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Quite generally, if $P(x)$ is the probability density of $n$ independent random variables and $F(x)=\int_{-\infty}^{x}P(x')dx'$ is their cumulative distribution function, then the joint distribution of the smallest and largest variables $x_{\rm min}<x_{\rm max}$ is given by

$$P(x_{\rm min},x_{\rm max})=n(n-1)P(x_{\rm min})P(x_{\rm max})[F(x_{\rm max})-F(x_{\rm min})]^{n-2}.$$

Several ways to prove this result are given here.

You ask for the probability ${\cal P}(d)$ that the largest spacing exceeds $d$, which follows from

$${\cal P}(d)=\int_{-\infty}^{\infty}dx_{\rm min}\int_{x_{\rm min}+d}^{\infty}dx_{\rm max}\,P(x_{\rm min},x_{\rm max}).$$

For the uniform distribution $P(x)=1$ and $F(x)=x$ for $0<x<1$, so

$$P(x_{\rm min},x_{\rm max})=n(n-1)(x_{\rm max}-x_{\rm min})^{n-2},$$

for $0<x_{\rm min}<x_{\rm max}<1$, and

$${\cal P}(d)=\int_{0}^{1-d}dx_{\rm min}\int_{x_{\rm min}+d}^{1}dx_{\rm max}\,n(n-1)(x_{\rm max}-x_{\rm min})^{n-2}=1-n d^{n-1}+(n-1) d^{n}.$$

Quite generally, if $P(x)$ is the probability density of $n$ independent random variables and $F(x)=\int_{-\infty}^{x}P(x')dx'$ is their cumulative distribution function, then the joint distribution of the smallest and largest variables $x_{\rm min}<x_{\rm max}$ is given by

$$P(x_{\rm min},x_{\rm max})=n(n-1)P(x_{\rm min})P(x_{\rm max})[F(x_{\rm max})-F(x_{\rm min})]^{n-2}.$$

Several ways to prove this result are given here.

You ask for the probability ${\cal P}(d)$ that the largest spacing exceeds $d$, which follows from

$${\cal P}(d)=\int_{-\infty}^{\infty}dx_{\rm min}\int_{x_{\rm min}+d}^{\infty}dx_{\rm max}\,P(x_{\rm min},x_{\rm max}).$$

For the uniform distribution $P(x)=1$ and $F(x)=x$ for $0<x<1$, so

$$P(x_{\rm min},x_{\rm max})=n(n-1)(x_{\rm max}-x_{\rm min})^{n-2},$$

for $0<x_{\rm min}<x_{\rm max}<1$, and

$${\cal P}(d)=\int_{0}^{1-d}dx_{\rm min}\int_{x_{\rm min}+d}^{1}dx_{\rm max}\,n(n-1)(x_{\rm max}-x_{\rm min})^{n-2}=1-n d^{n-1}+(n-1) d^{n}.$$

Two limits as a check: ${\cal P}(0)=1$, ${\cal P}(1)=0$ (assuming $n\geq 2$).

Quite generally, if $P(x)$ is the probability density of $n$ independent random variables and $F(x)=\int_{-\infty}^{x}P(x')dx'$ is their cumulative distribution function, then the joint distribution of the smallest and largest variables $x_{\rm min}<x_{\rm max}$ is given by

$$P(x_{\rm min},x_{\rm max})=n(n-1)P(x_{\rm min})P(x_{\rm max})[F(x_{\rm max})-F(x_{\rm min})]^{n-2}.$$

Several ways to prove this result are given here.

You ask for the probability ${\cal P}(d)$ that the largest spacing exceeds $d$, which follows from

$${\cal P}(d)=\int_{-\infty}^{\infty}dx_{\rm min}\int_{x_{\rm min}+d}^{\infty}dx_{\rm max}\,P(x_{\rm min},x_{\rm max}).$$

For the uniform distribution $P(x)=1$ and $F(x)=x$ for $0<x<1$, so

$$P(x_{\rm min},x_{\rm max})=n(n-1)(x_{\rm max}-x_{\rm min})^{n-2},$$

for $0<x_{\rm min}<x_{\rm max}<1$, and

$${\cal P}(d)=\int_{0}^{1-d}dx_{\rm min}\int_{x_{\rm min}+d}^{1}dx_{\rm max}\,n(n-1)(x_{\rm max}-x_{\rm min})^{n-2}=\frac{(d n-d-n) d^n+d}{d (n-1) n}.$$

Quite generally, if $P(x)$ is the probability density of $n$ independent random variables and $F(x)=\int_{-\infty}^{x}P(x')dx'$ is their cumulative distribution function, then the joint distribution of the smallest and largest variables $x_{\rm min}<x_{\rm max}$ is given by

$$P(x_{\rm min},x_{\rm max})=n(n-1)P(x_{\rm min})P(x_{\rm max})[F(x_{\rm max})-F(x_{\rm min})]^{n-2}.$$

Several ways to prove this result are given here.

You ask for the probability ${\cal P}(d)$ that the largest spacing exceeds $d$, which follows from

$${\cal P}(d)=\int_{-\infty}^{\infty}dx_{\rm min}\int_{x_{\rm min}+d}^{\infty}dx_{\rm max}\,P(x_{\rm min},x_{\rm max}).$$

For the uniform distribution $P(x)=1$ and $F(x)=x$ for $0<x<1$, so

$$P(x_{\rm min},x_{\rm max})=n(n-1)(x_{\rm max}-x_{\rm min})^{n-2},$$

for $0<x_{\rm min}<x_{\rm max}<1$, and

$${\cal P}(d)=\int_{0}^{1-d}dx_{\rm min}\int_{x_{\rm min}+d}^{1}dx_{\rm max}\,n(n-1)(x_{\rm max}-x_{\rm min})^{n-2}=1-n d^{n-1}+(n-1) d^{n}.$$

Quite generally, if $P(x)$ is the probability density of $n$ independent random variables and $F(x)=\int_{-\infty}^{x}P(x')dx'$ is their cumulative distribution function, then the joint distribution of the smallest and largest variables $x_{\rm min}<x_{\rm max}$ is given by

$$P(x_{\rm min},x_{\rm max})=n(n-1)P(x_{\rm min})P(x_{\rm max})[F(x_{\rm max})-F(x_{\rm min})]^{n-2}.$$

Several ways to prove this result are given here.

You ask for the probability ${\cal P}(d)$ that the largest spacing exceeds $d$, which follows from

$${\cal P}(d)=\int_{-\infty}^{\infty}dx_{\rm min}\int_{x_{\rm min}+d}^{\infty}dx_{\rm max}\,P(x_{\rm min},x_{\rm max}).$$

Quite generally, if $P(x)$ is the probability density of $n$ independent random variables and $F(x)=\int_{-\infty}^{x}P(x')dx'$ is their cumulative distribution function, then the joint distribution of the smallest and largest variables $x_{\rm min}<x_{\rm max}$ is given by

$$P(x_{\rm min},x_{\rm max})=n(n-1)P(x_{\rm min})P(x_{\rm max})[F(x_{\rm max})-F(x_{\rm min})]^{n-2}.$$

Several ways to prove this result are given here.

You ask for the probability ${\cal P}(d)$ that the largest spacing exceeds $d$, which follows from

$${\cal P}(d)=\int_{-\infty}^{\infty}dx_{\rm min}\int_{x_{\rm min}+d}^{\infty}dx_{\rm max}\,P(x_{\rm min},x_{\rm max}).$$

For the uniform distribution $P(x)=1$ and $F(x)=x$ for $0<x<1$, so

$$P(x_{\rm min},x_{\rm max})=n(n-1)(x_{\rm max}-x_{\rm min})^{n-2},$$

for $0<x_{\rm min}<x_{\rm max}<1$, and

$${\cal P}(d)=\int_{0}^{1-d}dx_{\rm min}\int_{x_{\rm min}+d}^{1}dx_{\rm max}\,n(n-1)(x_{\rm max}-x_{\rm min})^{n-2}=\frac{(d n-d-n) d^n+d}{d (n-1) n}.$$