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?




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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](https://projecteuclid.org/download/pdf_1/euclid.ojm/1200775827). However, I have been unable to find a comparison of the desired kind.



A comparison of diffusions
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.


