# On the largest and smallest spacings for the uniform distribution

Let $$Z_1,\dots,Z_n$$ be iid random variables (r.v.'s) each uniformly distributed on $$[0,1]$$. Let $$Z_{n:1}\le\cdots\le Z_{n:n}$$ be the corresponding order statistics. For $$i=1,\dots,n-1$$, let $$G_i:=Z_{n:i+1}-Z_{n:i}$$, the $$i$$th spacing/gap. Let $$$$A_n:=G_{n-1:1}=\min_{i\le n-1}G_i,\quad B_n:=G_{n-1:n-1}=\max_{i\le n-1}G_i.$$$$

In comments at Expected value ..., Brendan McKay asked if $$EB_n/EA_n\to\infty$$ (as $$n\to\infty$$), and Anthony Quas asked if $$med(B_n/A_n)\to\infty$$, where $$med$$ denotes the median. The purpose here is to answer these questions (affirmatively).

$$\newcommand{\eD}{\overset{\text{D}}\to} \newcommand{\D}{\overset{\text{D}}=}$$ As was noted on the linked MO page Expected value ..., the gaps $$G_1,\dots,G_{n-1}$$ between the adjacent points are jointly distributed as $$\frac{H_1}{H_1+\dots+H_{n+1}},\dots,\frac{H_{n-1}}{H_1+\dots+H_{n+1}}$$, where the $$H_i$$'s are iid standard exponential random variables (r.v.'s); see e.g. Theorem 6.6(c).

So,
$$\begin{equation*} B_n\D M_n:=\frac{H_{n-1:n-1}}{S_{n+1}}=\frac1{S_{n+1}}\,\max_{i\le n-1}H_i, \end{equation*}$$ where $$\D$$ means the equality in distribution, and $$S_{n+1}:=H_1+\dots+H_{n+1}$$. Next, for any real $$x$$ and all large enough natural $$n$$, $$\begin{multline*} P(H_{n-1:n-1}-\ln n for some r.v. $$Y$$, so that $$\begin{equation*} Y_n:=H_{n-1:n-1}-\ln n\eD Y, \end{equation*}$$ where $$\eD$$ means the convergence in distribution. Also, by the strong law of large numbers (SLLN) $$\frac n{S_{n+1}}\to1$$ almost surely and hence in distribution. So, $$\begin{equation*} \frac n{\ln n}\,B_n\D\frac n{\ln n}\,M_n=\frac n{\ln n}\,\frac{H_{n-1:n-1}}{S_{n+1}} =\frac{Y_n+\ln n}{\ln n}\,\frac n{S_{n+1}}\eD1. \tag{1} \end{equation*}$$ So, by the Fatou lemma, $$\begin{equation*} \liminf_n\frac n{\ln n}\,EB_n\ge1. \end{equation*}$$

On the other hand,
$$\begin{equation*} A_n\le G_1, \end{equation*}$$ and $$G_1$$ has the beta distribution with parameters $$1,n$$. So,
$$\begin{equation*} EA_n\le EG_1=\frac1{n+1}. \end{equation*}$$ So, $$\begin{equation*} \liminf_n\frac{EB_n}{EA_n}\ge\lim_n\frac{\ln n}n\,(n+1)=\infty. \end{equation*}$$ Thus, it is confirmed that $$EB_n/EA_n\to\infty$$.

Also, $$$$\frac{B_n}{A_n}\ge \frac{B_n}{G_1} \D\frac{nB_n}{H_1}\,\frac{S_{n+1}}n\eD\infty,$$$$ because, by (1), $$nB_n\eD\infty$$ and, by the SLLN, $$\frac{S_{n+1}}n\eD1$$. Thus, $$\frac{B_n}{A_n}\eD\infty$$ and hence indeed $$med(B_n/A_n)\to\infty$$.

• It’s interesting that you did the work on the length of the longest interval instead of the shortest one. You show the longest one is $\log n$ times the average gap; while it’s also true that the shortest one is about $1/n$ times the shortest gap (so I would have expected that to be the easier place to look for a big ratio). – Anthony Quas Aug 26 '19 at 2:43
• What you are saying is of course correct (I gather you meant "$1/n$ times the average gap"), and I did have that in mind. My preference, however, was to give an answer as simple and short as possible to both questions at once. Since I wanted to use the straightforward Fatou lemma, it looked like I had to work mainly with $B_n$ rather than $A_n$. – Iosif Pinelis Aug 26 '19 at 3:48
• Fair enough. I was surprised that the largest gap was that long, while the shortest gap feels like the birthday paradox. – Anthony Quas Aug 26 '19 at 4:34
• I too was surprised how short the shortest gap is. Was not intuitive to me at first, even though formulas tell me that right away. – Iosif Pinelis Aug 26 '19 at 13:01