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Let $B_{t}$ be a Brownian motion relative to a filtration $F_{t}$, is $(B_{t}+t)^{2}$ a Markov process? Thanks!

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Homework? If so, please tag as such. –  cardinal Mar 19 '11 at 20:02
    
This is a question to the OP. For every $t\ge0$, let $X_t=(B_t+t)^2$ and $F^X_t$ the sigma-algebra $F^X_t=\sigma(X_s,s\le t)$. Are you asking if $(X_t)$ is a Markov process in the usual sense, see en.wikipedia.org/wiki/Markov_process#The_Markov_property, that is, with respect to its own filtration $(F^X_t)$? Or, with respect to the filtration $(F_t)$ of the Brownian motion $(B_t)$? –  Did Mar 25 '11 at 22:55
    
With respect to the filtration $\mathcal{F}_t$ seems like the more natural question to me, but also the easiest. With respect to the filtration of $X$ is more interesting. The answer is the same though. It is not Markov. –  George Lowther Mar 25 '11 at 23:47
    
btw, I'd have posted this as an answer, but I'm just posting from my mobile at the moment, so it's a bit tricky. –  George Lowther Mar 25 '11 at 23:58
    
Thinking about it with a bit more care (doing this calculation in my head...) it looks like it is Markov under the filtration of $X$. Amazingly, when you calculate the distribution of $X_t$ conditional on $\mathcal{F}_s^X$ there is a whole lot of cancelation, and it ends up only depending on $X_s$. –  George Lowther Mar 26 '11 at 0:33

3 Answers 3

up vote 10 down vote accepted

There are two possible filtrations here -- the original filtration $\mathcal{F}_t$ generated by the Brownian motion and the one generated by the process $X$, which I'll denote by $\mathcal{F}^X_t$. So, there are two ways to interpret this question, (i) is $X$ Markov with respect to $\mathcal{F}_t$, and (ii) is $X$ Markov with respect to its own filtration $\mathcal{F}^X_t$? Unsurprisingly, the answer to (i) is no, $X$ is not Markov. To see this, it is straightforward to compute $\mathbb{E}[X_t\vert\mathcal{F}_s]$ for times $s < t$, and you get $$ \mathbb{E}[X_t\vert\mathcal{F}_s]=(B_s+t)^2+t-s=X_s + 2(t-s)B_s +(t-s)^2+t-s. $$ Due to the dependence on $B_s$, which is not uniquely detemined by $X_s$, this is not a function of $X_s$, so the process is not Markov (which is Byron's point in his answer).

The second question (ii) is a bit harder because computing the distribution of $X_t$ conditioned on $\mathcal{F}^X_s$ is tricky. The, perhaps surprising, answer to (ii) is yes, $X$ is Markov with respect to its own filtration! To prove this, it is necessary to show that $\mathbb{E}[f(X_t)\mid\mathcal{F}^X_s]$ is a function of $X_s$ (for times $s < t$ and bounded measurable function $f$). As a first step, note that, as the Brownian motion $B$ is Markov, $\mathbb{E}[f(X_t)\mid\mathcal{F}_s]=g(B_s)$ for a measurable function $g$. Apply the tower law for conditional expectations, $$ \mathbb{E}[f(X_t)\mid\mathcal{F}^X_s]=\mathbb{E}[\mathbb{E}[f(X_t)\mid\mathcal{F}_s]\mid\mathcal{F}^X_s]=\mathbb{E}[g(B_s)\mid\mathcal{F}^X_s]. $$ So, to show that $X$ is Markov under its own filtration we only have to show that the distribution of $B_s$ conditioned on $\mathcal{F}^X_s$ depends only on $X_s$. Also, given $X_s$ then $B_s$ can only take one of the two possible values $-s\pm\sqrt{X_s}$. We have to show that the probability of these two possibilities does not depend on the history of $X$. There are two main ways that I can think of showing this.

Direct Computation

We can directly compute the distribution of $B_s$ conditioned on $\mathcal{F}^X_s$ by breaking the time interval $[0,s]$ into $n$ discrete steps, which allows it to be computed as a ratio of probability density functions, then take the limit $n\to\infty$. Writing $h=s/n$ and $\delta w_k\equiv w_k-w_{k-1}$, the probability density function of $\hat B\equiv(B_{h},B_{2h},\ldots,B_{nh})$ can be written out by applying the independent Gaussian increments property as $$ p(w)=(2\pi h)^{-\frac{n}{2}}\exp\left(\frac{-1}{2h}\sum_{k=1}^n(\delta w_k)^2\right). $$ The distribution of $B_s$ conditional on $\hat X\equiv(X_h,X_{2h},\cdots,X_{nh})$ is simply given by a ratio of sums over the probability density function $p$, $$ \mathbb{P}\left(B_s=-s+\sqrt{X_s}\;\Big\vert\;\hat X\right)=\frac{\sum_{w\in P}p(w)}{\sum_{w\in P}p(w)+\sum_{w\in P^\prime}p(w)}. $$ Here, $P$ is the set of discrete paths for $\hat B$ agreeing with the values of $X$ and ending at $-s+\sqrt{X_s}$. Then, $P^\prime$ is the similar set of paths ending at $-s-\sqrt{X_s}$. If, as $w$ runs through $P$, we set $w^\prime_k\equiv-2kh-w_k$, then it can be seen that $w^\prime$ runs through $P^\prime$. So $\delta w^\prime_k=-2h-\delta w_k$ and, $$ \begin{align} \sum_{w\in P^\prime}p(w)&=\sum_{w\in P}p(w^\prime)=\sum_{w\in P}(2\pi h)^{-\frac{n}{2}}\exp\left(\frac{-1}{2h}\sum_{k=1}^n(-2h-\delta w_k)^2\right)\\ &=\sum_{w\in P}p(w)\exp\left(-2nh-2w_n\right) \end{align} $$ As $2nh+2w_n=2\sqrt{X_s}$, this can be plugged into the expression above, $$ \mathbb{P}\left(B_s=-s+\sqrt{X_s}\;\Big\vert\;\hat X\right)=\frac{1}{1+e^{-2\sqrt{X_s}}}. $$ This only depends on $X_s$ and, letting $n$ go to infinity, this expression also holds for the probability conditioned on $\mathcal{F}^X_s$.

Girsanov Transformations

The theory of Girsanov transformations tells us that, defining $U=\exp(-B_s-\frac12s)$ and the new measure $\mathbb{Q}=U\cdot\mathbb{P}$, then $\tilde B_u\equiv B_u+u$ is a standard $\mathbb{Q}$-Brownian motion on the interval $[0,s]$. Also write $V=U^{-1}=\exp(\tilde B_s-\frac12s)$, so that $\mathbb{P}=V\cdot\mathbb{Q}$. Under the measure $\mathbb{Q}$, symmetry on reflecting the Brownian motion about zero shows that $\tilde B_s$ takes the values $\pm\sqrt{X_s}$ each with probability 1/2, when conditioned on $\mathcal{F}^X_s$. The conditional expectation under the $\mathbb{P}$ measure can be converted to a conditional expectation under $\mathbb{Q}$, $$ \begin{align} \mathbb{E}[g(B_s)\mid\mathcal{F}^X_s]&=\mathbb{E}_{\mathbb{Q}}[Vg(-s+\tilde B_s)\mid\mathcal{F}^X_s]/\mathbb{E}_{\mathbb{Q}}[V\mid\mathcal{F}^X_s]\\ &=\left(e^{\sqrt{X_s}}g(-s+\sqrt{X_s})+e^{-\sqrt{X_s}}g(-s-\sqrt{X_s})\right)/(e^{\sqrt{X_s}}+e^{-\sqrt{X}_s}) \end{align} $$ This is a function of $X_s$, so $X$ is Markov. This method works because the change of measure "adding" a constrant drift to a Brownian motion only depends on the value of the the process at the end of the time interval, and is otherwise independent of the path taken.

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Thanks for this! –  Byron Schmuland Mar 27 '11 at 1:04

Hint: If the process were Markov, then for $0< s\leq t$ we'd have $$\mathbb{E}((B_t+t)^2 | F_s) = P_{s,t}((B_s+s)^2).$$

Calculate the conditional expectation on the left and try to write it as a function of $(B_s+s)^2$.

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Thanks for your reply, Byron. I computed the LHS and get $(t-s)+(B_{s}+t)^{2}$, i guess there doesn't exist a f such that $f((B_{s}+s)^{2})=(t-s)+(B_{s}+t)^{2}$, but how to prove this? –  user7762 Mar 20 '11 at 1:07
    
Hi user7762, How do you derive $(t-s)+(B_{s}+t)^{2}$, for the LHS ? –  TheBridge Mar 22 '11 at 10:45

If $X_t:=(B_t+t)^2$ were Markovian, then you could (at very least) determine the distribution of $X_t$ given $\{X_r\}_{r \leq s}$ using only the knowledge of $X_s$.

If you were given $B_s$ then this would certainly be enough to find the desired distribution, but crucially for a given $X_s$ there are two possible values for $B_s$. And these two values result in different distributions for $X_t$, which contradicts the process being Markovian.

The calculation would be something like this. Let $(B_s+s)^2 = a$ so that $B_s = -s \pm \sqrt{a}$. But then $$(B_t+t)^2 = (B_t -B_s +(t-s) + (B_s +s))^2$$ $$ = (B_t -B_s +(t-s))^2+2(B_t -B_s +(t-s))(\pm \sqrt{a}) + a^2 $$ The distribution of $B_t-B_s$ is of course known (it is a Brownian increment). But to uniquely determine the distribution of the above expression, you need to know the value of $\pm \sqrt{a}$, indeed, the two choices lead to different distributions. But this is unknown if you are only given the value of $(B_s+s)^2$.

This thread has a nice discussion about when functions of Markov processes are still Markovian.

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@dtbkelly Indeed this is the idea of the proof but I feel the argument is somehow not quite complete. For example, one should be able to see why the exact same argument fails when one tries to use it to prove that $Y_t=B_t^2$ is not Markov (since $Y$ is Markov, as we all know). –  Did Mar 25 '11 at 15:01
    
But in that case $\pm\sqrt{a} = \pm B_s$ and both choices have the same distribution, so it does work. At least, it doesn't lead to a contradiction. –  user8463 Mar 25 '11 at 15:14
    
@dtbkelly I know it does work, I know. But you have to have an argument saying that such "miracles" as the one you just described for $Y$ cannot happen with $X$ as well. –  Did Mar 25 '11 at 15:18
    
That miracle would be when the two choices of $\pm \sqrt{a}$ lead to $(B_t +t)^2$ having the same distribution. It is easy to check that this is not the case for $X$. You just check that $\pm(B_t - B_s + t-s)$ have different distributions. –  user8463 Mar 25 '11 at 15:25
    
@dtbkelly Note that the OP's question and my hint are about the Markov property of $(X_t)$ relative to the original filtration generated by $(B_t)$. Your first sentence suggests that you are tackling the (slightly?) harder problem of showing that $(X_t)$ in not even Markov with respect to its own natural filtration. –  Byron Schmuland Mar 25 '11 at 16:29

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