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I am pretty sure $(B_{t}^{2})$ not Markov because the squared random walk is not.

Showing the square of a Markov process is or isn't Markov

I guess I can repeat the method since to be Markov it must satisfy the discrete.

Thanks

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Thomas Kojar
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2 Answers2

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For every $t\geqslant0$, $(B_{t+s}^2)_{s\geqslant0}$ is distributed as $(X_s)_{s\geqslant0}$, where, for every $s\geqslant0$, $$X_s=B_t^2+2\sqrt{B_t^2}\cdot W_s+W_s^2,$$ where $(W_s)_{s\geqslant0}$ is a Brownian motion independent of $(B_u)_{0\leqslant u\leqslant t}$. Thus, indeed, $(B^2_t)_{t\geqslant0}$ is a Markov process.

The discrete analogue of this result is that, if $(X_n)_{n\geqslant0}$ is a random walk with $\pm1$ steps of equal probabilities $\frac12$, then $(|X_n|)_{n\geqslant0}$ is also Markov.

Did
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  • Thank you. Did you use that $2B_{t}W_{s}\stackrel{d} {\sim}2\sqrt{B^{2}{t}}W{s}$? – Thomas Kojar Nov 03 '14 at 23:13
  • Yes, provided one adds a missing factor 2 on the RHS and one passes to the level of identity in distribution of processes, not only of random variables. – Did Nov 03 '14 at 23:14
  • what do you mean by "level of identity in distribution of processes"? – Thomas Kojar Nov 03 '14 at 23:25
  • how can $B_{t}\stackrel{d}{\sim} |B_{t}|$? The latter is not even Gaussian. – Thomas Kojar Nov 03 '14 at 23:26
  • You do not need only the identity in distribution mentioned in your first comment to hold for each fixed $s$, but the identity $$(B_tW_s){s\geqslant0}\stackrel{d}{=}(|B_t|\cdot W_s){s\geqslant0}.$$ – Did Nov 03 '14 at 23:27
  • "The latter is not even Gaussian" It is not, and? (And who said that $B\stackrel{d}{=}|B|$?) – Did Nov 03 '14 at 23:27
  • By the way, what happened there? – Did Nov 03 '14 at 23:30
  • Why $B_{t}W_{s}\stackrel{d}{\sim}|B_{t}|W_{s}$ for all s? The $B_{t}$ doesn't have same distribution with $|B_{t}|$ so it is not obvious to me that the product with $W_{s}$ has the same distribution. – Thomas Kojar Nov 03 '14 at 23:31
  • Yeah, and? Note that if $U=\pm1$ is independent of $(W_s)_s$, then $(UW_s)_s$ is distributed like $(W_s)_s$. Apply this to $U=\mathrm{sign}(B_t)$. – Did Nov 03 '14 at 23:34
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Hint:

Use $B_{t_n}=sign(B_{t_n})|B_{t_n}|$ and $\sigma(B_s ^2)= \sigma(|B_{s}| )$ and independent increment property of B.M to show

$$P(B_t^2 \leq x | B_{t_1}^2 ,..., B_{t_n}^2)=P(B_t^2 \leq x | B_{t_n}^2) $$ where $0 <t_1<t_2 <...<t_n<t$ and $x \in \mathbb R$ .

In the process, you need to show $sign(B_{t_n})$ is independent with $\sigma(|B_{t_1}|,|B_{t_2}|,..., |B_{t_n}| )$ by calculating marginal distributions.

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