I have one question about the example of renewal process.

For a renewal process, $N(t)+1$ is a stopping time for the interarrival sequence $X_1,X_2,\cdots$. Show that $\mathbb{E}[N(t)] < \infty$, i.e, check that $m(t) < \infty$.

When I search for similar questions, I didn't notice anyone mentioned this idea. Below is the proof, but I am confused with one step.

  • Recall that $m(t)=\sum_{n=1}^{\infty} F_n(t)$, where $F_n(t)=\mathbb{P}\{S_n\leq t\}=\mathbb{E} [I_{\{S_n \leq t\}}]$ can always be bounded by $e^{-(S_n-t)}$ from 0 to t.
  • $\mathbb{E} [I_{\{S_n \leq t\}}] \leq \mathbb{E}[e^{-(S_n-t)}]=e^t (\mathbb{E}[e^{-X_1}])^n$
  • Now since $X_1 \geq 0$ and $\mathbb{P}\{X_1 > 0\} > 0$, so $\mathbb{E}[e^{-X_1}]=\zeta < 1$
  • So $F_n(t) \leq \zeta^n e^t$, and $m(t)\leq e^t \frac{\zeta}{1-\zeta} < +\infty$

I am confused with the step $\mathbb{P} \{ X_1 > 0 \} > 0$, why does it matter and how does it influence the subsequent proof?

What I try to convince myself is that

$\mathbb{E}(e^{-X_1}) = \int_0^{\infty}e^{-x}\mathrm{d}F(x) < \int_0^{\infty} 1 \cdot \mathrm{d} F(x) = 1$. Although the $e^0=1$ doesn't hold the inequality, but only finite number which have abnormal values makes no difference to the integral.

Does that part account for this integral to some extent?? Thank you for your reply.


$P(X_1>0)>0$ implies $\zeta<1$. If $P(X_1>0)=0$, then $\zeta=1$ giving you a divergent upper bound when summing $F_n(t)$.

In your integration, you are assuming $X_1$ has continuous density. On the other hand if $X_1$ has a jump, in particular if $X_1=0$ with probability 1,then the expectation of $e^{-X_1}$ is clearly 1 as well. You've written down a Stieltjes integral, not a Riemann one. $F(x)$ experiences a jump discontinuity when $X$ takes on a single value with positive probability.

  • $\begingroup$ thanks a lot, you are right. I seem to assume that F is continuous. So that inequality is the necessary requirement for any F of renewal process! $\endgroup$ – NiubilityDiu Jan 30 '15 at 6:53

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