If $X$ and $Y$ are Gaussian random variables with $\mathrm{Cov}(X,Y) = c$, then \begin{equation}\label{eq:20171005} \mathbb{E}\big[e^X f(Y)\big] = \mathbb{E}\big[e^X \big] \mathbb{E}\big[ f(Y + c)\big] \end{equation}

My proof goes as follows. It is sufficient to prove formula \eqref{eq:20171005} for functions of the form $f(x) = e^{i k x}$ because "all" functions can be expanded in $e^{i k x}$ by Fourier transformation. Then, the left hand side of \eqref{eq:20171005} is \begin{equation*} \mathbb{E}\big[e^{X + i k Y}\big] = \exp \left[ \mathbb{E} [ X + i k Y] + \frac{1}{2} \mathrm{Var}[ X + i k Y] \right] = \exp \left[ \mathbb{E} [ X] + i k\ \mathbb{E}[Y] + \frac{1}{2} \mathrm{Var}[ X ] + i k\ \mathrm{Cov}[X,Y] + \frac{1}{2} (ik)^2\ \mathrm{Var}[Y] \right] \end{equation*} The right hand side of \eqref{eq:20171005} is \begin{equation*}\mathbb{E}\big[e^{X} \big] \mathbb{E}\big[e^{i k (Y + c)}\big] = \exp \left[ \mathbb{E} [ X] + \frac{1}{2} \mathrm{Var}[ X ]\right] \exp\left[i k c + ik\ \mathbb{E}[Y] +\frac{1}{2} (ik)^2\ \mathrm{Var}[Y]\right] \end{equation*} which is clearly equal to the left hand side of \eqref{eq:20171005}.As an example of the use of \eqref{eq:20171005}, I calculate the price of an option to exchange one asset for another. I assume the reader is familiar with the Black-Scholes model. Suppose $S_1(T)$ and $S_2(T)$ are the prices of two risky assets at time $T$. The option gives the buyer the right, but not the obligation, to exchange the second asset for the first at the time of maturity $T$. In other words, the payoff is $\max(0, S_1(T) - S_2(T))$. I calculate \begin{equation*} \mathbb{E}\left[ \max( S_1 - S_2,0 )\right] =\mathbb{E}\left[ S_2 \max\left( \frac{S_1}{S_2} - 1,0 \right)\right] = \mathbb{E}\left[ S_2\right] \mathbb{E}\left[ \max\left( \frac{S_1}{S_2}e^c - 1,0 \right)\right] \end{equation*} with $ c = \mathrm{Cov}( \log S_2 , \log\frac{S_1}{S_2} )$. The calculation is now reduced to one dimension only. The Black formula gives \begin{equation*} \mathbb{E}\left[ \max(A -1,0 )\right] = \mathbb{E}\left[A \right] N(d_{+}) - N(d_{-}) \end{equation*} with $A = \dfrac{S_1}{S_2} e^c$ and \begin{equation*} d_{\pm} = \frac{\log \left( \mathbb{E}\left[A \right] \right)\pm \frac{1}{2} w}{ \sqrt{w}} \end{equation*} and $w = \mathrm{Var} \left[ \log A \right]$. Using \eqref{eq:20171005} again gives \begin{equation*} \mathbb{E}\left[ S_1 \right] = \mathbb{E}\left[ S_2 \right] \mathbb{E}\left[ \frac{S_1}{S_2} e^c\right] \end{equation*} thus \begin{equation*} \mathbb{E}\left[A\right] = \mathbb{E}\left[ \frac{S_1}{S_2} e^c\right] = \frac{\mathbb{E}\left[ S_1 \right]}{\mathbb{E}\left[ S_2 \right]} \end{equation*} The expectation value of the spot is the forward price, thus $\mathbb{E}\left[ S_1 \right] = F_1$ and $\mathbb{E}\left[ S_2 \right] = F_2$. The undiscounted price of the exchange option is therefore \begin{equation}\label{eq:20171006} F_1\ N(d_{+}) - F_2\ N(d_{-}) \end{equation} with \begin{equation*} d_{\pm} = \frac{\log \left( \dfrac{F_1}{F_2} \right)\pm \frac{1}{2} w}{ \sqrt{w}} \end{equation*} where \begin{equation*} w = \mathrm{Var}[ \log A ] = \mathrm{Var}\left[ \log S_1 - \log S_2\right] = (\sigma_1^2 - 2 \rho \sigma_1 \sigma_2 +\sigma_2^2 )T \end{equation*} Notice that I did not need to calculate $c$. Formula \eqref{eq:20171006} is known as Margrabe's formula.

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