Correlation of two ab(Normals) | ISI MStat 2016 PSB Problem 6

This problem is an interesting application of the moment generating function of normal random variable to see how the correlation behaves under monotone function. This is the problem 6 from ISI MStat 2016 PSB.

Problem

Suppose that random variables $X$ and $Y$ jointly have a bivariate normal distribution with $\mathrm{E}(X)=\mathrm{E}(Y)=0, {Var}(X)={Var}(Y)=1,$ and
correlation $\rho$ . Compute the correlation between $e^{X}$ and $e^{Y}$ .

Prerequisites

Correlation Coefficient
Moment Generating Function
Moment Generating Function $M_X(t)$ of Normal $X$ ~ $N(\mu, \sigma^2)$ is $e^{t\mu + \frac{\sigma^2t^2}{2}}$ .

Solution

$M_X(t) = E(e^{tX})$ is called the moment generating function.

Now, let's try to calculate $Cov(e^X, e^Y) = E(e^{X+Y}) - E(e^X)E(e^Y)$

For, that we need to have the following in our arsenal.

$X$ ~ $N(0, 1)$
$Y$ ~ $N(0, 1)$
$X+Y$ ~ $N(0, \sigma^2 = 2(1+\rho))$ [ We will calculate this $\sigma^2$ just below ].

$\sigma^2 = Var(X+Y) = Var(X) + 2Cov(X,Y) + Var(Y) = 1 + 2\rho + 1 = 2(1+\rho)$ .

Now observe the following:

$E(e^{X+Y}) = M_{X+Y}(1) = e^{1+\rho}$
$E(e^X) = M_{X}(1) = e^{\frac{1}{2}}$
$E(e^Y) = M_{Y}(1) = e^{\frac{1}{2}}$
$\Rightarrow Cov(e^X, e^Y) = E(e^{X+Y}) - E(e^X)E(e^Y) = e^{1+\rho} - e = e(e^{\rho} - 1)$

Important Observation

$Cor(e^X, e^Y)$ and $Cor(X,Y) = \rho$ always have the same sign. Can you guess why? There is, in fact, a general result, which we will mention soon.

Now, we are left to calculate $Var(e^X) = (Var(e^Y)$ .

$Var(e^X) = E(e^{2X}) - (E(e^{X}))^2 = M_X(2) - M_X(1)^2 = e^{\frac{4}{2}} - (e^{\frac{1}{2}})^2 = e^2 - e^1 = Var(e^Y)$ .

Therefore, $Cor(e^X, e^Y) = \frac{e(e^{\rho} - 1)}{e^2 - e^1} =\frac{e^{\rho} - 1}{e - 1}$ .

Observe that the mininum correlation of $e^X$ and $e^Y$ is $\frac{-1}{e}$ .

Back to the important observation

$Cor(e^X, e^Y)$ and $Cor(X,Y) = \rho$ always have the same sign. Why is this true?

Because, $f(x) = e^x$ is an increasing function. So, if $X$ and $Y$ are positively correlated then, as $X$ increases, $Y$ also increases in general, hence, $e^X$ also increases along with $e^Y$ hence, the result, which is quite intuitive.

Observe that in place of $f(x) = e^x$ if we would have taken, any increasing function $f(x)$ , this will be the case. Can you prove it?

Research Problem of the day ( Is the following true? )

Let $f(x)$ be an increasing function of $x$ , then

$Cor(f(X), f(Y)) > 0 \iff Cor(X, Y) > 0$
$Cor(f(X), f(Y)) = 0 \iff Cor(X, Y) = 0$
$Cor(f(X), f(Y)) < 0 \iff Cor(X, Y) < 0$