均匀点集间距 II

设最小两点距离为 $M$。对连续分布样本，$M>t$ 等价于“任意两点距离都 $>t$”。

对 $n$ 个点（这里 $n=101$），有经典结果（与上一题同一推导）：

$$\mathbb{P}(M>t)=(1-(n-1)t)^n,\quad 0\le t\le \frac{1}{n-1}.$$

因此

$$\mathbb{E}[M]=\int_0^{\infty}\mathbb{P}(M>t)\,dt =\int_0^{1/(n-1)}(1-(n-1)t)^n\,dt.$$

令 $u=1-(n-1)t$，则 $dt=-\frac{du}{n-1}$，得到

$$\mathbb{E}[M]=\frac{1}{n-1}\int_0^1 u^n\,du=\frac{1}{(n-1)(n+1)}=\frac{1}{n^2-1}.$$

代入 $n=101$ 得

$$\boxed{\mathbb{E}[M]=\frac{1}{10200}}.$$

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英文解析

Set the minimum distance between two points to $M$. For a continuous distribution sample, $M>t$is equivalent to “$>t$ for any two-point distance.”

For $n$ points (here $n=101$), there are classic results (same deduction as the previous question):

$$\mathbb{P}(M>t)=(1-(n-1)t)^n,\quad 0\le t\le \frac{1}{n-1}.$$

Therefore,

$$\mathbb{E}[M]=\int_0^{\infty}\mathbb{P}(M>t)\,dt =\int_0^{1/(n-1)}(1-(n-1)t)^n\,dt.$$

Orders $u=1-(n-1)t$, then $dt=-\frac{du}{n-1}$, get

$$\mathbb{E}[M]=\frac{1}{n-1}\int_0^1 u^n\,du=\frac{1}{(n-1)(n+1)}=\frac{1}{n^2-1}.$$

Substitute $n=101$ to get

$$\boxed{\mathbb{E}[M]=\frac{1}{10200}}.$$