Consider a renewal process whose holding times are given by a continuous random variable $X$ supported on $[0,1]$. It is known (e.g. Stone '65) that the renewal function $m(t)$ converges to $t/\mathbb{E} X + \mathbb{E}[X^2]/2\mathbb{E}^2[X]$ exponentially fast. I am looking for a bound on the rate of convergence which doesn't depend on the particular $X$, but holds for all continuous distributions supported on $[0,1]$ with, say, density bounded by some constant $C$ (if that helps).

Is such a uniform version of the renewal theorem with exponential convergence rate known?

In my particular application, we are given a "nice" random variable $X$ (we can choose what "nice" means), and we are interested in bounds for the conditional random variables $X|X \geq \alpha$ and (separately) $X|X \leq \beta$. In both cases $\alpha,\beta$ can be chosen "close" to the (properly defined) minimum and maximum values of $X$. This restriction corresponds to a bound on the density of the conditioned random variables.


If the density is bounded by a constant $C$ then I think the rate of convergence can be bounded as $\exp(-\delta t /C^2)$ for some fixed constant $\delta >0$. If the density is unbounded then I don't think there needs to be a uniform version of exponential convergence.

To elaborate on this, I will use my answer to the related question Error term for renewal function by Johan Wastlund. From the work there, we find that for any $c>0$ $$ m(t) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{e^{st}}{s(1-{\Bbb E}(e^{-sX}))}ds. $$ There is a double pole at $s=0$ and its residue is $t/{\Bbb E}(X) + {\Bbb E}(X^2)/(2{\Bbb E}(X)^2)$. Clearly all the other zeros lie in the half-plane Re$(s)<0$ . To show that there is a uniform exponential rate of convergence, we need to show that there is a half-plane $\text{Re}(s) < -\delta/C^2$ which is free of zeros of $1-{\Bbb E}(e^{-sX})$ (apart from the trivial zero at $s=0$).

First note that ${\Bbb E}(e^{-sX})= 1- s {\Bbb E}(X) + O(|s|^2 {\Bbb E}(X^2))$ from which it follows that there is some neighborhood $|s| <\delta$ which is has no zeros except for $s=0$.

Now suppose that $s=-x+iy$ is a zero, and we want a lower bound for $x$. We may assume that $|y| \ge \delta/2$. Let $f$ be the density function for the random variable $X$. Then we know that $$ \int_0^1 f(u) e^{ux} \cos(uy) du = \int_0^1 f(u) du. $$ Rearranging this means that $$ \int_0^1 f(u) (1-\cos(uy))du \le \int_0^1 f(u) (e^{ux}-1) du \le (e^x-1). $$ Now we get a lower bound on the left hand side, which will finish the proof. Since $y$ is bounded away from zero, if $\epsilon$ is small enough then the set of $u$ with $(1-\cos(uy))<\epsilon$ is bounded by some constant times $\sqrt{\epsilon}$. Therefore, since $f(u)\le C$, $$ \int_0^1 f(u) (1-\cos(uy)) du \ge \epsilon - \epsilon \int_{u:(1-\cos (uy))\le \epsilon} f(u) du \ge \epsilon (1- KC\sqrt{\epsilon}), $$ for some constant $K$. Thus taking $\epsilon = \delta/C^2$ for a suitably small $\delta$, we obtain that $$ (e^x-1) \ge \delta/C^2, $$ for some $\delta >0$. (In this write up, $\delta$ just denotes some positive constant which may change from line to line.) This completes the proof.

If the density is not bounded, then one can arrange for zeros to get arbitrarily close to the line $\text{Re }(s)=0$, and therefore there would be no uniform rate of exponential decay.

  • $\begingroup$ Great! I wonder why the literature doesn't bother with this sort of question. $\endgroup$ – Yuval Filmus Sep 10 '13 at 2:28
  • $\begingroup$ This shows that the error term behaves as $O(\exp(-\delta t/C^2))$, where the constant in the big O notation depends on the other solutions to $\mathbb{E}(e^{-sX})=1$, and so could depend on $X$. Can we get a uniform bound on the constant as well? (This is what my application actually requires.) $\endgroup$ – Yuval Filmus Sep 10 '13 at 2:41
  • $\begingroup$ The $O$-constant is absolute. You can move the line of integration staying within the zero free region above, and the denominator is uniformly bounded away from zero. You need to be a little careful with the integral -- truncating it from $c-iT$ to $c+iT$ and estimating errors etc, but the argument is standard. $\endgroup$ – Lucia Sep 10 '13 at 3:15
  • $\begingroup$ I would like to cite you - can you please send me your full name? (My e-mail is available on my user page.) $\endgroup$ – Yuval Filmus Sep 10 '13 at 19:15

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