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<C} 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.

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