I would like to ask for a more rigorous statement and proof of Lemma on page 5 of this paper. In essence, it states that two distinct sample paths of a Brownian motion does not strictly cross (meaning once they intersect at some time, they merge together from then on) with probability $1$. They claim this stems from the Markov property of the Brownian motion which stipulates that a path is uniquely determined by its initial position. This seems strange to me. All the sample path starts from $0$ at time $0$. If the statement is true, then there would have been only one path and the process would have been deterministic. Also Markov property in essence states that the conditional probability depends not on the historical but the current state. It does not seem to say two paths can not start from the same point.

Can someone resolve this confusion?

  • $\begingroup$ The statement holds and is actually relatively straightforward. Solutions $(\xi_t)_{t\geqslant0}$ of stochastic differential equations that have unique pathwise solutions are such that, at any time $t$, there exists some deterministic function $\Phi_t$ such that $$(\xi_s(\omega))_{s\geqslant t}=\Phi_t(\xi_t(\omega),(B_s(\omega)-B_t(\omega))_{s\geqslant t}).$$ Thus, if $\xi_t(\omega)=\xi'_t(\omega)$ for some $t$ then one knows that $\xi_t(\omega)=\xi'_t(\omega)$ for every $s\geqslant t$, which is what the Lemma formalizes. $\endgroup$ – Did May 25 '16 at 14:28
  • $\begingroup$ See Hairer's answer there. $\endgroup$ – Did May 25 '16 at 19:58
  • $\begingroup$ @Did: I misunderstood the premise of the Lemma. But it is not stated clearly in the paper either. The paper should write, $\xi_t(\omega)$, for a given sample $\omega$, exactly as you have written, or as Jay. H clarifies in his answer below, $\xi$ is conditioned on a given path of $B_t$. $r(x,t)$ and $\sigma(x,t)$ being Lipschitz continuous in $x$ and continuous in $t$ would enable us to extend Picard-Lindelof Theorem to this Stieltjes (the $dB$ integral) integral equation to obtain the existence and uniqueness of the pathwise solution. $\endgroup$ – Hans May 25 '16 at 20:08
  • $\begingroup$ Yeah, I got that the first time, why delete and repost? $\endgroup$ – Did May 25 '16 at 20:08
  • $\begingroup$ @Did: Immediately after I posted the comment, I checked the paper and saw they specified on top of page 4 that $r$ and $\sigma$ satisfy whatever regularity condition, including the Lipschitz and growth conditions. So I changed my comment. The link you just posted is very helpful. Thank you. $\endgroup$ – Hans May 25 '16 at 20:20

I think you misunderstood the meaning there. They talked about "contingent" claims, these are random variables given some other process. A simple example will be something like this: you have a simple SDE,

$dX = dB$, with $X_0 = a$,

the (strong) solution is, of course,

$X_t = a+ B_t$.

Now, you have another process on the same Brownian Filtration:

$dY = dB$, with $Y_0 = b$,

and so, $Y_t = b+B_t$

If $b>a$, then $Y_t>X_t$, for any given underline sample path $B_t(\cdot)$.

  • 1
    $\begingroup$ Thank you. Yes, I misunderstood the description in the paper. But the Lemma is not stated clearly. It should have written $\xi(\omega)$ for a given sample $\omega$, or as you have emphasized for the given sample path $B_t$. See my conversation with Did above. The proof of the Lemma is not clear either. It should have invoke the Lipschitz and growth condition of $r$ and $\sigma$ to prove the uniqueness. Instead it talks about the Markov property. How does that imply the uniqueness of an integral equation? $\endgroup$ – Hans May 25 '16 at 20:35

Here is an example in discrete time explaining why the Markov property is crucial to guarantee that solutions do not cross, in the sense that if two solutions coincide at some given time then they coincide at any later time.

Consider the AR2 process $(x_n)_{n\geqslant0}$ defined by some initial conditions $(x_0,x_1)$ and by the recursion $$x_n=2x_{n-1}-x_{n-2}+\epsilon_{n-2},$$ for every $n\geqslant2$, where $(\epsilon_n)_{n\geqslant0}$ is i.i.d. and independent of $(x_0,x_1)$. More specifically, let $(x_n)_{n\geqslant0}$ denote the solution when $x_0=1$ and $x_1=3$, and $(\bar x_n)_{n\geqslant0}$ the solution when $\bar x_0=4$ and $\bar x_1=5$.

Simple computations yield $x_2=5+\epsilon_0$, $\bar x_2=6+\epsilon_0$, $x_3=\bar x_3=7+2\epsilon_0+\epsilon_1$, hence, irrespectively of the realization of the process $(\epsilon_n)_{n\geqslant0}$, $$x_2\ne\bar x_2,\qquad x_3=\bar x_3,$$ while, interestingly in our context, $x_4=9+3\epsilon_0+2\epsilon_1+\epsilon_2$, $\bar x_4=8+3\epsilon_0+2\epsilon_1+\epsilon_2$, hence, again irrespectively of the realization of the process $(\epsilon_n)_{n\geqslant0}$, $$x_4\ne \bar x_4.$$ (And it happens that $x_n\ne \bar x_n$ for every $n\geqslant4$.) Thus, in this example, the fact that two trajectories meet at a given time does not imply that they coincide after this time, although they are both defined pathwise and run by the same process $(\epsilon_n)_{n\geqslant0}$. The missing piece of the argument is that, here, $x_n$ depends on the past $(x_k)_{k\leqslant n-1}$, not only through $x_{n-1}$ but also through $x_{n-2}$, in other words, the fact that the process $(x_n)_{n\geqslant0}$ is not Markov.

  • $\begingroup$ I missed your example earlier. Of course you are right. +1. $\endgroup$ – Hans Jun 8 '16 at 21:08

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