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# 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

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