# 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: $$$$\label{Wystup} \frac{1}{S} v(S, K, T, t, \sigma, r_d, r_f, \phi) = K v\left(\frac{1}{S}, \frac{1}{K}, T, t, \sigma, r_f, r_d, -\phi \right)$$$$ where: $$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.

• You are missing a measure change in the FX case, as both $S$ and $\frac{1}{S}$ are tradable we know what their respective drifts should be in their respective risk neutral measures. Jun 18 at 21:46
• To show that a Call on S = Put on 1/S you can look here. You literally just invert K and S, switch rates (as you now have the other quotation) and use same vol, and switch call to put. If you want to show it is identical, your notional (is in EUR for EURUSD but in USD for USDEUR) needs to be scaled by strike. That is not the same as paying in foreign though. You need to divide the flow by final spot $S_t$ (which is $N_{fgn}*max(S_t-K,0)/S_t$) so that it would stay the same (and not invert). Two different things really. Jun 18 at 22:26
• You may have a look of this question. Jun 18 at 22:52
• @Gordon Thank you very much. I had missed that highly relevant already-asked question. Somehow I find it strange that $dP^f/dP^d$ and $dP_S/dP^d$ (for stock numeraire measure change) are so similar - I need to upgrade my intuition on this stuff. Basically, I understand the answer to my question 2 (which way is correct) to be "Part 2" (pricing in the foreign r.n. measure). I plan to clean up the original question since it's too long/messy and to post a question on what I actually need, which is the foreign-paying Asian option.
– X Y
Jun 22 at 17:28
• I've posted the related Asian option question at quant.stackexchange.com/questions/65860/…. If anyone has any useful input, please post it there. Thanks.
– X Y
Jul 4 at 18:50