inordinatum

Physics and Mathematics of Disordered Systems

Posts Tagged ‘Fokker-Planck

Fokker-Planck equation for a jump diffusion process

with 8 comments

One of the simplest stochastic processes is a Brownian motion with drift, or a diffusion process:

\displaystyle   \begin{array}{rl}  \dot{x}(t) =& \mu + \sigma\,\xi(t), \\  x(t) =& \mu t + \sigma \,W(t).  \end{array}   (1)

Here, \xi(t) is Gaussian white noise with mean zero and variance \left\langle \xi(t_1)\xi(t_2)\right\rangle = \delta(t_1-t_2). Its integral W(t) = \int_0^t \mathrm{d} t'\,\xi(t') is a Brownian motion.
Continuous Itô stochastic processes such as eq. (1) are insufficient for applications where the random variable x may jump suddenly (such as in avalanches). A natural extension of (1) for modelling this is a so-called jump-diffusion process. Let us suppose that our jump sizes s are positive, independent and identically distributed with density Q(s). Then, the jump diffusion process y(t) is

\displaystyle   y(t) = x(t) + \sum_{i=0}^{N(t)} s_i = \mu t + \sigma\, W(t) + \sum_{i=0}^{N(t)} s_i,   (2)

where s_i, i\in \mathbb{N} are i.i.d. jump sizes as above, and N(t) is the number of jumps encountered up to time t. For simplicitly, let us assume that jumps occur independently with rate \lambda, i.e. that the probability to have a jump in a time interval \mathrm{d}t is \lambda\, \mathrm{d}t. Then, N(t) is a Poisson process with rate \lambda.

It is well-known that the diffusion process x(t) in (1) is equivalently described by a partial differential equation for the distribution P(x) of x, the Fokker-Planck equation (FPE)

\displaystyle \partial_t P_t(x) = \frac{\sigma^2}{2}\partial_x^2 P_t(x) - \mu \partial_x P_t(x).   (3)

This representation is useful e.g. for first-passage problems: they correspond to various boundaries introduced in the PDE (3). So how does one generalise the Fokker-Planck (3) to the case of the jump-diffusion process y(t) in (2)? I will explain in the following that the answer is

\displaystyle \partial_t P_t(y) = \frac{\sigma^2}{2}\partial_y^2 P_t(y) - \mu \partial_y P_t(y)- \lambda P_t(y) + \lambda\,\int_0^\infty \mathrm{d}s \,Q(s)P_t(y - s) ,   (4)

and then discuss a specific example.

1. Deriving the jump-diffusion FPE

Let us consider a time step from t to t + \mathrm{d}t. The probability for a jump to occur during this interval is \lambda\,\mathrm{d}t, so

\displaystyle P_{t+\mathrm{d}t}(y) = (1-\lambda \mathrm{d}t)\left\langle P_t\left(y-\mu \,\mathrm{d}t - \left[W(t+\mathrm{d}t)-W(t)\right] \right) \right\rangle_W + \lambda \mathrm{d}t\,\left\langle P_t(y-s) \right\rangle_s,   (5)

where \left\langle \cdot \right\rangle_W denotes averaging over all realizations of the Brownian motion W, and \left\langle \cdot \right\rangle_s denotes averaging over the distribution Q(s) of the jump size s. Since the jump term is already multiplied by the jump probability \propto \mathrm{d}t, the drift and noise contributions there are of higher order in \mathrm{d}t and were dropped.

The averaging over the jump size in (5) yields a convolution with the jump size distribution Q(s):

\displaystyle \left\langle P_t(y-s) \right\rangle_s = \int_0^\infty \mathrm{d}s \,Q(s)P_t(y - s).

The average over the noise W in (5) is the same as for standard diffusion. During the interval \mathrm{d}t, the increment of the noise term in (2) is

\displaystyle W(t+\mathrm{d}t)-W(t) = \int_t^{t+\mathrm{d}t}\mathrm{d}t'\,\xi(t') =: V\sqrt{\mathrm{d}t},

where the last equality is a definition for V.
Since W is a Brownian motion, V is normally distributed:

\displaystyle \mathcal{N}(V) = \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{V^2}{2}\right).

Thus, the average over W in (5) is

\displaystyle  \begin{array}{rl}  \displaystyle \left\langle P_t\left(y-\mu \,\mathrm{d}t - \left[W(t+\mathrm{d}t)-W(t)\right] \right) \right\rangle_W = & \displaystyle \int_{-\infty}^\infty \mathrm{d}V\,\mathcal{N}(V)\, P_t\left(y-\mu \,\mathrm{d}t - V \sqrt{\mathrm{d}t} \right) \\   = & \displaystyle P_t(y) - \mu\, \mathrm{d}t\, \partial_y P_t(y)- \sqrt{\mathrm{d}t} \partial_y P_t(y) \int_{-\infty}^\infty \mathrm{d}V\, \mathcal{N}(V)\, V  \\  & \displaystyle + \frac{1}{2}\mathrm{d}t \partial_y^2 P_t(y)\, \int_{-\infty}^\infty \mathrm{d}V\, \mathcal{N}(V)\, V^2 + \mathcal{O}(\mathrm{d}t)^2 \\  =& \displaystyle P_t(y) - \mu\, \mathrm{d}t\, \partial_y P_t(y) + \frac{1}{2}\mathrm{d}t\,\partial_y^2 P_t(y)  \end{array} .

In the last line we used the fact that the normal distribution has mean zero and variance 1, and dropped all terms of higher order than \mathrm{d}t. Inserting all this into (5), we obtain the jump-diffusion FPE (4).

2. Example: Exponential jumps

Now let us consider a simple example, where the jump sizes are distributed exponentially:

\displaystyle  Q(s) = \alpha\,\exp(-\alpha s),\quad\quad \alpha > 0  .

We will compute the distribution of y(t), starting from y(0)=0.
For this, we solve the jump-diffusion FPE (4) with the initial condition P_{t=0}(y) = \delta(y). Let us take a Fourier transform of (4) from y to k:

\begin{array}{rl}  \displaystyle  \tilde{P}_t(k) := & \displaystyle \int_{-\infty}^\infty \mathrm{d}y\,e^{i k y}\,P_t(y), \\  \displaystyle  \partial_t \tilde{P}_t(k) = & \displaystyle -\frac{\sigma^2}{2}k^2\,\tilde{P}_t(k) + i\mu k\, \tilde{P}_t(k) - \lambda \tilde{P}_t(k) + \lambda\tilde{Q}(k)\tilde{P}_t(k)  \end{array}  .

The different spatial modes decouple, and instead of a partial differential equation we obtain an ordinary differential equation for each value of the Fourier variable k.
For the above exponential form of the jump size distribution Q,

\displaystyle  \tilde{Q}(k) = \frac{\alpha}{\alpha - i k}  .

Furthermore, the initial condition P_{t=0}(y) = \delta(y) gives \tilde{P}_{t=0}(k) = 1. Hence, the solution $\tilde{P}$ reads

\displaystyle  \tilde{P}_t(k) = \exp\left[ \left( -\frac{\sigma^2}{2}k^2+  i\mu k + i\frac{\lambda k}{\alpha - i k} \right) t\right]  .

While it does not seem feasible to invert this Fourier transform to obtain a closed expression for P_t(y) (but if anyone has an idea, let me know!), \tilde{P}_t(y) is already enough to determine the moments of y(t). Taking derivatives, we obtain for example

\displaystyle  \begin{array}{rl}  \left\langle y(t) \right\rangle = & -i\partial_k \big|_{k=0} \tilde{P}_t(k) = \left(\frac{\lambda}{\alpha}+\mu\right)t \\  \left\langle y(t)^2 \right\rangle^c = & -\partial_k^2 \big|_{k=0} \tilde{P}_t(k) + \left[\partial_k \big|_{k=0} \tilde{P}_t(k)\right]^2  \\  = & \left(\frac{2\lambda}{\alpha^2}+\sigma^2\right)t, \\  ... &  \end{array}  .

Similarly, solving the FPE (4) with an absorbing boundary allows computing first-passage times (or at least their moments) for our jump-diffusion process.

Have fun, and do let me know if you have any comments!

Advertisements

Written by inordinatum

November 20, 2013 at 8:04 pm