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!



Absolute value of a diffusion
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.


