I have a random walk $$R(t)= \sum\_{n<t} X\_n,$$ with $X\_n \sim U(-\tfrac{1}{n^\alpha}, \tfrac{1}{n^\alpha}),$ where $X\_n$ are independant and $\alpha >0$.


I think that someone must have studied this before. I am interested in understanding the behavior of $R(t)$ for large $t$.


For example can we estimate the probability of $R(t) \in [1, x)$?


Obviously, $E(R(t))=0,$ and $Var(R)= \tfrac{1}{3}\sum\_{n<t} \tfrac{1}{n^{2\alpha}}.$


Therefore depends on $\alpha,$ the variance can grow with $t$. Any information regarding the behavior of $R$ is appreciated.



Random walk with decreasing steps
Let $a:=\alpha$. Note that
\begin{equation}
 Ee^{zX\_n}=\frac{\sinh(z/n^a)}{z/n^a}
\end{equation}
for real $z>0$. Using the inequality $\dfrac{\sinh u}u<e^{u^2/6}$ for real $u\ne0$ (see e.g. [this MathSE answer](https://math.stackexchange.com/a/1759418/96609)) and the independence of the $X\_n$'s, we get
\begin{equation}
 Ee^{zR(t)}\le e^{z^2 B\_{a,t}/6},
\end{equation}
where
\begin{equation}
 B\_{a,t}:=\sum\_{n<t}\frac1{n^{2a}}. 
\end{equation}
So, for any real $x>0$,
\begin{equation}
 P(R(t)\ge x)\le e^{-zx+z^2 B\_{a,t}/6}.
\end{equation}
The latter bound on $P(R(t)\ge x)$ is minimized at $z=3x/B\_{a,t}$. Thus,
\begin{equation}
 P(R(t)\ge x)\le e^{-3x^2/(2B\_{a,t})}.
\end{equation}


