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I am studying renewal theory.

In the text book, $\{X_n, n=1, 2, \cdots\}$ denotes a sequence of non-negative i.i.d. with a common distribution $F$, and to avoid trivialities suppose that $F(0)<1$. We shall interpret $X_n$ as the time between the $(n-1)$st and $n$th event(renewal).

I do not understand what $F(0)<1$ means. Why is it possible to avoid trivialities if supposing like that?

$S_n$ denotes the time of the $n$th event. $$S_0=0, \quad S_n=\sum_{i=1}^{n}{X_i}, \quad n\ge1$$

So, we have that $N(t)$, the number of events by time $t$, is given by $$N(t)=\sup\{n: S_n\le t\}.$$

Now, the question is whether an infinite number of renewals can occur in a finite time. It cannot and will be proved like the following:

$$ \frac{S_n}{a}\to\mu \mbox{ as } n\to\infty $$ where $$\mu=E[X_n]=\int_0^\infty xdF(x).$$

Since $\mu\gt 0$, $S_n$ must be going to infinity as $n$ goes to infinity. Thus, $S_n$ can be less than or equal to $t$ for at most a finite number of values of $n$. Hence, $N(t)$ must be finite, and we can write $$N(t) = \max\{n: S_n\le t\}.$$

Why is $N(t)$ changed from $\sup$ to $\max$?

I copied my text book with slightly shortening. If my question has some problems, please tell me. Then, I will delete my question. Sorry for asking with vague question.

epimorphic
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Danny_Kim
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  • $F(0)<1$ just means that $\mathbb P(X_1 = 0) < 1$, in which case the $X_n$ are almost surely zero. In most cases we have $\mathbb P(X_1 = 0)=0$, though. – Math1000 Jan 18 '16 at 00:43
  • @Math1000 Ah... thank you. I got it. So, its purpose is telling that $\mu$ is greater than 0. – Danny_Kim Jan 18 '16 at 00:46
  • Sup changed to max: This is because unbounded subsets of $\mathbb N$ have no maximum, only bounded nonempty ones have. Since one knows that $0\in{n\mid S_n\leqslant t}$ for every $t\geqslant0$, this proves that $N(t)$ is indeed a maximum, not only a supremum. – Did Jan 18 '16 at 07:48

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