A Poisson process $N_t$ has a jump size $dN$ and a rate $\lambda$. Can a Poisson process be viewed as a diffusion process in the infinitesimal limit, i.e. as $dN \rightarrow 0$? Here is a related question, but my question is about scaling $dN$ in a finite time frame $t\in [0,T]$ in which case I don't think $\lambda$ would need to go to infinity.
The mean and variance of a Poisson process are both $\lambda$, so I was thinking the diffusion would be
$$dN_t = \lambda dt + \sqrt{\lambda} dB_t.$$
I don't know whether this form holds or how to prove/disprove it. The Poisson process transition probability is
$$ P(N(t+dt) – N(t) = k) = \frac{(\lambda dt)^k e^{-\lambda dt}}{k!}$$
How can we extend this to the continuous case? Would it be, for $x\in \mathbb{R}$,
$$ P(N(t+dt) – N(t) = x) = \frac{(\lambda dt)^x e^{-\lambda dt}}{\Gamma(x)}?$$
Best Answer
We have $$ P(N(t+\Delta t)- N(t) = k) = \frac{(\lambda \Delta t)^k \mathrm{e}^{- \lambda \Delta t}}{k !} $$ as you noted. Over large time intervals $\Delta t$ we thus have a convergence to a normal distribution $$ \frac{N(t+\Delta t)- N(t) - \lambda \Delta t}{\sqrt{\lambda \Delta t}} \quad \xrightarrow[]{d} \quad \text{Normal}(0,1). $$ as is standard for a Poisson distribution.
Now consider a random walk $$ x(t+ d t) = \mu d t + \sigma d W_t $$ for which we have $$ \frac{x(t+\Delta t) - x(t) - \mu \Delta t}{\sigma \sqrt{\Delta t}} \sim \text{Normal(0,1)} $$ Thus if one restricts to looking stroboscopically at time intervals $t \in \Delta t \mathbb{N}$ we see the two processes exhibit the same convergence in distribution if we set $\sigma = \sqrt{\lambda}$, and $\mu = \lambda$ in the limit of large $\Delta t$.
More coarsely, a Poisson process "looks" like a random walk as long as it is observed on timescales much larger than than the inverse rate constant $$ \Delta t \gg \lambda^{-1}. $$