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

  • $\begingroup$ Did you lose the right parenthesis on the exponents? $\endgroup$
    – Hans
    May 4, 2016 at 3:46
  • 1
    $\begingroup$ @Jim, there was a problem with my previous answer. It is now fixed. $\endgroup$
    – Quantuple
    May 4, 2016 at 9:31

1 Answer 1


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)

  • $\begingroup$ For the line when you calculate the mean, how did you get the $r_f$ to cancel out. $\endgroup$
    – Jim
    May 4, 2016 at 10:29
  • $\begingroup$ 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? $\endgroup$
    – Jim
    May 4, 2016 at 10:57
  • $\begingroup$ @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 $\mu_{\tilde{X}} = \ln(\tilde{X_0})+(r_f-r_d-\sigma_{\tilde{X}}^2/2)t$ $\endgroup$
    – Quantuple
    May 4, 2016 at 11:01
  • $\begingroup$ @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 $X_t$. $\endgroup$
    – Quantuple
    May 4, 2016 at 11:03
  • $\begingroup$ perfect explanations with excellent examples. $\endgroup$
    – Jim
    May 4, 2016 at 11:15

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