# inordinatum

Physics and Mathematics of Disordered Systems

## Fokker-Planck equation for a jump diffusion process

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!