correlationcovariancenormal distributionself-study
Consider jointly normally distributed random variables $X,Y\sim N(0,1)$ that have $\text{Corr}(X,Y)=\rho$.
Show that $\text{Corr}(X^2,Y^2)=\rho^2$.
(Hint: Consider $X,U\sim N(0,1)$ where they are independent then we have $Y=\rho X+\sqrt{1-\rho^2}U$).
I sense that a transform is required for both $X$ and $Y$ but not sure where to go next.
edit: I am more interested in knowing how the hint is used.
You cannot use the linear representation of the correlation in discrete support distributions.
In the special case of the Binomial distribution, the representation $$X=\sum_{i=1}^8 \delta_i\quad Y=\sum_{i=1}^{18} \gamma_i\quad \delta_i,\gamma_i\sim \text{B}(1,2/3)$$ can be exploited since $$\text{cov}(X,Y)=\sum_{i=1}^8\sum_{j=1}^{18}\text{cov}(\delta_i,\gamma_j)$$ If we choose some of the $\delta_i$'s to be equal to some of the $\gamma_j$'s, and independently generated otherwise, we obtain $$\text{cov}(X,Y)=\sum_{i=1}^8\sum_{j=1}^{18}\mathbb{I}(\delta_i:=\gamma_j)\text{var}(\gamma_j)$$ where the notation $\mathbb{I}(\delta_i:=\gamma_j)$ indicates that $\delta_i$ is chosen identical to $\gamma_j$ rather than generated as a Bernoulli $\text{B}(1,2/3)$.
Since the constraint is$$\text{cov}(X,Y)=0.5\times\sqrt{8\times 18}\times\frac{2}{3}\times\frac{1}{3}$$we have to solve $$\sum_{i=1}^8\sum_{j=1}^{18}\mathbb{I}(\delta_i:=\gamma_j)=0.5\times\sqrt{8\times 18}=6$$This means that if we pick 6 of the 8 $\delta_i$'s equal to 6 of the 18 $\gamma_j$'s we should get this correlation of 0.5.
The implementation goes as follows:
We can check this result with an R simulation
> z=rbinom(10^8,6,.66)
> y=z+rbinom(10^8,12,.66)
> x=z+rbinom(10^8,2,.66)
cor(x,y)
> cor(x,y)
[1] 0.5000539
Comment
This is a rather artificial solution to the problem in that it only works because $8\times 18$ is a perfect square and because $\text{cor}(X,Y)\times\sqrt{8\times 18}$ is an integer. For other acceptable correlations, randomisation would be necessary, i.e. $\mathbb{I}(\delta_i:=\gamma_j)$ would be zero or one with some probability $\varrho$.
Addendum
The problem was proposed and solved years ago on Stack Overflow with the same idea of sharing Bernoullis.
I don't think the formula given in the question can be correct in all cases, it is developed using joint normality. Without joint normality we can use copulas. For $X,Y$ random variables with joint distribution with cumulative distribution function $F(x,y)$ and joint density $f(x,y)$ define the transformed random variables (rv) $U=F_X(X), V=F_Y(Y)$ where $F_X, F_Y$ denotes the marginal cumulative distribution functions (cdf). Then the joint distribution of $U;V$ $$ \DeclareMathOperator{\P}{\mathbb{P}} \P(U \le u, V \le v)=C(u,v) $$ is the copula of $X$ and $Y$, with copula density $c(u,v)$ (when it exists). So, in your setup let us assume that the copula density exists, and for simplicity I will take both $X$ and $Y$ as standard normals. So what possibility exists for the conditional expectation of $X$ given $Y=y$ ? Using Sklar's theorem we can write the joint density as $$ f(x,y) = c(\Phi(x),\Phi(y)) \phi(x) \phi(y) $$ where $\Phi, \phi$ are the standard normal cdf, pdf, respectively. Then the conditional density is given by $$ f(x \mid y) = \frac{f(x,y)}{f(y)}=\frac{c(\Phi(x),\Phi(y))\phi(x)\phi(y)}{\phi(y)}= c(\Phi(x),\Phi(y))\phi(x) $$ Then we can look at this with various copula functions, see https://en.wikipedia.org/wiki/Copula_(probability_theory) .
There is a general inequality for copulas $$ W(u,v)=\max(u+v-1,0) \le C(u,v) \le M(u,w)=\min(u,v) $$ where both upper and lower limits are copulas (This is the Frechet-Hoeffding bounds). The upper limit isn't very interesting, since it gives $\P(U=V)=1$ so gives correlation equal 1. The lower limit similarly corresponds to $U=1-V$ with probability one, so correlation is -1. But for these two extremal copula the conditional expectation function certainly is linear!
Lets look at some intermediate cases. I will use the R package copula and some numerical integration to find the conditional expectation function, for the case of the gumbel copula. The code can be simply adapted for other copulas.
library(copula)
C <- gumbelCopula(2)
make_cond <- Vectorize(function(y) {
function(x) dCopula( cbind(pnorm(x), pnorm(y)), C)*dnorm(x)
} )
the last command makes a function representing the conditional density given $Y=y$. Let us look at how this looks like for three different values of $y$:
cond_dens <- make_cond(c(-1, 0, 1))
plot(cond_dens[[1]], from=-3, to=3, col="blue", ylim=c(0, 0.7))
plot(cond_dens[[2]], from=-3, to=3, col="orange", add=TRUE)
plot(cond_dens[[3]], from=-3, to=3, col="red", add=TRUE)
title("conditional densities for y=-1, 0, 1")
showing clearly that the conditional distributions now are non-normal. We can also see clearly that the conditional variance is non-constant.
For more examples using the copula package see https://www.r-bloggers.com/modelling-dependence-with-copulas-in-r/ and Generating values from copula using copula package in R . Then we can find the conditional expectation function using numerical integration:
plot(function(y) sapply(make_cond(y), FUN=function(fun)
integrate(function(x) x*fun(x) ,
lower=-Inf, upper=Inf)$value),
from=-3, to=3,
ylab="conditional expectation given y", xlab="y")
title("conditional expectation of X given Y=y")
and it is quite clear that the conditional expectation function is not linear!
Best Answer
Hint:
Let $(X,Y)$ be jointly normal variables with zero means and unit variances and $\text{Corr}(X,Y)=\rho$.
By definition, \begin{align}\text{Corr}(X^2,Y^2)=\frac{\text{Cov}(X^2,Y^2)}{\sqrt{\text{Var}(X^2)\text{Var}(Y^2)}}\end{align}
where $\text{Cov}(X^2,Y^2)=\mathbb E(X^2Y^2)-\mathbb E(X^2)\mathbb E(Y^2)$, and
$\text{Var}(X^2)=\mathbb E(X^4)-(\mathbb E(X^2))^2=\text{Var}(Y^2)$.
For finding $\mathbb E(X^2Y^2)$ quickly, note that $\mathbb E(X^2Y^2)=\mathbb E(\mathbb E(X^2Y^2\mid X))=\mathbb E(X^2\,\mathbb E(Y^2\mid X))$.
And we know that $Y\mid X\sim\mathcal{N}(\rho X,1-\rho^2)$.
So, $\mathbb E(Y^2\mid X)=\text{Var}(Y\mid X)+(\mathbb E(Y\mid X))^2=\cdots$.
I think you can find the moments now.