Consider the Heston model given by the following set of stochastic differential equations: $$\frac{dS_{t}}{S_{t}}=\mu_{t}dt+\sqrt{V_{t}}dW_{t}, S_{0}>0,$$ $$dV_{t}=\kappa(\theta-V_{t})dt+\xi\sqrt{V_{t}}dZ_{t}, V_{0}=v_{0}>0,$$ $$d<W,Z>_{t}=\rho dt$$ where $W_{t}$ and $Z_{t}$ are two brownian motion, $\kappa,\theta,\xi>0, \rho\in(-1,1).$ I don't understand what impact would $\rho$ have on the shape of volatility smile when it's negative or positive. In addition, How the volatility smile would change if $\xi$ increases? Thank you.


1 Answer 1


Intuition: You can think of the vol smile as a reflection of the risk neutral distribution (compared to the Black Scholes Gaussian density). A fat tailed distribution creates the smile: fat tail -> higher prob of exercise than Gaussian with constant stdev -> higher option price than BS with ATM vol -> higher implied vol for given strike. Skewed distributions cause skewed smile: neg skew -> left tail thicker than right -> OTM put impl vol higher than OTM call impl vol.

Also you can think of a stoch vol model as mixture of Gaussians, each with different volatility. The way that this mix is produced generates fat tailed and skewed distributions.

Non-zero Rho will produce asymmetric volatility smiles (that look more like skews). Negative correlation means that a negative spot shock is more likely accompanied by a positive vol shock -> as spot goes down we mix with higher vols -> will produce a fatter left tail than right -> negatively skewed risk neutral density -> downward sloping skew.

Large Xi produces a steeper, more pronounced smile/skew, as we increase the vol-of-vol potentially mixing with higher volatilities which generate more leptokurtic risk neutral distributions. Zero Xi would correspond to the Black Scholes case where the smile becomes flat.

To be more rigorous you can take the Heston char function, take derivatives to compute moments and show the relation of Rho/Xi with Skewness/Kurtosis.

  • $\begingroup$ Thanks. However, I don't know how to prove rigorously why negative correlation will produce a downward sloping skew and the opposite for positive $\rho$, Is there an intuitive explanation? In terms of $\xi$, I'm also wondering how to explain why the price of the at-the-money option would decrease while the price of the in-the-money or out-of-the-money option would increase when $\xi$ becomes larger. $\endgroup$
    – cmd1991
    May 10, 2015 at 2:12
  • $\begingroup$ Edited to answer your further questions $\endgroup$
    – Kiwiakos
    May 10, 2015 at 8:13

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