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Suppose $B_t$ is a standard Brownian motion on a filtered probability space $\langle \Omega, \mathcal F, \{\mathcal F_t\}_t, \mathbb P\rangle$. Consider two SDEs below.

Suppose, $X_0 = Y_0 = 0$ \begin{align*} dX_t =& sign(X_t) dt + d B_t\\ dY_t =& \alpha_t dt + dB_t \end{align*} where $\alpha_t$ is some $[-1,1]$-valued, $\mathcal F_t$ measurable process.

I am wondering if the following is true. And if yes, how might one approach this thing? $$\mathbb E[\vert X_T\vert] \ge \mathbb E[\vert Y_T \vert]$$ for all $T \ge 0$.

Intuitively, the reason for this conjecture is the following: Both the processes start from $0$. Say, at some time they are equal to some $x > 0$. Now, $X_t$ has an upward drift of $1$ while the upward drift of $Y_t$ is weakly less than $1$. So, the moment the two processes have the same sign, it seems that $X_t$ runs farther away compared to $Y_t$. But, I have no idea how to make this precise, even if true.

Thanks!

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  • $\begingroup$ What happens in the case $\alpha_t =1$ for all $t$? $\endgroup$
    – John Dawkins
    Commented Apr 14, 2020 at 15:34
  • $\begingroup$ Do you conjecture that with $\alpha_t = 1$ for all $t$, $\mathbb E \vert X_T \vert \le \mathbb E \vert Y_T \vert$? I would be surprised. Heuristically, if we get a negative shock at $t=0$ of size $\sqrt{d t}$, then, $\alpha_t = - 1$ there makes it $-d t - \sqrt{ dt}$, while $\alpha_t = +1$ makes it $dt - \sqrt{ dt}$. So, on the negative side, negative push via drift should be better, and on the positive a positive shock is a vague reasoning I am employing. $\endgroup$
    – avk255
    Commented Apr 14, 2020 at 18:17
  • $\begingroup$ The comparison theorem of N. Ikeda & S. Watanabe (projecteuclid.org/download/pdf_1/euclid.ojm/1200770674) tells us that $X_t \le Y_t$ for all $t$, a.s., when $\alpha_t =1$. $\endgroup$
    – John Dawkins
    Commented Apr 17, 2020 at 17:22

1 Answer 1

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Yes, square the processes then $$ dX_t^2 = 2X_t dX_t + (dX_t)^2 = 2|X_t| dt + 1 + 2X_t dB_t$$ and similarly for the other. Then you see the the drift on this process must be bigger than that of the $Y_t$ process at the same level. Then you can finish it off with a comparison theorem that says $X_t^2$ will be stochastically larger than $Y_t^2$ and the same is therefore true about the absolute value as well. There is a chapter on the comparison theorems in Ikeda and Watanabe , I don't know where else they can be found. If this is true, the I acknowledge that I learned this trick from Ionnis Karatzis about 1986, but if I'm muddled, i's just me.

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  • $\begingroup$ A very neat trick! Actually, the comparison theorem yields that here $|X_t|\ge |Y_t|$ pathwise, not just stochastically. $\endgroup$
    – zhoraster
    Commented Apr 15, 2020 at 16:18
  • $\begingroup$ Pathwise? Is this under an additional Markovian assumption? $\endgroup$
    – e.lipnowski
    Commented Apr 15, 2020 at 17:19
  • $\begingroup$ @e.lipnowski, the comparison theorem is also valid for random (adapted) coefficients (however, I do not know any classic reference for that, so I had to prove it for our paper, see Theorem 8). $\endgroup$
    – zhoraster
    Commented Apr 15, 2020 at 17:57
  • $\begingroup$ Thanks @mike This is great, and very useful! $\endgroup$
    – avk255
    Commented Apr 16, 2020 at 6:15

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