1
$\begingroup$

Suppose $X$ and $Y$ follow $N(0,1)$ and $ \operatorname{Corr}(X,Y)=\rho$. Find $ \operatorname{Cov}(X^2,Y^2)$.

Here is what I know $$ \operatorname{Cov}(X^2,Y^2)=E[X^2Y^2]-E[X^2]E[Y^2]$$
Since $E[X^2]= \operatorname{Var}(X)+E^2[X]= \operatorname{Var}(X)$, $\enspace E[X^2]E[Y^2]= \operatorname{Var}(X) \operatorname{Var}(Y)=1$.
But I don't know how to deal with $E[X^2Y^2]$.
Am I in the right track?

$\endgroup$
  • $\begingroup$ To say that $X\sim N(0,1)$ and $Y\sim N(0,1)$ and $\operatorname{corr}(X,Y) = \text{a particular number}$ is not enough information to specify the distribution of the pair $(X,Y).$ However, it is enough if you add an additional bit of information: that the pair $(X,Y)$ is JOINTLY normally distributed. There is this simple way of getting $X\sim N(0,1)$ and $Y\sim N(0,1)$ and $\operatorname{corr}(X,Y) = \text{your preferred number}$ without $(X,Y)$ being jointly normal: Let $Y=\begin{cases} \phantom{+}X&\text{if } |X|\le c, \\ -X&\text{if }|X| > c, \end{cases}\quad$ and then$\,\ldots \qquad$ $\endgroup$ – Michael Hardy Sep 22 '18 at 21:21
  • $\begingroup$ $\ldots\,$ choose the value of $c$ to to make the correlation what you want it to be. $\qquad$ $\endgroup$ – Michael Hardy Sep 22 '18 at 21:23
  • $\begingroup$ I don't know whether or not the information given, without the assumption of JOINT normality, is enough to determine the correlation between $X^2$ and $Y^2. \qquad$ $\endgroup$ – Michael Hardy Sep 22 '18 at 21:25
2
$\begingroup$

Assuming you mean $(X,Y)$ is jointly normal where $X$ and $Y$ have zero means and unit variances and $\operatorname{Corr}(X,Y)=\rho$, we know the conditional distribution of $Y\mid X$, namely

$$Y\mid X\sim N(\rho X,1-\rho^2)$$

Then using the law of total expectation,

\begin{align} E(X^2Y^2)&=E\left[E(X^2Y^2\mid X)\right] \\&=E\left[X^2E(Y^2\mid X)\right] \\&=E\left[X^2\left(\operatorname{Var}(Y\mid X)+(E(Y\mid X))^2\right)\right] \\&=\quad\cdots \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.