How would you price \begin{equation*} \mathbb{E}^{Q} \left[ e^{-\int_{0}^{T}r_{s}ds} f \left( S_{T_f}^1, S_{T_f}^2 \right) | \mathcal{F}_{0} \right]\end{equation*} with $T_{f} \le T$ and $S^{1}, S^{2}$ the assets.

by static replication , is using a copula the only way ?

  • $\begingroup$ Does the payoff function $f$ have any specific property? Such as linearity or homogeneity for example. $\endgroup$ Commented Jul 1, 2020 at 16:39
  • $\begingroup$ Sadly not , the objective would be to integrate with respect to the density with the right adjustment so that any payoffs can be replicate using the information available on the market. ( here for example would be deriving the price from spread option for all strikes ) to get the joint density. $\endgroup$
    – Kupoc
    Commented Jul 1, 2020 at 16:44
  • 1
    $\begingroup$ are $S_1$ and $S_2$ independent assets? $\endgroup$
    – will
    Commented Jul 1, 2020 at 19:54
  • $\begingroup$ No , there are the least assumptions possible made on the assets. $\endgroup$
    – Kupoc
    Commented Jul 2, 2020 at 6:33

1 Answer 1


As long as you are able to generate a joint terminal distribution, any model will do the job. Copula is only one such approach.

Now, in theory, you cannot completely statically replicate this payoff in general. To see this, know that all you have is vanillas, and the most you can do is imply the marginal distribution from them (the usual risk neutral density). However, your exotic is also sensitive to the conditional distribution, which the vanillas cannot capture.

Even if you do assume that you have spread options (like you say in the comment), it is still not enough to determine the joint distribution completely; as you can only determine the distribution of their difference (so you now have 3 marginals). To see why, note that you cannot hope to find the moment generating function of the joint at many points (say [1,5] for instance). So you can't pin down the MGF (equivalently, the joint distribution).

One can say something stronger: You cannot find the generating function for any random vector f except the spread itself.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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