Probability Distribution of the Heston Model, Part II

Starting point for a semi-analytical solution of the Fokker-Planck forward equation for the Heston model is the exact sampling algorithm of Broadie and Kaya [1] (for the notation please see [2])

\begin{array}{rcl} x_t &=& x_0 + m(t) + \sigma(t)Z \nonumber \\ \sigma^2(t) &=& (1-\rho^2)\int_0^t \nu_s ds \nonumber \\ m(t) &=& (r_t-q_t)t - \frac{1}{2}\int_0^t \nu_s ds + \rho\int_0^t \sqrt{\nu_s}dW_s^{(1)} \nonumber \end{array}

The probability distribution function can be described as

p(x_t, \nu_t, t) = p(\nu_t, t) p(x_t,t\mid \nu = \nu_t)

and p(\nu_t, t) is  given by a noncentral chi squared distribution. The distribution p(x_t, t \mid \nu = \nu_t) can be calculated using the exact simulation algorithm. In this algorithm the variable x_t is given as a function of two random variables \int_0^t \nu_s ds and Z.

The distribution of x_t can now be derived using the general transformation theorem for random variables: Let be a random variable with probability density function f. The transformed random variable Y=h(X) has the probability density function

p(y) = f(h^{-1}(y)) \left| \det \left( \frac{\partial h^{-1}_i(y)}{\partial y_j} \right)\right|

First step is now to rewrite the exact simulation method in terms of the two random variables

X_1 = \int_0^t \nu_s ds \ , X_2=Z .

The simulation scheme then becomes

\begin{array}{rcl} x_t &=& x_0+a(t) -\frac{1}{2}X_1 + \frac{\rho\kappa}{\sigma}X_1+\sigma(t) X_2 \nonumber \\\sigma^2(t)&=&(1-\rho^2)X_1 \nonumber \\a(t)&=&(r_t-q_t)t+\frac{\rho}{\sigma}\left( \nu_t-\nu_0-\kappa\theta t \right)\end{array}

or in terms of the transformed random variable

\begin{array}{rcl} Y_1 =h_1(X_1,X_2) &=& x_0+a(t)-\frac{1}{2}X_1 + \frac{\rho\kappa}{\sigma}X_1+\sqrt{\left(1-\rho^2\right)X_1}X_2 \nonumber \\ Y_2=h_2(X_1,X_2)&=& X_1 \nonumber\end{array}.

Let \phi(x_1) be the density function of X_1

\phi(x_1)=\frac{2}{\pi}\int_0^\infty \cos ux_1 \mathrm{Re}(\Phi(u))\mathrm{d}u

and X_2 follows by definition a normal distribution. The joint probability density function of (X_1, X_2) is then

f\left( \begin{matrix} x_1 \\ x_2 \end{matrix}\right)=\phi(x_1) \frac{1}{\sqrt{2\pi}}e^{-\frac{x_2^2}{2}}

with

\begin{array}{rcl}f\left(h^{-1}(y)\right)&=&f\left(\begin{matrix}y_2 \\ \frac{1}{\sqrt{\left(1-\rho^2\right)y_2}}\left(y_1-x_0-a(t)+\frac{1}{2}y_2-\frac{\rho\kappa}{\sigma}y_2\right)\end{matrix}\right)  \nonumber \\  \left|\det \left( \frac{\partial h^{-1}_i(y)}{\partial y_j} \right)\right| &=& \frac{1}{\sqrt{\left(1-\rho^2\right)y_2}}  \end{array}.

This yields to the semi-analytical formula for the solution of the Fokker-Planck equation because by definition p(x_t,t \mid \nu = \nu_t) is the distribution density function of Y_1, which is given by

p(x_t,t \mid \nu = \nu_t) = \int_0^\infty \mathrm{d}y_2 p(y_1, y_2)\mid_{y_1=x_t} = \int_0^\infty \mathrm{d}y_2 \left[f\left(h^{-1}(y)\right)\left|\det \left( \frac{\partial h^{-1}_i(y)}{\partial y_j} \right)\right| \right]_{y_1=x_t}

The integration over y_2 can be carried out using e.g. the Simpson integral rule together with the Cornish-Fisher expansion, which gives an upper bound for the truncation of the upper limit of the integration.

The contour plots below show the probability density function of the Heston model for some example parametrisations.

plot
\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline {\rm Parameters} & x_0 & \nu_0 & r & q & \kappa & \theta & \sigma & \rho \\ \hline \hline  a & 4.6052 & 0.4 & 5\% & 2.5\% & 1.0 & 0.4 & 0.8 & -75\%  \\ \hline  b & 4.6052 & 0.4 & 5\% & 2.5\% & 1.0 & 0.4 & 0.8 & \ \ 75\%  \\ \hline  c & 4.6052 & 0.4 & 5\% & 2.5\% & 1.0 & 0.4 & 0.4 & -75\%  \\ \hline  d & 4.6052 & 0.4 & 5\% & 2.5\% & 1.0 & 0.4 & 0.4 & \ \ \ \ 0\%  \\ \hline  \end{array}

The example code is available here and depends on the upcoming QuantLib version 1.4.

[1] M. Broadie, Ö. Kaya, Exact Simulation of Stochastic Volatility and other Affine Jump Diffusion Processes

[2] K. Spanderen, Probability Distribution of the Heston Model, Part I

[3] R.U. Seydel, Tools for Computational Finance, pp 86

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s