Poisson-like random walk expressed as Bernoulli-like random walks (splitting scheme)

Question

In our problem we have the transition density for $x,y\in \mathbb{Z}$ and $t\in \mathbb{N}$

$$R_{t}(x,y):=e^{-t}\frac{t^{x-y}}{(x-y)!}1_{x\geq y},$$

which is the Poisson distribution pdf. (This is also in semigroup form of the operator $I+\nabla^{-}$:

$$R_{t}(x,y)=e^{-t(I+\nabla^{-})}(x,y),$$

where $If(x)=f(x)$ and $\nabla^{-}f(x)=f(x)-f(x-1)$).

Q: I am curious if we can write this as a convolution of unit step operators (i.e. $P_{t,k}(x,y):=c(x,y,t,k)1_{1\geq x-y\geq 0}$).

By convolution we mean $$A\ast B(x,y):=\sum_{z\in \mathbb{Z}}A(x,z)B(z,y).$$

So I am curious if we can decompose a Poisson jump into Bernoulli jumps

$$R_{t}(x,y)=P_{t,1}(x,y)\ast \cdots \ast P_{t,n}(x,y)$$

for some n, possibly infinite. The $R_{t}(x,y)$ is a probability a random walk of jumping to the left from $x$ to $y$ with Poisson jumps. So heuristically I am wondering if I can "split" this Poisson-random walk into Bernoulli-random walks. In a way I am trying to run over all possible unit step paths from from $x$ to $y$.

Same operator

Let $P_{t}(x,y)=c(x,y,t)1_{1\geq x-y\geq 0}$ for some function $c$ ,ideally of the form $c(x,y,t)=c(x-y,t)$. These are unit steps transition and so we ask to find some n (possibly infinite) to have the n-convolution equal $R_{t}$: $$R_{t}(x,y)=P_{t}\ast\cdots \ast P_{t}(x,y),$$

where $A\ast B(x,y):=\sum_{z\in \mathbb{Z}}A(x,z)B(z,y)$. That doesn't seem to work because by a simple induction argument we get

$$P_{t}\ast\cdots \ast P_{t}(x,y)=\sum_{k=0}^{n}\binom{n}{k}c(1,t)^{n-k}c(0,t)^{k}c(k,t) 1_{1\geq x-y-n+k\geq 0} $$ $$= \binom{n}{n-(x-y)+1}c(1,t)^{x-y-1}c(0,t)^{n-(x-y)+1}c(n-(x-y)+1,t)+\binom{n}{n-(x-y)}c(1,t)^{x-y}c(0,t)^{n-(x-y)}c(n-(x-y),t)$$

$$= (c(1,t)/c(0,t))^{x-y}c(0,t)^{n}[\binom{n}{n-(x-y)+1}\frac{c(1,t)}{c(0,t)}c(n-(x-y)+1,t)+\binom{n}{n-(x-y)}c(n-(x-y),t)].$$

Here are some cases that show it is not true if we use the same operator.

$\bullet$ If $\frac{c(1,t)}{c(0,t)}c(n-(x-y)+1,t)=c(n-(x-y),t)$, then using the summation identity for the binomial we get $$= (c(1,t)/c(0,t))^{x-y}c(0,t)^{n}\binom{n+1}{n-(x-y)+1}c(n-(x-y),t).$$

So to get $\frac{1}{(x-y)!}$ requires $\binom{n+1}{n-(x-y)+1}=\frac{1}{(x-y)!}$, which implies $(n-(x-y)+1)!$ and so $x-y=0$, which is not necessarily true for general $x\geq y$.

$\bullet$ If we let $n=x-y$, we get $$=\binom{x-y}{1}c(1,t)^{x-y}c(0,t)+\binom{x-y}{0}c(1,t)^{x-y}c(0,t)=c(1,t)^{x-y}c(0,t)(x-y+1),$$

which is not equal to $(x-y)!$.

Different operators

Using the property

$$e^{-t(I+\nabla^{-})}(x,y)=e^{-\sum_{k=1}\frac{t}{2^{k}}(I+\nabla^{-})}(x,y)=\ast_{k=1}^{\infty}R_{t/2^{k}}(x,y)=\lim_{n\to +\infty}R_{t/2}\ast\dots\ast R_{t/2^{n}}(x,y),$$

we could use $P_{t,k}(x,y):=R_{t/2^{k}}(x,y)=e^{-t}\frac{(t/2^{k})^{x-y}}{(x-y)!}1_{x\geq y}$. But these are not unit steps.

Splitting scheme point of view

One point of view is from the splitting scheme for semigroups literature (eg. see here ). Here they decompose semigroup operators into smaller steps.

Answer 1

If you insist on $P_t(x,y) = c(t,x-y) \mathbb{1}_{1\geqslant x-y \geqslant 0}$, then it is not possible to factorise $R_t$ as $P_t Q_t$ for whatever "reasonable" $Q_t$. Indeed, this would mean that the corresponding Z-transforms satisfy $$ \sum_{x=0}^\infty R_t(x,0) z^n = \sum_{x=0}^\infty P_t(x,0) z^n \times \sum_{x=0}^\infty Q_t(x,0) z^n $$ for all $z$. However $$ \sum_{x=0}^\infty P_t(x,0) z^n = c(t,0) + c(t,1) z $$ has a zero at some negative $z$, while $$ \sum_{x=0}^\infty R_t(x,0) z^n = e^z $$ is non-zero everywhere.

(Here "reasonable" means, for example, non-negative. Indeed, in this case $Q_t(x,0)$ necessarily vanishes exponentially fast as $x \to \infty$, and all Z-transforms converge well for all $z$.)