}+d^{2})} = (\rho_{X,Y})^{2}$$

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Shoul I compute covariance of $a^2M^2 + b^2N^ + 2abMN$ and $c^2M^2 + d^2N^ + 2cdMN$ ? – Anna Dec 12 '12 at 15:35
Yes. The terms with MN might need conditional expectations. I ll write it up in a couple of minutes – madprob Dec 12 '12 at 16:00
Thanks! So I was wrong thinking $ρ_{X^2,Y^2}=(ρ_{X,Y})^2$. Are you sure about $Var(X^2)$? Isn't is $2a^4+2b^4+4a^2b^2$? – Anna Dec 12 '12 at 18:51
Indeed, you are right! – madprob Dec 12 '12 at 19:25