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CS50ai: week2 Uncertainty我的笔记A版 - 实践

news 2026/6/15 6:30:47

该篇文章为随手笔记，如想要阅读更系统与细致的笔记可阅读B版

ω：概率中的一个世界，即一种可能的事件状态组合

ω发生的概率为：0 <= P(ω) <=1
Σ(ω∈Ω)P(ω) = 1 所有世界发生的概率为1

unconditional probability

degree of belief in a proposition in the absence of any other evidence

conditional probability

degree of belief in a proposition given some evidence that has already been revealed

符号：P(a|b) a given b 在b已知基础上a的概率
Eg. 如P(disease | test results)，基于实践结果的得病概率
P(a|b) =Р(а ^ b)/P(b)我们可以忽略b概率不存在的情况，他是基础。也可以写作Р(а ^ b)=P(b)P(a|b)=P(a)P(b|a)

random variable

a variable in probability theory with a domain of possible values it can take on

Eg. Weather {sun, cloud, rain, wind, snow} 后面这些就是对于Weather可以选择的变量
或者 Flight {on time, delayed, cancelled}则又有

probability distribution概率分布

P(Flight = on time) = 0.6
P(Flight = delayed) = 0.3
P(Flight = cancelled) = 0.1

另一种形式：（P要粗体）P(Flight) = <0.6, 0.3, 0.1>（概率的顺序很重要，要和上面的三个状态顺序对应哦）

probability theory

independence

the knowledge that one event occurs does not affect the probability of the other event

并不绝对，支持的例子如蓝红骰子，反对的如雨和云，根据实际分析

对于P(a ۸ b) = P(a)P(b|a)因为独立，所以a发生和a下b发生相乘得到a^b，因此是相互独立的
不独立的情况：投骰子如果为6则不可能为4，不成立

Bayes' Rule

贝叶斯规则，公式由上面的conditional probability转化而来

$$P(b|a)=\frac {P(a|b)P(b)}{P(a)}$$

Eg.

Given clouds in the morning,what's the probability of rain in the afternoon?
80% of rainy afternoons start with cloudy mornings.
40% of days have cloudy mornings.
10% of days have rainy afternoons.

$$P(rain | clouds) = \frac{P(clouds | rain)P(rain)}{P(clouds)$$

则我们知道

Knowing

$$P(\text{visible effect} | \text{unknown cause}$$

we can calculate

$$P(\text{unknown cause} | \text{visible effect}$$

Joint Probability

联合概率

通过此时能够看到，雨和云并非是独立的，因此当两个同时发生可能看到概率并非是两者相乘

因此$$P(\text{C} | \text{rain})$$可得公式：

$$P(\text{C} | \text{rain}) = \frac{P(\text{C}, \text{rain})}{P(\text{rain})} = \alpha P(\text{C}, \text{rain}) = \alpha\langle0.08, 0.02\rangle = \langle0.8, 0.2\rangl$$

(C, rain)表示C和rain一起发生
乘某个常数让其总值为1

Probability Rules

Negation

很简单的公式

Inclusion-Exclusion

减去求和的重复项

Marginalization

合并两个式子

也能够写为，不一定只合并两个

Conditioning

P(a)=P(a∣b)P(b)+P(a∣¬b)P(¬b)

等效概率，a概率等于b发生和不发生时a发生的概率

Bayesian network

data structure that represents the dependencies among random variables

directed graph
each node represents a random variable
each node X has probability distribution$$\mathbf{P}(X \mid \text{Parents}(X))$$

Eg. 下面是四个节点。是否能赴约往上取决

如下图，下雨程度影响了是否需要维护的概率，其他的节点之间也是一样的关系

因此我们可以得到如下式子：

P(Appointment | light, no) = α * P(Appointment, light, no)

P(Appointment, light, no) = P(Appointment, light, no, on time) + P(Appointment, light, no, delayed)

即——Inference by Enumeration

X is the query variable.
e is the evidence.
y ranges over values of hidden variables.
$$\alph$$ normalizes the result.

在python的实现

使用pomegranate库，让我们能自己创建节点，将分布的概率填入等等

Sampling采样(一种Approximate Inference)

根据最初项的概率分布随机选一个

后面继续列概率随机选一项继续列

......

得到贝叶斯可能的取值，如果重复进行好多次，那么会生成各种样本

Eg. 我们假设了10000个样本，让他们寻路，符合条件sample["train"] == "delayed"的记录下来：

# Rejection sampling
# Compute distribution of Appointment given that train is delayed
N = 10000
data = []
for i in range(N):sample = generate_sample()if sample["train"] == "delayed":data.append(sample["appointment"])
print(Counter(data))

Rejection Sampling

根据特定条件选取，丢弃样本。但对于一些很苛刻的条件效率很低，因此我们有下面新的采样方法

Likelihood Weighting

Start by fixing the values for evidence variables.
Sample the non-evidence variables using conditional probabilities in the Bayesian Network.
Weight each sample by its likelihood: the probability of all of the evidence.

此时对于上面那个例子来说，train delayed是一个固定量，不计入加权。

对于变量的加权：