MathProblem 35 Light bulb problem #1
作者：互联网
You have ten light bulbs. Five have an average life of 100 hours, and the other five have a average life of 200 hours. These light bulbs have a memoryless property in that their current age (measured in how long they have already been on) has no bearing on their future life expectancy. Assuming they are all already on what is the expected number of hours before the first one burns out?
Hint: The density function for this kind of light bulb with average life of \(n\) hours is \(f(x)=\frac{e^{-x/n}}{n}\).
Solution
题目中的分布即为指数分布。概率密度函数：
\[f(x;\beta) = \frac{1}{\beta}e^{-\frac{1}{\beta}x} \]其期望以及方差：
\[E[X] = \beta, Var[X] = \beta^2 \]那什么是无记忆性呢？即：
\[P(T>s+t|T>t)=P(T>s) \]回到题目，一种灯泡的寿命概率密度为
\[f_1(x;100)=\frac{1}{100}e^{-\frac{1}{100}x} \]另一种则为
\[f_1(x;200)=\frac{1}{200}e^{-\frac{1}{200}x} \]因此累积分布函数为：
\[P(X\leq x) = 1-e^{-\frac{1}{\beta}x} \]因此存活的概率为：
\[P(X\ge x) = e^{-\frac{1}{\beta}x} \]所以对于所有的灯泡：
\[P(x)=(e^{-\frac{1}{100}x})^5(e^{-\frac{1}{200}x})^5=(e^{-\frac{3}{40}x}) \]所以至少有一个损坏：\(1-P(x)\)，其概率密度函数：\((1-P(x))'\)，那么本题的答案就是：
\[E[x] = \int_{x=0}^{\infin} (1-P(x))'\cdot x dx \]标签：200,life,frac,Light,MathProblem,hours,beta,problem,100 来源： https://www.cnblogs.com/xinyu04/p/16590282.html