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](http://A%20Semigroup%20Point%20Of%20View%20On%20Splitting%20Schemes%20For%20Stochastic%20(Partial)%20Differential%20Equations) ). Here they decompose semigroup operators into smaller steps.



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


