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Consider two SDEs (stochastic differential equations) as follows:

$$dX_t=b^-(t,X_t) \, dt+a(t,X_t) \, dW_t;\quad dY_t = b^+(t,Y_t)\,dt+a(t,Y_t)\,dW_t,$$

where $b^-,b^+,a$ are Lipschitz such that $b^-<b^+$ pointwise. For any $x\le y$, it is known that, see e.g. Theorem 1.1 of https://www.sciencedirect.com/science/article/pii/0304414994900558

$$X^x_t\le Y^y_t,\quad \forall t\ge 0, $$

where $X^x, Y^y$ denote the solutions to the above SDEs with initial conditions $X^x_0=x, Y^y_0=y$. My question is as follows: Let $X,Y$ be two arbitrary solution to the above SDEs (note that such $X,Y$ are not unique), does

$$\mathbb P\big[X_s\ge Y_s, \forall 0\le s\le t \mid X_t=z=Y_t\big]=1$$

hold for (almost) all $t$ and $z$?

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  • $\begingroup$ did you mean to write $X_{s}\leq Y_{s}$? $\endgroup$
    – Thomas Kojar
    Commented Aug 3, 2023 at 20:06
  • $\begingroup$ So then if I understand correctly, your concern is whether this bridging process still satisfies the comparison? $\endgroup$
    – Thomas Kojar
    Commented Aug 3, 2023 at 20:06
  • $\begingroup$ @ThomasKojar No. Knowing $X_t=Y_t$, I wish to show $X_s\ge Y_s$ for $s\le t$. Roughly speaking, knowing the present, does the comparison principle say something about the history? $\endgroup$
    – Fawen90
    Commented Aug 3, 2023 at 20:08
  • $\begingroup$ I can understand maybe $X_{s}=Y_{s}$, but the drifts $b_{x}<b_{y}$, would likely just give again $X_{s}\leq Y_{s}$. $\endgroup$
    – Thomas Kojar
    Commented Aug 3, 2023 at 20:13

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