Let $T\in\mathbb{R}_{>0}$, $[0,T]$ be a time-index set and $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space. We endow the latter with a filtration $\mathbb{F}$ which is the augmented filtration generated by a Brownian motion $W$. We define:
- A random variable $A$ such that $A(\omega)>1$ for every $\omega\in\Omega$.
- A stochastic process $B:=(B_{t})_{t\in[0,T]}$ where $B_{t}:=\mathbb{E}[A\,\vert\,\mathcal{F}_{t}]$.
Is this reasoning correct: By the monotonicity of conditional expectation we can deduce that $B_{t}\neq 0$ for every $t\in[0,T]$, $\mathbb{P}$-a.s. However, we cannot deduce that $B_{t}\neq 0$, $\mathbb{P}$-a.s. for every $t\in[0,T]$, since an uncountable union of null-sets is not necessarily a null-set. In particular, we might have a situation where each path of $B$ is at least once equal to zero.
Is there a possibility to get the implication mentioned above? Maybe via continuity of the martingale $B$?
A standard assumption is to assume that the process has (almost all )paths that are continuous from the right (or from the left). If we assume that (with probability $1$) $B_s \to B_t$ as $ s$ decreases to $t$ (for any $t$) then the we get $B_t \geq 1$ for all rational $t$ with probability $1$ which implies $B_t \geq 1$ for all $t$ with probability $1$.