Currency Black–Scholes Pricing and Inverting the Exchange Rate

# Black-Scholes formula for currency exchange rates


# Black-Scholes formula for currency exchange rates












Exercise 17.6 of Options, Futures, And Other Derivatives by John Hull (Page 397):

> Show that the formula in equation (17.12) for a put option to sell one unit of currency A for currency B at strike price K gives the same value as equation (17.11) for a call option to buy K units of currency B for currency A at strike price 1>K.

Formulas 17.12 and 17.11 that were mentioned are, respectively:

$$c =S_0e^{-r_fT}N(d_1) - Ke^{-rT}N(d_2) $$ $$c =Ke^{-rT}N(-d_2)- S_0e^{-r_fT}N(-d_1) $$ where $$d_1 = \frac{ln(S_0/K)+(r-r_f+\sigma^2/2)T}{\sigma\sqrt{T}}$$ $$d_2 = \frac{ln(S_0/K)+(r-r_f-\sigma^2/2)T}{\sigma\sqrt{T}}$$ $S_0$, $r$ and $r_f$ are the value of one unit of A in terms of B, domestic and foreign risk-free rate, respectively.

The way the book wants us to solve this practice is not that much challenging, indeed the solution manual solved the exercise in this manner: A put option to sell one unit of currency A for K units of currency B is worth: $$ Ke^{-r_BT}N(-d_2)-S_0e^{-r_AT}N(-d_1)\,\,\,\,\,\,\,\,\,(1)$$ Factoring $KS_0$ we have: $$KS_0[S_0^*e^{-r_BT}N(d_1^*)-K^*e^{-r_AT}N(d_2^*)]\,\,\,\,\,\,\,\,\,(2)$$ where the letters with asterisk are duals of the same letters in (1), but from point of view of the A currency (i.e $S_0^* = 1/S_0$ etc.). With (2), we solved the exercise. But here emerges my problem; The book's approach to prove Black-Scholes formulas (17.11 and 17.12) is to use risk-neutrality to derive $E[max(S_T-K.0)]e^{-rT}$. The distribution of the random variable $S_T$ is assumed to be lognormal, that is: $$ln(S_T)\sim\phi[ln(S_0)+(r_B-r_A-\sigma^2/2)T,\sigma^2T]\,\,\,\,\,\,\,\,\,\,(3)$$ that is, $ln(S_T)$ is normally distributed with mean and variance as above. Now if we accept that one of the currencies' distribution (say the domestic one) conforms with (3), then necessarily we accepted that the other exchange rate, that is the value of domestic currency in terms of foreign currency at time T is a RV with lognormal distribution with mean and variance as follows: $$ln(1/S_T) = ln(S_t^*)\sim\phi[ln(S_0^*)+(r_A-r_B+\sigma^2/2)T,\sigma^2T]\,\,\,\,\,\,\,\,\,\,(4)$$

Now, there is an evident difference between (3) and (4); The sign of variance term in the mean is flipped. Hence, this is not that form of lognormal distribution derived for geometric brownian motion upon which the Black-Scholes formula derivation, at least in this book, was explicitly founded. By risk-neutrality approach we proceed to derive $E(max(S_t^*-k,0))$ to obtain c, the value of the call option with strike price $K^*$ (we assume $S_T^*$'s P.D.F is $f(x)$): $$c = E(max(S_t^*-K^*,0))e^{-r_AT} = \int_{K^*}^\infty x f(x)dx - K^* \int_{K^*}^\infty f(x) dx$$ $$= \frac{1}{\sqrt{2\pi} \sigma \sqrt{T}} \int_{ln(K^*)}^\infty e^u e^{-\frac{(u-\mu)^2}{2 \sigma^2T}}du-K^*N(\frac{ln(S^*/K^*)+(r_A-r_B+\sigma^2/2)T}{\sigma\sqrt{T}})\,\,\,\,(5)$$ where $\mu$ is the mean in (4). The first integral is equal to: $$=\frac{1}{\sqrt{2\pi} \sigma \sqrt{T}}\int_{ln(K^*)}^\infty Exp (-\frac{(u-\mu-\sigma^2T)^2-(\sigma^4T^2+2\mu\sigma^2T)}{2\sigma^2T})$$ $$=Exp[\sigma^2T/2+ln(S_0^*)+(r_A-r_B+\sigma^2/2)T].N(\frac{ln(S_0^*/K_0^*)+(r_A-r_B+3\sigma^2/2)T}{\sigma\sqrt{T}})\,\,\,\,\,(6)$$ Accordingly the c is obtained as: $$c = E(max(S_T^*-K^*,0))e^{-r_AT}=S_0^*e^{(-r_B+\sigma^2)T}N(\frac{ln(S_0^*/K_0^*)+(r_A-r_B+3\sigma^2/2)T}{\sigma\sqrt{T}}) -K^*e^{-r_AT}N(\frac{ln(S^*/K^*)+(r_A-r_B+\sigma^2/2)T}{\sigma\sqrt{T}})\,\,\,\,\,\,\,(7) $$ Unless I have done a huge and embarrassing blunder, this formula for the price of the call is not equivalent to the segment inside bracket in (2). If my reasoning is right, then if we assume a currency exchange rate follows a geometric brownian motion, then its inverse, that is the exchange rate in terms of the other currency does not follow the geometric brownian motion with the same growth rate as that of the canonical one, hence cannot be used in Black-Scholes formula to derive its options' values. Thus, it is meaningless to talk about Black-Scholes formula for that call option in the exercise above, unless, of course, we accept that the currency value is expected to grow with rate$(r_A-r_B+\sigma^2)T$ instead of $(r_A-r_B)T$ and substitute the terms in (17.11) and (17.12) accordingly. Where is the error in my reasoning or calculation?