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.
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 \(e^{Y}\).
\(M_X(t) = E(e^{tX})\) 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.
\( \sigma^2 = Var(X+Y) = Var(X) + 2Cov(X,Y) + Var(Y) = 1 + 2\rho + 1 = 2(1+\rho) \).
Now observe the following:
\( Cor(e^X, e^Y)\) 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 \(\frac{-1}{e}\).
\( Cor(e^X, e^Y)\) 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