Let $(X_n)_{n\ge 0}$ be a simple asymmetric random walk on states $0,1,\dots,M$, where $0$ and $M$ are absorbing. Initial state is $i\neq 0,M$. Let $(X_n^*)_{n\ge 0}$ be the process $(X_n)$ conditional on event $A:=\{X_m=M\text{ finally }\}$. That is $$P((X_0^*,X_1^*,\dots,X_k^*)\in B)=P((X_0,X_1,\dots,X_k)\in B|A)=\frac{P(\{(X_0,X_1,\dots,X_k)\in B\}\cap A)}{P(A)}$$ for any $k\ge 0 $ and $B\subset \{0,1,\dots,M\}^{k+1}$.

I need to show that $(X^*_n)$ is a Markov chain.

It's clear to me intuitively why it is so: on the event $A$ none of $X_k$'s should be equal 0, and then the Markov property is inherited from $(X_n)$, but I'm struggling with showing it rigorously.


1 Answer 1


We can prove this by using the Strong Markov Property. So, the Strong Markov Property says that (Theorem 1.4.2 from the book Markov Chain, by Norris'):

Let $(X_n)_{n\geq0}$ be $Markov(\lambda,p)$ and let $T$ be a stopping time of $(X_n)_{n\geq0}$. Then, conditional on $T<\infty$ and $X_T = i, (X_{T+n})_{n\geq0}$ is $Markov(\delta_i,P)$ and independent of $X_0,X_1,...,X_T$, where $\delta_i$ is the unit mass at $i$.

Now, we only have to prove that the time time to hit $M$ for the first time is a stopping time. The definition of stopping time is:

A random variable $T:\Omega\to \{{0,1,2,...\}}\cup\{{\infty\}}$ is called stopping time if the event $\{{T=n\}}$ depends only on $X_0,X_1,...,X_n$ for $n=0,1,2,...$.

Well, the time to hit $M$ for the first time is clearly a stopping time, since:

$$\{{T_M=m\}} = \{X_0\neq M,X_1\neq M,...,X_{m-1}\neq M,X_m=M\}$$

Therefore, $(X^*_n)=(X_{T_M+n})$ is a Markov chain.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.