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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 \(e^{Y}\).

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

\(M_X(t) = E(e^{tX})\) 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:

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

\( Cor(e^X, e^Y)\) 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

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

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.

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 \(e^{Y}\).

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

\(M_X(t) = E(e^{tX})\) 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:

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

\( Cor(e^X, e^Y)\) 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

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

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