Convergence rate of maximum interval of Poisson process Let $\{t_0,t_1,\ldots,t_{N}\}\subset [0,1]$ be a collection of points from a Poisson process with intensity $\lambda$.
Here, for convenience, assume that $t_0=0$.
Then what is the convergence rate of maximum interval of these points,
\begin{align*}
 \max_{i:t_i\in(0,1]} (t_i-t_{i-1}) = O_p(?)
\end{align*}
as $\lambda\rightarrow \infty$?
Here, what I have done.
Let $\max \Delta t_i = \max_{i:t_i\in(0,1]} (t_i-t_{i-1})$.
We have $N\sim \operatorname{Pois}(\lambda)$, and "without considering" the restriction $t_N \le 1$, each interval follows
\begin{align*}
  t_i - t_{i-1} \sim \operatorname{Exp}(\lambda),
\end{align*}
independently.
So, for some fixed $n$, we have
\begin{align*}
\def\bbP{{\mathbb{P}}}
\def\eps{{\varepsilon}}
  \bbP\{ \max_{i\le n} \Delta t_i > \eps \} &= \bbP \left\{ \Delta t_1 > \eps, \ldots, \Delta t_n > \eps\right\}  \\
&= \prod_{i=1}^n e^{-\lambda \eps} \\
&= (e^{-\lambda \eps})^n.
\end{align*}
Since $N$ is not a constant, but a random, we have to sum up with probability $\bbP\{N=n\}$ as follows:
\begin{align*}
  \bbP\{ \max_{i\le N} \Delta t_i > \eps \} &= \sum_{n=0}^{\infty} (e^{-\lambda \eps})^n \bbP\{N=n\}  \\
&= \sum_{n=0}^{\infty} (e^{-\lambda \eps})^n \frac{\lambda^n e^{-\lambda}}{n!}  \\
&= e^{-\lambda} \sum_{n=0}^{\infty}  \frac{\left(e^{-\lambda \eps}\lambda\right)^n}{n!}  \\
&= e^{e^{-\lambda \eps}\lambda - \lambda}  \\
&= e^{\lambda \left( e^{-\lambda \eps} - 1\right)}.
\end{align*}
Then we have $\max_{i\le n} \Delta t_i \xrightarrow{\bbP} 0$ by
\begin{align*}
  \lim_{\lambda \rightarrow \infty} \lambda \left( e^{-\lambda \eps} - 1\right) = \lim_{\lambda \rightarrow \infty} -\lambda=-\infty,
\end{align*}
and the definition of in probability convergence:
\begin{align*}
  \bbP\{ \max_{i\le n} \Delta t_i > \eps \} \rightarrow  \lim_{\lambda \rightarrow \infty} e^{\lambda \left(e^{-\lambda \eps}-1\right)} = e^{-\infty} = 0.
\end{align*}
Thus, for some small $c>0$,
\begin{align*}
  e^{\lambda \left( e^{-\lambda \eps} - 1\right)} &= e^{-c}, \\
   \lambda \left( e^{-\lambda \eps} - 1\right) &= -c, \\
   e^{-\lambda \eps} &= 1- \frac{c}{\lambda}, \\
  \lambda \eps &= \log \left( \frac{\lambda}{\lambda-c} \right),  \\
  \eps &= \frac{\log \frac{\lambda}{\lambda-c}}{\lambda},
\end{align*}
and we have the convergence rate (exactly) as
\begin{align*}
  \max_{i:t_i\in(0,1]} (t_i-t_{i-1}) = O_p\left(\frac{\log \frac{\lambda}{\lambda-c}}{\lambda}\right).
\end{align*}
However, I have no idea how to handle the restriction $t_N \le 1$.
Of course the convergence rate may be similar to this, but I don't know how to rigorously prove it.
Thanks,
 A: I don't have a solution, but I have a couple of ideas:
1.
Conditional on $N$, the distribution of $(t_1,...,t_N)$ is the same as the distribution of the order statistics of a random sample of size $N$ from the Unif(0,1) distribution.
Now, let $M_N$ be the maximum spacing from a random sample from the Unif(0,1) distribution of size $N$.
$$P[N \times M_N -\log{N}<x|N]\rightarrow e^{-e^x}$$
$$E[N \times M_N -\log{N}|N]\rightarrow \gamma =0.5772157..$$
$$Var[N \times M_N|N]\rightarrow \frac{\pi^2}{6}$$
So, the unconditional mean is
$$E[M]\approx  E\left[\frac{\gamma+\log{N}}{N}\right]\approx \frac{\gamma+\log{\lambda}}{\lambda}$$
and the unconditional variance is
$$Var[M] \approx \frac{\pi^2}{6}E\left[\frac{1}{N^2}\right]+Var\left[\frac{\gamma+\log{N}}{N}\right]
\approx \frac{\pi^2}{6}\frac{1}{\lambda^2}+\frac{\gamma+\log{\lambda}-1}{\lambda^3}$$
Devroye, Luc. "Laws of the iterated logarithm for order statistics of uniform spacings." The Annals of Probability (1981): 860-867.
2.
This seems like a way to eliminate the restriction that $t_N<1$.
Let $N^*=ceiling(\lambda+\sqrt{\lambda}\log{\lambda})$
Then,
$$P\left[\max_{i\le N}\Delta t_i>x\right]$$
$$=P\left[\max_{i\le N}\Delta t_i>x,N<N^*\right]+P\left[\max_{i\le N}\Delta t_i>x,N\ge N^*\right]$$
$$\le P\left[\max_{i\le N^*}\Delta t_i>x,N<N^*\right]+P\left[\max_{i\le N}\Delta t_i>x,N\ge N^*\right]$$
$$\le P\left[\max_{i\le N^*}\Delta t_i>x\right]+P\left[N\ge N^*\right]$$
$N$ has mean and variance $\lambda$ and is asymptotically normal, so $P[N \ge N^*]\approx 1-\Phi(\log{\lambda})\rightarrow 0$ where $\Phi$ is the standard normal cdf.
For large $\lambda$, $\lambda-\sqrt{\lambda}\log{\lambda}<N<\lambda+\sqrt{\lambda}\log{\lambda}$ with probability approaching 1.
Therefore, $P\left[\max_{i\le N}\Delta t_i>x\right]$ is bounded between the similar probabilities with these lower and upper bounds replacing the $N$.
