Pricing an fx option in the same currency

Question

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?

Answer 1

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*}