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Prove, or give an explicit counterexample to refute, the following assertion: if $\{X_n\}$ is a Markov chain, then $\{X_n^2\}$ is also a Markov chain.

It's easy to show that $$P\{X_{n+1}^2=j^2|X_0=i_0,\ldots,X_{n}=i_n\}=P\{X_{n+1}^2=j^2|X_n=i_n\}$$ thus I think if I can also show that $$P\{X_{n+1}^2=j^2|X_0=\{\pm i_0\},\ldots,X_n=\{\pm i_n\}\}=P\{X_{n+1}^2|X_n=\{\pm i_n\}\}$$ then this problem is done. But it seems incorrect. So now I have no idea how to proof it.

(Additional question: What properties should $f(x)$ satisfies to make $\{f(X_n)\}$ a Markov chain?)

user95640
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1 Answers1

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For a counterexample consider the Markov chain $\{X_n\}$ described here by Did. Assume everything to be the same as in the example, but instead of states $0, 1$ and $2$ consider the states $-1, 1$ and $2$.

Now the function $g(x)=x^2$ lumps together states $-1$ and $1$. That is, we have that $g(-1)=g(1)=1$ and $g(2)=4$, exactly as described in the aforementioned example. Then you have directly from the result that is proved there, that the process $g(X_n)=X_n^2$ does not have the markov property.

Concerning your additional question, you can read this article or see here for more details.

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Jimmy R.
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