Show that the following process based on exponential waiting time is Markov wrt natural filtration For a parameter $\lambda>0$ consider an exponential random variable: $T\sim \operatorname{Exp}(\lambda)$ and define a process $X_{t}=(0)1_{T> t}+(t-T)1_{T\leq t}$. Show that $X_{t}$ is $\mathcal{F}_{t}^{X}$ - Markov process with transition function:
$P_{t}(0,A)=\exp(-\lambda t)1_{A}(0)+\int_{0}^{t}1_{A}(t-s)\lambda\exp(-\lambda s)ds$
and for $x>0$:
$P_{t}(x,A)=1_{A}(x+t)$
Basically all I need to show is that for $t,h\geq 0, \; A \in \mathcal{B}((0,\infty))$ and $n\in \mathbb N$ with $0=t_{0}<...<t_{n}=t$ and $A_{i}\in \mathcal{B}((0,\infty))$ for all $i = 0,...,n$ that:
$E[1_{\{X_{t+h}\in A\}}\prod\limits_{i=0}^{n}1_{\{X_{t_{i}}\in A_{i}\}}]=E[P_{h}(X_{t},A)\prod\limits_{i=0}^{n}1_{\{X_{t_{i}}\in A_{i}\}}]\; (*)$
Now in order to evaluate $(*)$ I will consider the two cases $\{T\leq t\}$ and $\{T> t\}$, thus considering
$E[P_{h}(X_{t},A)\prod\limits_{i=0}^{n}1_{\{X_{t_{i}}\in A_{i}\}},T\leq t]=E[1_{A}(X_{t}+h)\prod\limits_{i=0}^{n}1_{\{X_{t_{i}}\in A_{i}\}},T\leq t]=E[1_{A}(t+h-T)\prod\limits_{i=0}^{n}1_{\{X_{t_{i}}\in A_{i}\}},T\leq t]=E[1_{A}(X_{t+h})\prod\limits_{i=0}^{n}1_{\{X_{t_{i}}\in A_{i}\}},T\leq t]$ and so we are done in the case that $T\leq t$. But now in the case that $T> t$:
$E[P_{h}(X_{t},A)\prod\limits_{i=0}^{n}1_{\{X_{t_{i}}\in A_{i}\}},T> t]=E[P_{h}(0,A)\prod\limits_{i=0}^{n}1_{\{X_{t_{i}}\in A_{i}\}},T> t]=\prod\limits_{i=0}^{n}\delta_{0}(A_{i})P_{h}(0,A)\mathbb P(T> t)$
So I supppose this problem would reduce itself to showing that $P_{h}(0,A)\mathbb P(T> t)=1_{A}(X_{t+h})$, so now clearly:
$P_{h}(0,A)\mathbb P(T> t)=\exp(-\lambda t)P_{h}(0,A)=\exp(-\lambda t)(\exp(-\lambda h)1_{A}(0)+\int_{0}^{h}1_{A}(h-s)\lambda\exp(-\lambda s)ds)$
This is where I am unsure of how to proceed, to show that the above is equal to $1_{A}(X_{t+h})$, any ideas?
 A: I think there is a typo in your formula. I feel like we should have $P_t(0,A)=e^{\lambda t}1_A(0)+\int_0^t1_A(t-s)e^{\lambda s}\textrm{d}s$ (or, indeed, the calculation below should show it!).
Well, $P_h(0,A) \mathbb{P}(T>t)$ is definitely equal to neither $0$ nor $1$, so that's not gonna be equal to an indicator function. However, we have
\begin{align}
&\mathbb{P}[X_{t+h}\in A, X_i\in A_i\; \forall i, T>t]=\mathbb{P}[X_{t+h}\in A, 0\in A_i \; \forall i, T>t]\\
=&\mathbb{E}[1_A(X_{t+h}) 1_{T>t} \prod_i 1_{A_i}(0)]=\prod_{i} 1_{A_i}(0) \mathbb{E}[1_A(X_{t+h}) 1_{T>t}]
\end{align}
And, furthermore,
\begin{align}
\mathbb{E}[1_A(X_{t+h}) 1_{T>t}] = & \mathbb{E}[ 1_A(t+h-T)1_{T\in (t,t+h]}+1_A(0)1_{T\geq t+h}] \\
=& \int_0^{\infty} e^{-\lambda s} \left(1_A(t+h-s)1_{[t,t+h]}(s)+1_A(0)1_{(t+h,\infty)}(s)\right)\textrm{d}s\\
=& \int_{t}^{t+h} e^{-\lambda s} 1_A(t+h-s)\textrm{d}s+\int_{t+h}^{\infty} e^{-\lambda s}1_A(0)\textrm{d}s \\
=& e^{-\lambda t}\int_0^h e^{-\lambda s}1_A(h-s)\textrm{d}s+1_A(0) e^{-\lambda(t+h)}\\
=& P_h(0,A)e^{-\lambda t}\\
=& P_h(0,A)\mathbb{P}(T>t),
\end{align}
which should finish the proof for you.
