I need to price a "foreign-paying" fixed-strike Asian (i.e., average) option. Thus, the payoff is:

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

where $A_T$ is the arithmetic average, at expiry $T$, of some spot FX rates. E.g, $A_T = \sum_{i=1}^N w_i S_{t_i}$ for spot FX rates $S_{t_i}$ and weights $w_i$. In the reduced case $N = 1$, we simply have a European vanilla option, which we can price via domestic-foreign symmetry, as in https://quant.stackexchange.com/a/44541/43050.

My question is: Can we use the same domestic-foreign symmetry for the Asian option, e.g. if we approximate the distribution of $A_T$ as lognormal via moment-matching? Following the logic for the vanilla case, we would set $\frac{dP^f}{dP^d}\big|_t = \frac{A_t B^f_t}{A_0 B^d_t}$, where $B^d_t$ and $B^f_t$ are the respective domestic and foreign money-market account values at time $t$, and assuming we can prove that the expectation is $1$ so it's an actual Radon-Nikodym derivative. Then we'd have:

\begin{align*} E^f\bigg( \frac{1}{B^f_T A_T} (A_T-K)^+\bigg) &= E^d\bigg(\frac{A_T B^f_T}{A_0 B^d_T} \frac{1}{B^f_T A_T} (A_T-K)^+\bigg)\\ &= \frac{1}{A_0}E^d\bigg(\frac{1}{B^d_T} (A_T-K)^+\bigg), \end{align*}

thus reducing to the domestic-payout case, which we know how to price (via moment-matching). Does this make any sense, and, if not, why not? If the symmetry approach does not work, can we use moment-matching on the harmonic average directly? I have only seen moment-matching discussed in the arithmetic case. Many thanks in advance.


1 Answer 1


Regarding the second question, if we assume that $A$ is lognormal (shifted lognormal) and imply its two (three) moments from respective spot rate components under the appropriate probability measure, then we have its probability density function $f_A$ which allows us to compute the expectation of the payoff the usual way:

$$ E\left[\left(\frac{1}{K}-\frac{1}{A}\right)^+ \right] =\int_K^\infty \left(\frac{1}{K}-\frac{1}{x}\right)f_A(x) dx. $$

  • $\begingroup$ So: $E^f[S^k_T] = E^d\left[\frac{S_T B^f_T}{S_0 B^d_T} S^k_T \right] = \frac{B^f_T}{S_0 B^d_T} E^d[S^{k+1}_T]$. Use these identities to compute $M_k := E^f[A^k_T]$. Then let $f_A$ be the pdf of the lognormal distribution with moments $M_k$ and compute $\int_K^\infty \left(\frac{1}{K}-\frac{1}{x}\right)f_A(x) dx$. Am I understanding your answer correctly? $\endgroup$
    – X Y
    Commented Jul 5, 2021 at 17:07
  • $\begingroup$ You need the (distribution) moments of $S_{t_i}$, then $A$'s can be calculated ($S_T$ might not even be one of the fixings). You can stick with doing all calculations under $Q^f$, where $S$'s dynamics would be $dS/S=(r_d-r_f -\sigma^2)dt + \sigma dW^f$. (Girsanov gives: $dW^f = dW^d -\sigma dt$.) $\endgroup$
    – ir7
    Commented Jul 5, 2021 at 18:23

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