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
 A: 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,
$$
\mathrm{Corr}(X^2,Y^2)=\frac{\mathbb E(X^2Y^2)-\mathbb E(X^2)\mathbb E(Y^2)}{\sqrt{\mathrm{var}(X^2)\mathrm{var}(Y^2)}}=\frac{1+2\varrho^2-1}{\sqrt{2\cdot2}}=\varrho^2.
$$
A: 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,
$$\rho_{X^{2},Y^{2}} = \frac{2a^{2}c^{2}+2b^{2}d^{2} +4abcd}{\sqrt{2a^{4} + 4a^{2}b^{2} + 2b^{4}} \sqrt{2c^{4} + 4c^{2}d^{2} + 2d^{4}}} =$$
$$= \frac{(ac+bd)^2}{(a^{2}+b^{2})(c^{2}+d^{2})} = (\rho_{X,Y})^{2}$$
