Let $(X_n)$ be an iid. sequence of real, integrable random variables with $EX_1=a>0$. Let $S_n=X_1+...+X_n$, $n=1,2,...$ and $N_t:=\sup\{n\geq 1|S_1,...,S_n\leq t\}$, $t\geq 0$ where $\sup\emptyset=0$. Then the following holds:

  1. $P(N_t<\infty)=1\,\forall t\geq 0$
  2. $\lim_{t\to\infty} N_t=\infty$ $P$-almost surely
  3. $\lim_{t\to\infty} N_t/t=1/a$ $P$-almost surely

My textbook says this follows immediately from the strong law of large numbers, but I don't see this. In fact, I don't even know how I could prove this. Can anyone help me out here? Thanks.

  • 1
    $\begingroup$ picture the SLLN as saying that that $(n,S_n)$ is in the cone $\lbrace (x,y) : (a- \epsilon) x < y < (a + \epsilon ) x$ for large n, $\endgroup$
    – mike
    Jan 8, 2014 at 17:47
  • $\begingroup$ okay, so basically $N_t$ lies in this cone, too. By intuition I think I understand that because $N_t$ is in the cone we have $P(N_t<\infty)=1$. But how can I show this mathematically? $\endgroup$
    – dinosaur
    Jan 8, 2014 at 21:09
  • $\begingroup$ Got something from the answer below? $\endgroup$
    – Did
    Jul 1, 2016 at 11:06

1 Answer 1


$$\forall n\geqslant n_\varepsilon,\ a(1-\varepsilon)\leqslant\frac{S_n}n\leqslant a(1+\varepsilon)\implies\forall t\geqslant a(1+\varepsilon)n_\varepsilon,\ \frac1{a(1+\varepsilon)}\leqslant\frac{N_t}t\leqslant\frac1{a(1-\varepsilon)}$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .