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Consider two diffusions given by $$X_j(t)=\int_0^t a_j(s,X_j(s))\,dW_s$$ for $j=1,2$ and $t\ge 0$, where $W_\cdot$ is a standard Wiener process/Brownian motion and the $a_j$'s are smooth enough functions such that $0\le a_1\le a_2$.

Does it then follow that $P(|X_1(1)|>x)\le P(|X_2(1)|>x)$ for all real $x$?

At least, does this comparison hold when $a_2$ is a constant?

There are a number of results comparing two diffusions with the same diffusion coefficients but with the drift of one of the two diffusions greater than the drift of the other one -- see e.g. Nakao. However, I have been unable to find a comparison of the desired kind.

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The inequality is not true in general — additional assumptions are needed. I think some kind of monotonicity of $a_1$ and $a_2$ should help, but this is merely a guess.

Here is a counterexample. Consider $a_2 = 1$, so that $X_2(1) = W(1)$. Let $a_1(s, x) = 1$ when $|x| < 1$ and $a_1(s, x) = 0$ otherwise. Then $X_2(1) = W(1 \wedge \tau)$, where $\tau$ is the hitting time of $\{-1, 1\}$ for $W$. It is fairly straightforward to see that if $|X_2(1)| \geqslant 1$, then $|X_1(1)| = 1$, while if $|X_2(1)| < 1$, then still $|X_1(1)| = 1$ with positive probability. Thus, the desired inequality fails to hold for $x = 1 - \epsilon$ with $\epsilon > 0$ small enough.

The question asks for smooth $a_1$ and $a_2$, so one should mollify them appropriately, but this is just a technicality, I hope the idea is clear.

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This is just a comment on a related issue.

You don't have stochastic comparison of processes at a given time but you DO have stochastic comparison of hitting times.

To see this, you can couple the two processes $j=1,2$ AFTER having performed the (respective) time change transformations into standard BM.

As a consequence, the two processes will have the SAME path but with different time parametizations, one FASTER than the other.

This implies stochastic comparison of hitting times.

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    $\begingroup$ It is not clear to me why one may expect the comparison of hitting times. Clearly one has $a_1(t,x)\le a_2(t,x)$ for all $(t,x)$, but why this implies that $a_1(t,X_1(t))\le a_2(t,X_2(t))$? $\endgroup$
    – GJC20
    Commented Feb 26, 2022 at 16:04
  • $\begingroup$ You are perfectly correct, this is not as direct as my answer misleadingly suggested. For some reason I found it quite intuitive, but one might need some form of comparison theorem for one dimensional ODE similar to math.stackexchange.com/questions/912468/…. $\endgroup$
    – Mathias Rousset
    Commented Mar 1, 2022 at 10:59
  • $\begingroup$ The comparison should apply to $\tau_j(t)$, the time change that transform $W$ into the $X_j$'s. This should imply that $\tau_1 \leq \tau_2$, hence the comparison of hitting times. $\endgroup$
    – Mathias Rousset
    Commented Mar 1, 2022 at 11:01
  • $\begingroup$ @GJC20 I'll be delighted were you able to check that what I am suggesting is correct. $\endgroup$
    – Mathias Rousset
    Commented Mar 1, 2022 at 11:14
  • $\begingroup$ Thanks for your reply. For most comparison results, the SDEs share the same volatlity term. It is not clear to me why $a_1\le a_2$ yields larger quadratic variation. Indeed, my question is related to my post mathoverflow.net/questions/416766/… $\endgroup$
    – GJC20
    Commented Mar 1, 2022 at 11:45

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