For the purpose of this question a local vol model is a 1d SDE which specifies the price process and we have a contingent claim that depends on those prices (in general, at multiple times). e.g. $dX_t = \sigma(X_t, t)dW_t$. A stochastic vol model is an at least 2d SDE where one of the equations is for the aforementioned prices process, but the additional equations specify other variables that the price process is not independent of. e.g. $dX_t = X_t Y_t dW^1_t, dY_t = \nu Y_t dW^2_t$.

Do you agree that in general, given a stochastic vol model, there is no equivalent local vol model in the following sense: The joint density across all times of the price process $X_t$ in both models can be made the same. In other words, given a set of prices of contingent claims on $X_t$(that depend on multiple dates) a stochastic vol model determines, there is no local vol model that gives the same set of prices. If so, can you point me in the direction of a proof of this? If no proof is available, a nice counterexample will suffice. A local vol and stochastic vol model which give the same vanilla options price, but have at least one different joint density.


1 Answer 1


The local vol model has exactly enough freedom to match the individual densities $X_t.$ There is no additional freedom in the local vol model to match even a joint density for a pair of times $(X_t,X_s).$

When you ask about the joint density across the continuum of times $t \in [0,T]$ it is pretty easy to show that any local vol model differs from any stochastic vol model provided the stochastic volatility model has nonzeeo vol-of-vol. That is because the joint density along the continuum of times gives full information about the measure on paths $X_t$.

For a local vol model, the instantaneous variance of a path $X_t$ at time $t$ is almost surely $\sigma^2(X_t,t).$ For a stochastic volatility model with nontrivial vol-of-vol this is not true for $t>0$: the instantaneous variance at time $t$ conditioned on $X_t$ is a random variable with mean $\sigma^2(X_t,t)$ but posive variance.

  • $\begingroup$ Although I appreciate your comment, you have basically just restated my question. The easy proof is what I am looking for. $\endgroup$
    – muaddib
    Commented May 21, 2015 at 11:23
  • $\begingroup$ Expanded answer. $\endgroup$
    – q.t.f.
    Commented May 21, 2015 at 11:50
  • $\begingroup$ Yes they do look different at first glance. But these two variances have been chosen in such a way that the marginals for each time are exactly the same. In particular, the transformation is unique and given by Dupire's formula. The question is then why given that, their joint densities can be different. A proof might look something like, assume p1(x, t) = p2(x, t), then there exists x, y, s, t such that p1(x, y, s, t) <> p2(x, y, s, t). $\endgroup$
    – muaddib
    Commented May 21, 2015 at 12:15

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