FX Option Symmetry for Foreign-Currency Payoffs

# Equivalence of Call Option on $S_T$ and Put Option on $\frac{1}{S_T}$ in FX Markets


# Equivalence of Call Option on $S_T$ and Put Option on $\frac{1}{S_T}$ in FX Markets












Part 1: I am trying to price an option in the FX world. It naturally pays in the domestic currency, but in this case the payout currency must be the foreign currency. For example, consider the payoff:

$$\left(\frac{S_T - K}{S_T}\right)^{+} = \left(1 - \frac{K}{S_T}\right)^{+} = K \left(\frac{1}{K}-\frac{1}{S_T}\right)^{+}$$

The inclusion of $S_T$ as the denominator turns the plain vanilla call on $S_T$ into a "foreign-payout" option. I've done the computation for $\mathbb{E}\left[\left(\frac{1}{K}-\frac{1}{S_T}\right)^{+} \right]$, Black-Scholes-style, using the equation for $\frac{1}{S_T}$, and got the same answer as in Pricing an Option with payoff $\left(1-\frac{K}{S_t}\right)^{+}$. That is, ignoring the interest rates for now (i.e. setting both interest rates to zero): $$\mathbb{E}\left[\left(\frac{1}{K}-\frac{1}{S_T}\right)^{+} \right] = N(d_2) - \frac{K}{S_0} e^{\sigma^2 T} N(d_3)$$

Part 2: In the FX world, a call on one currency is a put on the other currency. For example, in Uwe Wystup's 2008 `Foreign Exchange Symmetries' working paper (https://core.ac.uk/download/pdf/6671934.pdf), Section 4.5, we have: \begin{equation}\label{Wystup} \frac{1}{S} v(S, K, T, t, \sigma, r_d, r_f, \phi) = K v\left(\frac{1}{S}, \frac{1}{K}, T, t, \sigma, r_f, r_d, -\phi \right) \end{equation} where: $r_d$ is the domestic interest rate, $r_f$ is the foreign interest rate, and $\phi = 1$ for a call and $-1$ for a put. (The SDE is $dS_t = (r_d - r_f) S_t dt + \sigma S_t dW_t$.)

Quoting from the aforementioned paper, ``We consider the example of $S_t$ modeling the exchange rate of EUR/USD. In New York, the call option $(S_T - K)^{+}$ costs $v(S, K, T, t, \sigma, r_{usd}, r_{eur}, 1)$ USD and hence $v(S, K, T, t, \sigma, r_{usd}, r_{eur}, 1) / S$ EUR. This EUR-call option can also be viewed as a USD put option with payoff $K \left(\frac{1}{K}-\frac{1}{S_T}\right)^{+}$. This option costs $K v\left(\frac{1}{S}, \frac{1}{K}, T, t, \sigma, r_{eur}, r_{usd}, -1 \right)$ EUR in Frankfurt, because $S_t$ and $\frac{1}{S_t}$ have the same volatility. Of course, the New York value and the Frankfurt value must agree, which leads to [equation above].''

We now apply the above equality to obtain an alternate solution to our computation: $$\mathbb{E}\left[ K \left(\frac{1}{K}-\frac{1}{S_T}\right)^{+} \right] = \frac{1}{S_0} (S_0 N(d1) - K N(d2)) = N(d1) - \frac{K}{S_0} N(d2)$$ where we just plugged in the Black-Scholes formula for the call option $v(S, K)$, again ignoring the interest rates for simplicity.

Clearly the two answers (Part 1 and Part 2) don't match, and I'm at a loss as to why.

Question 1: Is the expectation in Part 2 under a different measure than that in Part 1, and is this the reason for the discrepancy?

Question 2: Which is the correct way to price an option as first described (payout in the foreign currency)?

Remark: While the option as first described (payout in the foreign currency) may be artificial, there are Asian call options with this feature, e.g. $\left(\frac{A_T - K}{A_T}\right)^{+}$, where $A_T$ is the average, and I believe they are equivalent to put options on $\frac{1}{A_T}$ (with strike $\frac{1}{K}$). I'm hoping to figure out how to treat these by first looking at the simpler case of $S_T$ instead of $A_T$.

Any help is greatly appreciated.