# [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

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

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.

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 $b < 1$.

• Dear George, so grateful for your answer and you really understand the subject. Indeed, I did not have counter-example, and always believed it was true.

• Fixed, by the way, what is $X$ in your definition of $U$?

• My X should be a Y. Also, the b inside the max and min should be squared (and you may as well delete my comments pointing out typos once they’re fixed.

• Dear George, Thanks. I will fix it and delete the comments after.

• Dear George, After my calculation, I got different SDE for $B$, that is
$d B = 2 \sqrt{(1 + b^2) B} dZ + (1+b^2) dt.$
I am sorry, but would you verify once more and confirm it?

• I think I am correct. Itos formula gives $dB=2Y^\prime \sigma dW + {\rm tr}(\sigma\sigma^\prime)dt$. Then use $Y^\prime\sigma =Y^\prime$. So the first term on the rhs has quadratic variation $4Y^\prime Y dt$, and you can take it from there.

• Dear George, Yes you are right. I made a very simple mistake.

2. Dear George, Yes you are right. I made a very simple mistake.