Has anyone ever formally proved that Markets are incomplete under the stochastic volatility model?

I know that if there are more random sources than traded assets, then the market is incomplete but does there exist a proof of that meta-fact?

If there is, could you please give me the reference to the paper(s)?

  • 4
    $\begingroup$ Can you show that there are multiple different equivalent martingale measures? Completeness is equivalent to the existence of a unique martingale measure. This is called 2nd Fundamental Theorem of Asset Pricing sometimes. $\endgroup$
    – quasi
    May 15, 2013 at 22:42
  • $\begingroup$ It might have been discussed in "Foundation of Stochastic Financial Mathematics" by A. Shiryaev $\endgroup$
    – SBF
    May 17, 2013 at 11:02

4 Answers 4


It's really quite simple. It's just a matter of the fact that we can change measure on the stochastic volatility while not changing the fact that the stock is a martingale. Once we can do this, we have payoffs that have different values under different measures, so the market can't be complete.

For clarity, just consider a stock S, a money market account M and a Brownian motions B and W. Let dS = a dt + v dB, dM = r dt, and dv = b dt + w dW, where our filtration is generated by B and W, and a, r, b, and w are adapted processes. Using M as numeraire (i.e. - divide by M so as to express all prices in terms of shares of M or equivalently, assuming interest rates are zero), then under an equivalent martingale measure, assets and admissible strategies are martingales, so now S is a martingale, so dS' = v' dB', and dv' = b' dt + w' dW'.

The point is we can change measure to add a drift to W. Since dS' = v' dB', this doesn't change the fact that S' has zero drift, so S' remains a martingale. But it does change the distribution of v', and thus the distribution of S'.

If the market was complete, all contingent claims on S would be replicable. Their prices would be equal to the initial value of the replicating strategy, and hence fixed. However, since the distribution of S' changes when we change the drift of W, there are contingent claims on S which will have different values depending on the drift we add to W. Hence the market is incomplete.

  • $\begingroup$ The volatility change of $dS$ does not make process of $S$ non martingale. It needs to add a drift to $dS$. I think you need to clarify that point $\endgroup$
    – adam
    Mar 7, 2014 at 8:53
  • $\begingroup$ Noted and edited. $\endgroup$
    – hjs
    Mar 12, 2014 at 19:02
  • $\begingroup$ Great answer! :) $\endgroup$ Feb 6, 2019 at 16:19

I think a sketch of the proof would look like this

Let's say you start from $$ dS_t = S_t \odot (\mu_t dt + \sigma_t dW_t) $$ where $S$ is an vector valued process of your $n$ risky assets prices, $W$ a standard $k$-dimensionnal brownian motion under the historic probability, $\sigma_t$ an $n \times k$ matrix valued process and $\odot$ is the Hadamard coordinate by coordinate product of vectors. Add $S^0_t = e^{\int_0^t r_u du}$ the cash (risk free asset).

A risk premium process $\lambda$ is a solution of $\sigma_t\lambda_t = \mu_t - r_t \mathbf 1$ (with $\mathbf 1$ the $n$-dimensional vector of $1$'s). For each such solution you get a risk neutral probability $\mathbb{Q}^\lambda|_{\mathcal{F}_t} = \exp( - \int_0^t \lambda_s dW_s - \frac{1}{2} \int_0^t |\lambda_s|^2 ds) \mathbb{P}|_{\mathcal{F}_t}$. Indeed you can write $$ dS_t = S_t \odot ( r_t \mathbf 1 dt + \sigma_t (\underbrace{dW_t + \lambda_t dt}_{dW^\lambda_t}) ) $$
and Girsanov theorem tells you that $dW_t + \lambda_t dt$ is a $\mathbb{Q}^\lambda$-brownian motion. So $S/S^0$ is a local martingale under $\mathbb{Q}^\lambda$ (for the historic brownian filtration).

If $k> n$ (for exemple $k=2$ and $n= 1$ in the Heston model) and $\sigma_t$ is of full rank then you get an infinite number of risk premiums so you have an infinite number of risk neutral probability measures so your market is incomplete.

I guess the devil is in the details like the integrability condition for $\lambda$ to actually define a probability measure but I think that's the basic idea.

  • 1
    $\begingroup$ The key point your missing is that \sigma_t can be degenerate, in which case the market could still be complete. $\endgroup$
    – hjs
    Mar 7, 2014 at 7:09
  • $\begingroup$ Yes I should have mentionned the rank of $\sigma$. $\endgroup$
    – AFK
    Mar 7, 2014 at 10:16
  • $\begingroup$ That should be $dS_t = S_t ⊙ (\mu_t dt + \sigma_t dW_t)$, etc, yes? $\endgroup$
    – hjs
    Mar 12, 2014 at 19:06

The paper by Marc Romano and Nizar Touzi, Section 3, contains a general proof that a stochastic volatility model cannot be complete in the sense that the addition of the option completes the market (in the sense of Harrison and Pliska) generated by the underlying and risk-free borrowing/lending: http://citeseerx.ist.psu.edu/viewdoc/download?doi=

The paper only deals with a bivariate diffusion with correlated Wiener noise.


You can have a look at Heston (1993) who gives a closed form solution for options under stochastic volatility.


Keep in mind that an investor demands extra compensation for volatility risk. Breeden's consumption based model offers a way of determining this premium. When calibrating a stochastic volatility model those extra premiums are already contained in market prices.


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