# Proving a Markov Chain (Random Walk) is Time-Homogeneous

Let $$Y_0, Y_1,Y_2,...$$ be a sequence of independent and identically distributed random variables. Then we define $$X_n = \displaystyle\sum_{j=0}^{n} Y_j$$ , $$n=0,1,2,...$$.

This is a Random Walk process. I would like to get help to prove that this is Time-homogeneous.

For the Markov property, I considered increments of this process and proved that they are independent and then used that to deduce the Markov property. But I am unable to prove that the transition probabilities do not depend on time. Could anyone help me with this please?

Thank You

Hi: For time homogeneity, you need $$P(X_{n+1} = I_{n+1}| X_{n} = I_{n})$$ is independent of the time, namely $$n$$.
$$P(X_{n+1} = I_{n+1}| X_{n} = I_{n}) = P(Y_{n+1} = I_{n+1} - I_{n})$$.
But $$Y_{n}$$ is iid, so $$P(Y_{n+1} = I_{n+1} - I_{n})$$ is not dependent on $$n$$ which means that the chain is time homogeneous.
• @Tejo Gedela: Your welcome but reading this again, I'm not 100 percent sure it's correct because $I_{n+1} - I_{n}$ is still a function of $n$. Could someone comment on whether this answer is correct ? or correct it. Thanks. Commented Sep 21, 2019 at 13:45
• I just looked at this now and it should be okay because $P(I_{n+1} - I_{n} ) = P(Y_{n+1})$. But $Y_{i}$ is independent and identically distributed so $P(Y_{n+1})$ equals $P(Y_{i}) ~\forall i~$ so it's not a function of $n$. Commented Oct 28, 2023 at 2:25