Let $N_1(t)$ and $N_2(t)$ be two random processes with $t\geq 0$ such that $\mathbb{P}(N_1(t)\geq b) \geq \mathbb{P}(N_2(t)\geq b) $ for all $b \in \mathbb{R}$. Can I conclude the following?: $$\mathbb{P}(\forall t\in[0,a]: N_1(t)\le 0 ) \leq \mathbb{P}(\forall t\in[0,a]: N_2(t)\le 0 )$$
This is a part of a bigger problem. In one step, I conclude the above. While it is very clear intuitively, I was not able to prove it. Is that right? How can I prove it?
It is incorrect so long as we do not know the time dependence of these processes; here is a counter-example:
$N_1(t) = X$ for all time $t$, where $X$ is a Rademacher random variable (i.e. takes values $\pm 1$ with equal probability).
$N_2(t) = X_1$ for time $t \in [0,a/2]$, $X_2$ otherwise, where $X_1, X_2$ are also Rademacher random variables.
Now, for a fixed time, you can tell that the condition is satisfied (they have the same distribution function)
However, if you look at $N_1(t)$, it has 1/2 probability of always being below 0, whereas $N_2(t)$ will only be uniformly below 0 with probability 1/4, thus violating your result.