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I'm reading about stochastic volatility models - the ones which resulted after Wiggins proposed in 1986/7 that $\sigma$ in Black-Scholes should be a stochastic process rather than a constant.

In particular, I am looking at 3 models:

  • Hull & White
  • Stein & Stein
  • Heston

The book says, after introducing the last one (Heston), that it is an incomplete market because the model has 2 Brownian motions but we only have one risk asset for replication. What I don't understand is, whether this statement about the market completeness relates to just the last model or all 3. To me, they all seem to be incomplete markets.

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  • $\begingroup$ You are right, they are all incomplete. $\endgroup$
    – user39119
    Commented Jun 3, 2020 at 21:44
  • $\begingroup$ They are incomplete until they are complete(d). $\endgroup$
    – user34971
    Commented Jun 4, 2020 at 9:09

2 Answers 2

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Any stochastic volatility model will be incomplete.

Asset price under no arbitrage satisfies an SDE $dS_t = r(t, S_t) dt + \sigma(t, S_t) dW_t$, where $r(t, S_t)$ and $\sigma(t, S_t)$ can be stochastic. Second Fundamental Theorem of Asset Pricing states that a model is incomplete if and only if the associated equivalent martingale measure is not unique. In stochastic volatility model we can change the law of the volatility process $\sigma_t$ (which changes the law of $S_t$) without affecting the martingality of the discounted asset price $\exp\left(-\int _0 ^t r_s ds\right) S_t$. This yields another martingale measure and thus incompleteness.

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You generally need one tradable asset per source of risk in your model that is someway dependent on that noise. So in a world where you can trade a single stock, but have two sources of variance your model would be incomplete as there would be no way to fix the market price of volatility risk. However, if some other asset was tradable (say some reference option) it would complete the model and let you calibrate the market price of risk.

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