# Tag Info

0

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 ...

2

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 ...

0

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: ...

6

To recover the Black-Scholes pricing equation, you should first express the standard normal cdf in terms of its characteristic function analogous to the Heston solution: $$N(x) = \frac{1}{2} - \frac{1}{\pi} \int_0^{\infty} Re [\frac{e^{-i\phi x} f(\phi)}{i\phi}] d\phi$$ where $f(\phi)$ is the characteristic function of the standard normal distribution:  ...

Top 50 recent answers are included