Let $T_n$ be a random variable with $T_n=X_1+...+X_n$ where the $X_i$'s are iid. Further we set $N(t)=max\{ n: T_n\leq n\}$ with the property $\Pr(N(t)<\infty)=1$.
I want to prove that $\mathbb{E}[N(t)]<\infty$.
I have done this so far. Let $F_n$ be the distribution function of $T_n$. $$\mathbb{E}[N(t)]=\sum_{n=1}^{\infty} n P(N(t)=n)=\sum_{n=1}^{\infty} n (F_n(t)-F_{n+1}(t))=\sum_{n=1}^{\infty} F_n(t)$$
Further I know that $0 \leq F_n \leq 1$. But does this implies $\mathbb{E}[N(t)]<\infty$ ? I don't think so... I don't know what I can do further...
Can anybody help me how I can proof it?
Thank you for any help!!!
Define a new renewal process $\{\bar N(t), t\geq 0\}$ and interarrival times, $\bar X_n$, are defined as follows where $\alpha$ is a positive number \begin{equation} \bar X=\begin{cases} 0 \text{ if } X_n<\alpha\\ \alpha \text{ if } X_n\geq\alpha \end{cases} \end{equation} and $\bar N(t)=\sup\{n: \bar X_1+\bar X_2+...+\bar X_n\leq t\}$.
In other words, the new renewal process has a gap between the renewals only if the original process has an interarrival time larger than $\alpha$. Draw a sample path for the original process, then draw the corresponding new process, I believe it will help you to understand.
It is obvious that the new process has renewals only at times $t=\alpha n$. Think of an arbitrary interval, $(k\alpha, (k+1)\alpha]$. The expected number of renewals in the interval is $1/P(X_n\geq \alpha)$. This might not be obvious at first. Convince yourself that number of renewals in the interval is distributed with a geometric distribution with $P(X_n\geq \alpha)$.
Then, \begin{align} E[\bar N(t)]\leq\frac{t/\alpha+1}{P(X_n\geq \alpha)}<\infty \end{align}
Here, again drawing a sample path helps a lot. For a given $t$, $t/\alpha+1$ is the smallest $\alpha$ value that is greater than $t$. So, the expected number of renewals up to such $\alpha$ is greater than or equal to the expected number of renewals up to $t$.
Also, $N(t)\leq \bar N(t)<\infty$. So, you have your result.