# Girsanov Theorem for Quanto/Compo adjustment

Assume that I have a foreign asset $$Y_t = Y_0 \exp \left((r_f-\frac{1}{2}\sigma^2_Y)t+\sigma_Y W_t^1\right)$$ and an exchange rate $$X_t = X_0 \exp\left((r_d-r_f-\frac{1}{2}\sigma^2_X)t+\sigma_X W_t^2\right)$$

I would like to compute the expectation of $Y_tX_t$ under the domestic rsik-neutral market measure. I know I would like to use Girsanov, but am not sure how to approach this.

My ultimate goal would then be to extend the workings to $Y_t^2 X_t$ or $X_t^2 Y_t$ or $X_t^2 Y_t^2$ etc. so this change of measure would be useful to me

• Did you lose the right parenthesis on the exponents? – Hans May 4 '16 at 3:46
• @Jim, there was a problem with my previous answer. It is now fixed. – Quantuple May 4 '16 at 9:31

Assume deterministic and constant interest rates.

For an investor in the foreign economy i.e. a market participant that can only trade assets delivering a payout in the foreign currency, let us define

$$\tilde{X}_t = \tilde{X}_0 \exp \left(\left(r_f-r_d-\frac{\sigma_\tilde{X}^2}{2}\right)+\sigma_\tilde{X} W_t^{\tilde{X},\mathbb{Q}^f} \right)$$

$$Y_t =Y_0\exp \left(\left(r_f-\frac{\sigma_Y^2}{2}\right)t+\sigma_Y W_t^{Y,\mathbb{Q}^f} \right)$$ where

• $$\mathbb{Q}^f$$ figures the foreign risk-neutral measure (risk-free MMA $$B^f_t = \exp(r_f\ t)$$ is the numéraire).
• $$\tilde{X}_t$$ representing the instantaneous DOM/FOR exchange rate. $$\tilde{X}_t = \text{x}$$ means that, at time $$t$$, 1 unit of domestic currency equals $$\text{x}$$ units of foreign currency.
• $$Y_t$$ an equity underlying denominated in the foreign currency.

Let's further assume that the 2 Brownian motions $$W_t^{\tilde{X},\mathbb{Q}^f}$$ and $$W_t^{Y,\mathbb{Q}^f}$$ are correlated $$d\langle W^{\tilde{X},\mathbb{Q}^f}, W^{Y,\mathbb{Q}^f} \rangle_t = \rho dt$$

Notice how I have used $$\tilde{X}_t$$ (DOM/FOR) and not $$X_t$$ (FOR/DOM) as you propose, because in the foreign economy, the only tradable assets are: $$Y_t$$, $$B^f_t$$ and $$B^d_t \tilde{X}_t$$ as hinted above (and these should be all $$\mathbb{Q}^f$$-martingales when expressed under the numéraire $$B_t^f$$). We do have the relationship, $$\tilde{X}_t = 1/X_t$$.

Thanks to the fundamental theorem of asset pricing, for any tradable asset $$V_t$$ denominated in the foreign currency, we have that, under the foreign risk-neutral measure $$\mathbb{Q}^f$$

$$\frac{V_t}{B^f_t} \text{ is a } \mathbb{Q}^f \text{- martingale} \iff \frac{V_0}{B^f_0} = E^{\mathbb{Q}^f}_0 \left[ \frac{V_t}{B^f_t} \right]$$

Under the domestic risk-neutral measure $$\mathbb{Q}^d$$ (risk-free MMA $$B^d_t = \exp(r_d\ t)$$ is the numéraire)

$$\frac{V_t/\tilde{X}_t}{B^d_t} \text{ is a } \mathbb{Q}^d \text{- martingale} \iff \frac{V_0/\tilde{X}_0}{B^d_0 } = E^{\mathbb{Q}^d}_0 \left[ \frac{V_t /\tilde{X}_t^d}{B^d_t} \right]$$ in words, the foreign asset value converted to domestic currency units is a martingale under the domestic risk-neutral measure.

From the above, we see that the Radon-Nikodym derivative writes $$\left. \frac{d\mathbb{Q}^d}{d\mathbb{Q}^f} \right\vert_{\mathcal{F}_0} = \frac{B_0^f B_t^d \tilde{X}_t}{B_t^f B_0^d \tilde{X}_0}$$ yet because \begin{align} \tilde{X}_t &= \tilde{X}_0\exp \left(\left(r_f-r_d-\frac{1}{2}\sigma_\tilde{X}^2\right)t+\sigma_\tilde{X} W_t^{\tilde{X},\mathbb{Q}^f} \right) \\ &= \tilde{X}_0 \frac{B^f_t}{B^f_0}\frac{B^d_0}{B^d_t}\exp \left(-\frac{1}{2}\sigma_\tilde{X}^2t+\sigma_\tilde{X} W_t^{\tilde{X},\mathbb{Q}^f} \right) \end{align}

this deriative also writes \begin{align} \left. \frac{d\mathbb{Q}^d}{d\mathbb{Q}^f} \right\vert_{\mathcal{F}_0} &= \exp\left(\sigma_\tilde{X} W_t^{\tilde{X},\mathbb{Q}^f}-\frac{1}{2}\sigma_\tilde{X}^2t\right) \\ &= \mathcal{E}\left(\sigma_\tilde{X} W_t^{\tilde{X},\mathbb{Q}^f} \right) \end{align} which is indeed a well-behaved Doléans-Dade exponential where we've used the notation $$\mathcal{E}(M_t) = \exp \left( M_t - \frac{1}{2}\langle M \rangle_t \right)$$ to denote the stochastic exponential.

Hence Girsanov theorem can be applied to transform Brownian motions under $$\mathbb{Q}^f$$ as Brownian motions under $$\mathbb{Q}^d$$. How does it work?

Girsanov Theorem (non rigourous version) - Let $$W_t^{\mathbb{Q^f}}$$ represent a standard Brownian motion under $$\mathbb{Q^f}$$ and assume the Radon-Nikodym derivative can be written as: $$\left. \frac{d\mathbb{Q}^d}{d\mathbb{Q}^f} \right\vert_{\mathcal{F}_0} = \mathcal{E}(L_t)$$ In that case, the process $$W_t^{\mathbb{Q^d}}$$ defined as $$W_t^{\mathbb{Q^d}} = W_t^{\mathbb{Q^f}} - \langle W^{\mathbb{Q^f}}, L \rangle_t$$ is a standard a Brownian motion under $$\mathbb{Q^d}$$.

In our particular example, we see that $$L_t := \sigma_\tilde{X} W_t^{\tilde{X},\mathbb{Q}^f}$$

Applying Girsanov theorem then allows us to write \begin{align} W_t^{\tilde{X},\mathbb{Q}^d} &= W_t^{\tilde{X},\mathbb{Q}^f} - \langle W^{\tilde{X},\mathbb{Q}^f}, \sigma_\tilde{X} W^{\tilde{X},\mathbb{Q}^f} \rangle_t \\ &= W_t^{\tilde{X},\mathbb{Q}^f} - \sigma_\tilde{X}t \\ W_t^{Y,\mathbb{Q}^d} &= W_t^{Y,\mathbb{Q}^f} - \langle W^{Y,\mathbb{Q}^f}, \sigma_\tilde{X} W^{\tilde{X},\mathbb{Q}^f} \rangle_t \\ &= W_t^{Y,\mathbb{Q}^f} - \rho \sigma_\tilde{X} t \end{align} meaning that, to move from $$\mathbb{Q}^f$$ to $$\mathbb{Q}^d$$ one can just replace \begin{align} W_t^{\tilde{X},\mathbb{Q}^f} = W_t^{\tilde{X},\mathbb{Q}^d} + \sigma_\tilde{X} t \\ W_t^{Y,\mathbb{Q}^f} = W_t^{Y,\mathbb{Q}^d} + \rho \sigma_\tilde{X} t \\ \end{align} in the expressions for $$\tilde{X}_t$$ and $$Y_t$$ to obtain: \begin{align} \tilde{X}_t = \tilde{X}_0 \exp \left(\left(r_f - r_d + \frac{\sigma_\tilde{X}^2}{2}\right) t + \sigma_\tilde{X} W_t^{\tilde{X},\mathbb{Q}^d} \right) \\ Y_t = Y_0 \exp \left(\left(r_f + \rho \sigma_\tilde{X} \sigma_Y - \frac{\sigma_Y^2}{2}\right) t + \sigma_Y W_t^{Y,\mathbb{Q}^d} \right) \end{align}

Now assume we want to compute the expectation of $$Y_tX_t = Y_t/\tilde{X}_t$$ under $$\mathbb{Q}^d$$. The random variable $$Y_t/\tilde{X}_t$$ being lognormally distributed (ratio of two lognormals) with mean $$\mu = \ln(Y_0/\tilde{X}_0) + \left(r_d - \frac{\sigma^2_X - 2\rho\sigma_\tilde{X}\sigma_Y + \sigma_Y^2}{2}\right)t$$ and variance $$\sigma^2 = \left(\sigma_\tilde{X}^2 - 2 \rho \sigma_\tilde{X} \sigma_Y + \sigma_Y^2 \right)t$$ applying the usual formula gives \begin{align} E^{\mathbb{Q}^d}[Y_t/\tilde{X}_t] &= \exp \left(\mu+\frac{\sigma^2}{2} \right) \\ &= Y_0/\tilde{X}_0 \exp \left(r_d t \right) \\ &= Y_0/\tilde{X}_0 B_t^d \end{align} hence $$E^{\mathbb{Q}^d} \left[ \frac{Y_t/\tilde{X}_t}{B_t^d} \right] = \frac{Y_0/\tilde{X}_0}{B_0^d}$$ as it should since we already knew that $$\frac{Y_t/\tilde{X}_t}{B^d_t} \text{ was a } \mathbb{Q}^d \text{- martingale}$$

For quanto derivatives we prefer to express the equity/forex dynamics in terms of $$X_t$$ the FOR/DOM exchange rate instead of the DOM/FOR exchange rate $$\tilde{X}_t$$. This can be done through a simple application of Itô's lemma noticing that $$\tilde{X}_t = 1/X_t$$. This would typically yield: \begin{align} \frac{dX_t}{X_t} = (r_d - r_f) dt + \sigma_X dW_t^{X,\mathbb{Q}^d} \\ \frac{dY_t}{Y_t} = (r_f - \rho_{XY}\sigma_X\sigma_Y) dt + \sigma_Y dW_t^{Y,\mathbb{Q}^d} \end{align} where we have introduced $$W_t^{X,\mathbb{Q}^d} = -W_t^{\tilde{X},\mathbb{Q}^d}$$ such that $$\langle W_t^{X,\mathbb{Q}^d}, W_t^{Y,\mathbb{Q}^d} \rangle_t = \rho_{XY} t = -\rho t$$ and we used $$\sigma_X = \sigma_{\tilde{X}}$$ for clarity.

Hence finally, under $$\mathbb{Q}^d$$ we can write:

$$X_t = X_0 \exp \left(\left(r_d-r_f-\frac{\sigma_X^2}{2}\right)+\sigma_X W_t^{X,\mathbb{Q}^d} \right)$$

$$Y_t = Y_0\exp \left(\left(r_f - \rho_{XY}\sigma_X\sigma_Y -\frac{\sigma_Y^2}{2}\right)t + \sigma_Y W_t^{Y,\mathbb{Q}^d} \right)$$

where the quantity $$F(0,t) = E^{\mathbb{Q}^d}_0 \left[ Y_t \right] = Y_0\exp \left(\left(r_f - \rho_{XY}\sigma_X\sigma_Y\right)t\right)$$ is known as the quanto forward.

and it is once again easy to show that $$\frac{Y_tX_t}{B_t^d} \text{ is a } \mathbb{Q}^d \text{- martingale}$$ using the fact that $$Z=Y_t X_t$$ is a product of lognormals (and not a ratio as before)

• For the line when you calculate the mean, how did you get the $r_f$ to cancel out. – Jim May 4 '16 at 10:29
• Ah I got it. What would happen if I didn't use $\tildeX$ and just did the whole calculations with $X_t$ and the new $Y_t$. Wouldn't you still get the same answer? – Jim May 4 '16 at 10:57
• @Jim, Sorry I had left some typos. Basically, because $Z = Y_t/\tilde{X}_t$ is a ratio of lognormally-distributed variables, $\ln(Z) = \ln(Y_t) - \ln(\tilde{X}_t)$ is normally distributed with mean $\mu = \mu_{Y} - \mu_{\tilde{X}}$ where $\mu_Y = \ln(\tilde{Y_0})+r_f+\rho\sigma_\tilde{X} \sigma_Y - \sigma_Y^2/2)t$ and $\mu_{\tilde{X}} = \ln(\tilde{X_0})+(r_f-r_d-\sigma_{\tilde{X}}^2/2)t$ – Quantuple May 4 '16 at 11:01
• @Jim, you can but be careful under what measure you write your dynamics... The $X_t$ you provide is the arbitrage free solution of a GBM-like dynamics specified under $\mathbb{Q}^d$ (see second part of my answer), while the $Y_t$ you propose is the arbitrage free solution of a GBM-like dynamics under $\mathbb{Q}^f$. You need to work under the same probability measure. So either you work with a quanto adjusted $Y_t$ under $\mathbb{Q}^d$ (see second part of my post), or you work under $\mathbb{Q}^f$ with $\tilde{X}_t$ instead of $X_t$. – Quantuple May 4 '16 at 11:03
• perfect explanations with excellent examples. – Jim May 4 '16 at 11:15