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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!!!

mr_T
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1 Answers1

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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.

marvinthemartian
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    Thank you marvin so far!!! I understand it until the last steps. We have so far: $\mathbb{E}[\bar{N}(t)]=\frac{1}{P(X_n \geq \alpha)}$. Since $0 \leq P(X_n \geq \alpha) \leq 1$ and $t/\alpha \geq 0$ , we can conclude that: $\mathbb{E}[\bar{N}(t)] \leq \frac{1+t/\alpha }{P(X_n \geq \alpha)}$. But why is this finite? I don't understand what u wrote later on. Can you please explain that step a little bit closer to me? I would really appreciate that!! – mr_T Jun 18 '14 at 13:30