离散均匀分布中未替换的样本之间的最大间隙

这个问题与我实验室对机器人覆盖率的研究有关：

随机绘制$n$从组数字$\{1,2,\ldots,m\}$无需更换，并以升序排序的数字。 。$1\le n\le m$

从此排序的数字，生成连续数字和边界之间的差：。这给出了间隙。$\{a_{(1)},a_{(2)},…,a_{(n)}\}$$g = \{a_{(1)},a_{(2)}−a_{(1)},\ldots,a_{(n)}−a_{(n-1)},m+1-a_{(n)}\}$$n+1$

最大差距的分布是什么？

$P(\max(g) = k) = P(k;m,n) = ?$

可以使用订单统计信息来构架： $P(g_{(n+1)} = k) = P(k;m,n) = ?$

有关间隙的分布，请参见链接，但是此问题要求最大间隙的分布。

我会对平均值\ mathbb {E} [g _ {（n + 1）}]感到满意$\mathbb{E}[g_{(n+1)}]$。

如果$n=m$所有间隙的大小均为1。如果$n+1 = m$则存在一个大小为2的间隙$2$，并且$n+1$可能的位置。最大间隙大小为$m-n+1$，并且该间隙可以放置在$n$ 数字中的任意一个之前或之后，总共$n+1$可能的位置。最小的最大间隙大小为$\lceil\frac{m-n}{n+1}\rceil$。定义任何给定组合T = {m \ choose n} ^ {-1}的概率$T= {m \choose n}^{-1}$。

我已经部分解决了概率质量函数，如 $P(g_{(n+1)} = k) = P(k;m,n) = \begin{cases} 0 & k < \lceil\frac{m-n}{n+1}\rceil\\ 1 & k = \frac{m-n}{n+1} \\ 1 & k = 1 \text{ (occurs when $m=n$)} \\ T(n+1)& k = 2 \text{ (occurs when $m=n+1$)} \\ T(n+1)& k = \frac{m-(n-1)}{n} \\ ? & \frac{m-(n-1)}{n} \le k \le m-n+1 \\ T(n+1)& k = m-n+1\\ 0 & k > m-n+1 \end{cases} \tag{1}$

当前工作（1）： 第一个间隙的方程很简单： 期望值具有一个简单值： 。通过对称性，我希望所有间隙都具有这种分布。也许可以通过从该分布中绘制次来找到解决方案。$a_{(1)}$

$$P(a_{(1)} = k) = P(k;m,n) = \frac{1}{{m \choose n}} \sum_{k=1}^{m-n+1} {m-k-1 \choose n-1}$$ $\mathbb{E}[P(a_{(1)})] = \frac{1}{ {m \choose n}} \sum_{k=1}^{m-n+1} {m-k-1 \choose n-1} k = \frac{m-n}{1+n}$$n$$n$

当前的工作（2）：容易进行蒙特卡洛模拟。

simMaxGap[m_, n_] := Max[Differences[Sort[Join[RandomSample[Range[m], n], {0, m+1}]]]];
m = 1000; n = 1; trials = 100000;
SmoothHistogram[Table[simMaxGap[m, n], {trials}], Filling -> Axis,
Frame -> {True, True, False, False},
FrameLabel -> {"k (Max gap)", "Probability"},
PlotLabel -> StringForm["m=``,n=``,smooth histogram of maximum map for `` trials", m, n, trials]][![enter image description here][1]][1]

令f （g ; n ，m ）为最小值a （1 ）等于g的机会；也就是说，样本由g和n − 1子集{ g + 1 ，g + 2 ，… ，m }组成。有（ m − $f(g;n,m)$$a_{(1)}$$g$$g$$n-1$$\{g+1,g+2,\ldots,m\}$ñ - 1）个这种子集出的（$\binom{m-g}{n-1}$n）同样可能的子集$\binom{m}{n}$

$$\Pr(a_{(1)}=g = f(g;n,m) = \frac{\binom{m-g}{n-1}}{\binom{m}{n}}.$$

对于所有大于g的k的可能值，将f （k ; n ，m ）相加得出生存函数$f(k;n,m)$$k$$g$

$$\Pr(a_{(1)} \gt g) = Q(g;n,m)= \frac{(m-g)\binom{m-g-1}{n-1}}{n \binom{m}{n}}.$$

令G n ，m为最大差距给定的随机变量：$G_{n,m}$

$$G_{n,m} = \max\left(a_{(1)}, a_{(2)}-a_{(1)}, \ldots, a_{(n)}-a_{(n-1)}\right).$$

（此响应的问题如最初框架，它被修改为包括之间的间隙前一个（Ñ ）和米）。$a_{(n)}$$m$ 我们将计算其生存函数P （克; Ñ ，米）= 镨（ģ Ñ ，米 > 克），从中可以轻松得出G n ，m的整个分布。该方法是一个从n = 1开始的动态程序，对于

$$P(g;n,m)=\Pr(G_{n,m}\gt g),$$ $G_{n,m}$$n=1$

$$P(g;1,m) = \Pr(G_{1,m} \gt 1) = \frac{m-g}{m},\ g=0, 1, \ldots, m.\tag{1}$$

For larger $n\gt 1$, note that the event $G_{n,m}\gt g$ is the disjoint union of the event

$$a_{1} \gt g,$$

for which the very first gap exceeds $g$, and the $g$ separate events

$$a_{1}=k\text{ and } G_{n-1,m-k} \gt g, \ k=1, 2, \ldots, g$$

for which the first gap equals $k$ and a gap greater than $g$ occurs later in the sample. The Law of Total Probability asserts the probabilities of these events add, whence

$$P(g;n,m) = Q(g;n,m) + \sum_{k=1}^g f(k;n,m) P(g;n-1,m-k).\tag{2}$$

Fixing $g$ and laying out a two-way array indexed by $i=1,2,\ldots,n$ and $j=1,2,\ldots,m$, we may compute $P(g;n,m)$ by using $(1)$ to fill in its first row and $(2)$ to fill in each successive row using $O(gm)$ operations per row. Consequently the table can be completed in $O(gmn)$ operations and all tables for $g=1$ through $g=m-n+1$ can be constructed in $O(m^3n)$ operations.

These graphs show the survival function $g\to P(g;n,64)$ for $n=1,2,4,8,16,32,64$. As $n$ increases, the graph moves to the left, corresponding to the decreasing chances of large gaps.

Closed formulas for $P(g;n,m)$ can be obtained in many special cases, especially for large $n$, but I have not been able to obtain a closed formula that applies to all $g,n,m$. Good approximations are readily available by replacing this problem with the analogous problem for continuous uniform variables.

Finally, the expectation of $G_{n,m}$ is obtained by summing its survival function starting at $g=0$:

$$\mathbb{E}(G_{n,m}) = \sum_{g=0}^{m-n+1} P(g;n,m).$$

This contour plot of the expectation shows contours at $2, 4, 6, \ldots, 32$, graduating from dark to light.