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.


  • $\begingroup$ What happens in the case $\alpha_t =1$ for all $t$? $\endgroup$ – John Dawkins Apr 14 '20 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 Apr 14 '20 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 Apr 17 '20 at 17:22

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.

  • $\begingroup$ A very neat trick! Actually, the comparison theorem yields that here $|X_t|\ge |Y_t|$ pathwise, not just stochastically. $\endgroup$ – zhoraster Apr 15 '20 at 16:18
  • $\begingroup$ Pathwise? Is this under an additional Markovian assumption? $\endgroup$ – e.lipnowski Apr 15 '20 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 Apr 15 '20 at 17:57
  • $\begingroup$ Thanks @mike This is great, and very useful! $\endgroup$ – avk255 Apr 16 '20 at 6:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.