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随机价格购买：剩余金额期望

What is the expected amount of money that remaining

专题
Probability / 概率
难度
L4

题目详情

概率题：随机价格购买：剩余金额期望。

英文原题

Assume that M>0M > 0 is a fixed real number and that XX is a random variable with uniform distribution on the interval (0,M)(0,M) . A price of an item is exactly SX\mathbb{S}X . A person with a total wealth of SM\mathbb{S}M has decided to buy as many of these items as possible. What is the expected amount of money that the person will have remaining?

解析

设 X∼U(0,M)X\sim U(0,M)，能买 ⌊M/X⌋\lfloor M/X\rfloor 件，剩余

R=M−⌊M/X⌋X.R=M-\lfloor M/X\rfloor X.

令 Y=X/M∼U(0,1)Y=X/M\sim U(0,1)，则

E[R]=M(1−∑k=1∞∫1/(k+1)1/kky dy) =M(1−12∑k=1∞k(1k2−1(k+1)2)).\mathbb{E}[R]=M\left(1-\sum_{k=1}^\infty \int_{1/(k+1)}^{1/k}k y\,dy\right)\ = M\left(1-\frac12\sum_{k=1}^\infty k\left(\frac{1}{k^2}-\frac{1}{(k+1)^2}\right)\right).

该和为 π2/6\pi^2/6，因此

E[R]=M(1−π212)≈0.178M.\boxed{\mathbb{E}[R]=M\left(1-\frac{\pi^2}{12}\right)}\approx 0.178M.

英文解析

R=M−⌊M/X⌋X.R=M-\lfloor M/X\rfloor X. E[R]=M(1−∑k=1∞∫1/(k+1)1/kky dy) =M(1−12∑k=1∞k(1k2−1(k+1)2)).\mathbb{E}[R]=M\left(1-\sum_{k=1}^\infty \int_{1/(k+1)}^{1/k}k y\,dy\right)\ = M\left(1-\frac12\sum_{k=1}^\infty k\left(\frac{1}{k^2}-\frac{1}{(k+1)^2}\right)\right). E[R]=M(1−π212)≈0.178M.\boxed{\mathbb{E}[R]=M\left(1-\frac{\pi^2}{12}\right)}\approx 0.178M.