Consider the random walk on $\mathbb R$ with $X_0 = a >0$ and
$$X_{n+1} = X_n + U_n,$$
where $U_0, U_1, U_2,\ldots $ is an i.i.d. sequence of uniform random numbers in $[-1,1]$.
How does the hitting time to $(-\infty,0]$; i.e.
$$\tau_a:=\min\{n : X_n\leq 0\}$$
behave for large $a$?
In need to prove $\tau_a$ is of order $a^2$; i.e. for any $\epsilon>0$ there are constants $C$ and $C'$ such that for large enough $a$ we have
$$\mathbb P[Ca^2<\tau_a<C'a^2]>1-\epsilon.$$
Best Answer
We assume that $U$ is centered, square integrable and we denote by $\sigma^2>0$ its variance. Given $a\geq 0$, I denote by $\tau_a$ the hitting time of $[a,+\infty)$ by $X$ starting from $X_0=0$ (this formulation is of course equivalent but more natural when using the following method). Fix $\varepsilon>0$ and let us prove that there exists $C,C'$ such that, for all $a$ large enough, \begin{align*} \mathbb{P}(C a^2< \tau_a \leq C' a^2+1)\geq 1-\varepsilon. \end{align*} We will use Donsker invariance principle, although there is no need to carefully check the law of hitting times for the Brownian motion, nore to quantify the speed of convergence to the Brownian motion.
Let $B$ be a standard one dimensional Brownian motion and choose $C>0$ and $C'>C$ such that \begin{align*} \mathbb{P}(C < T_{1/\sigma}\leq C')\geq 1-\varepsilon/3, \end{align*} where $T_{1/\sigma}$ is the hitting time of $1/\sigma$ by the Brownian motion $B$.
Let $(\varphi_k)_{k\in\mathbb{N}}$ (resp. $(\psi_k)_{k\in\mathbb{N}}$) be an increasing (resp. bounded decreasing) sequence of continuous functions converging pointwisely to $\mathbf{1}_{\cdot < 1/\sigma}$. We define the continous function $f_k$ and $g_k$ on $C([0,\infty[)$ (with the topology defined p. 60 of Karatzas-Shreve) by \begin{align*} f_k(\omega)=\varphi_k(\max_{t\in[0,C]} \omega_t) \text{ and }g_k(\omega)=\psi_k(\max_{t\in[0,C']} \omega_t). \end{align*} We thus have, almost surely, \begin{align*} \mathbf{1}_{C < T_{1/\sigma}\leq C'}=\lim_{k\rightarrow\infty} f_k(B)-g_k(B). \end{align*} Hence, by the dominated convergence theorem, we can choose $k_0$ such that \begin{align*} \mathbb{E}(f_{k_0}(B)-g_{k_0}(B))\geq 1-2\varepsilon/3. \end{align*}
For any $n\in\mathbb{N}$, let us define the affine process starting from $0$ and such that \begin{align*} X_t^{(n)}=\frac{1}{\sigma \sqrt{n}}Y_{nt},\text{ with } Y_t=\sum_{n=1}^{\lfloor t\rfloor}U_n+(t-\lfloor t\rfloor)U_{\lfloor t\rfloor+1}. \end{align*} Denoting by $T^{(n)}_{1/\sigma}$ the first hitting time of $1/\sigma$ by $X^{(n)}$, it is clear that $a^2 T^{(a^2)}_{1/\sigma}\leq \tau_a < a^2 T^{(a^2)}_{1/\sigma}+1$. Hence \begin{align*} \mathbb{P}(C a^2< \tau_a \leq C' a^2+1)&\geq \mathbb{P}(C a^2< a^2 T^{(a^2)}_{1/\sigma}\leq C' a^2)\\ &= \mathbb{P}(C < T^{(a^2)}_{1/\sigma}\leq C')\\ &\geq \mathbb{E}(f_{k_0}(X^{a^2})-g_{k_0}(X^{a^2})) \end{align*} We know that the law of $(X_t^{(n)})_{t\geq 0}$ converges weakly to the Brownian motion on $C([0,\infty))$ when $n\rightarrow\infty$ (see for instance Theorem~4.20 p.71 in Karatzas-Shreve), hence \begin{align*} \mathbb{E}(f_{k_0}(X^{a^2})-g_{k_0}(X^{a^2}))\xrightarrow[a\rightarrow\infty]{} \mathbb{E}(f_{k_0}(B)-g_{k_0}(B))\geq 1-2\varepsilon/3. \end{align*} As a consequence, there exists $a_0$ such that, for all $a\geq a_0$, \begin{align*} \mathbb{P}(C a^2< \tau_a \leq C' a^2+1)\geq 1-\varepsilon. \end{align*}