Limits of Static Replication for Two-Asset Equity Payoffs

Summary

The document considers pricing a discounted payoff that depends jointly on two equity prices at a future date. Its central point is that pricing requires a joint risk-neutral distribution; a copula is one way to construct one, but it is not the only possible model. Any approach that supplies a suitable joint terminal distribution can be used.

Vanilla options reveal the risk-neutral marginal distribution of each asset, but those marginals do not determine how the assets move together. Consequently, they generally cannot statically replicate a payoff sensitive to their joint outcomes. Adding spread options supplies information about the distribution of the price difference, yet still does not fully identify the joint distribution. The discussion is conceptual and offers no empirical pricing comparison or particular calibration procedure, so the quality of a price depends on the assumptions used to complete the joint distribution.

Key ideas

• A two-asset payoff generally requires a model for the joint risk-neutral distribution.
• Copulas are one method for specifying dependence, but other joint-distribution models are possible.
• Vanilla option prices identify marginal distributions but do not reveal all dependence information.
• Even adding spread-option information does not generally determine the full joint distribution.

Tags

# Pricing multidimensional equity


# Pricing multidimensional equity












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 ?

## Answer by Arshdeep (score 2)

https://quant.stackexchange.com/a/55372

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.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.