三段中段长度期望

Expected Breakup Length

专题: Probability / 概率
难度: L6
来源: Brainstellar

题目详情

把一根长度为 1 的木棍在 $[0,1]$ 上随机选取两个切点并切开，得到三段。

求中间那段（位于两切点之间）的长度期望。

提示：用期望定义（积分）。

A stick is broken into 3 pieces, by randomly choosing two points along its unit length, and cutting it. What is the expected length of the middle part?

Hint

Selecting the random point from a small 'dt' length element is dt , as length of stick=1. Now use the definition of Expectation.

解析

答案是 $\frac{1}{3}$ 。

设两切点为 $X,Y\sim\mathrm{Unif}(0,1)$ 且独立。中段长度为 $|X-Y|$ 。

因此

E[|X-Y|]=\int_0^1\int_0^1 |x-y|\,dx\,dy=\frac{1}{3}.

Original Explanation

1/3

Solution

Double integral of |x-y|dxdy gives 1/3 as answer. This is same as one would expect from a broken pencil.

Palak's Solution: Integrate from 0 to 1, x * x/2 + (1-x) * (1-x)/2 = 1/3 logic: if one cut is at distance x from left, with probability x, the second cut comes before it, and expected length of middle piece is x/2.. Similarly with prob (1-x) it, middle piece is expected to have length (1-x)/2. Thus adding and integrating from 0 to 1.