If $\{X_n\}$ is a homogeneous Markov chain, is it true that ${X_{n^2}}$ ($n$ is of the power $2$ not $X_n$) is also a homogeneous Markov chain? And why?
Asked
Active
Viewed 95 times
1
-
What have you tried? – Alex Vong Mar 28 '19 at 20:47
-
I know that its not true, but I dont know the counter example. – Jennyfer Lahoud Mar 28 '19 at 20:50
1 Answers
2
Here is a simple example. Consider the random walk on $\mathbb Z$ that stays at its position, goes to right or left with probability $1/3$ each. We consider the a particular transition probability \begin{align} p_n(0,2):=\mathbb P(X_{n^2}=2 \mid X_{(n-1)^2}=0 ) \end{align} which, if $X_n$ is a homoegeneous Markov chain, should not depend on $n$. But guess what? It depends on $n$, we have \begin{align} p_1(0,2)=\mathbb P(X_1=2\mid X_0=0)=0 \end{align} while \begin{align} p_2(0,2)=\mathbb P(X_4=2\mid X_1=0)>0 \end{align} Now it remains to think about why we have $p_2(0,2)>0$... Do you see why?
Shashi
- 8,738