# Absolute value of a diffusion

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!

• What happens in the case $\alpha_t =1$ for all $t$? Apr 14, 2020 at 15:34
• 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. Apr 14, 2020 at 18:17
• 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 $\alpha_t =1$. Apr 17, 2020 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.
• A very neat trick! Actually, the comparison theorem yields that here $|X_t|\ge |Y_t|$ pathwise, not just stochastically. Apr 15, 2020 at 16:18