# FX Binary Option and differences in prices based on different numeraires?

I suppose, but I can't figure it out by myself mathematically, that the Exchange-Rate put call duality (see Section 9.3.3 of Stochastic Calculus For Finance 2 by Shreve is related with these criticalities of currency bet (binary, digital) option stated by N. Taleb in the book Dynamic Hedging:

1. A bet in dollars for a dollar based person on USD-DEM is different in price from the translation into German marks of a bet in German marks on USD-DEM of the same strike and expiration (p. 286).
2. The price of a bet for a dollar-based person, which is N(d2), is different from the price of the bet for the person based in DEM (for example) as this latter will be N(d1) (p. 286).

Could someone exaplain to me if these concepts are effectively related to the duality principle and give me some explanation with the math involved with the pricing of these bets?

Please, let me know if more details are needed. Thanks for the help.

Edit: Extended answer for 2) thanks to @Rylan

Considering a general bet option with strike K that pays 1$or 1€ depending on the location of the operator: 1. "American" side: $$N(d2)=E^u[1_{C(T)>K}|F_t]$$ 2. "European" side: convert its payoff in EUR to dollar using the new numeraire $$C(t)$$. The Radon-Nikodym derivative is $$C(T)/C(t)$$, assuming no interesest rates. 3. Final equalities: $$E^e[1_{C(T)>K}|F_t]=\frac{1}{C(t)}E^e[C(T)1_{C(T)>K}]=\frac{1}{C(t)}N(d1)=N(d2)$$ ## 1 Answer I've read through the pages where he discusses this a few times. I think the key is this, although I am not 100% sure if I am interpreting his phrases correctly. (Note: All options/payoffs I'm describing below are European in nature; if I refer to European or American it just refers to currency. Notation and argments build off this discussion. Rates are zero as in the reference text.) Let's call $$C(t)$$ the price in euros of one dollar at time $$t$$, and note that it's around 0.9 today. Let's imagine making bets around a strike of 1, ie if 1 EUR becomes worth the same as 1 USD. For both the European and the American, a "bet" is a binary option, defined as $$N(d_2)$$. In point 1., it looks like the American's bet means he is paid $$\\\1$$ if $$C(T) > 1$$ else $$0$$, while the European's bet means he is paid $$\text{EUR}1$$ if $$C(T) > 1$$. This already points to a difference -- the European will be paid in a currency that has fallen while the American will be paid in a currency that has risen. Setting both currency's rates to zero for simplicity, we have: $$\text{The American's bet} = V^u(t) = \mathbb{E}^u(\mathbb{1}_{\{C(T) < 1\}}| F(t))$$ $$\text{The European's bet} = V^e(t) = \mathbb{E}^e(\mathbb{1}_{\{C(T) < 1\}}| F(t))$$ We note that $$\frac{C(T)}{C(t)}$$ is an $$e-$$martingale, and similar to the other answer it defines the change of numeraire to switch from the $$e-$$measure to the $$u-$$measure. Note that $$\mathbb{E}^e\Big(\frac{C(T)}{C(t)}\mathbb{1}_{\{C(T) < 1\}}| F(t)\Big) = \mathbb{E}^u(\mathbb{1}_{\{C(T) < 1\}}| F(t))$$ And therefore $$V^u(t) = \mathbb{E}^u(\mathbb{1}_{\{C(T) < 1\}}| F(t)) = \frac{1}{C(t)}\mathbb{E}^e(C(T)\mathbb{1}_{\{C(T) < 1\}}| F(t))$$ Where we pay specific attention to the fact that the "European measure" way of expressing the bet, $$\frac{1}{C(t)}\mathbb{E}^e(C(T)\mathbb{1}_{\{C(T) < 1\}}| F(t))$$, differs from the European version of the bet $$V^e(t)$$. It has a scalar of $$\frac{1}{C(t)}$$ out front (which is I think what he meant by the translation; an easy fix either way) but more fundamentally it has a term $$C(T)$$ multiplying the indicator. To see why "one is $$N(d_1)$$ and the other is $$N(d_2)$$" you can revisit the derivation of the Black-Scholes formula where the expectation $$E^{\mathbb{Q}}[(S(T) - K)^+]$$ is computed directly as an integral. • thanks. I have to study better the change of numeraire at this point, I do not get the intuition behind the idea of the e-martingale$C(T)/C(t)$. Commented Aug 31 at 12:47 • The numeraire chosen is C(T), correct? So we are converting EUR to USD. Commented Sep 1 at 11:42 • For 2, am I on the right path? I start from the American side$N(d2)=E^u[1_{C(T)>K}]$, change the numeraire$1/C(T)$, to express this 1USD in EUR, to compute this integral$E^e[C(t)/C(T)1_{C(T)>K}]$. From here I can't see how this is$N(d1)$. Thanks for the help. Commented Sep 2 at 19:33 • I'd suggest reviewing the "standard" Black-Scholes formula derivation for a call option on a stock, where we calculate$V(0) = \mathbb{E}^\mathbb{Q}[D(T)(S(T) - K)^+] = \mathbb{E}^\mathbb{Q}[D(T)S(T)\mathbb{1}_{S(T) > K} - D(T)K\mathbb{1}_{S(T) > K}]$by direct integration. You will see how "a scalar multiple of the indicator variable" integrates to "something times$N(d_2)$while "a stock or currency price times the indicator variable" integrates to "something times$N(d_1)\$. Commented Sep 3 at 8:02
• If it makes sense to you the last equality in my second comment I think I can close my question. It's a little strange to me. As always, thank you. Commented Sep 3 at 13:25