I understand that stochastic volatility models should be used when the exotic option payoff is volatility dependent (such as variance swaps and volatility swaps).

Stochastic volailtiy models should also be used when the exotic option has forward starting features (such as with cliquet options). Can someone explain why should we use stochastic volatility models for forward starting options?

Also, given an exotic option, how would you decide whether to use a local volatility or stochastic volatility model for pricing it?

  • $\begingroup$ If anyone has an idea ? $\endgroup$
    – glork
    Aug 31, 2016 at 13:44

1 Answer 1


Whenever you use any model to price anything, all you need to do is make sure you model the underlying dynamics that the product you're pricing actually depends on.

Any product will be dependent on numerous facets, to varying degrees - this is the same with modelling anything.

The modelling that happens in pricing financial derivatives is an integration over the space of possible outcomes, i.e. calculating the expectation. Now, one key aspect of the expectation is Linearity - that is, if we have $\mathrm{E}[f(X)]$ where $f(X)$ is a linear function (i.e. $f(X) = mX + c$) then $f(\mathrm{E}[X])$ - that is, we don't actually care about the distribution of the variable $X$, only its mean.

Why is this important though? Well, if you have a product which does not really depend on some variable, then we can say it is approximately linear, and we do not have to model its dynamics correctly, since we only care about the expected value of the underlying.

Now, to relate this back to your question, say we have a product which depends on the distribution of the volatility of the underlying - for example an option on the realised volatility, and lets think about what the distribution of the volatility will look like in the local vol. and stoch. vol models:

  1. local vol: here, each simulated path will randomly move from its initial price to some future price, and the volatility at each time step will be determined deterministically from the local volatility function. It will have some random distribution, coming from the random variation of the spot process across the local volatility surface - and we have no guarantee it will be the correct distribution of the volatility. Furthermore, local volatility surfaces tend to flatten out at farther tenors, so the volatility of volatility will converge to zero - that is clearly wrong.
  2. A stoch. vol process however has an extra process for the volatility, following its own dynamics (governed by the chosen model) - these dynamics should be chosen to correctly model (/approximate) the dynamics of volatility, such that the realised volatility of the paths accurately represents the driving volatility of the underlying.

And now another explanation using the cliquet from your question, and some fairly handwavey explanations:

  1. The local volatility is the instantaneous volatility dependent on the spot and point in time the process happens to be at.
  2. To create the term volatility surface (i.e. recreate the black scholes vol surface / option prices used to create it) we basically have to integrate the realised volatility from $S=S_0, t=t_0$ up to $S=K, t=\tau$, where the integration is over all possible paths from start to end, weighted by the probability of that path occurring (this isn't actually what's done, but it's a useful analogy).
  3. If we want to get the local volatility implied vol surface from some future time, conditional on being at $S=K_0$, then we need to do the above integration to all points on the vol surface (in reality, what you do is start at $S=K_0, t=\tau$ and then diffuse the underlying and price options on some grid to reconstruct the surface).
  4. As we mentioned above, the local vol surface tends to flatten out for larger tenors. A result of this is that when you perform the creation of the hypothetical vol surface in the future, it is much flatter than one would expect, such that the skew converges to zero. This means that something like a cliquet where you are buying an option at 90% atm, on a negatively skewed underlying (i.e. most equities / equity indices) will be underpriced, since the conditional implied vol on the down side will be underestimated, underpricing all of the fwd start options*.
  5. Even for atm cliquets, the options will be underpriced, since they are a little bit dependent on convexity/kurtosis - but the effect is less pronounced than for otm options.

One of the properties of stoch vol models is that when you calculate the conditional vol. in the future, they still produce a skew, meaning that you're not habitually mispricing all of the fwd start options in the cliquet.

On the subject of the question "when do i use model X, and when model Y?" the answer is "if you don't already know which to use, price with both. If you get a different answer, then try to understand why".

  • $\begingroup$ I cant see how stochastic vol produces a skew in the future but local vol doesnt? I think that the key thing is I cant see how this works mathematically $\endgroup$
    – Trajan
    Mar 25, 2018 at 15:30
  • 1
    $\begingroup$ In stock vol it comes from the correlation between the spot process and the vol process. Because of this, where the (instantaneous) correlation is non zero, you will have a skew. You do not have this in local vol though, so all skew must come from the input surface. And since option prices' skew tends to flatten with extended maturities, the forward skew is under pronounced. $\endgroup$
    – will
    Mar 25, 2018 at 15:46
  • $\begingroup$ Yeah I cant see why non zero correlation would cause skew, nor why skew would flatten in T large. Mathematically that is $\endgroup$
    – Trajan
    Mar 25, 2018 at 15:53
  • $\begingroup$ This is something that should be more obvious intuitively before having some mathematical proof - if there is a negative spot/vol correlation, then when the spot has experienced negative returns, you would expect the vol process to have experienced positive returns - ie where the spot has been decreasing, the vol has been increasing, and vice versa. When you translate this into a smile you get skew. $\endgroup$
    – will
    Mar 25, 2018 at 15:56
  • 1
    $\begingroup$ In local volatility there is no random process for the volatility, so there is no correlation. If you were to measure the realized volatility of each path, then you would end up with a correlation which should be related to the skew, but it is not a correlation between random processes, because there isn't one. $\endgroup$
    – will
    Mar 25, 2018 at 16:02

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