# Probability density function of simple equation, compound Poisson noise

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.

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.

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.

• 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. Apr 7, 2016 at 10:13
• 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 Apr 7, 2016 at 13:28
• 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. Apr 7, 2016 at 13:38
• 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. Apr 9, 2016 at 14:50