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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

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Hi: For time homogeneity, you need $P(X_{n+1} = I_{n+1}| X_{n} = I_{n})$ is independent of the time, namely $n$.

The process is a random walk so

$P(X_{n+1} = I_{n+1}| X_{n} = I_{n}) = P(Y_{n+1} = I_{n+1} - I_{n})$.

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.

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  • $\begingroup$ Thank you! This was very helpful. $\endgroup$
    – Tejo Gedela
    Commented Sep 20, 2019 at 10:29
  • $\begingroup$ @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. $\endgroup$
    – mlofton
    Commented Sep 21, 2019 at 13:45
  • $\begingroup$ 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 $n$. $\endgroup$
    – mlofton
    Commented Oct 28, 2023 at 2:25

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