I found the following proposition and I want to prove it.
Let $S_n$ be a discrete-time random walk with increment distribution p and $N_t$ be a Poisson process with parameter $1$.Then the process $T_t:=S_{N_t}$ has the distribution of a continuous-time random walk with increment distribution p.
The same literature gives the following definition of a continuous-time random walk:
A continuous-time random walk with increment distribution $p$ is the continuous-time Markov chain $T_t$ with rates $p$, i.e.:
- $\mathbb{P}[T_{t+\Delta t}=y\mid T_t=x]=p(y-x)\Delta t+o(\Delta t)$ if $x\neq y$
- $\mathbb{P}[T_{t+\Delta t}=x\mid T_t=x]=1-[\sum_{y\neq x}p(y-x)]\Delta t+o(\Delta t)$ if $x=y$.
A discrete-time random walk with increment distribution $p$ can be seen as a time-homogeneous Markov chain $S_n$ with transition probabilities:
- $p(y,z)=\mathbb{P}[S_{n+1}=z\mid S_n=y]=p(z-y)$
and n-th step distribution
- $p_n(x,y)=\mathbb{P}[S_{n}=y\mid S_0=x]$
So I want now to use the discrete definition to prove the continuous case. Note that $\mathbb{P}[N_{t+\Delta t}-N_t=k]=\frac{e^{-\Delta t}(\Delta t)^k}{k!}$. I want to use this condition so I thought the following:
$\mathbb{P}[S_{N_{t+\Delta t}}=y\mid S_{N_t}=x]=\frac{\sum_{k\geq 0}{\mathbb{P}[S_{N_{t+\Delta t}}=y,S_{N_t}=x,N_{t+\Delta t}-N_t=k]}}{\mathbb{P}[S_{N_t}=x]}=\frac{\sum_{k\geq 0}{\mathbb{P}[S_{N_{t}+k}=y,S_{N_t}=x,N_{t+\Delta t}-N_t=k]}}{\mathbb{P}[S_{N_t}=x]}=\sum_{k\geq 0}{\mathbb{P}[S_{N_{t}+k}=y,N_{t+\Delta t}-N_t=k\mid S_{N_t}=x]}$.
But now I don't see how to go on from there and honestly it does not looks that good..
UPDATE
So I tried the following. But I think I still need some help to conclude the proof.
Want to show: $T_t:=S_{N_t}$ is a continuous-time Markov chain with transition probability $p$.
To check: $\mathbb{P}[T_{t+\Delta t}=y\mid T_t=x]=p(y-x)\Delta t+o(\Delta t)$
Using that the discrete-time random walk is a Markov chain with transition probability $p$ we have: $\mathbb{P}[S_{n+1}=z\mid S_n=y]=p(z-y)$
Now from the cmputation we have: $\mathbb{P}[T_{t+\Delta t}=y\mid T_t=x]=\mathbb{P}[S_{N_{t+\Delta t}}=y\mid S_{N_t}=x]=\frac{\sum_{k\geq 0}{\mathbb{P}[S_{N_{t+\Delta t}}=y,S_{N_t}=x,N_{t+\Delta t}-N_t=k]}}{\mathbb{P}[S_{N_t}=x]}=\frac{\sum_{k\geq 0}{\mathbb{P}[S_{N_{t}+k}=y,S_{N_t}=x,N_{t+\Delta t}-N_t=k]}}{\mathbb{P}[S_{N_t}=x]}=\sum_{k\geq 0}{\frac{\mathbb{P}[S_{N_{t}+k}-S_{N_t}=y-x,N_{t+\Delta t}-N_t=k,S_{N_t}=x]}{\mathbb{P}[S_{N_t}=x]}}$
Now since $S_{N_{t}+k}-S_{N_t}\overset{L}{=}S_k$ and $S_k$ is independent of $N_{\Delta t}\overset{L}{=}N_{t+\Delta t}-N_{t}$, we have $S_{N_{t}+k}-S_{N_t}$ is independent of $N_{t+\Delta t}-N_{t}$ and also independent of $S_{N_t}$. And finally since the Poisson process has independent increments we also have that $N_{t+\Delta t}-N_{t}$ is independent of $N_{\Delta t}$ and so (since by assumption $N_t$ is independent of $S_n$) is also independent of $S_{N_t}$. So we get:
$=\sum_{k\geq 0}{\frac{\mathbb{P}[S_{N_{t}+k}-S_{N_t}=y-x]\mathbb{P}[N_{t+\Delta t}-N_t=k]\mathbb{P}[S_{N_t}=x]}{\mathbb{P}[S_{N_t}=x]}}=\sum_{k\geq 0}{\mathbb{P}[S_{k}=y-x]\mathbb{P}[N_{\Delta t}=k]}=\sum_{k\geq 0}{\mathbb{P}[S_{k}=y-x]\frac{e^{-\Delta t}(\Delta t)^{k}}{k!}}$
So now I think I should use that $S_n$ is a Markov chain to rewrite the term $\mathbb{P}[S_{k}=y-x]$. I was thinking to rewrit it as $\sum_{z\in\mathbb{Z}}{\mathbb{P}[S_{k}=y-x\mid S_{k-1}=z]}=\sum_{z\in \mathbb{Z}}{p(y-x-z)}$, but again this does not looks very helpful...