Let imagine we have an option from EUR to USD priced in EUR, therefore the payoff for a call is:

$$\frac{(S - K)^{+}}{S} = K (1/K - 1/S)^{+}$$ This is basically the payoff of a price of a put on 1/S multiply by K.

The discounted expectation gives the price. What is the way to price this option with Black Scholes?


Let $P^d$ and $P^f$ denote the respective USD and EUR risk-neutral measures. We assume that, under the USD risk-neutral measure, \begin{align*} dS_t = S_t \Big(\big(r^d-r^f \big)dt +\sigma dW_t \Big), \end{align*} where

  • $r^d$ and $r^f$ denote respectively the USD and EUR interest rates,
  • $\sigma$ is the constant volatility, and
  • $W_t$ is a standard Brownian motion.

Then \begin{align*} \frac{dP^f}{dP^d}\big|_t &= \frac{S_t B^f_t}{S_0 B^d_t}\\ &=\exp\left(-\frac{\sigma^2}{2}t +\sigma W_t\right), \end{align*} where $B^d_t$ and $B^f_t$ are the respective USD and EUR money-market account values at time $t$. Consequently, \begin{align*} \hat{W}_t = W_t -\sigma t \end{align*} is a standard Brownian motion under the EUR risk-neutral measure. Moreover, \begin{align*} d\frac{1}{S_t} &= \frac{1}{S_t}\Big(\big(r^f-r^d +\sigma^2\big)dt -\sigma dW_t \Big)\\ &=\frac{1}{S_t}\Big(\big(r^f-r^d \big)dt +\sigma d(-\hat{W}_t) \Big) \\ &=\frac{1}{S_t}\Big(\big(r^f-r^d \big)dt +\sigma d\tilde{W}_t \Big), \end{align*} where $\tilde{W}_t = -\hat{W}_t$ is also a standard Brownian motion under the foreign risk-neutral measure.

Let $E^d$ and $E^f$ be expectations corresponding to USD and EUR risk-neutral measures $P^d$ and $P^f$. Then \begin{align*} E^f\bigg( \frac{1}{B^f_TS_T} (S_T-K)^+\bigg) &= E^d\bigg(\frac{S_T B^f_T}{S_0 B^d_T} \frac{1}{B^f_TS_T} (S_T-K)^+\bigg)\\ &= \frac{1}{S_0}E^d\bigg(\frac{1}{B^d_T} (S_T-K)^+\bigg)\\ &= e^{-r^d T}\frac{1}{S_0}\left[F\Phi(d_1) - K\Phi(d_2) \right], \end{align*} where

  • $F= S_0 e^{(r^d-r^f)T}$,
  • $\Phi$ is the cumulative distribution function of a standard normal random variable,
  • $d_1 = \Big[\ln \frac{S_0}{K} + \big(r^d-r^f + \frac{1}{2}\sigma^2 \big)T\Big]/(\sigma \sqrt{T})$, and
  • $d_2 = d_1 -\sigma \sqrt{T}$.

Similarly, \begin{align*} E^f\bigg(\frac{K}{B^f_T}\Big(\frac{1}{K} - \frac{1}{S_T}\Big)^+\bigg) &= e^{-r^f T} K\bigg[\frac{1}{K}\Phi(-\hat{d}_2) -\hat{F}\Phi(-\hat{d}_1) \bigg], \end{align*} where

  • $\hat{F}= \frac{1}{S_0}e^{(r^f-r^d)T}$,
  • $\hat{d}_1 = \Big[\ln \frac{K}{S_0} + \big(r^f-r^d + \frac{1}{2}\sigma^2 \big)T\Big]/(\sigma \sqrt{T}) = -d_2$, and
  • $\hat{d}_2 = \hat{d}_1 - \sigma \sqrt{T} = -d_1$.

Then, \begin{align*} E^f\bigg(\frac{K}{B^f_T}\Big(\frac{1}{K} - \frac{1}{S_T}\Big)^+\bigg) &= e^{-r^f T} K\bigg[\frac{1}{K}\Phi(-\hat{d}_2) -\hat{F}\Phi(-\hat{d}_1) \bigg]\\ &=e^{-r^f T} K\bigg[\frac{1}{K}\Phi(d_1) -\hat{F}\Phi(d_2) \bigg]\\ &= e^{-r^d T}\frac{1}{S_0}\left[F\Phi(d_1) - K\Phi(d_2) \right]. \end{align*}

| improve this answer | |
  • $\begingroup$ Thanks. I do not really get the first part on the change of measure but at the end the point is just to see that the expectation can be in the form of the typical call price divided by the spot. Once you have that the option can be priced. One thing that is surprising that the price is just the option price in USD divided by the spot, intuitively I would have thought there would have been more than simply converting USD to EUR. $\endgroup$ – user26616 Mar 13 '19 at 10:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.