3
$\begingroup$

Consider the ito integral of the sign of the Brownian motion $W_s$ from $0$ to $t$: $$\int_0^t \operatorname{sign}(W_s)\,dW_s$$ This appears for instance in the Tanaka formula. I think this is a Brownian motion, by Levy's characterization, since this is a continuous Martingale, according to the theory of stochastic integration, whose quadratic variation is $t$. However, I don't quite understand it intuitively.

For instance, it seems that the above ito integral is always positive, at least bounded form below. Say, $W_s$ stays positive, then the above ito integral is $W_t$, which is positive, since the sign of $W_s$ is 1. If $W_s$ stays negative, then the ito integral is $|W_s|$, which is also positive, since the sign of $W_s$ is $-1$. So it seems the above ito integral is positive or bounded from below. In fact, the above ito integral appears in the semi-Martingale representation of $|W_t|$, which is given by Tanaka formula: $$|W_t|=\int_0^t \operatorname{sign}(W_s)\,dW_s +L_t$$ where $L_t$ is the Brownian local time.

So, my question is how can the ito integral be a Brownian motion, if it seems that the integral is positive. Maybe, my intuition has something wrong. Can someone explain to me what goes wrong here?

$\endgroup$
6
  • 3
    $\begingroup$ I think most textbooks on stochastic integration discuss this in detail, but to get some intuition think about the discrete case: let $S_n = X_1 + \ldots + X_n$ be a random walk with i.i.d. symmetric $X_j$, and define the `discrete stochastic integral' $$\tilde{S}_n = \sum_{j = 1}^n \operatorname{sign}(S_{j-1}) \Delta S_j = \sum_{j = 1}^n \operatorname{sign}(S_{n-1}) X_j .$$ Now it is much easier to see that $\tilde{S}_n$ is the same random walk, right? (Note: you had better define $\operatorname{sign}(0) = 1$ or $-1$ rather than $0$ here). $\endgroup$ – Mateusz Kwaśnicki Jul 8 '18 at 19:25
  • $\begingroup$ Thanks, Mateusz. Since $\tilde{S}_n$ has independent increment at every $n$ having the same identical distribution, it is a random walk. Of course, sign of 0 has to be defined as 1 or -1, otherwise it will have 0 increment when $S_{n-1}=0$, if sign of 0 is defined to be 0. However, it will not be so obvious for $W_t$. Let say the Ito integral is denoted by $X_t$. Then $X_s$ will behave like Brownian motion when $W_s$ is away from $0$. The problem is when $W_s$ crosses 0. During this time, it's not obvious it's a Brownian motion. For r. walk, it's still rand. walk when $S_{n-1}$ crosses 0. $\endgroup$ – Wai Jul 9 '18 at 12:11
  • $\begingroup$ Mateusz, can you show me some references which discuss this stochastic integral? I am not aware of any such references. Thank you. $\endgroup$ – Wai Jul 9 '18 at 12:19
  • $\begingroup$ Karatzas and Shreve, Section 3.6, for example. $\endgroup$ – Mateusz Kwaśnicki Jul 9 '18 at 19:50
  • $\begingroup$ Wai, the first name of losif Pinelis starts with l not L. $\endgroup$ – Mahdi Jul 10 '18 at 16:39
1
$\begingroup$

Let $X_t:=\int_0^t \operatorname{sign}(W_s)\,dW_s$ for $t\ge0$. Intuitively, because of the independent increments and symmetry of the Brownian motion, for $t\ge0$, small $h>0$, and Borel sets $A$ we have \begin{align*} P(X_{t+h}\in A\mid X_t=x)&\approx P((\operatorname{sign}W_t)(W_{t+h}-W_t)\in A-x) \\ &=P(W_{t+h}-W_t\in A-x)=P(W_{t+h}\in A\mid W_t=x), \end{align*} so that $(X_t)$ is also a Brownian motion.

Now, when you say "$W_s$ stays positive", you apparently mean $W_s\ge0$ for some $t>0$ and all $s\in[0,t]$. But this event has probability $0$, and so, whatever happens in this event is of no consequence.

$\endgroup$
4
  • $\begingroup$ Losif, let me make some remarks concerning your explanation first. 1.$W_t=x$ and $X_t=x$ are 2 things. $W_t=x$ is more important. $\endgroup$ – Wai Jul 9 '18 at 18:52
  • $\begingroup$ Sorry,let me continue. One calculates: $$P(\int_t^{t+h} sign(W_s)dW_s\epsilon A|W_t=x)=P(W_{t+h} -W_t \epsilon A|W_t=x)$$ The argument is similar as yours. Also note that they are only approximately equal, as you indicated. 2.the calculation works only when $x$ is not $0$. That is $W_t$ is not 0. Because when $W_t$ is away from 0, $sign(W_t)$ stays constant and the above approximation works. So $X_t$ is a Brownian motion when $W_t$ is away from 0 and does not cross 0. Now, the problem is to explain why $X_t$ is still a Brownian motion even when $W_t$ crosses 0. Concerning this, I'm not sure. $\endgroup$ – Wai Jul 9 '18 at 19:08
  • $\begingroup$ Iosif, let me explain better why I feel $X_t$ is quite strange and $W_t$ and $X_t$ are different Brownian motions. 1.When $W_s$ stays positive on $[a,b]$ or stays negative on $[a,b]$, then if $t \epsilon [a,b]$, then $$\int_a^t sign(W_s)dW_s > 0$$ Clearly this event has positive probability. That means $X_t$ decreases only when $W_t=0$ and crosses 0. 2.For convenience, define $sign(0)=0$, then for the 2 Brownian paths $W$ and $-W$, the corresponding integrals are equal for all t: $$X_t (W)=X_t (-W)$ $\endgroup$ – Wai Jul 9 '18 at 19:28
  • $\begingroup$ Sorry, some (typing) errors. 1.That means $X_t$ "finds new minimum" only when $W_t=0$ and crosses 0. 2. $X_t(W)=X_t(-W)$ $\endgroup$ – Wai Jul 9 '18 at 19:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.