Concentration of a modified random walk 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 $.5$, $X^{(t+1)} = X^{(t)} + (1+\varepsilon)$

*If $X^{(t)} = 0$, then we do the same thing as the original random walk: with probability $.5$, $X^{(t+1)} = X^{(t)} + 1$, and with probability $.5$, $X^{(t+1)} = X^{(t)} - 1$.


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 variance $t$ (thus a deviation of $\sqrt{t}$) in the case of $\varepsilon=0$?
 A: Let me try an answer.
[Edit: simplified and (hopefully) corrected.]
Let $\alpha$ be the only positive solution of $\mathrm{ch}\alpha = \exp(\varepsilon\alpha)$, so that
$$ \exp(\alpha x) = \frac12\left(\exp(\alpha(x+1-\varepsilon))+\exp(\alpha(x - 1 - \varepsilon))\right)\text. $$
The purpose of the definition lies in the fact that $\mathbb E[\mathbf e^{\alpha|X^{(n+1)}|}|\mathcal F_n]=\mathbf e^{\alpha|X^{(n)}|}$ whenever $|X^{(n)}|\geq 1+\varepsilon$. I claim that:

For $C>0$ large enough, $$\mathbf e^{\alpha|X^{(n)}|}-Cn$$ is a supermartingale.

Indeed, because of the remark above, it is enough to show that for some constant $C>0$, $$\mathbb E[\mathbf e^{\alpha|X^{(n+1)}|}|\mathcal F_n]\leq \mathbf e^{\alpha|X^{(n)}|} + C$$ whenever $|X^{(n)}|< 1+\varepsilon$. This is true for $C=\max\{\mathbf e^{\alpha x},0\leq x\leq 2\}=\mathbf e^{2\alpha}$, for example.
This means that
$\mathbb E[\mathbf e^{\alpha |X^{(n)}|}]\leq 1+Cn$, so $X^{(n)}$ cannot explode that much. For example, about the variance, for any positive exponent $p\in\mathbb N$,
$$   \mathbb E[|X^{(n)}|^2]
\leq \mathbb E[|X^{(n)}|^{2p}]^{1/p}
\leq K\mathbb E[\mathbf e^{\alpha|X^{(n)}|}]^{1/p}
\leq K'(1+n)^{1/p}\text,$$
so the variance increases slower that any root.

As Martin pointed out below (thank you!), $\mathbb E[\exp(a|X^{(n)}|)]\leq C'$ whenever $a<\alpha$. It can be seen as a consequence of the modified claim:

For $C\geq1$ large enough and $\lambda=\mathrm{ch}(a)\exp(-\varepsilon a)$,
  $$\mathbb E[\mathbf e^{a|X^{(n+1)}|}|\mathcal F_n]\leq \lambda\mathbf e^{a|X^{(n)}|} + C\text.$$

When $X^{(n)}\geq 1+\varepsilon$, this is a consequence of
\begin{align*}
\mathbb E[\mathbf e^{a|X^{(n+1)}|}|\mathcal F_n]
= & \frac12\left(\exp(a(X^{(n)}+1-\varepsilon))+\exp(a(X^{(n)} - 1 - \varepsilon))\right)\\
= & \mathbf e^{a|X^{(n)}|}\mathrm{ch}(a)\exp(-\varepsilon a)\text,
\end{align*}
and similarly when $X^{(n)}\leq -1-\varepsilon$; when $|X^{(n)}|<1+\varepsilon$,
$$\mathbb E[\mathbf e^{a|X^{(n+1)}|}|\mathcal F_n]\leq\mathbf e^{2a}\text,$$
and the claim follows.
From the claim, and noting that $\lambda<1$, we see that
$$   \mathbb E[\mathbf e^{a|X^{(n)}|}|]
\leq C(1+\lambda+\cdots+\lambda^n)
\leq \frac C{1-\lambda}$$
so $\mathbb E[\exp(a|X^{(n)}|)]$ (hence the variance) is uniformly bounded.
