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Let $B = \{B_t\}_{t\geq 0}$ be a standard brownian motion. I'm trying to show that the process $X = \{X_t\}_{t\geq 0}$, where $$ X_t := \int_0^t B_s \,\mathrm{d}s, \tag{1} $$ is not a Markov process with respect to the natural filtration $\{\mathscr{F}_t^B\}_{t\geq 0}$ of $B$. I understand that, by definition, $X$ is a Markov process if for every Borel mesurable set $A \subset \mathbb{R}$ and for every $s, t > 0$, $$ \mathbb{P}[X_{t+s} \in A \mid \mathscr{F}_s^B] = \mathbb{P}[X_{t+s} \in A \mid X_s]. $$ However, I'm having trouble to figure out how to work with this definition, especially to prove that the process is not a Markov process.

I found similar questions about this process, such as this one. However, the proof given there that it is not a Markov process relies on other results about gaussian processes, which I don't know about.

Is there a way to prove that the process $X$ defined by $(1)$ is not a Markov process, directly from the definition?

Vincent
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    See also this question – saz Apr 25 '20 at 07:15
  • @saz Thanks, these are two very interesting approaches to the problem. However there has been no feedback from others, do they seem like good proofs to you? I found both to be satisfying, but I'm still new to the subject so I might be missing something. – Vincent Apr 25 '20 at 13:55

1 Answers1

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Observe that $X_{t+s} = tB_s+ X_s+\int_0^t [B_{u+s}-B_s]\,du$. The first two terms on the right side of this expansion are $\mathscr F^B_s$-measurable, while the third is independent of $\mathscr F^B_s$ and has the same distribution as $Xt$ (because of the stationary independent increments of Brownian motion). It follows that the conditonal distribution of $X_{t+s}$, given $\mathscr F^B_s$ is the same as that of $X_t$ translated by the fixed amount $tb+x$, where $b=B_s$ and $x=X_s$. In particular, that conditional distribution depends not just on $X_s$ but also on $B_s$.

John Dawkins
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