How can we derive expectation of two dependent normal distribution?

Question

$\mathbf{X}$ and $\mathbf{Y}$ are each dependent normal random variable, then how can we derive like this one? $$\mathbf{E}\{e^{\mathbf{X}}e^{\mathbf{Y}}\}$$ I know the each first moment is $\mathbf{E}\{e^{\mathbf{X}}\}=exp(m_{\mathbf{X}}+\frac12\sigma^2_{\mathbf{X}})$ so if their are independent(no correlation) that equation changes like $$\mathbf{E}\{e^\mathbf{X}e^\mathbf{Y}\}=\mathbf{E}\{e^\mathbf{X}\}\mathbf{E}\{e^\mathbf{Y}\}=exp(m_\mathbf{X}+m_\mathbf{Y})exp(\frac12(\sigma^2_\mathbf{X}+\sigma^2_\mathbf{Y})).$$ But if their are dependent each other with correlation coefficient $\rho$, that equation changes like $$\mathbf{E}\{e^\mathbf{X}e^\mathbf{Y}\}=exp(m_\mathbf{X}+m_\mathbf{Y})exp(\frac12(\sigma^2_\mathbf{X}+\sigma^2_\mathbf{Y}+2\rho\sigma_\mathbf{X}\sigma_\mathbf{Y})).$$ I don't know how to explain why the component $2\rho\sigma_\mathbf{X}\sigma_\mathbf{Y}$ is added because derivation of that equation is too long. I hope you advise me just appropriate textbook or papers including that problem. Thank you.

Answer 1

Consider $Z = a X + b Y$ for some yet unknown coefficients $a$ and $b$. $Z$ is Gaussian random variable as a linear combination of jointly Gaussian random variables $X$ and $Y$.

Moreover $X$ and $Z$ are jointly Gaussian, since $(X,Z)^T = \begin{pmatrix} 1 & 0 \cr a & b \end{pmatrix} (X,Y)^T$. Since $X$ and $Z$ are jointly Gaussian, their distribution is determinated by their means, and covariance matrix. We will attempt to choose indeterminates $a$ and $b$ so that $X$ and $Z$ are independent, i.e. $\operatorname{Cov}(X,Z)=0$: $$ \mathbb{Cov}(X,Z) = a \sigma_X^2 + b \rho \sigma_X \sigma_Y = \sigma_X \left( a \sigma_X + b \rho \sigma_Y \right) $$ Pick a solution, for example $b = 1$ and $a = -\rho \frac{\sigma_Y}{\sigma_X}$.

Then $Y = \frac{Z-a X}{b}$ and the expectation you seek becomes easy $$ \mathbb{E}\left(\mathrm{e}^X \mathrm{e}^Y\right) = \mathbb{E}\left(\mathrm{e}^{X (1-a/b)} \mathrm{e}^{Z/b}\right) = \mathbb{E}\left(\mathrm{e}^{X (1-a/b)} \right) \mathbb{E}\left( \mathrm{e}^{Z/b}\right) = \exp\left( \mathbb{E}\left(X(1-a/b)\right) + \frac{1}{2} \mathbb{Var}\left(X(1-a/b)\right)\right) \cdot \exp\left( \mathbb{E}\left(Z/b\right) + \frac{1}{2} \mathbb{Var}\left(Z/b\right)\right) $$ It remains to fill in rather tedious algebra: $$ \mathbb{E}\left(X(1-a/b)\right) = \mathbb{E}\left(X\right) (1-a/b) = m_X + \rho \frac{\sigma_Y}{\sigma_X} m_X $$ $$ \mathbb{E}(Z/b) = \frac{a}{b} \mathbb{E}(X) + \mathbb{E}(Y) = m_Y - \rho \frac{\sigma_Y}{\sigma_X} m_X $$ $$ \mathbb{Var}\left(X(1-a/b)\right) = (1-a/b)^2 \mathbb{Var}(X) = \sigma_X^2 \frac{(\sigma_X + \rho \sigma_Y)^2}{\sigma_X^2} = (\sigma_X + \rho \sigma_Y)^2 $$ $$ \mathbb{Var}\left(Z/b\right) = \frac{1}{b^2} \mathbb{Var}(Z) = \frac{1}{b^2} \left( a^2 \sigma_X^2 + b^2 \sigma_Y^2 + 2 \rho a b \sigma_X \sigma_Y \right) = \sigma_Y^2 \left(1 - \rho^2\right) $$

Combining these terms give the desired result:

In[13]:= ((1 - a/b) mX + 1/2 (1 - a/b)^2 sX^2) + ((a/b mX + mY) + 
     1/2 1/b^2 (a^2 sX^2 + b^2 sY^2 + 2 r a b sX sY)) /. {b -> 1, 
   a -> -r sY/sX} // Expand

Out[13]= mX + mY + sX^2/2 + r sX sY + sY^2/2