# Quanto/Compo adjustments - Product of two geometric brownian motion

Let's say I have two processes $X_t =X_0 \exp((a-\frac{1}{2}\sigma_X^2)t +\sigma_X dW_t^1)$ and $Y_t=Y_0 \exp((b-\frac{1}{2}\sigma_Y^2)t +\sigma_Y dW_t^2)$ and I then multiply them together (like converting a foreign asset into domestic currency). I arrive at

$X_tY_t = X_0Y_0 \exp((a+b-\frac{1}{2}\sigma_X^2-\frac{1}{2}\sigma_Y^2)t +\sigma_X dW_t^1+\sigma_Y dW_t^2)$

If I then assume the $W_t$ have a correlation of $\rho$, how would I obtain the expectation of $X_tY_t$. The thing that is troubling me is because I have 2 Brownian motions and I'm not sure how to compute from there.

Edit: I would like to make this question a little bit more general. What would I do if I wanted to compute $E[X_t Y_s]$, where say $s < t$. So something to do with independent increments would play a factor.

Just use the fact that $$\sigma_X W_t^1 + \sigma_Y W_t^2 = \sqrt{ \sigma_X^2 + \sigma_Y^2 + 2\rho\sigma_X\sigma_Y } W_t$$ holds in probability assuming that $W_t^1$ and $W_t^2$ are 2 correlated Brownian motions with $$d\langle W_t^1, W_t^2 \rangle_t = \rho dt$$ and $W_t$ is a new standard Brownian motion defined over the same probability space.

Simply put, just replace your sum of two correlated Gaussians (LHS above) by a single Gaussian (RHS above) exhibitting the exact same statistical properties (for a Gaussian identical mean/variance is enough). Doing so, you can now use the formulas you are accustomed to.

Applying this shows $$X_t Y_t = X_0Y_0 \exp((a+b-\frac{1}{2}\sigma_X^2-\frac{1}{2}\sigma_Y^2)t +\sqrt{ \sigma_X^2 + \sigma_Y^2 + 2\rho\sigma_X\sigma_Y } W_t)$$ is lognormally distributed with mean $$\mu = \ln(X_0Y_0)+(a+b-\frac{1}{2}\sigma_X^2-\frac{1}{2}\sigma_Y^2)t$$ and variance $$\sigma^2 = (\sigma_X^2 + \sigma_Y^2 + 2\rho\sigma_X\sigma_Y)t$$

hence applying the usual formula for the mean of a lognormally distributed variable $$E[X_t Y_t] = e^{\mu + \sigma^2/2}$$ is a function of $\rho$.

I think you are completely mixing up two different problems (here two different probability measures).

I understand from your comment that you define the dynamics of the FOR/DOM instantaneous exchange rate $X_t$ (i.e. $X_t = x$ meaning that, at time $t$, 1 unit of foreign currency = x units of domestic currency) under the foreign risk-neutral measure $\mathbb{Q}^f$ (or rather the probability space $(\Omega,\mathcal{F},\mathbb{Q}^f)$ along with the price process of an equity underlying denominated in the foreign currency.

In that case, under the domestic risk-neutral measure $\mathbb{Q}^d$ (or rather the probability space $(\Omega,\mathcal{F},\mathbb{Q}^d)$), one should have, by absence of arbitrage opportunities and assuming market completeness: $$\frac{Y_t X_t}{B_t^d} \text{ is a } \mathbb{Q}^d \text{- martingale}$$

with $B^d_t$ representing the time-$t$ value of a risk-free money market account in the domestic economy in which 1 unit of currency has been invested at $t=0$. Using the martingale property it then entails that: $$\frac{Y_0 X_0}{B_0^d} = Y_0 X_0 = E^{\mathbb{Q}^d} \left[ \frac{Y_t X_t}{B_t^d} \vert \mathcal{F}_0 \right]$$ and further assuming deterministic rates: $$E ^{\mathbb{Q}^d} \left[ X_t Y_t \vert \mathcal{F}_0 \right] = e^{-r_d t} X_0 Y_0$$ which is independent of $\rho$.

This is known as a compo adjustment in deriatives lingo.

That being said, under the measure $\mathbb{Q}^f$ where you have originally defined $X_t$ and $Y_t$, the expectation $E^{\mathbb{Q}^f} [ Y_t X_t ]$ does dependent on $\rho$ as hinted above.

• Thanks for that, but I guess I might be having trouble applying this to my problem. My problem has $X_t =X_0\exp((r_d-r_f-\frac{1}{2}\sigma_X^2)t+\sigma_X W_t^1)$ and $Y_t =Y_0\exp((r_f-\frac{1}{2}\sigma_Y^2)t+\sigma_Y W_t^2)$. What would be the expectation of $X_tY_t$ because my answer involves $\rho$, when I know for certain that it should not. – Jim May 3 '16 at 11:36
• Do you agree that if you manage to show that $Y_t X_t / B_t^d$ is a $\mathbb{Q}^d$-martingale then you are done? – Quantuple May 3 '16 at 13:07