If $(M_n)$ is a symmetric random walk then $(M_n^2)$ is a Markov process (or not?)

Question

Let $\{X_n:n=0,1,2,\ldots\}$ be an i.i.d. sequence of random variables with $$\mathbb P(X_1=1)=\mathbb P(X_1=-1)=\frac12 $$ and consider the process $\{M_n:n=0,1,2\ldots\}$ with $M_0=0$ and $$M_n = \sum_{j=1}^n X_j,\; n\geqslant 1. $$ It is clear that $M_n$ is a martingale, so the discrete stochastic integral $I_0=0$, $$I_n=\sum_{j=0}^{n-1}M_j(M_{j+1}-M_j) $$ exists and is a martingale, and by induction we find that $I_n = \frac12 M_n^2 - \frac12n$. Now, by definition, $\{I_n\}$ is a Markov process iff for each nonnegative integer $n$ and bounded measurable function $f$, there exists a measurable function $g_n$ such that $$\mathbb E[I_{n+1}\mid\mathcal F_n] = g_n(I_n). \tag 1$$ (where $\{\mathcal F_n\}$ is the canonical filtration of $\{I_n\}$). Writing $$I_{n+1}=\frac12\left(M_n+X_{n+1}\right)^2 - \frac12(n+1)$$ and conditioning on $X_{n+1}$, I found that $g_n:\mathbb Z\to\mathbb R$ with $$g_n(k) = \frac12 f\left( k + \sqrt{2k+n}\right) + \frac12 f\left(k-\sqrt{2k+n}\right) $$ satisfies $(1)$, but the computation (and the result!) are not particularly enlightening. Is there a more intuitive way to show that $\{I_n\}$ is a Markov process?

Answer 1

Let's eliminate the annoying $1/2$ by considering the martingale $J_n:=2I_n=M_n^2-n$. The process $\{J_n\}$ is Markov iff and only if $\{I_n\}$ is Markov. But $$ J_{n+1}=J_n+2X_{n+1}\sqrt{J_n+n}\cdot sign(M_n), $$ where $sign(M_n)$ is $+1$ if $M_n>0$ and $-1$ if $M_n<0$ and $0$ if $M_n=0$. This shows that the conditional distribution of $J_{n+1}$ given the history of $\{M_n\}$ depends on both $J_n$ and (the sign of) $M_n$, casting into doubt the Markov property of the process $\{J_n\}$. It is true that the pair $(J_n,M_n)$ is a Markov process with respect to the filtration $\{\mathcal G_n\}$ generated by $\{M_n\}$.