I don't know if it is allowed here, to repost again his own question. I hope it is ok... I already asked this question here: Finite expectation of renewal process But I don't understand the last steps and I didn't get any answer in 4 months... And I really want to understand it! I post again all here:
Let $T_n$ be a random variable with $T_n=X_1+...+X_n$ where the $X_i$'s are iid. Further we set $N(t)=max\{ n: T_n\leq t\}$ with the property $\Pr(N(t)<\infty)=1$.
Claim: $\mathbb{E}[N(t)]<\infty$.
Proof:
I define a new renewal process $\{\bar N(t), t\geq 0\}$ with interarrival times $\bar X_n$. Let $\alpha$ be a positive number: \begin{equation} \bar X_n=\begin{cases} 0 \text{ if } X_n<\alpha\\ \alpha \text{ if } X_n\geq\alpha \end{cases} \end{equation} and $\bar N(t)=\sup\{n: \bar X_1+\bar X_2+...+\bar X_n\leq t\}$.
So the new renewal process has a gap between the renewals only if the original process has an interarrival time larger than $\alpha$. If one draws a sample path for the original and the new process, then it is obvious that the new process has renewals only at times $t=\alpha n$. From the drawing it follows also that the number of renewals in an arbitrary interval $(k\alpha, (k+1)\alpha]$ is geometric distributed with $p=P(X_n\geq \alpha)$. Hence the expected number of renewals in the interval is $1/P(X_n\geq \alpha)$: $$\mathbb{E}[\bar N(t)]=\frac{1}{P(X_n\geq \alpha)}.$$ Since $N(t) \leq \bar N(t)$, we have $\mathbb{E}[ N(t)] \leq \mathbb{E}[\bar N(t)]$.
But unfortunately I don't know how can I conclude that $\frac{1}{P(X_n\geq \alpha)}<\infty$... Can anybody help me out here? I appreciate any help, idea or advice!