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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.

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

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

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