[Q]: Does non-degenerate martingale hit a given specific point by some chance?

Let {W^{x}} be a two dimensional standard Brownian motion with respect to a filtered probability space {(\Omega, \mathcal F, \mathbb P, \mathbb F := \{\mathcal F_{t}\}_{t\ge 0})}, starting from a point {x}.

It is well known that 2-D Brownian motion {W^{x}} is neighborhood recurrent, but point polar. In particular, following statement holds: for any given point {y\in \mathbb R^{2}}

\displaystyle \mathbb P\{ W^{x}(t) = y \hbox{ for some } t\in (0, 1)\} = 0.

Can we extend this fact to continuous non-degenerate martingale of the form below?

\displaystyle Y(t) = \int_{0}^{t} \sigma(s) d W(s)

under uniform non-degenerate condition? More precisely,

Assumption 1 {\sigma(t,\omega)} is a {\mathbb F}-progressively measurable 2 by 2 matrix process satisfying

\displaystyle \lambda \|\xi\|^{2} \le \xi' \sigma_{t} \sigma_{t}' \xi \le \Lambda \|\xi\|^{2}, \hbox{ for all } (\xi, t) \in \mathbb R^{2} \times (0,1) \hbox{ with some } \lambda, \Lambda >0.

[Q] With Assumption 1, can we prove

\displaystyle \mathbb P\{ Y(t) = (1,1) \hbox{ for some } t\in (0, 1)\} = 0? \ \ \ \ \ (1)

So far, we do not have the answer.

At this stage, we can prove that (1) is true if {\sigma} is further assumed piecewise constant, that is

[Claim] Suppose {\sigma} satisfies Assumption 1 and admits the form of

\displaystyle \sigma(t) = \sum_{i=1}^{n} \sigma(t_{i-1}) I(t_{i-1}\le t <t_{i})

for some partition {0 = t_{0} < t_{1}< \cdots < t_{n} = 1} and {n}. Then, (1) is true.

Proof: It based on the fact on point polar property of 2-D Brwonian motion, and induction. \Box

Next, the original attempt to answer [Q] was the following: assuming non-piecewise constant {\sigma} of Asssumption 1

  1. Let {\sigma^{(n)}} be a piecewise constant generate by {\sigma} with uniform {n}-equal-mesh:

    \displaystyle \sigma^{(n)}(t) = \sum_{i=1}^{n} \sigma\Big(\frac{i-1}{n}\Big) I((i-1)/n\le t <i/n).

  2. Let {Y^{(n)}} be

    \displaystyle Y^{(n)}(t) = \int_{0}^{T} \sigma^{(n)}(t) d W(t)

    Note that, {Y^{(n)}} converges to {Y} in distribution by the definition of Ito integral, denoted by {Y^{(n)} \Rightarrow Y}.

  3. By [Claim] above, we know {\mathbb P\{ Y^{(n)}(t) = (1,1) \hbox{ for some } t\in (0, 1)\} = 0}

However, it is not sufficient to draw the conclusion of (1) from the facts {Y^{(n)} \Rightarrow Y} and {\mathbb P\{ Y^{(n)}(t) = (1,1) \hbox{ for some } t\in (0, 1)\} = 0}, by Portmanteau theorem. We tried to prove the set of processes hitting {(1,1)} is a {\mathbb P}-continuity set in some process space under some appropriate topology, but we failed to cover this gap.

Any suggestion will be greatly appreciated.

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9 comments

  1. Hi. As this post is quite old now, maybe you have worked out a solution. However, I can answer the question – no, statement Q is false.

    To construct a counterexample, it is more convenient to move the point (1,1) to the origin, so the question is whether Y(t) = Y(0)+\int_0^t\sigma(s)\,dW_s with Y(0)\not=0 can ever hit the origin with positive probability. To construct a counterexample, set

    \displaystyle\sigma(s) = UU^\prime+ b VV^\prime

    where U = Y/\lVert Y\rVert and V is equal to U with a 90 degree rotation applied (i.e., it is a unit vector orthogonal to Y). Then, dY = \sigma\,dW is a stochastic differential equation. As \sigma is a smooth function of Y (for Y\not=0), it has a unique solution up until the first time Y hits 0 (once Y hits zero, we have the counterexample, so don’t really care any more and can take \sigma to be the identity if we like). Your condition is also satisfied with \lambda=\min(1,b^2) and \Lambda=\max(1,b^2). Setting B = \lVert Y\rVert^2 it can be seen that this satisfies the SDE

    \displaystyle dB = 2\sqrt{B}dZ+(1+b^2)dt

    This is a {\rm BES}^2_{1+b^2} process, which hits zero eventually with probability 1, and hits zero by any time t with positive probability, so long as b < 1.


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