Hi I have a question about the following definition. We want to define a random waldlk $S$ on $\mathbb{Z}^d$ in continuous time. For this let $p$ be the increment distribution of $S$. Then for each $x,y\in \mathbb{Z}^d$ we have
- $\mathbb{P}[S_{t+\Delta t}=y\mid S_t=x]=p(y-x)\Delta t+o(\Delta t)$
- $\mathbb{P}[S_{t+\Delta t}=x\mid S_t=x]=1-(\sum_{y\neq x}{p(y-x)})\Delta t+o(\Delta t)$
Now define:
- $p_t'(x,y)=\mathbb{P}[S_t=y\mid S_0=x]$ and $p_t'(y)=p'_t(0,y)=p_t'(x+x+y)$ Then all this should imply that
$\frac{d}{dt}p_t'(x)=\sum_{y\in\mathbb{Z}^d}{p(y)[p_t'(x-y)-p_t'(x)]}$
How this should imply that result?
I'm sorry I can't show it to you because I don't see it either. But you seem to study the same script like me.
If I read the definitions correctly, the following should hold: $$p'_t(x)=p'_t(0,x)=\mathbb P[S_t=x|S_0=0]=p(x)t.$$ This implies $\frac{\mathrm d}{\mathrm d t}p'_t(x)=p(x)$ and $$\sum_{y\in\mathbb{Z}^d}{p(y)[p_t'(x-y)-p_t'(x)]}=\sum_{y\in\mathbb{Z}^d}p(y)[p(x-y)t-p(x)t]=t\sum_{y\in\mathbb{Z}^d}p(y)[p(x-y)-p(x)].$$ I have no idea how the last term should work out to $p(x)$, so I must be terribly wrong at some point...