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I keep searching but I can't find any place that gives a good method of showing a process is NOT Markov. The definition I am using is that for every $s<t$ and $g$ bounded borel there is $f$ borel such that $E[g(X_t)|\mathcal{F}_s] = f(X_s)$ a.s.

I was told that $(B_t+1)^2$ is not a Markov process relative to the filtration generated by $B_t$ (standard Brownian Motion), I would like to show this.

Negating the definition, I need to find $s<t$ and $g$ bounded borel such that there is no function $f$ with $E[g((B_t+1)^2)|\mathcal{F}_s] = f((B_s+1)^2)$ a.s.

By the Markov property for $B_t$ i see that I can write $$ E[g((B_t+1)^2)|\mathcal{F}_s] = h(B_s) $$ where $$ h(x) = E[g((B_t-B_s+x+1)^2)] $$ but I don't see how I can show that for a particular $g$ this can't be rewritten as $f((B_s+1)^2)$.

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  • $\begingroup$ You may find this useful: math.stackexchange.com/questions/27994/… $\endgroup$
    – user940
    Mar 30, 2013 at 16:06
  • $\begingroup$ That doesn't really help as the proof that it gives is just due to the dependence on $B_s$ it doesn't work out. But that is exactly what I'm trying to show. How can I show $B_s$ is not $(B_s+1)^2$-measurable? Intuitively, the function isn't injective so it's "obvious", but I have been sitting here trying to prove it for a while and I can't figure it out. $\endgroup$
    – nullUser
    Mar 30, 2013 at 17:05
  • $\begingroup$ I apologize. I didn't realize your problem and the linked problem were that different. I just thought the other answer might give you some useful ideas. I've provided a fuller answer below which seems to show that $(B_t+1)^2$ is, in fact, Markov. $\endgroup$
    – user940
    Mar 30, 2013 at 22:44

1 Answer 1

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A bounded Borel function $h$ on $\mathbb{R}$ is a function of $(x+1)^2$ if and only if $h(x)=g(x+1)$ for some symmetric function $g$, i.e., $g(x)=g(-x)$.

Let $f$ be any bounded Borel function and define the symmetric function $g(x)=f(x^2)$ and let $h(x)=g(x+1)=f((x+1)^2)$. Since the transition function $p_t$ for Brownian motion is symmetric, we have for convolution $(p_t*g)(x)=(p_t*g)(-x)$ and since it is also shift invariant, we get $(p_t*h)(x)=(p_t*g)(x+1)$ so $(p_t*h)(x)=\phi_t((x+1)^2)$ for some bounded Borel $\phi_t$.

Let $(B_t)$ be standard Brownian motion with filtration $({\cal F}_t)$, and define $X_t:=(B_t+1)^2$. For $s<t$ we have $$\mathbb{E}\left[f(X_t)\,\big|\,{\cal F}_s\right] = \mathbb{E}\left[h(B_t)\,\big|\,{\cal F}_s\right] =(p_{t-s}*h)(B_s)=\phi_{t-s}(X_s).$$
So in fact, $(X_t)$ is a Markov process.

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  • $\begingroup$ ... This can't be true. $B_t$ is not $\sigma(X_t)$ measurable, so when $f$ is the identity function the statement cannot hold. $\endgroup$
    – nullUser
    Mar 30, 2013 at 23:21
  • $\begingroup$ @nullUser I don't claim that $B_s$ is $\sigma(X_s)$ measurable. But I do claim that $\mathbb{E}[f(X_t)\,|\,{\cal F}_s]$ is $\sigma(X_s)$ measurable. $\endgroup$
    – user940
    Mar 30, 2013 at 23:52
  • $\begingroup$ $E[X_1|\mathcal{F}_s] = X_s + 1-s+2(B_s+s)(1-s)+(1-s)^2$, no? $\endgroup$
    – nullUser
    Mar 31, 2013 at 1:03
  • $\begingroup$ $$X_t=(B_t+1)^2=(B_t-B_s)^2+2(B_t-B_s)(B_s+1)+(B_s+1)^2$$ Condition with respect to ${\cal F}_s$ to get $(t-s)+0+(B_s+1)^2=(t-s)+X_s$. $\endgroup$
    – user940
    Mar 31, 2013 at 1:08
  • $\begingroup$ Oh okay I got it now, arithmetic error. $\endgroup$
    – nullUser
    Mar 31, 2013 at 2:14

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