概率论快速学习04：概率公理全概率贝叶斯事件独立性

文章由LinuxBoy分享于2019-03-27 04:03:19热评（53）

概率论快速学习04：概率公理全概率贝叶斯事件独立性

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Written In The Font

数学和生活是技术之本, 有了数学,加上生活,才会开心.

今天继续概率论:

全概率
贝叶斯
事件独立性

Content

The total probability

In the Set :

The law of total probability is the proposition that if $\left\{{B_n : n = 1, 2, 3, \ldots}\right\}$ is a finite or countably infinitepartition of a sample space (in other words, a set of pairwise disjoint events whose union is the entire sample space) and each event $B_n$ is measurable, then for any event $A$ of the same probability space:

$\Pr(A)=\sum_n \Pr(A\mid B_n)\Pr(B_n),\,$

example:

例. 甲、乙两家工厂生产某型号车床，其中次品率分别为20%, 5%。已知每月甲厂生产的数量是乙厂的两倍，现从一个月的产品中任意抽检一件，求该件产品为合格的概率？

设A表示产品合格，B表示产品来自甲厂

Bayes

for some partition {B_j} of the event space, the event space is given or conceptualized in terms of P(B_j) and P(A|B_j). It is then useful to compute P(A) using the law of total probability:

example:

An entomologist spots what might be a rare subspecies of beetle, due to the pattern on its back. In the rare subspecies, 98% have the pattern, or P(Pattern|Rare) = 98%. In the common subspecies, 5% have the pattern. The rare subspecies accounts for only 0.1% of the population. How likely is the beetle having the pattern to be rare, or what is P(Rare|Pattern)?

From the extended form of Bayes' theorem (since any beetle can be only rare or common),

$\begin{align}P(\text{Rare}|\text{Pattern}) &=\frac{P(\text{Pattern}|\text{Rare})P(\text{Rare})} {P(\text{Pattern}|\text{Rare})P(\text{Rare}) \, + \, P(\text{Pattern}|\text{Common})P(\text{Common})} \\[8pt]&= \frac{0.98 \times 0.001} {0.98 \times 0.001 + 0.05 \times 0.999} \\[8pt]&\approx 1.9\%. \end{align}$

One more example:

Independence

Two events

Two events A and B are independent if and only if their joint probability equals the product of their probabilities:

$\mathrm{P}(A \cap B) = \mathrm{P}(A)\mathrm{P}(B)$ .

Why this defines independence is made clear by rewriting with conditional probabilities:

$\begin{align}\mathrm{P}(A \cap B) = \mathrm{P}(A)\mathrm{P}(B) &\Leftrightarrow \mathrm{P}(A) = \frac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)} \\&\Leftrightarrow \mathrm{P}(A) = \mathrm{P}(A\mid B)\end{align}$

how about Three events

sometimes , we will see the Opposition that can be used to make the mess done. We will use the rule of independence such as : $P(A^c)=1-P(A)\,$

Editor's Note

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概率论快速学习04：概率公理全概率贝叶斯事件独立性