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Hi I am trying to show that if $X_n$ is a markov process, whether or not $X_n^2$ is a markov process.

$X_n$ is a markov process if $P\{X_k = a_k|X_{k-1} = a_{k-1}, X_{k-2} = a_{k-2}, ..., X_k = a_1 \} = P\{X_k = a_k|X_{k-1} = a_{k-1}\}$.

Then $P\{X_k^2 = a_k|X_{k-1}^2 = a_{k-1}, X_{k-2}^2 = a_{k-2}, ..., X_k^2 = a_1 \} = P\{X_k = \pm \sqrt{a_k} |X_{k-1} = \pm \sqrt{a_{k-1}}, X_{k-2} = \pm \sqrt{a_{k-2}}, ..., X_k = \pm \sqrt{a_1} \} = P\{X_k = \pm \sqrt{a_k} |X_{k-1} = \pm \sqrt{a_{k-1}}\}$

may not be true since this knowing the probability of $X_{k-1} = - \sqrt{a_{k-1}}$ does not tell you anything about the probability of $X_{k} = + \sqrt{a_{k}}$

Am I on the right path here? Thanks!

Erroldactyl
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    This is not Markov in general. Hence the simplest approach is to provide a counterexample. – Did Apr 14 '14 at 21:26

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Here is a natural example: random walk on the numbers $-1,0,1,2$ which moves left or right with probability $1/2$, except that transitions $-1\to 0$ and $2\to 1$ have probability $1$. It's a Markov process, but its square is not.

Indeed, suppose we observed $X_1^2=4$ and $X_2^2=1$. Then we know $X_1=2$, hence $X_2=1$. Conclusion: $P(X_3^2=0) = 1/2$.

On the other hand, if we observed $X_1^2=0$ and $X_2^2=1$, then $$P(X_3^2=0) = \frac12 P(X_3=0 | X_2=1) + \frac12 P(X_3=0 | X_2=-1) = \frac12\cdot \frac12 + \frac12\cdot 1 = \frac34 $$