[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

Let ${\sigma^{(n)}}$ be a piecewise constant generate by ${\sigma}$ with uniform ${n}$ -equal-mesh:
$\displaystyle \sigma^{(n)}(t) = \sum_{i=1}^{n} \sigma\Big(\frac{i-1}{n}\Big) I((i-1)/n\le t <i/n).$
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 ${Y^{(n)} \Rightarrow Y}$ .
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.

9 comments

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 $b < 1$ .

kenneth says:|

August 23, 2016 at 7:54 am

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.

Reply
kenneth says:|

August 23, 2016 at 7:55 am

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

Reply
George Lowther says:|

August 23, 2016 at 8:21 am

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.

Reply
- kenneth says:|
  
  August 23, 2016 at 10:48 am
  
  Dear George, Thanks. I will fix it and delete the comments after.
kenneth says:|

August 23, 2016 at 10:52 am

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?

Reply
- George Lowther says:|
  
  August 23, 2016 at 3:41 pm
  
  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.
- kenneth says:|
  
  August 26, 2016 at 8:55 am
  
  Dear George, Yes you are right. I made a very simple mistake.