Stochastic inequality, true? Consider two stochastic processes $X$ and $Y$ satisfying the following SDEs (with the same drift!):
$$X_t = x + \int_0^t b(X_s)ds + B_t$$
$$Y_t = y + \int_0^t b(Y_s)ds + B_t.$$
If $0<x<y$, is it then true that the sample paths of $Y$ are a.s. bigger than the sample paths of $X$? In other words.. is it true that
$$P(\forall~ t \ Y_t>X_t)=1?$$
If it is true, I don't see how to prove it! :( Btw. the drifts are not Lipschitz so I'm not allowed to use Gronwall here.
Thanks a lot for the help.
 A: The claim holds true if there exists a unique solution to the SDE
$$dZ_t = b(Z_t) \, dt + dB_t, \qquad Z_0 = z \tag{1}$$
e.g. if $b$ is locally Lipschitz continuous.
Proof: Let $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$ be solutions to the SDEs
$$\begin{align*} dX_t &= b(X_t) \, dt + dB_t \qquad X_0 = x \tag{2} \\ dY_t &= b(Y_t) \, dt + dB_t \qquad Y_0 = y \tag{3}\end{align*}$$ for $x<y$. We define a stopping time $\tau$ by
$$\tau := \inf\{t>0; X_t > Y_t\}$$
where $\inf \emptyset := \infty$. Note that, by the continuity of the solutions, $X_{\tau} =Y_{\tau}$. Now set
$$\tilde{X}_t := \begin{cases} X_t, & t \leq \tau, \\ Y_t, & t>\tau. \end{cases}$$
Obviously, $\tilde{X}_0 = x$ and, moreover, 
$$d\tilde{X}_t = b(\tilde{X}_t) \, dt + dB_t.$$
(Intuitively, it should be clear that the last equation holds. Making this rigorous  requires some more effort; note that we do not need any further assumptions on $b$ to do so.) This means that $(\tilde{X}_t)_{t \geq 0}$ solves $(2)$. Consequently, we conclude from the uniqueness of the solution that $$\mathbb{P}(\tau<\infty)=0.$$
Hence,
$$\mathbb{P}(\forall t: X_t \leq Y_t)=1.$$
(Note that in general the equality $\mathbb{P}(\forall t: X_t<Y_t)=1$ does not hold.)
Remark: This statement is a very particular case of a so-called "comparison principle". 
