Consider $X_1, X_2, X_3$ ... random variables i.i.d. such that $P(X_i=1)=p$ and $P(X_i=-1)=1-p$. Consider the random walk $(S_n)_{n\ge 0} $ with $S_0=0$ and for $n\ge 1 $, $S_n = \displaystyle\sum^{n}_{i=1} X_i$. Let $d=2p-1=E X_i$ be the drift of $S_n $. Assume that $d>0.$
Define $T_x= \inf \{n\ge0; S_n=x\} $. I want to prove that $\displaystyle\sup_{y>0}\{T_y-\frac {2(y-1)}{d}\}$ is finite. It's clear that $T_y $ is finite, because $S_n$ is transient to right, but I can not control the difference. I thought of using the fact that the time the walker needs to go from $0$ to $y$ is of the ordem of $\frac {y}{d}$ since d is intuitively the speed of the random walk. Help?
$S_{T_x + n} - x = S_{T_x + n} - S_{T_x}$ are independent and distributed and $S_n = S_n - S_0$. Thus, $T_{x+1} - T_x$ are independent and distributed as $T_1$. Thus $\frac{T_{y}}{y}$ converges almost surely to $E[T_1]=\frac{1}{d}$ almost surely, according to the Law of Large Numbers. Thus $T_y - \frac{2(y - 1)}{d} = \frac{y}{d} - \frac{2(y-1)}{d} +o(y) = o(y) - \frac{y+1}{d}$ almost surely. Thus $\forall c \in \mathbb{N}$ almost surely $\exists y_0 \in \mathbb{N}$ such that $\forall y > y_0$ $T_y - \frac{2(y - 1)}{d} < c$. That results in $T_y - \frac{2(y - 1)}{d}$ being bounded.