# Probability question on non-degenerate martingale

Let ${(\Omega, \mathcal{F}, P)}$ be a probability space, on which ${\mathcal{F}_t}$ is filtration satisfying general conditions. ${W_t}$ is a standard Brownian motion. Let ${Y_t}$ be a martingale given by

$\displaystyle Y_t = \int_0^t \sigma_r d W_r$

where ${\sigma_t}$ is bounded ${\mathcal{F}_t}$ measurable process.

Proposition 1 Assume ${c< \sigma_t for some positive constants ${c}$ and ${C}$, then

$\displaystyle P\{Y_1 = b\} = 0$

for all constant ${b}$.

To prove the above proposition, we will use following two facts. We define ${f: \mathbb{R}^+ \times \mathbb{R}^+ \times \mathbb{R} \mapsto [0,1]}$ by

$\displaystyle f(x,y,u) = \mathbb{P}\{W_t = u \hbox{ for some } t\in (x,y)\}.$

1. By direct computation, one can have

$\displaystyle \sup_{u\in \mathbb{R}} f(x,y,u) = f(x,y,0) < 1.$

2. By time change argument, we have

$\displaystyle f(\lambda x, \lambda y, u) = f(x,y, \frac{u}{\sqrt \lambda}), \quad \forall \lambda>0.$

Now we are ready to present the proof of Proposition~1.

Proof: Since ${Y_t}$ is continuous process,

$\displaystyle \{Y_1 = b\} \in \mathcal{F}_{1^-} = \sigma(\{\mathcal{F}_t: t<1\}).$

By Levy’s zero one law, we have

$\displaystyle I_{\{Y_1 = b\}} = \lim_{t\uparrow 1} \mathbb{P}\{Y_1 = b |\mathcal{F}_t\}, \quad a.s.$

Therefore, it is enough to show that there exists ${a\in (0,1)}$ such that

$\displaystyle \mathbb{P}\{Y_1 = b|\mathcal{F}_t\} < a < 1, \forall t\in (0,1).$

Note that, the martingale ${(Y_s|Y_t = u:s>t)}$ has the same distribution as a time-changed Brownian motion starting from state ${u}$. Together with ${c^2 (1-t) \le \int_t^1 \sigma_r^2 dr \le C^2 (1-t)}$, we have for some standar Brownian motion ${B}$,

$\displaystyle \mathbb{P}\{Y_1 = b | Y_t = u\} = \mathbb{P}\{B_r = b - u, \hbox{ for some } r \in ( c^2(1-t), C^2(1-t)) \} = f(c^2, C^2, \frac{b - u}{\sqrt{1 - t}}) \le f(c^2, C^2, 0).$

Since ${f(c^2, C^2, 0)}$ is independent to ${t}$, and strictly less than ${1}$, we can simply take ${a = f(c^2, C^2, 0)}$. $\Box$

To this end, one may ask if the condition of ${\sigma}$ can be relaxed to ${\sigma_t>0}$ without affecting the result? The answer is unfortunately NO. See counter-example here.