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Let $\{X_i\}$ be a collection of i.i.d. Exp$(1)$ random variables. For $k \in \mathbb{Z}_{>0}$, define $$S_k = \sum_{i=1}^k X_i$$ and note that $\mathbb{E}[S_k] = k$.

I was wondering if there is any way or reference for studying the asymptotic of $$\mathbb{P}\Big(\min_{1\leq k \leq n} \tfrac{S_k - k}{\sqrt{k}} \geq \sqrt n\Big)?$$

In particular, I suspect that this probably decays like $e^{-Cn}$. When $n$ is large, this probability for fixed small $k\ll n$ decays like $e^{-C\sqrt n}$ because the steps are exponential. For large $n$ and $k$, because of CLT, we have $\mathbb{P}(\mathcal{N}(0,1) \geq \sqrt n) \sim e^{-Cn}$. Thus, $e^{-Cn}$ will dominate the decay.

I am interested in obtaining the leading order constant in the exponential, i.e. $$\lim_{n \to \infty} \frac{1}{n} \log \, \mathbb{P}\Big(\min_{1\leq k \leq n} \tfrac{S_k - k}{\sqrt{k}} \geq \sqrt n\Big).$$

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  • $\begingroup$ By the general invariance principle, for $t\in[0,1]$ and $n\to\infty$ $$ \frac{S_{\lfloor nt\rfloor}-\lfloor nt\rfloor}{\sqrt{n}}\to B_t $$ where $B_t$ is a Brownian motion. Therefore, $\frac{S_{k}-k}{\sqrt k}\asymp\frac{B_t}{\sqrt{t}}$ for $k\le n$ and $t=k/n$. So we are interested in $$ \sup_{t\in(0,1]} \frac{B_t}{\sqrt t} $$ So this is relevant to math.stackexchange.com/questions/1856443/… but alas the discrete time answer must be different... $\endgroup$ Commented Apr 18 at 10:52

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