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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# 1.随机事件与概率"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## 1.1 基础概念 \n",
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"随机试验：\n",
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"试验是指为了察看某事的结果或某物的性能而从事的某种活动. 在概率论与数理统计中，一个试验如果具有以下3个特点: \n",
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" - (1) 可重复性: 在相同条件下可以重复进行：\n",
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" - (2) 可观察性: 每次试验的可能结果不止一个，并且能事先明确试验的所有可能结果:\n",
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" - (3) 不确定性: 一次试验之前，不能预知会出现哪一个结果。 就称这样的试验是一个随机试验，也简称为试验。\n",
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"\n",
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"样本点和样本空间： \n",
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"每次试验的每一个结果称为基本事件，也称作样本点，记作 $w_{1}, w_{2}, \\cdots$ 全部样本点的集合称为样本空间，记作 $\\Omega,$ 则 $\\Omega=$ $\\left\\{w_{1}, w_{2}, \\cdots\\right\\}$ \n",
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"例子: \n",
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"投郑一颗均匀股子，观察出现的点数。这是一个随机试验。样本空间 $\\Omega=\\{1,2,3,4,5,6\\} .$ \n",
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"随机事件： \n",
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"基本事件是不可再分解的、最基本的事件，其他事件均可由它们复合而成，由基本事件复合而成的事件称为随机事件或简称事件。 常用大写字母 $A, B, \\quad C$ 等表示事件。比如 $A=\\{$ 出现的点数为偶数 $\\}=\\{2,4,6\\} .$ \n",
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"\n",
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"## 1.2 随机事件与概率 \n",
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"随机事件在一次试验中是否发生虽然不能确定，但让人感兴趣的是随机事件在一次试验中发生的可能性有多大。概率就是用来描述随机事件发生的可能性大小的。比如抛硬币的试验，拖得次数越多，出现正面的次数与投郑次数之间的比例（也叫频率 )愈加趋 于0.5。\n",
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"它的数学定义为: = : \n",
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"在多次重复试验中，若事件A发生的频率稳定在确定常数 $p$ 附近摆动，且随着试验次数的增加，这种摆动的幅度是很微小的。 则称确定常数p为事件A发生的概率，记作 $P(A)=p .$ \n",
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"思考题 \n",
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"设一年有365天，求下列事件 $A, B$ 的概率: \n",
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"$A=\\{$n个人中没有2人同一天生日 $\\}$ \n",
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"$B=\\{$ n个人中有2人同一天生日 $\\}$ \n",
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"解 \n",
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"显然事件 $A, \\quad B$ 是对立事件, $P(B)=1-P(A)$. 由于每个人的生日可以是365天的任意一天，因此， $n$ 个人的生日有 $365^{n}$ 种可能结果，而且每种结果是等可能的，因而是古典概型，事件A的发生必须是 $n$ 个不同的生日，因而 $A$ 的样本点数为从365中取 $n$ 个的排列数 $P_{365}^{n},$ 于是 \n",
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"$$\n",
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"\\begin{array}{c}\n",
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"P(A)=\\frac{P_{365}^{n}}{365^{n}} \\\\\n",
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"P(B)=1-P(A)=1-\\frac{P_{365}^{n}}{365^{n}}\n",
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"\\end{array}\n",
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"$$ \n",
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"\n",
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"## 1.3 条件概率\n",
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"定义\n",
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"设 $A, B$ 是两个事件，且 $P(A)>0,$ 则称 \n",
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"$$\n",
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"P(B \\mid A)=\\frac{P(A B)}{P(A)}\n",
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"$$\n",
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"为在事件A发生的条件下，事件B的条件概率。 \n",
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"例子 \n",
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"某种元件用满6000h未坏的概率是3/4，用满10000h未坏的概率是1 $/ 2$ ，现有一个此种元件，已经用过 $6000 h$ 未坏，试求它能用到10000h的概率。 \n",
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"解 \n",
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"设A表示\\{用满10000h未坏\\}, $B$ 表示\\{用满 $6000 h$ 未坏\\}, 则 \n",
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"$$\n",
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"P(B)=3 / 4, P(A)=1 / 2\n",
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"$$ \n",
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"由于 $A \\subset B, A B=A,$ 因而 $P(A B)=1 / 2,$ 故 \n",
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"$$\n",
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"P(A \\mid B)=\\frac{P(A B)}{P(B)}=\\frac{P(A)}{P(B)}=\\frac{\\frac{1}{2}}{\\frac{3}{4}}=\\frac{2}{3}\n",
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"$$ \n",
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"\n",
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"## 1.4 事件的独立性\n",
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"定义\n",
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"如果事件B发生的可能性不受事件A发生与否的影响，即\n",
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"$$\n",
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"P(B \\mid A)=P(B)\n",
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"$$\n",
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"则称事件B对于事件A独立.显然，若B对于A独立，则 $A$ 对于 $B$ 也一定独立，称事件 $A$ 与事件 $B$ 相互独立.\n",
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"例子.\n",
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"口袋里装有5个黑球与3个白球，从中有放回地取2次，每次取一个，设事件A表示第一次取到黑球，事件B表示第二次取到黑球，则\n",
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"有\n",
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"$$\n",
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"P(A)=\\frac{5}{8}, P(B)=\\frac{5}{8}, P(A B)=\\frac{5}{8} \\times \\frac{5}{8}=\\frac{25}{64}\n",
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"$$\n",
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"因而\n",
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"$$\n",
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"P(B \\mid A)=\\frac{P(A B)}{P(A)}=\\frac{5}{8}\n",
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"$$\n",
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"因此， $P(B \\mid A)=P(B),$ 事实上还可以算出 $P(B \\mid \\bar{A})=P(B)$ 。这表明不论 $A$ 发是不发生，都对 $B$ 发生的概率没有影响。即B 与A独立. \n",
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"\n",
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"性质： \n",
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"事件 $A$ 和事件 $B$ 相互独立的充分必要条件是 \n",
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"$$\n",
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"P(A B)=P(A) P(B)\n",
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"$$"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# 2.全概率公式和贝叶斯公式"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## 2.1 全概率公式\n",
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"定义： \n",
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"如果事件 $A_{1}, A_{2}, \\cdots, A_{n}$ 是一个完备事件组，并且都具有正概率，则有： \n",
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"$$\n",
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"\\begin{array}{c}\n",
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"P(B)=P\\left(A_{1}\\right) P\\left(B \\mid A_{1}\\right)+P\\left(A_{2}\\right) P\\left(B \\mid A_{2}\\right)+\\cdots+P\\left(A_{n}\\right) P\\left(B \\mid A_{n}\\right) \\\\\n",
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"=\\sum_{i=1}^{n} P\\left(A_{i}\\right) P\\left(B \\mid A_{i}\\right)\n",
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"\\end{array}\n",
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"$$ \n",
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"对于任何事件 $B,$ 事件 $A \\bar{A}$ 构成最简单的完备事件组，根据全概率公式得 \n",
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"$$\n",
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"\\begin{aligned}\n",
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"P(B) &=P(A B+\\bar{A} B)=P(A B)+P(\\bar{A} B) \\\\\n",
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"&=P(A) P(B \\mid A)+P(\\bar{A}) P(B \\mid \\bar{A})\n",
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"\\end{aligned}\n",
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"$$ \n",
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"\n",
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"## 2.2 贝叶斯公式 \n",
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"定义： \n",
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"设 $A_{1}, A_{2}, \\cdots, A_{n}$ 是一完备事件组，则对任一事件 $B, P(B)>0,$ 有 \n",
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"$$\n",
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"P\\left(A_{i} \\mid B\\right)=\\frac{P\\left(A_{i} B\\right)}{P(B)}=\\frac{P\\left(A_{i}\\right) P\\left(B \\mid A_{i}\\right)}{\\sum_{i=1}^{n} P\\left(A_{i}\\right) P\\left(B \\mid A_{i}\\right)}\n",
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"$$ \n",
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"以上公式就叫贝叶斯公式，可由条件概率的定义及全概率公式证得。 \n",
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"\n",
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"例子: \n",
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"市场上供应的某种商品只由甲、乙、丙3个厂生产，甲厂占45%，乙厂占35%，丙厂占20%。如果各厂的次品率依次为 4%, 2%, 5% 。现从市场上购买1件这种商品，发现是次品，试判断它是由甲厂生产的概率。 \n",
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"设事件 $A_{1}, A_{2}, A_{3},$ 分别表示“商品为甲、乙、丙厂生产的\"，事件 $B$ 表示“商品为次品”，由题意得到概率 \n",
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"$$\n",
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"\\begin{array}{c}\n",
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"P\\left(A_{1}\\right)=45 \\%, P\\left(A_{2}\\right)=35 \\%, P\\left(A_{3}\\right)=20 \\% \\\\\n",
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"P\\left(B \\mid A_{1}\\right)=4 \\%, P\\left(B \\mid A_{2}\\right)=2 \\%, P\\left(B \\mid A_{3}\\right)=5 \\%\n",
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"\\end{array}\n",
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"$$ \n",
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"根据贝叶斯公式，可得:\n",
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"$$\n",
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"\\begin{aligned}\n",
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"P\\left(A_{1} \\mid B\\right) &=\\frac{P\\left(A_{1}\\right) P\\left(B \\mid A_{1}\\right)}{P\\left(A_{1}\\right) P\\left(B \\mid A_{1}\\right)+P\\left(A_{2}\\right) P\\left(B \\mid A_{2}\\right)+P\\left(A_{3}\\right) P\\left(B \\mid A_{3}\\right)} \\\\\n",
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"&=\\frac{45 \\% \\times 4 \\%}{45 \\% \\times 4 \\%+35 \\% \\times 2 \\%+20 \\% \\times 5 \\%} \\approx 0.514\n",
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"\\end{aligned}\n",
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"$$\n",
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"在“购买一件商品”这个试验中， $P\\left(A_{i}\\right)$ 是在试验以前就已经知道的概率，所以习惯地称为先验概率。试验结果出现了次品（即 $B$ 发 生），这时条件概率 $P\\left(A_{i} \\mid B\\right)$ 反映了在试验以后对 $B$ 发生的“来源”(即次品的来源）的各种可能性的大小，通常称为后验概率。"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# 3.随机变量"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## 3.1 随机变量\n",
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"把试验的结果与实数对应起来，随试验结果的不同而变化的量就是随机变量，包含离散型随机变量和连续型随机变量。\n",
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"例子： \n",
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"郑一枚匀称的硬币，观察正面、背面的出现情况。这一试验的样本空间为 $\\Omega=\\{H, T\\}$, 其中，H表示\"正面朝上\", $T$ 表示“背面 朝上”。如果引入变量 $X$ ，对试验的两个结果进行数值化，将 $X$ 的值分别规定为1和0，即\n",
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"$$\n",
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"X=\\left\\{\\begin{array}{ll}\n",
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"1 & \\text { if } \\text { 出现 } H \\\\\n",
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"0 & \\text { if } & \\text { 出现 } T\n",
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"\\end{array}\\right.\n",
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"$$\n",
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"这里的 $X$ 就叫随机变量，因为它能取的值是离散的，我们就叫它离散型随机变量。"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## 3.2 概率分布\n",
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"定义：设离散型随机变量 $X$ 的所有可能取值为 $x_{1}, x_{2}, \\cdots, x_{n},$ 称为 $X$ 的概率分布。 \n",
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"离散型随机变量 $X$ 的分布律具有下列基本性质： \n",
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"1. $p_{k} \\geq 0, k=1,2, \\cdots$ \n",
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"2. $\\sum_{i=1}^{+\\infty} p_{k}=1$ \n",
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"二项分布： \n",
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"二项分布是一种离散型的概率分布。二项代表它有两种可能的结果：成功或者不成功。每次试验必须相互独立，重复n次，并且每 次试验成功的概率是相同的，为 $p $; 失败的概率也相同，为 $1-p$。掷硬币就是一个典型的二项分布。当我们要计算抛硬币吻，恰巧有 $x$ 次正面朝上的概率，可以使用二项分布的公式: \n",
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"$$\n",
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"P\\{X=k\\}=C_{n}^{k} p^{k}(1-p)^{n-k}\n",
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"$$\n",
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"\n",
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" \n",
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"\n",
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"泊松分布： \n",
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"如果随机变量 $X$ 的概率分布为\n",
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"$$\n",
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"P\\{X=k\\}=\\frac{\\lambda^{k}}{k !} e^{-\\lambda}, k=0,1,2, \\cdots\n",
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"$$\n",
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"式中， $\\lambda>0$ 为常数，则称随机变量 $X$ 服从参数为 $\\lambda$ 的泊松(Possion)分布，记为 $X \\sim P(\\lambda)$. \n",
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"\n",
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"\n",
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"\n"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## 3.3 概率密度函数\n",
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"定义：若存在非负函数 $f(x)$, 使一个连续型随机变量 $X$ 取值于任一区间 $(a, b]$ 的概率可以表示为\n",
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"$$\n",
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"P\\{a<X \\leq b\\}=\\int_{a}^{b} f(x) d_{x}\n",
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"$$\n",
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"\n",
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"正态分布： \n",
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"正态分布是概率论中最重要的连续型分布，在19世纪前叶由德国数学家高斯（Gauss）加以推广，故又常称为高斯分布。 正态分布的概率密度函数曲线呈钟形，概率密度函数为\n",
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"$$\n",
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"f(x)=\\frac{1}{\\sqrt{2 \\pi} \\sigma} e^{-\\frac{(x-\\mu)^{2}}{2 \\sigma^{2}}}\n",
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"$$\n",
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"具有两个参数 $\\mu$ 和 $\\sigma^{2}$ 。第一参数 $\\mu$ 是代表服从正态分布的随机变量的均值，第二个参数 $\\sigma^{2}$ 是此随机变量的方差。 \n",
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"如果一个随机变量 服从均值为 $\\mu,$ 标准差为 $\\sigma$ 的正态分布，数学上记作\n",
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"$$\n",
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"X \\sim N\\left(\\mu, \\sigma^{2}\\right)\n",
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"$$\n",
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"我们通常所说的标准正态分布均值为0, 标准差为1的正态分布。 \n",
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"\n",
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" \n",
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"\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": []
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## 3.4 随机变量的期望\n",
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"对于一个随机变量，时常要考虑它的平均取什么，期望就是概率中的平均值，对随机变量中心位置的一种度量。\n",
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"例子.\n",
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"经过长期观察积累，某射手在每次射击中命中的坏数 $X$ 服从分布：\n",
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"$$\n",
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"\\begin{array}{|c|c|c|c|c|c|c|c|}\n",
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"\\hline \\mathrm{X} & 0 & 5 & 6 & 7 & 8 & 9 & 10 \\\\\n",
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"\\hline p_{i} & 0 & 0.05 & 0.05 & 0.1 & 0.1 & 0.2 & 0.5 \\\\\n",
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"\\hline\n",
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"\\end{array}\n",
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"$$\n",
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"求这个射手平均命中的坏数是多少? \n",
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"解 \n",
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"一种很自然的考虑是：假定该射击手进行了100次射击，那么，约有5次命中5坏，5次命中6坏，10次命中7坏，10次命中8坏，20 次命中9坏，50次命10，没有脱革，从而在一次射击中，该射手平均命中的坏数为： \n",
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"$\\frac{1}{100}(10 \\times 50+9 \\times 20+8 \\times 10+7 \\times 10+6 \\times 5+5 \\times 5+0 \\times 0)=8.85$ \n",
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"所以，我们可以看到离散型的随机变量的期望值可以表示为： \n",
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"$E(X)=\\sum_{i=1}^{+\\infty} x_{i} p_{k}$ \n",
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"\n",
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"期望的性质： \n",
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"1. $E(c)=c$ \n",
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"2. $E(X+c)=E(X)+c$ \n",
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"3. $E(k X)=k E(X)$ \n",
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"4. $E(k X+c)=k E(X)+c$ \n",
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"5. $E(X+Y)=E(X)+E(Y)$ "
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
|
||||
"source": [
|
||||
"## 3.5 随机变量的方差：\n",
|
||||
"方差表示了随机变量的变异性，方差越大，随机变量的结果越不稳定。 \n",
|
||||
"定义：随机变量，若\n",
|
||||
"$$\n",
|
||||
"E[X-E(X)]^{2}\n",
|
||||
"$$\n",
|
||||
"存在，则称其为 $X$ 的方差，记为 $D(X),$ 即\n",
|
||||
"$$\n",
|
||||
"D(X)=E[X-E(X)]^{2}\n",
|
||||
"$$\n",
|
||||
"而称 $\\sqrt{D(X)}$ 为 $X$ 的标准差或均方差。\n",
|
||||
"由方差的定义和数学期望的性质，可以推出方差的计算公式:\n",
|
||||
"$$\n",
|
||||
"D(X)=E\\left(X^{2}\\right)-[E(X)]^{2}\n",
|
||||
"$$\n",
|
||||
"\n",
|
||||
"方差的性质 \n",
|
||||
"(1) $D(c)=0$ \n",
|
||||
"(2) $D(X+c)=D(X)$ \n",
|
||||
"(3) $D(c X)=c^{2} D(X)$ \n",
|
||||
"\n",
|
||||
"例子： \n",
|
||||
"甲、乙两车间生产同一种产品，设1000件产品中的次品数分别为随机变量 $X, Y,$ 已知他们的分布律如下: \n",
|
||||
"$$\n",
|
||||
"\\begin{array}{ccccc}\n",
|
||||
"\\mathbf{X} & \\mathbf{0} & 1 & \\mathbf{2} & \\mathbf{3} \\\\\n",
|
||||
"\\hline p_{i} & 0.2 & 0.1 & 0.5 & 0.2\n",
|
||||
"\\end{array}\n",
|
||||
"$$\n",
|
||||
"与\n",
|
||||
"$$\n",
|
||||
"\\begin{array}{|c|c|c|c|c|}\n",
|
||||
"\\hline \\mathrm{Y} & 0 & 1 & 2 & 3 \\\\\n",
|
||||
"\\hline p_{i} & 0.1 & 0.3 & 0.4 & 0.2 \\\\\n",
|
||||
"\\hline\n",
|
||||
"\\end{array}\n",
|
||||
"$$\n",
|
||||
"试讨论甲乙两者车厢的质量。 \n",
|
||||
"解 \n",
|
||||
"先计算均值 \n",
|
||||
"$$\n",
|
||||
"\\begin{array}{l}\n",
|
||||
"E(X)=0 \\times 0.2+1 \\times 0.1+2 \\times 0.5+3 \\times 0.2=1.7 \\\\\n",
|
||||
"E(Y)=0 \\times 0.1+1 \\times 0.3+2 \\times 0.4+3 \\times 0.2=1.7\n",
|
||||
"\\end{array}\n",
|
||||
"$$\n",
|
||||
"得到：甲、乙两车间次品数的均值相同。 再计算方差： \n",
|
||||
"$D(X)=(0-1.7)^{2} \\times 0.2+(1-1.7)^{2} \\times 0.1+(2-1.7)^{2} \\times 0.1+(2-1.7)^{2} \\times 0.5+(3-1.7)^{2} \\times 0.2=1.01$\n",
|
||||
"\n",
|
||||
"$D(Y)=(0-1.7)^{2} \\times 0.1+(1-1.7)^{2} \\times 0.3+(2-1.7)^{2} \\times 0.4+(2-1.7)^{2} \\times 0.4+(3-1.7)^{2} \\times 0.2=0.81$\n",
|
||||
"\n",
|
||||
"得到 $D(X)>D(Y),$ 说明乙车间的产品质量较稳定。"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## 随机模拟1：$\\pi$的估值\n",
|
||||
"\n",
|
||||
" "
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 61,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"name": "stdout",
|
||||
"output_type": "stream",
|
||||
"text": [
|
||||
"pi的估计值为 2.8\n",
|
||||
"pi的估计值为 2.8\n",
|
||||
"pi的估计值为 2.96\n",
|
||||
"pi的估计值为 3.016\n",
|
||||
"pi的估计值为 3.156\n",
|
||||
"pi的估计值为 3.156\n",
|
||||
"pi的估计值为 3.1208\n",
|
||||
"pi的估计值为 3.13528\n",
|
||||
"pi的估计值为 3.14612\n",
|
||||
"pi的估计值为 3.137168\n",
|
||||
"pi的估计值为 3.1410684\n"
|
||||
]
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"# pi的估计问题\n",
|
||||
"import numpy as np \n",
|
||||
"def pi_estimate(n):\n",
|
||||
" '''\n",
|
||||
" n为投点的数量\n",
|
||||
" '''\n",
|
||||
" n_rand_X = np.random.uniform(-1.0,1.0,n)\n",
|
||||
" n_rand_Y = np.random.uniform(-1.0,1.0,n)\n",
|
||||
" ## 判断是否在圆内\n",
|
||||
" distance = np.sqrt(n_rand_X**2 + n_rand_Y**2)\n",
|
||||
" dis_n = float(len(distance[distance<=1.0]))\n",
|
||||
" return 4 * (dis_n / n)\n",
|
||||
"\n",
|
||||
"for i in [10,50,100,500,1000,5000,10000,50000,100000,500000,10000000]:\n",
|
||||
" print(\"pi的估计值为\",pi_estimate(i))\n"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 41,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"name": "stdout",
|
||||
"output_type": "stream",
|
||||
"text": [
|
||||
"\u001b[1;31mDocstring:\u001b[0m\n",
|
||||
"uniform(low=0.0, high=1.0, size=None)\n",
|
||||
"\n",
|
||||
"Draw samples from a uniform distribution.\n",
|
||||
"\n",
|
||||
"Samples are uniformly distributed over the half-open interval\n",
|
||||
"``[low, high)`` (includes low, but excludes high). In other words,\n",
|
||||
"any value within the given interval is equally likely to be drawn\n",
|
||||
"by `uniform`.\n",
|
||||
"\n",
|
||||
".. note::\n",
|
||||
" New code should use the ``uniform`` method of a ``default_rng()``\n",
|
||||
" instance instead; please see the :ref:`random-quick-start`.\n",
|
||||
"\n",
|
||||
"Parameters\n",
|
||||
"----------\n",
|
||||
"low : float or array_like of floats, optional\n",
|
||||
" Lower boundary of the output interval. All values generated will be\n",
|
||||
" greater than or equal to low. The default value is 0.\n",
|
||||
"high : float or array_like of floats\n",
|
||||
" Upper boundary of the output interval. All values generated will be\n",
|
||||
" less than or equal to high. The default value is 1.0.\n",
|
||||
"size : int or tuple of ints, optional\n",
|
||||
" Output shape. If the given shape is, e.g., ``(m, n, k)``, then\n",
|
||||
" ``m * n * k`` samples are drawn. If size is ``None`` (default),\n",
|
||||
" a single value is returned if ``low`` and ``high`` are both scalars.\n",
|
||||
" Otherwise, ``np.broadcast(low, high).size`` samples are drawn.\n",
|
||||
"\n",
|
||||
"Returns\n",
|
||||
"-------\n",
|
||||
"out : ndarray or scalar\n",
|
||||
" Drawn samples from the parameterized uniform distribution.\n",
|
||||
"\n",
|
||||
"See Also\n",
|
||||
"--------\n",
|
||||
"randint : Discrete uniform distribution, yielding integers.\n",
|
||||
"random_integers : Discrete uniform distribution over the closed\n",
|
||||
" interval ``[low, high]``.\n",
|
||||
"random_sample : Floats uniformly distributed over ``[0, 1)``.\n",
|
||||
"random : Alias for `random_sample`.\n",
|
||||
"rand : Convenience function that accepts dimensions as input, e.g.,\n",
|
||||
" ``rand(2,2)`` would generate a 2-by-2 array of floats,\n",
|
||||
" uniformly distributed over ``[0, 1)``.\n",
|
||||
"Generator.uniform: which should be used for new code.\n",
|
||||
"\n",
|
||||
"Notes\n",
|
||||
"-----\n",
|
||||
"The probability density function of the uniform distribution is\n",
|
||||
"\n",
|
||||
".. math:: p(x) = \\frac{1}{b - a}\n",
|
||||
"\n",
|
||||
"anywhere within the interval ``[a, b)``, and zero elsewhere.\n",
|
||||
"\n",
|
||||
"When ``high`` == ``low``, values of ``low`` will be returned.\n",
|
||||
"If ``high`` < ``low``, the results are officially undefined\n",
|
||||
"and may eventually raise an error, i.e. do not rely on this\n",
|
||||
"function to behave when passed arguments satisfying that\n",
|
||||
"inequality condition. The ``high`` limit may be included in the\n",
|
||||
"returned array of floats due to floating-point rounding in the\n",
|
||||
"equation ``low + (high-low) * random_sample()``. For example:\n",
|
||||
"\n",
|
||||
">>> x = np.float32(5*0.99999999)\n",
|
||||
">>> x\n",
|
||||
"5.0\n",
|
||||
"\n",
|
||||
"\n",
|
||||
"Examples\n",
|
||||
"--------\n",
|
||||
"Draw samples from the distribution:\n",
|
||||
"\n",
|
||||
">>> s = np.random.uniform(-1,0,1000)\n",
|
||||
"\n",
|
||||
"All values are within the given interval:\n",
|
||||
"\n",
|
||||
">>> np.all(s >= -1)\n",
|
||||
"True\n",
|
||||
">>> np.all(s < 0)\n",
|
||||
"True\n",
|
||||
"\n",
|
||||
"Display the histogram of the samples, along with the\n",
|
||||
"probability density function:\n",
|
||||
"\n",
|
||||
">>> import matplotlib.pyplot as plt\n",
|
||||
">>> count, bins, ignored = plt.hist(s, 15, density=True)\n",
|
||||
">>> plt.plot(bins, np.ones_like(bins), linewidth=2, color='r')\n",
|
||||
">>> plt.show()\n",
|
||||
"\u001b[1;31mType:\u001b[0m builtin_function_or_method\n"
|
||||
]
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"?np.random.uniform"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## 随机模拟2:电子元件寿命问题\n",
|
||||
"\n",
|
||||
" \n",
|
||||
" "
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": null,
|
||||
"metadata": {},
|
||||
"outputs": [],
|
||||
"source": []
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 39,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
"text/plain": [
|
||||
"0.9124"
|
||||
]
|
||||
},
|
||||
"execution_count": 39,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"# 电子元件寿命问题\n",
|
||||
"import numpy as np \n",
|
||||
"\n",
|
||||
"def ele_life(n,c,h,t,lamb):\n",
|
||||
" \"\"\"\n",
|
||||
" 参数n:模拟实验的次数\n",
|
||||
" 参数c:每次试验中的c个元件\n",
|
||||
" 参数t:每c个元件中规定的合格品数量\n",
|
||||
" 参数h:小时数\n",
|
||||
" \"\"\"\n",
|
||||
" times = 0.0\n",
|
||||
" for i in range(n):\n",
|
||||
" c_rand = np.random.exponential(1/lamb,c)\n",
|
||||
" c_rand_t = len(c_rand[c_rand>h])\n",
|
||||
" if c_rand_t > t:\n",
|
||||
" times = times + 1\n",
|
||||
" return times / n \n",
|
||||
"\n",
|
||||
"ele_life(10000,1000,18,20,0.2)"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 34,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"name": "stdout",
|
||||
"output_type": "stream",
|
||||
"text": [
|
||||
"\u001b[1;31mDocstring:\u001b[0m\n",
|
||||
"exponential(scale=1.0, size=None)\n",
|
||||
"\n",
|
||||
"Draw samples from an exponential distribution.\n",
|
||||
"\n",
|
||||
"Its probability density function is\n",
|
||||
"\n",
|
||||
".. math:: f(x; \\frac{1}{\\beta}) = \\frac{1}{\\beta} \\exp(-\\frac{x}{\\beta}),\n",
|
||||
"\n",
|
||||
"for ``x > 0`` and 0 elsewhere. :math:`\\beta` is the scale parameter,\n",
|
||||
"which is the inverse of the rate parameter :math:`\\lambda = 1/\\beta`.\n",
|
||||
"The rate parameter is an alternative, widely used parameterization\n",
|
||||
"of the exponential distribution [3]_.\n",
|
||||
"\n",
|
||||
"The exponential distribution is a continuous analogue of the\n",
|
||||
"geometric distribution. It describes many common situations, such as\n",
|
||||
"the size of raindrops measured over many rainstorms [1]_, or the time\n",
|
||||
"between page requests to Wikipedia [2]_.\n",
|
||||
"\n",
|
||||
".. note::\n",
|
||||
" New code should use the ``exponential`` method of a ``default_rng()``\n",
|
||||
" instance instead; please see the :ref:`random-quick-start`.\n",
|
||||
"\n",
|
||||
"Parameters\n",
|
||||
"----------\n",
|
||||
"scale : float or array_like of floats\n",
|
||||
" The scale parameter, :math:`\\beta = 1/\\lambda`. Must be\n",
|
||||
" non-negative.\n",
|
||||
"size : int or tuple of ints, optional\n",
|
||||
" Output shape. If the given shape is, e.g., ``(m, n, k)``, then\n",
|
||||
" ``m * n * k`` samples are drawn. If size is ``None`` (default),\n",
|
||||
" a single value is returned if ``scale`` is a scalar. Otherwise,\n",
|
||||
" ``np.array(scale).size`` samples are drawn.\n",
|
||||
"\n",
|
||||
"Returns\n",
|
||||
"-------\n",
|
||||
"out : ndarray or scalar\n",
|
||||
" Drawn samples from the parameterized exponential distribution.\n",
|
||||
"\n",
|
||||
"See Also\n",
|
||||
"--------\n",
|
||||
"Generator.exponential: which should be used for new code.\n",
|
||||
"\n",
|
||||
"References\n",
|
||||
"----------\n",
|
||||
".. [1] Peyton Z. Peebles Jr., \"Probability, Random Variables and\n",
|
||||
" Random Signal Principles\", 4th ed, 2001, p. 57.\n",
|
||||
".. [2] Wikipedia, \"Poisson process\",\n",
|
||||
" https://en.wikipedia.org/wiki/Poisson_process\n",
|
||||
".. [3] Wikipedia, \"Exponential distribution\",\n",
|
||||
" https://en.wikipedia.org/wiki/Exponential_distribution\n",
|
||||
"\u001b[1;31mType:\u001b[0m builtin_function_or_method\n"
|
||||
]
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"?np.random.exponential"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## 随机模拟3：三门问题\n",
|
||||
"蒙提霍尔问题：假如你参与一个有主持人的游戏，你会看见三扇关闭了的门，其中一扇的后面有一辆汽车，另外2扇门后面各是一只山羊，你看不见门后面的情况，但主持人知道一切。你被主持人要求在三扇门中选择一扇，但不能打开，在你选定之后主持人开启了另一扇后面有山羊的门，然后你可以坚持原来选定的门，也可以改主意重新选择。问题是：改与不改对选中汽车的概率有影响吗？请使用模拟实验的方法回答该问题。"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 64,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"name": "stdout",
|
||||
"output_type": "stream",
|
||||
"text": [
|
||||
"在100000次测试中,坚持原则第一次就选中的次数是33063,改变决定选择另一扇门中奖的次数是66937\n",
|
||||
"概率分别是0.33063和0.66937，改变决定选择另一扇门中奖几率是坚持选择的2.02倍\n"
|
||||
]
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"# 三门问题\n",
|
||||
"import numpy.random as random\n",
|
||||
"def MontyHallProblem(n_test):\n",
|
||||
" #测试次数\n",
|
||||
" \n",
|
||||
" winning_door = random.randint(0,3,n_test)\n",
|
||||
" first_get = 0\n",
|
||||
" change_get = 0\n",
|
||||
" for winning_doors in winning_door:\n",
|
||||
" act_door = random.randint(0,3)\n",
|
||||
" if winning_doors == act_door:\n",
|
||||
" first_get += 1\n",
|
||||
" else :\n",
|
||||
" change_get += 1\n",
|
||||
" first_pro = first_get / n_test\n",
|
||||
" change_pro = change_get / n_test\n",
|
||||
" compar1 = round(change_get / first_get,2)\n",
|
||||
" print (\"在%d次测试中,坚持原则第一次就选中的次数是%d,改变决定选择另一扇门中奖的次数是%d\"% (n_test,first_get,change_get))\n",
|
||||
" print (\"概率分别是{0}和{1}，改变决定选择另一扇门中奖几率是坚持选择的{2}倍\".format(first_pro,change_pro,compar1))\n",
|
||||
"MontyHallProblem(100000)"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": null,
|
||||
"metadata": {},
|
||||
"outputs": [],
|
||||
"source": []
|
||||
}
|
||||
],
|
||||
"metadata": {
|
||||
"kernelspec": {
|
||||
"display_name": "Python 3",
|
||||
"language": "python",
|
||||
"name": "python3"
|
||||
},
|
||||
"language_info": {
|
||||
"codemirror_mode": {
|
||||
"name": "ipython",
|
||||
"version": 3
|
||||
},
|
||||
"file_extension": ".py",
|
||||
"mimetype": "text/x-python",
|
||||
"name": "python",
|
||||
"nbconvert_exporter": "python",
|
||||
"pygments_lexer": "ipython3",
|
||||
"version": "3.9.2"
|
||||
}
|
||||
},
|
||||
"nbformat": 4,
|
||||
"nbformat_minor": 4
|
||||
}
|
||||
|
|
@ -1,40 +0,0 @@
|
|||
# MCMC方法之Metropolis-Hastings算法
|
||||
|
||||
import numpy as np
|
||||
import random
|
||||
import matplotlib.pyplot as plt
|
||||
|
||||
## 设置参数
|
||||
mu = 0.5
|
||||
sigma = 0.1
|
||||
skip = 700 ## 设置收敛步长
|
||||
num = 100000 ##采样点数
|
||||
|
||||
def Gussian(x,mu,sigma):
|
||||
return 1/(np.sqrt(2*np.pi)*sigma)*np.exp(-np.square((x-mu))/(2*np.square(sigma)))
|
||||
|
||||
def M_H(num):
|
||||
x_0 = 0
|
||||
samples = []
|
||||
j = 1
|
||||
while(len(samples) <= num):
|
||||
while True:
|
||||
x_1 = random.random() # 转移函数(转移矩阵)
|
||||
q_i = Gussian(x_0,mu,sigma)
|
||||
q_j = Gussian(x_1,mu,sigma)
|
||||
alpha = min(1,q_i/q_j)
|
||||
u = random.random()
|
||||
if u <= alpha:
|
||||
x_0 = x_1
|
||||
if j >= skip:
|
||||
samples.append(x_1)
|
||||
j = j + 1
|
||||
break
|
||||
return samples
|
||||
|
||||
norm_samples = M_H(num)
|
||||
|
||||
X = np.array(norm_samples)
|
||||
px = Gussian(X,mu,sigma)
|
||||
plt.scatter(X,px)
|
||||
plt.show()
|
||||
|
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|
|
@ -1,668 +0,0 @@
|
|||
{
|
||||
"cells": [
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"# 1.随机事件与概率"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## 1.1 基础概念 \n",
|
||||
"随机试验：\n",
|
||||
"试验是指为了察看某事的结果或某物的性能而从事的某种活动. 在概率论与数理统计中，一个试验如果具有以下3个特点: \n",
|
||||
" - (1) 可重复性: 在相同条件下可以重复进行：\n",
|
||||
" - (2) 可观察性: 每次试验的可能结果不止一个，并且能事先明确试验的所有可能结果:\n",
|
||||
" - (3) 不确定性: 一次试验之前，不能预知会出现哪一个结果。 就称这样的试验是一个随机试验，也简称为试验。\n",
|
||||
"\n",
|
||||
"样本点和样本空间： \n",
|
||||
"每次试验的每一个结果称为基本事件，也称作样本点，记作 $w_{1}, w_{2}, \\cdots$ 全部样本点的集合称为样本空间，记作 $\\Omega,$ 则 $\\Omega=$ $\\left\\{w_{1}, w_{2}, \\cdots\\right\\}$ \n",
|
||||
"例子: \n",
|
||||
"投郑一颗均匀股子，观察出现的点数。这是一个随机试验。样本空间 $\\Omega=\\{1,2,3,4,5,6\\} .$ \n",
|
||||
"随机事件： \n",
|
||||
"基本事件是不可再分解的、最基本的事件，其他事件均可由它们复合而成，由基本事件复合而成的事件称为随机事件或简称事件。 常用大写字母 $A, B, \\quad C$ 等表示事件。比如 $A=\\{$ 出现的点数为偶数 $\\}=\\{2,4,6\\} .$ \n",
|
||||
"\n",
|
||||
"## 1.2 随机事件与概率 \n",
|
||||
"随机事件在一次试验中是否发生虽然不能确定，但让人感兴趣的是随机事件在一次试验中发生的可能性有多大。概率就是用来描述随机事件发生的可能性大小的。比如抛硬币的试验，拖得次数越多，出现正面的次数与投郑次数之间的比例（也叫频率 )愈加趋 于0.5。\n",
|
||||
"它的数学定义为: = : \n",
|
||||
"在多次重复试验中，若事件A发生的频率稳定在确定常数 $p$ 附近摆动，且随着试验次数的增加，这种摆动的幅度是很微小的。 则称确定常数p为事件A发生的概率，记作 $P(A)=p .$ \n",
|
||||
"思考题 \n",
|
||||
"设一年有365天，求下列事件 $A, B$ 的概率: \n",
|
||||
"$A=\\{$n个人中没有2人同一天生日 $\\}$ \n",
|
||||
"$B=\\{$ n个人中有2人同一天生日 $\\}$ \n",
|
||||
"解 \n",
|
||||
"显然事件 $A, \\quad B$ 是对立事件, $P(B)=1-P(A)$. 由于每个人的生日可以是365天的任意一天，因此， $n$ 个人的生日有 $365^{n}$ 种可能结果，而且每种结果是等可能的，因而是古典概型，事件A的发生必须是 $n$ 个不同的生日，因而 $A$ 的样本点数为从365中取 $n$ 个的排列数 $P_{365}^{n},$ 于是 \n",
|
||||
"$$\n",
|
||||
"\\begin{array}{c}\n",
|
||||
"P(A)=\\frac{P_{365}^{n}}{365^{n}} \\\\\n",
|
||||
"P(B)=1-P(A)=1-\\frac{P_{365}^{n}}{365^{n}}\n",
|
||||
"\\end{array}\n",
|
||||
"$$ \n",
|
||||
"\n",
|
||||
"## 1.3 条件概率\n",
|
||||
"定义\n",
|
||||
"设 $A, B$ 是两个事件，且 $P(A)>0,$ 则称 \n",
|
||||
"$$\n",
|
||||
"P(B \\mid A)=\\frac{P(A B)}{P(A)}\n",
|
||||
"$$\n",
|
||||
"为在事件A发生的条件下，事件B的条件概率。 \n",
|
||||
"例子 \n",
|
||||
"某种元件用满6000h未坏的概率是3/4，用满10000h未坏的概率是1 $/ 2$ ，现有一个此种元件，已经用过 $6000 h$ 未坏，试求它能用到10000h的概率。 \n",
|
||||
"解 \n",
|
||||
"设A表示\\{用满10000h未坏\\}, $B$ 表示\\{用满 $6000 h$ 未坏\\}, 则 \n",
|
||||
"$$\n",
|
||||
"P(B)=3 / 4, P(A)=1 / 2\n",
|
||||
"$$ \n",
|
||||
"由于 $A \\subset B, A B=A,$ 因而 $P(A B)=1 / 2,$ 故 \n",
|
||||
"$$\n",
|
||||
"P(A \\mid B)=\\frac{P(A B)}{P(B)}=\\frac{P(A)}{P(B)}=\\frac{\\frac{1}{2}}{\\frac{3}{4}}=\\frac{2}{3}\n",
|
||||
"$$ \n",
|
||||
"\n",
|
||||
"## 1.4 事件的独立性\n",
|
||||
"定义\n",
|
||||
"如果事件B发生的可能性不受事件A发生与否的影响，即\n",
|
||||
"$$\n",
|
||||
"P(B \\mid A)=P(B)\n",
|
||||
"$$\n",
|
||||
"则称事件B对于事件A独立.显然，若B对于A独立，则 $A$ 对于 $B$ 也一定独立，称事件 $A$ 与事件 $B$ 相互独立.\n",
|
||||
"例子.\n",
|
||||
"口袋里装有5个黑球与3个白球，从中有放回地取2次，每次取一个，设事件A表示第一次取到黑球，事件B表示第二次取到黑球，则\n",
|
||||
"有\n",
|
||||
"$$\n",
|
||||
"P(A)=\\frac{5}{8}, P(B)=\\frac{5}{8}, P(A B)=\\frac{5}{8} \\times \\frac{5}{8}=\\frac{25}{64}\n",
|
||||
"$$\n",
|
||||
"因而\n",
|
||||
"$$\n",
|
||||
"P(B \\mid A)=\\frac{P(A B)}{P(A)}=\\frac{5}{8}\n",
|
||||
"$$\n",
|
||||
"因此， $P(B \\mid A)=P(B),$ 事实上还可以算出 $P(B \\mid \\bar{A})=P(B)$ 。这表明不论 $A$ 发是不发生，都对 $B$ 发生的概率没有影响。即B 与A独立. \n",
|
||||
"\n",
|
||||
"性质： \n",
|
||||
"事件 $A$ 和事件 $B$ 相互独立的充分必要条件是 \n",
|
||||
"$$\n",
|
||||
"P(A B)=P(A) P(B)\n",
|
||||
"$$"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"# 2.全概率公式和贝叶斯公式"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## 2.1 全概率公式\n",
|
||||
"定义： \n",
|
||||
"如果事件 $A_{1}, A_{2}, \\cdots, A_{n}$ 是一个完备事件组，并且都具有正概率，则有： \n",
|
||||
"$$\n",
|
||||
"\\begin{array}{c}\n",
|
||||
"P(B)=P\\left(A_{1}\\right) P\\left(B \\mid A_{1}\\right)+P\\left(A_{2}\\right) P\\left(B \\mid A_{2}\\right)+\\cdots+P\\left(A_{n}\\right) P\\left(B \\mid A_{n}\\right) \\\\\n",
|
||||
"=\\sum_{i=1}^{n} P\\left(A_{i}\\right) P\\left(B \\mid A_{i}\\right)\n",
|
||||
"\\end{array}\n",
|
||||
"$$ \n",
|
||||
"对于任何事件 $B,$ 事件 $A \\bar{A}$ 构成最简单的完备事件组，根据全概率公式得 \n",
|
||||
"$$\n",
|
||||
"\\begin{aligned}\n",
|
||||
"P(B) &=P(A B+\\bar{A} B)=P(A B)+P(\\bar{A} B) \\\\\n",
|
||||
"&=P(A) P(B \\mid A)+P(\\bar{A}) P(B \\mid \\bar{A})\n",
|
||||
"\\end{aligned}\n",
|
||||
"$$ \n",
|
||||
"\n",
|
||||
"## 2.2 贝叶斯公式 \n",
|
||||
"定义： \n",
|
||||
"设 $A_{1}, A_{2}, \\cdots, A_{n}$ 是一完备事件组，则对任一事件 $B, P(B)>0,$ 有 \n",
|
||||
"$$\n",
|
||||
"P\\left(A_{i} \\mid B\\right)=\\frac{P\\left(A_{i} B\\right)}{P(B)}=\\frac{P\\left(A_{i}\\right) P\\left(B \\mid A_{i}\\right)}{\\sum_{i=1}^{n} P\\left(A_{i}\\right) P\\left(B \\mid A_{i}\\right)}\n",
|
||||
"$$ \n",
|
||||
"以上公式就叫贝叶斯公式，可由条件概率的定义及全概率公式证得。 \n",
|
||||
"\n",
|
||||
"例子: \n",
|
||||
"市场上供应的某种商品只由甲、乙、丙3个厂生产，甲厂占45%，乙厂占35%，丙厂占20%。如果各厂的次品率依次为 4%, 2%, 5% 。现从市场上购买1件这种商品，发现是次品，试判断它是由甲厂生产的概率。 \n",
|
||||
"设事件 $A_{1}, A_{2}, A_{3},$ 分别表示“商品为甲、乙、丙厂生产的\"，事件 $B$ 表示“商品为次品”，由题意得到概率 \n",
|
||||
"$$\n",
|
||||
"\\begin{array}{c}\n",
|
||||
"P\\left(A_{1}\\right)=45 \\%, P\\left(A_{2}\\right)=35 \\%, P\\left(A_{3}\\right)=20 \\% \\\\\n",
|
||||
"P\\left(B \\mid A_{1}\\right)=4 \\%, P\\left(B \\mid A_{2}\\right)=2 \\%, P\\left(B \\mid A_{3}\\right)=5 \\%\n",
|
||||
"\\end{array}\n",
|
||||
"$$ \n",
|
||||
"根据贝叶斯公式，可得:\n",
|
||||
"$$\n",
|
||||
"\\begin{aligned}\n",
|
||||
"P\\left(A_{1} \\mid B\\right) &=\\frac{P\\left(A_{1}\\right) P\\left(B \\mid A_{1}\\right)}{P\\left(A_{1}\\right) P\\left(B \\mid A_{1}\\right)+P\\left(A_{2}\\right) P\\left(B \\mid A_{2}\\right)+P\\left(A_{3}\\right) P\\left(B \\mid A_{3}\\right)} \\\\\n",
|
||||
"&=\\frac{45 \\% \\times 4 \\%}{45 \\% \\times 4 \\%+35 \\% \\times 2 \\%+20 \\% \\times 5 \\%} \\approx 0.514\n",
|
||||
"\\end{aligned}\n",
|
||||
"$$\n",
|
||||
"在“购买一件商品”这个试验中， $P\\left(A_{i}\\right)$ 是在试验以前就已经知道的概率，所以习惯地称为先验概率。试验结果出现了次品（即 $B$ 发 生），这时条件概率 $P\\left(A_{i} \\mid B\\right)$ 反映了在试验以后对 $B$ 发生的“来源”(即次品的来源）的各种可能性的大小，通常称为后验概率。"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"# 3.随机变量"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## 3.1 随机变量\n",
|
||||
"把试验的结果与实数对应起来，随试验结果的不同而变化的量就是随机变量，包含离散型随机变量和连续型随机变量。\n",
|
||||
"例子： \n",
|
||||
"郑一枚匀称的硬币，观察正面、背面的出现情况。这一试验的样本空间为 $\\Omega=\\{H, T\\}$, 其中，H表示\"正面朝上\", $T$ 表示“背面 朝上”。如果引入变量 $X$ ，对试验的两个结果进行数值化，将 $X$ 的值分别规定为1和0，即\n",
|
||||
"$$\n",
|
||||
"X=\\left\\{\\begin{array}{ll}\n",
|
||||
"1 & \\text { if } \\text { 出现 } H \\\\\n",
|
||||
"0 & \\text { if } & \\text { 出现 } T\n",
|
||||
"\\end{array}\\right.\n",
|
||||
"$$\n",
|
||||
"这里的 $X$ 就叫随机变量，因为它能取的值是离散的，我们就叫它离散型随机变量。"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## 3.2 概率分布\n",
|
||||
"定义：设离散型随机变量 $X$ 的所有可能取值为 $x_{1}, x_{2}, \\cdots, x_{n},$ 称为 $X$ 的概率分布。 \n",
|
||||
"离散型随机变量 $X$ 的分布律具有下列基本性质： \n",
|
||||
"1. $p_{k} \\geq 0, k=1,2, \\cdots$ \n",
|
||||
"2. $\\sum_{i=1}^{+\\infty} p_{k}=1$ \n",
|
||||
"二项分布： \n",
|
||||
"二项分布是一种离散型的概率分布。二项代表它有两种可能的结果：成功或者不成功。每次试验必须相互独立，重复n次，并且每 次试验成功的概率是相同的，为 $p $; 失败的概率也相同，为 $1-p$。掷硬币就是一个典型的二项分布。当我们要计算抛硬币吻，恰巧有 $x$ 次正面朝上的概率，可以使用二项分布的公式: \n",
|
||||
"$$\n",
|
||||
"P\\{X=k\\}=C_{n}^{k} p^{k}(1-p)^{n-k}\n",
|
||||
"$$\n",
|
||||
"\n",
|
||||
" \n",
|
||||
"\n",
|
||||
"泊松分布： \n",
|
||||
"如果随机变量 $X$ 的概率分布为\n",
|
||||
"$$\n",
|
||||
"P\\{X=k\\}=\\frac{\\lambda^{k}}{k !} e^{-\\lambda}, k=0,1,2, \\cdots\n",
|
||||
"$$\n",
|
||||
"式中， $\\lambda>0$ 为常数，则称随机变量 $X$ 服从参数为 $\\lambda$ 的泊松(Possion)分布，记为 $X \\sim P(\\lambda)$. \n",
|
||||
"\n",
|
||||
"\n",
|
||||
"\n"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## 3.3 概率密度函数\n",
|
||||
"定义：若存在非负函数 $f(x)$, 使一个连续型随机变量 $X$ 取值于任一区间 $(a, b]$ 的概率可以表示为\n",
|
||||
"$$\n",
|
||||
"P\\{a<X \\leq b\\}=\\int_{a}^{b} f(x) d_{x}\n",
|
||||
"$$\n",
|
||||
"\n",
|
||||
"正态分布： \n",
|
||||
"正态分布是概率论中最重要的连续型分布，在19世纪前叶由德国数学家高斯（Gauss）加以推广，故又常称为高斯分布。 正态分布的概率密度函数曲线呈钟形，概率密度函数为\n",
|
||||
"$$\n",
|
||||
"f(x)=\\frac{1}{\\sqrt{2 \\pi} \\sigma} e^{-\\frac{(x-\\mu)^{2}}{2 \\sigma^{2}}}\n",
|
||||
"$$\n",
|
||||
"具有两个参数 $\\mu$ 和 $\\sigma^{2}$ 。第一参数 $\\mu$ 是代表服从正态分布的随机变量的均值，第二个参数 $\\sigma^{2}$ 是此随机变量的方差。 \n",
|
||||
"如果一个随机变量 服从均值为 $\\mu,$ 标准差为 $\\sigma$ 的正态分布，数学上记作\n",
|
||||
"$$\n",
|
||||
"X \\sim N\\left(\\mu, \\sigma^{2}\\right)\n",
|
||||
"$$\n",
|
||||
"我们通常所说的标准正态分布均值为0, 标准差为1的正态分布。 \n",
|
||||
"\n",
|
||||
" \n",
|
||||
"\n"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": null,
|
||||
"metadata": {},
|
||||
"outputs": [],
|
||||
"source": []
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## 3.4 随机变量的期望\n",
|
||||
"对于一个随机变量，时常要考虑它的平均取什么，期望就是概率中的平均值，对随机变量中心位置的一种度量。\n",
|
||||
"例子.\n",
|
||||
"经过长期观察积累，某射手在每次射击中命中的坏数 $X$ 服从分布：\n",
|
||||
"$$\n",
|
||||
"\\begin{array}{|c|c|c|c|c|c|c|c|}\n",
|
||||
"\\hline \\mathrm{X} & 0 & 5 & 6 & 7 & 8 & 9 & 10 \\\\\n",
|
||||
"\\hline p_{i} & 0 & 0.05 & 0.05 & 0.1 & 0.1 & 0.2 & 0.5 \\\\\n",
|
||||
"\\hline\n",
|
||||
"\\end{array}\n",
|
||||
"$$\n",
|
||||
"求这个射手平均命中的坏数是多少? \n",
|
||||
"解 \n",
|
||||
"一种很自然的考虑是：假定该射击手进行了100次射击，那么，约有5次命中5坏，5次命中6坏，10次命中7坏，10次命中8坏，20 次命中9坏，50次命10，没有脱革，从而在一次射击中，该射手平均命中的坏数为： \n",
|
||||
"$\\frac{1}{100}(10 \\times 50+9 \\times 20+8 \\times 10+7 \\times 10+6 \\times 5+5 \\times 5+0 \\times 0)=8.85$ \n",
|
||||
"所以，我们可以看到离散型的随机变量的期望值可以表示为： \n",
|
||||
"$E(X)=\\sum_{i=1}^{+\\infty} x_{i} p_{k}$ \n",
|
||||
"\n",
|
||||
"期望的性质： \n",
|
||||
"1. $E(c)=c$ \n",
|
||||
"2. $E(X+c)=E(X)+c$ \n",
|
||||
"3. $E(k X)=k E(X)$ \n",
|
||||
"4. $E(k X+c)=k E(X)+c$ \n",
|
||||
"5. $E(X+Y)=E(X)+E(Y)$ "
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## 3.5 随机变量的方差：\n",
|
||||
"方差表示了随机变量的变异性，方差越大，随机变量的结果越不稳定。 \n",
|
||||
"定义：随机变量，若\n",
|
||||
"$$\n",
|
||||
"E[X-E(X)]^{2}\n",
|
||||
"$$\n",
|
||||
"存在，则称其为 $X$ 的方差，记为 $D(X),$ 即\n",
|
||||
"$$\n",
|
||||
"D(X)=E[X-E(X)]^{2}\n",
|
||||
"$$\n",
|
||||
"而称 $\\sqrt{D(X)}$ 为 $X$ 的标准差或均方差。\n",
|
||||
"由方差的定义和数学期望的性质，可以推出方差的计算公式:\n",
|
||||
"$$\n",
|
||||
"D(X)=E\\left(X^{2}\\right)-[E(X)]^{2}\n",
|
||||
"$$\n",
|
||||
"\n",
|
||||
"方差的性质 \n",
|
||||
"(1) $D(c)=0$ \n",
|
||||
"(2) $D(X+c)=D(X)$ \n",
|
||||
"(3) $D(c X)=c^{2} D(X)$ \n",
|
||||
"\n",
|
||||
"例子： \n",
|
||||
"甲、乙两车间生产同一种产品，设1000件产品中的次品数分别为随机变量 $X, Y,$ 已知他们的分布律如下: \n",
|
||||
"$$\n",
|
||||
"\\begin{array}{ccccc}\n",
|
||||
"\\mathbf{X} & \\mathbf{0} & 1 & \\mathbf{2} & \\mathbf{3} \\\\\n",
|
||||
"\\hline p_{i} & 0.2 & 0.1 & 0.5 & 0.2\n",
|
||||
"\\end{array}\n",
|
||||
"$$\n",
|
||||
"与\n",
|
||||
"$$\n",
|
||||
"\\begin{array}{|c|c|c|c|c|}\n",
|
||||
"\\hline \\mathrm{Y} & 0 & 1 & 2 & 3 \\\\\n",
|
||||
"\\hline p_{i} & 0.1 & 0.3 & 0.4 & 0.2 \\\\\n",
|
||||
"\\hline\n",
|
||||
"\\end{array}\n",
|
||||
"$$\n",
|
||||
"试讨论甲乙两者车厢的质量。 \n",
|
||||
"解 \n",
|
||||
"先计算均值 \n",
|
||||
"$$\n",
|
||||
"\\begin{array}{l}\n",
|
||||
"E(X)=0 \\times 0.2+1 \\times 0.1+2 \\times 0.5+3 \\times 0.2=1.7 \\\\\n",
|
||||
"E(Y)=0 \\times 0.1+1 \\times 0.3+2 \\times 0.4+3 \\times 0.2=1.7\n",
|
||||
"\\end{array}\n",
|
||||
"$$\n",
|
||||
"得到：甲、乙两车间次品数的均值相同。 再计算方差： \n",
|
||||
"$D(X)=(0-1.7)^{2} \\times 0.2+(1-1.7)^{2} \\times 0.1+(2-1.7)^{2} \\times 0.1+(2-1.7)^{2} \\times 0.5+(3-1.7)^{2} \\times 0.2=1.01$\n",
|
||||
"\n",
|
||||
"$D(Y)=(0-1.7)^{2} \\times 0.1+(1-1.7)^{2} \\times 0.3+(2-1.7)^{2} \\times 0.4+(2-1.7)^{2} \\times 0.4+(3-1.7)^{2} \\times 0.2=0.81$\n",
|
||||
"\n",
|
||||
"得到 $D(X)>D(Y),$ 说明乙车间的产品质量较稳定。"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## 随机模拟1：$\\pi$的估值\n",
|
||||
"\n",
|
||||
" "
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 61,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"name": "stdout",
|
||||
"output_type": "stream",
|
||||
"text": [
|
||||
"pi的估计值为 2.8\n",
|
||||
"pi的估计值为 2.8\n",
|
||||
"pi的估计值为 2.96\n",
|
||||
"pi的估计值为 3.016\n",
|
||||
"pi的估计值为 3.156\n",
|
||||
"pi的估计值为 3.156\n",
|
||||
"pi的估计值为 3.1208\n",
|
||||
"pi的估计值为 3.13528\n",
|
||||
"pi的估计值为 3.14612\n",
|
||||
"pi的估计值为 3.137168\n",
|
||||
"pi的估计值为 3.1410684\n"
|
||||
]
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"# pi的估计问题\n",
|
||||
"import numpy as np \n",
|
||||
"def pi_estimate(n):\n",
|
||||
" '''\n",
|
||||
" n为投点的数量\n",
|
||||
" '''\n",
|
||||
" n_rand_X = np.random.uniform(-1.0,1.0,n)\n",
|
||||
" n_rand_Y = np.random.uniform(-1.0,1.0,n)\n",
|
||||
" ## 判断是否在圆内\n",
|
||||
" distance = np.sqrt(n_rand_X**2 + n_rand_Y**2)\n",
|
||||
" dis_n = float(len(distance[distance<=1.0]))\n",
|
||||
" return 4 * (dis_n / n)\n",
|
||||
"\n",
|
||||
"for i in [10,50,100,500,1000,5000,10000,50000,100000,500000,10000000]:\n",
|
||||
" print(\"pi的估计值为\",pi_estimate(i))\n"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 41,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"name": "stdout",
|
||||
"output_type": "stream",
|
||||
"text": [
|
||||
"\u001b[1;31mDocstring:\u001b[0m\n",
|
||||
"uniform(low=0.0, high=1.0, size=None)\n",
|
||||
"\n",
|
||||
"Draw samples from a uniform distribution.\n",
|
||||
"\n",
|
||||
"Samples are uniformly distributed over the half-open interval\n",
|
||||
"``[low, high)`` (includes low, but excludes high). In other words,\n",
|
||||
"any value within the given interval is equally likely to be drawn\n",
|
||||
"by `uniform`.\n",
|
||||
"\n",
|
||||
".. note::\n",
|
||||
" New code should use the ``uniform`` method of a ``default_rng()``\n",
|
||||
" instance instead; please see the :ref:`random-quick-start`.\n",
|
||||
"\n",
|
||||
"Parameters\n",
|
||||
"----------\n",
|
||||
"low : float or array_like of floats, optional\n",
|
||||
" Lower boundary of the output interval. All values generated will be\n",
|
||||
" greater than or equal to low. The default value is 0.\n",
|
||||
"high : float or array_like of floats\n",
|
||||
" Upper boundary of the output interval. All values generated will be\n",
|
||||
" less than or equal to high. The default value is 1.0.\n",
|
||||
"size : int or tuple of ints, optional\n",
|
||||
" Output shape. If the given shape is, e.g., ``(m, n, k)``, then\n",
|
||||
" ``m * n * k`` samples are drawn. If size is ``None`` (default),\n",
|
||||
" a single value is returned if ``low`` and ``high`` are both scalars.\n",
|
||||
" Otherwise, ``np.broadcast(low, high).size`` samples are drawn.\n",
|
||||
"\n",
|
||||
"Returns\n",
|
||||
"-------\n",
|
||||
"out : ndarray or scalar\n",
|
||||
" Drawn samples from the parameterized uniform distribution.\n",
|
||||
"\n",
|
||||
"See Also\n",
|
||||
"--------\n",
|
||||
"randint : Discrete uniform distribution, yielding integers.\n",
|
||||
"random_integers : Discrete uniform distribution over the closed\n",
|
||||
" interval ``[low, high]``.\n",
|
||||
"random_sample : Floats uniformly distributed over ``[0, 1)``.\n",
|
||||
"random : Alias for `random_sample`.\n",
|
||||
"rand : Convenience function that accepts dimensions as input, e.g.,\n",
|
||||
" ``rand(2,2)`` would generate a 2-by-2 array of floats,\n",
|
||||
" uniformly distributed over ``[0, 1)``.\n",
|
||||
"Generator.uniform: which should be used for new code.\n",
|
||||
"\n",
|
||||
"Notes\n",
|
||||
"-----\n",
|
||||
"The probability density function of the uniform distribution is\n",
|
||||
"\n",
|
||||
".. math:: p(x) = \\frac{1}{b - a}\n",
|
||||
"\n",
|
||||
"anywhere within the interval ``[a, b)``, and zero elsewhere.\n",
|
||||
"\n",
|
||||
"When ``high`` == ``low``, values of ``low`` will be returned.\n",
|
||||
"If ``high`` < ``low``, the results are officially undefined\n",
|
||||
"and may eventually raise an error, i.e. do not rely on this\n",
|
||||
"function to behave when passed arguments satisfying that\n",
|
||||
"inequality condition. The ``high`` limit may be included in the\n",
|
||||
"returned array of floats due to floating-point rounding in the\n",
|
||||
"equation ``low + (high-low) * random_sample()``. For example:\n",
|
||||
"\n",
|
||||
">>> x = np.float32(5*0.99999999)\n",
|
||||
">>> x\n",
|
||||
"5.0\n",
|
||||
"\n",
|
||||
"\n",
|
||||
"Examples\n",
|
||||
"--------\n",
|
||||
"Draw samples from the distribution:\n",
|
||||
"\n",
|
||||
">>> s = np.random.uniform(-1,0,1000)\n",
|
||||
"\n",
|
||||
"All values are within the given interval:\n",
|
||||
"\n",
|
||||
">>> np.all(s >= -1)\n",
|
||||
"True\n",
|
||||
">>> np.all(s < 0)\n",
|
||||
"True\n",
|
||||
"\n",
|
||||
"Display the histogram of the samples, along with the\n",
|
||||
"probability density function:\n",
|
||||
"\n",
|
||||
">>> import matplotlib.pyplot as plt\n",
|
||||
">>> count, bins, ignored = plt.hist(s, 15, density=True)\n",
|
||||
">>> plt.plot(bins, np.ones_like(bins), linewidth=2, color='r')\n",
|
||||
">>> plt.show()\n",
|
||||
"\u001b[1;31mType:\u001b[0m builtin_function_or_method\n"
|
||||
]
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"?np.random.uniform"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## 随机模拟2:电子元件寿命问题\n",
|
||||
"\n",
|
||||
" \n",
|
||||
" "
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": null,
|
||||
"metadata": {},
|
||||
"outputs": [],
|
||||
"source": []
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 39,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
"text/plain": [
|
||||
"0.9124"
|
||||
]
|
||||
},
|
||||
"execution_count": 39,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"# 电子元件寿命问题\n",
|
||||
"import numpy as np \n",
|
||||
"\n",
|
||||
"def ele_life(n,c,h,t,lamb):\n",
|
||||
" \"\"\"\n",
|
||||
" 参数n:模拟实验的次数\n",
|
||||
" 参数c:每次试验中的c个元件\n",
|
||||
" 参数t:每c个元件中规定的合格品数量\n",
|
||||
" 参数h:小时数\n",
|
||||
" \"\"\"\n",
|
||||
" times = 0.0\n",
|
||||
" for i in range(n):\n",
|
||||
" c_rand = np.random.exponential(1/lamb,c)\n",
|
||||
" c_rand_t = len(c_rand[c_rand>h])\n",
|
||||
" if c_rand_t > t:\n",
|
||||
" times = times + 1\n",
|
||||
" return times / n \n",
|
||||
"\n",
|
||||
"ele_life(10000,1000,18,20,0.2)"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 34,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"name": "stdout",
|
||||
"output_type": "stream",
|
||||
"text": [
|
||||
"\u001b[1;31mDocstring:\u001b[0m\n",
|
||||
"exponential(scale=1.0, size=None)\n",
|
||||
"\n",
|
||||
"Draw samples from an exponential distribution.\n",
|
||||
"\n",
|
||||
"Its probability density function is\n",
|
||||
"\n",
|
||||
".. math:: f(x; \\frac{1}{\\beta}) = \\frac{1}{\\beta} \\exp(-\\frac{x}{\\beta}),\n",
|
||||
"\n",
|
||||
"for ``x > 0`` and 0 elsewhere. :math:`\\beta` is the scale parameter,\n",
|
||||
"which is the inverse of the rate parameter :math:`\\lambda = 1/\\beta`.\n",
|
||||
"The rate parameter is an alternative, widely used parameterization\n",
|
||||
"of the exponential distribution [3]_.\n",
|
||||
"\n",
|
||||
"The exponential distribution is a continuous analogue of the\n",
|
||||
"geometric distribution. It describes many common situations, such as\n",
|
||||
"the size of raindrops measured over many rainstorms [1]_, or the time\n",
|
||||
"between page requests to Wikipedia [2]_.\n",
|
||||
"\n",
|
||||
".. note::\n",
|
||||
" New code should use the ``exponential`` method of a ``default_rng()``\n",
|
||||
" instance instead; please see the :ref:`random-quick-start`.\n",
|
||||
"\n",
|
||||
"Parameters\n",
|
||||
"----------\n",
|
||||
"scale : float or array_like of floats\n",
|
||||
" The scale parameter, :math:`\\beta = 1/\\lambda`. Must be\n",
|
||||
" non-negative.\n",
|
||||
"size : int or tuple of ints, optional\n",
|
||||
" Output shape. If the given shape is, e.g., ``(m, n, k)``, then\n",
|
||||
" ``m * n * k`` samples are drawn. If size is ``None`` (default),\n",
|
||||
" a single value is returned if ``scale`` is a scalar. Otherwise,\n",
|
||||
" ``np.array(scale).size`` samples are drawn.\n",
|
||||
"\n",
|
||||
"Returns\n",
|
||||
"-------\n",
|
||||
"out : ndarray or scalar\n",
|
||||
" Drawn samples from the parameterized exponential distribution.\n",
|
||||
"\n",
|
||||
"See Also\n",
|
||||
"--------\n",
|
||||
"Generator.exponential: which should be used for new code.\n",
|
||||
"\n",
|
||||
"References\n",
|
||||
"----------\n",
|
||||
".. [1] Peyton Z. Peebles Jr., \"Probability, Random Variables and\n",
|
||||
" Random Signal Principles\", 4th ed, 2001, p. 57.\n",
|
||||
".. [2] Wikipedia, \"Poisson process\",\n",
|
||||
" https://en.wikipedia.org/wiki/Poisson_process\n",
|
||||
".. [3] Wikipedia, \"Exponential distribution\",\n",
|
||||
" https://en.wikipedia.org/wiki/Exponential_distribution\n",
|
||||
"\u001b[1;31mType:\u001b[0m builtin_function_or_method\n"
|
||||
]
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"?np.random.exponential"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"## 随机模拟3：三门问题\n",
|
||||
"蒙提霍尔问题：假如你参与一个有主持人的游戏，你会看见三扇关闭了的门，其中一扇的后面有一辆汽车，另外2扇门后面各是一只山羊，你看不见门后面的情况，但主持人知道一切。你被主持人要求在三扇门中选择一扇，但不能打开，在你选定之后主持人开启了另一扇后面有山羊的门，然后你可以坚持原来选定的门，也可以改主意重新选择。问题是：改与不改对选中汽车的概率有影响吗？请使用模拟实验的方法回答该问题。"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 64,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"name": "stdout",
|
||||
"output_type": "stream",
|
||||
"text": [
|
||||
"在100000次测试中,坚持原则第一次就选中的次数是33063,改变决定选择另一扇门中奖的次数是66937\n",
|
||||
"概率分别是0.33063和0.66937，改变决定选择另一扇门中奖几率是坚持选择的2.02倍\n"
|
||||
]
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"# 三门问题\n",
|
||||
"import numpy.random as random\n",
|
||||
"def MontyHallProblem(n_test):\n",
|
||||
" #测试次数\n",
|
||||
" \n",
|
||||
" winning_door = random.randint(0,3,n_test)\n",
|
||||
" first_get = 0\n",
|
||||
" change_get = 0\n",
|
||||
" for winning_doors in winning_door:\n",
|
||||
" act_door = random.randint(0,3)\n",
|
||||
" if winning_doors == act_door:\n",
|
||||
" first_get += 1\n",
|
||||
" else :\n",
|
||||
" change_get += 1\n",
|
||||
" first_pro = first_get / n_test\n",
|
||||
" change_pro = change_get / n_test\n",
|
||||
" compar1 = round(change_get / first_get,2)\n",
|
||||
" print (\"在%d次测试中,坚持原则第一次就选中的次数是%d,改变决定选择另一扇门中奖的次数是%d\"% (n_test,first_get,change_get))\n",
|
||||
" print (\"概率分别是{0}和{1}，改变决定选择另一扇门中奖几率是坚持选择的{2}倍\".format(first_pro,change_pro,compar1))\n",
|
||||
"MontyHallProblem(100000)"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": null,
|
||||
"metadata": {},
|
||||
"outputs": [],
|
||||
"source": []
|
||||
}
|
||||
],
|
||||
"metadata": {
|
||||
"kernelspec": {
|
||||
"display_name": "Python 3",
|
||||
"language": "python",
|
||||
"name": "python3"
|
||||
},
|
||||
"language_info": {
|
||||
"codemirror_mode": {
|
||||
"name": "ipython",
|
||||
"version": 3
|
||||
},
|
||||
"file_extension": ".py",
|
||||
"mimetype": "text/x-python",
|
||||
"name": "python",
|
||||
"nbconvert_exporter": "python",
|
||||
"pygments_lexer": "ipython3",
|
||||
"version": "3.9.2"
|
||||
}
|
||||
},
|
||||
"nbformat": 4,
|
||||
"nbformat_minor": 4
|
||||
}
|
||||
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