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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

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

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

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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