2
$\begingroup$

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.

$\endgroup$
  • 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 '14 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 '14 at 21:09
  • $\begingroup$ Got something from the answer below? $\endgroup$ – Did Jul 1 '16 at 11:06
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.