Let $\varepsilon$ be a number in $(0, 1)$, consider the following random walk on the real line $X^{(0)}, X^{(1)}, \dots$, where

• $X^{(0)}=0$
• If $X^{(t)} > 0$, then with probability $.5$, $X^{(t+1)} = X^{(t)} + (1-\varepsilon)$, and with probability $.5$, $X^{(t+1)} = X^{(t)} - (1+\varepsilon)$
• If $X^{(t)} < 0$, then with probability $.5$, $X^{(t+1)} = X^{(t)} - (1-\varepsilon)$, and with probability $.5$, $X^{(t+1)} = X^{(t)} + (1+\varepsilon)$

In other words, if $X^{(t)}$ is biased from $0$, we want to "drag it towards $0$," and the strength is controlled by $\varepsilon$.

Is it the case that for nontrivial setting of $\varepsilon$ (for example, $\varepsilon = .3$), that $X^{(t)}$ is well concentrated around $0$ with smaller variance, compared to a $\sqrt{t}$ variance in the case of $\varepsilon=0$?