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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?

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    $\begingroup$ 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
    Commented Jul 8, 2018 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
    Commented Jul 9, 2018 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
    Commented Jul 9, 2018 at 12:19
  • $\begingroup$ Karatzas and Shreve, Section 3.6, for example. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jul 9, 2018 at 19:50
  • $\begingroup$ Wai, the first name of losif Pinelis starts with l not L. $\endgroup$
    – Mahdi - Free Palestine
    Commented Jul 10, 2018 at 16:39

1 Answer 1

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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.

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  • $\begingroup$ 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
    Commented Jul 9, 2018 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
    Commented Jul 9, 2018 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
    Commented Jul 9, 2018 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
    Commented Jul 9, 2018 at 19:39

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