A Note on First Passages for $S_n/n^{\frac{1}{2} 1}$

Abstract

Let $X_1, X_2, \cdots$ be independent random variables defined on a probability space $(\Omega, F, P)$ and with mean 0 and variance1. Let $F_0 = \{\Omega, \phi\}$, and for $n \geqq 1$ let $S_n = X_1 + X_2 + \cdots + X_n$ and $F_n = \sigma(X_1, X_2,\cdots, X_n)$. Denote by $T$ the class of all stopping rules with respect to $\{F_n\}$, i.e., the class of all $t: \Omega \rightarrow \{1, 2,\cdots, \infty\}$ such that $\{t = k\}\in F_k$ for $k = 1, 2,\cdots$, and let $T_n = \{t \in T: t\geqq n \operatorname{a.s.}\}$. If $t \in T$ we adopt the convention of the author of [3] that $|S_t|/t = \lim \sup_n |S_n|/n$ if $t = \infty$. We remind the reader that for this sequence of random variables $X_1, X_2,\cdots, \lim \sup_n |S_n|/n = 0 \operatorname{a.s}.$ Since $E(\sup_n|S_n|/n) < \infty$ (Lemma 9 [1]), it follows at once from the results of [3] that LEMMA 1. If $f_n = \operatorname{ess} \sup_{t\in T_n} E(|S_t\mid/t| F_n)$, then $f_n = \max (|S_n\mid/n, E(f_{n+1} |F_n))$ a.s. and $\lim \sup_nf_n = \lim \sup_n|S_n|/n = 0$ a.s. For each $c > 0$ define $t(c) =$ first $n \geqq 1$ for which $|S_n| > cn^{\frac{1}{2}}, = \infty$ if no such $n$ exists. In this note we prove THEOREM 1. If for each $c > 0, P(t(c) < \infty) = 1$, then $P(|S_n| = nf_n \operatorname{i.o.}) = 1$. Although this theorem will not be a surprise to readers of the recent literature on optimal stopping problems related to $S_n/n$, this proof may be of some interest.