# Showing a process is not markov

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 $f((B_s+1)^2)$.

• You may find this useful: math.stackexchange.com/questions/27994/…
– user940
Mar 30, 2013 at 16:06
• 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. Mar 30, 2013 at 17:05
• 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.
– user940
Mar 30, 2013 at 22:44

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 $$s we have $$\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.

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