Pay in currency : cur

The FX is : $FX^{cur_2/cur_1}$

European options on the FX (and itself) are quoted in currency cur 1.

I'm looking for the price of \begin{equation*} \mathbb{E}^{Q} \left[ e^{-\int_{0}^{T}r_{s}^{cur}ds} f \left( FX_{T_f}^{cur_2/cur_1} \right) | \mathcal{F}_{0} \right] = ? \end{equation*}

If i integrate with respect to the FX_rate density $\phi_{T_{f}}$, is $\frac{B(0,T)^{cur}}{B(0,T_{f})^{cur1}}\int_{0}^{\infty}f(x)\phi_{T_{f}}(x)dx$ the right answer?

  • $\begingroup$ You mean you have a quanto payoff? $\endgroup$
    – ir7
    Jul 2, 2020 at 14:51
  • $\begingroup$ Yes , I think you're right , but i'm quite unsure of the result . I would need to introduce FX cur1/cur no ? $\endgroup$
    – Kupoc
    Jul 2, 2020 at 15:03

1 Answer 1


If $X$ is FOR-DOM exchange rate (asset always on the left, numeraire on the right), then its dynamics in the QUANTO currency measure (currency different from FOR and DOM; let $Y$ be the DOM-QUANTO exchange rate) is:

$$ dX/X = \left(r_{\rm DOM}-r_{\rm FOR}-\rho_{XY}\sigma_X \sigma_Y \right) dt + \sigma_X dW $$

Terminal distribution:

$$ \ln (X_T/X_0) \sim \phi \left((r_{\rm DOM}-r_{\rm FOR}-\rho_{XY}\sigma_X \sigma_Y - 0.5 \sigma_X^2)T, \sigma^2_X T\right) $$

with $\phi$ normal density.


$$ {\rm e}^{-r_{\rm QUANTO} T} \int_{-\infty}^{\infty} g(x)\phi(x)dx $$

(we get Black-Scholes formula under my assumptions here and $g(x)=(X_0{\rm e}^x-K)^+$).

(See this resource for further details and proofs on quanto FX options.)

  • $\begingroup$ But without supposing any model , just using the implied risk neutral density , how can i do ? $\endgroup$
    – Kupoc
    Jul 2, 2020 at 15:12
  • $\begingroup$ You still need mean and variance for the terminal distribution. The point is the mean needs to be adjusted to account for the switch from the natural measures (FOR or DOM) to QUANTO measure. $\endgroup$
    – ir7
    Jul 2, 2020 at 15:32
  • $\begingroup$ I added the terminal distribution (for standard dynamics). $\endgroup$
    – ir7
    Jul 2, 2020 at 15:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.