# Transforming joint Gaussians to independent random variables

I saw this question in UIUC's probability class homework and it got me curious on how to solve it.

Suppose $X$ and $Y$ are jointly Gaussian such that $X \sim N(0, 9)$, $Y \sim N(0, 4)$; and the correlation coefficient is denoted by $\rho$. The solutions to the questions below may depend on $\rho$ and may fail to exist for some values of $\rho$.

For what values of $d$ is $X + dY$ independent of $(X - dY)^3$?

However I wasn't able to find a solution. How do you obtain the covariance of $(X - dY)^3$? I am trying to do $Cov(Z-X,X^3)$ turns into $Cov(Z,X^3) - Cov(X,X^3)$ but I don't know what $Cov(X,X^3)$ solves to?

• Note that there is no such thing as "covariance of $(X-dY)^3$. Also, proving that $\text{cov}(X+dY,(X-dY)^3) = 0$ does not suffice to prove that $X+dY$ and $(X-dY)^3$ are independent random variables, though if it turns out that $\text{cov}(X+dY,(X-dY)^3) \neq 0$, then you can be sure that $X+dY$ and $(X-dY)^3$ are dependent random variables. – Dilip Sarwate Dec 9 '12 at 17:20
• I added a link to the actual problem set from which this problem is taken (I taught that course for more than 35 years, and recognized the problem....) – Dilip Sarwate Dec 9 '12 at 19:34

There are two arguments you need to make. First, if $A$ and $B$ are independent random variables, then so are $g(A)$ and $h(B)$ independent random variables for all (measurable) functions $g(\cdot)$ and $h(\cdot)$. So, the question can be transformed into

For what values of $d$ are $X+dY$ and $X-dY$ independent random variables?

Second, since $X$ and $Y$ are jointly normal, so are $X+dY$ and $X-dY$ jointly normal random variables, and so they are independent if $\text{cov}(X+dY,X-dY)$ equals $0$. Now,

\begin{align*}\text{cov}(X+dY,X-dY) &= \text{var}(X) -d^2\cdot\text{var}(Y) + d\cdot\text{cov}(Y,X) -d\cdot\text{cov}(X,Y)\\ &= \text{var}(X) -d^2\cdot\text{var}(Y) \end{align*}

since the covariance terms cancel, and so $\text{cov}(X+dY,X-dY) = 0$ for $d = \pm\sqrt{\frac{9}{4}}=\pm\frac{3}{2}$.

As a special case, note that if $\rho=\pm 1$, then $Y=\rho\frac{\sigma_Y}{\sigma_X}X = \pm\frac{2}{3}X$, and in this case, one of the random variables $X+\frac{3}{2}Y$ and $X-\frac{3}{2}Y$ is zero with probability $1$.

The answer is $d=\sqrt{\frac{9}{4}}$ and relies on a characterization of the normal distribution:

• for two jointly normal random variable $X$and $Y$ with identical variance, $(X+Y)$ and $(X-Y)$ are independent normal random variables

user603 is specifically requested to not delete any of the boldfaced text in the above statement since deleting either part makes the statement demonstrably incorrect.

• -1 The statement "for two standard normal random variables $X$ and $Y$, $(X+Y)$ and $(X-Y)$ are independent gaussian random variables" is false. It does not hold if $X$ and $Y$ are marginally normal but not jointly normal random variables. It does hold for jointly normal (not necessarily independent) random variables with identical variance. – Dilip Sarwate Dec 9 '12 at 21:20
• @DilipSarwate: oula, yes, indeed: corrected. – user603 Dec 10 '12 at 0:11
• Your rollback of the edit that I made makes you answer incorrect again. It is NOT true that if $X$ and $Y$ are jointly normal random variables, then $X+Y$ and $X-Y$ are independent (normal) random variables. In fact, $$\operatorname{cov}(X+Y,X-Y) = \operatorname{var}(X) - \operatorname{var}(Y) \neq 0$$ when $\operatorname{var}(X)$ and $\operatorname{var}(Y)$ have different values, and so $X+Y$ and $X-Y$ are manifestly dependent random variables. That is, you need identical variance (as you had originally with standard normals) PLUS joint normality for the result to hold. – Dilip Sarwate Oct 16 '16 at 15:57
• @DilipSarwate: thanks, I didn't read the edit with enough care, missed the main point of it! – user603 Oct 16 '16 at 16:19

You probably have to construct covariance matrix $\left(\begin{array}{cc} 9 & 6\rho \\ 6\rho & 4 \end{array}\right)$ and to find its eigenvalues and eigenvectors (both depending on $\rho$). Eigenvectors will led to the independent random variables expressed as linear combination of $X$ and $Y$: $Z_1 = e_{11} X + e_{12} Y$ and $Z_2 = e_{21} X + e_{22} Y$.