I would like to find the probability density function (at stationarity) of the random variable $X_t$, where: \begin{equation*} dX_t = -aX_t dt + d N_t, \end{equation*} $a$ is a constant and $N_t$ is a compound Poisson process with Poisson jump size distribution.

In other words, $X_t$ solves the ordinary differential equation $\frac{d X_t}{dt} + a X_t=0$, but at times $t_i$ say, where the $t_i$ are exponentially distributed with mean $1/k$, $X_t$ increases by an integer drawn from $M\sim Poi(m)$ (i.e. $X_t$ gets a Poisson-distributed "kick" upwards at exponentially distributed intervals).

Is there a way of obtaining the pdf for this random variable $X$? If I have understood things correctly, the Kramers-Moyal equation for the pdf of $X$ is of infinite order because it is a jump Markov process. I have also tried looking at the Master Equation but I get lost. However, I am new to this literature and was wondering if the solution is easy for those in the know, since it is such a simple system.

Many thanks for your help!


In the following paper, an expression for the Fourier transform of the probability density function is provided for general jump size distribution (see Section 5.2):

Generalized Fokker-Planck equation: Derivation and exact solutions

or in the ArXiv: http://arxiv.org/pdf/0808.0274.pdf

For the case I am interested in (Poisson jump size distribution), I don't think the Fourier transform can be inverted analytically. However, the paper gives the exact solution for exponentially-distributed jump sizes as an example.

Thank you to those who have answered or commented so far. It is actually very helpful for me to know that there isn't a straightforward way to obtain the pdf (if at all). Nevertheless, if anyone else out there does know of a way I'd be interested to hear about it.


1 Answer 1


I don't think you can have an explicit form.

Let $Y_t= e^{at}X_t$ then :

$$ Y_t -Y_0 =\sum_{i=1}^{N_t}e^{aT_i} $$ where $(T_i)_{i=1...N_t}$ are the jump times of your poisson process.

then $$P(Y_t\leq x)=\sum_{n\geq 0}\frac{(mt)^n}{n!}e^{-mt}P(\sum_{i=1}^{N_t}e^{aT_i}\leq x|N_t=n)$$

$$P(\sum_{i=1}^{N_t}e^{aT_i}\leq x|N_t=n) =\int_{[0,+\infty]^n}\mathbf{1}_{\sum_{i=1}^n e^{at_i}\leq x}\mathbf{1}_{t_1<t_2<...<t_n}m^ne^{-mt_n} dt_1dt_2\dots dt_n$$

and then it becomes difficult.

  • 1
    $\begingroup$ As I understand it from the OP: $N_t := \sum_{i=1}^{P_t} J_i$ where $P_t$ denotes a Poisson counting process and $J_i$ i.i.d. Poisson jump sizes. In your answer, didn't you assimilate $N_t$ to a standard Poisson process (as it is the usual notation for Poisson processes in most articles) ? This doesn't change much to your argument though. $\endgroup$
    – Quantuple
    Commented Apr 7, 2016 at 10:13
  • 1
    $\begingroup$ you're right with $N_t=\sum_{i=1}^{P_t}\xi_i$ it will give us : $$Y_t-Y_0 = \sum_{i=1}^{P_t}e^{aT_i}\xi_i$$ where $T_i$ are the jump times of $N$ and it remains difficult $\endgroup$ Commented Apr 7, 2016 at 13:28
  • $\begingroup$ I wonder if this process does indeed reaches some sort of stationarity... but I agree with you, difficult to come up with a closed-form expression. $\endgroup$
    – Quantuple
    Commented Apr 7, 2016 at 13:38
  • 2
    $\begingroup$ Thank you both! It's helpful to know that there isn't a simple solution (if any). I found a paper that provides the Fourier transform of the solution for general jump size distribution, but it isn't invertible (as far as I can tell) for the Poisson distribution. I will add an update to my question now in case others are interested. $\endgroup$ Commented Apr 9, 2016 at 14:50

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