2026-08-06-Preview

2026-08-06-Preview · Probability Theory Ch3 Conditional Probability and Independence（条件概率与独立性）

Ref: Lecture slides Ch3 · Ross Ch3 Conditional Probability and Independence

Key Concepts

Term	Definition
Conditional probability（条件概率）	$P(E\Vert F)=\frac{P(E\cap F)}{P(F)}$ ($P(F)>0$); once F occurs, F is the new sample space
Multiplication rule（乘法公式）	$P(E_1\cap \cdots \cap E_n)=P(E_1)P(E_2\Vert E_1)\cdots P(E_n\Vert E_1\cap \cdots \cap E_{n-1})$
Total probability（全概率公式）	If $F_i$ partition $S$: $P(E)={\sum }_{i}P(E\Vert F_i)P(F_i)$
Bayes’ formula（贝叶斯公式）	$P(F_i\Vert E)=\frac{P(E\Vert F_i)P(F_i)}{{\sum }_{j}P(E\Vert F_j)P(F_j)}$
Independence（独立性）	$P(E\cap F)=P(E)P(F)$, equivalently $P(E\Vert F)=P(E)$
Conditional independence（条件独立性）	$P(T_{i_1}\cap \cdots \cap T_{i_r}\Vert E)={\prod }_{j}P(T_{i_j}\Vert E)$, and same given $E^c$

Per-Subsection: Definition & Examples

3.2 Conditional Probability（条件概率）

Definition

If $P(F)>0$, the conditional probability of $E$ given $F$ is

$P(E|F)=\frac{P(E\cap F)}{P(F)}$

Once $F$ has occurred, $F$ is the new sample space: the occurrence must be in $E\cap F$.

Multiplication rule

$P(E_1\cap E_2\cap \cdots \cap E_n)=P(E_1)P(E_2|E_1)P(E_3|E_1\cap E_2)\cdots P(E_n|E_1\cap \cdots \cap E_{n-1})$

Proof (multiplication rule)

• Apply $P(A|B)=P(A\cap B)/P(B)$ repeatedly with $A=E_1\cap \cdots \cap E_i$, $B=E_1\cap \cdots \cap E_{i-1}$;
• telescoping（连乘相消）: each step’s numerator is the previous step’s denominator, so only $P(E_1)$ remains in the denominator.

Example: Smith family（史密斯家庭）— 2 children, 4 configurations equally likely; probability of two girls?

• No information: $P(\{G,G\})=\frac{1}{4}$;
• Elder child is a girl: $F=\{(G,G),(G,B)\}$, $P(\{G,G\}|F)=\frac{1/4}{1/2}=\frac{1}{2}$;
• At least one child is a girl: $F=\{(G,G),(G,B),(B,G)\}$, $P(\{G,G\}|F)=\frac{1/4}{3/4}=\frac{1}{3}$.
• Key idea: conditioning changes the denominator; “one is a girl”（信息更弱）≠ “the elder is a girl”.

3.3 Bayes’ Formula（贝叶斯公式）

Definition

If $F_1,\dots ,F_n$ partition $S$（两两不交且并为 $S$）:

Total probability: $P(E)={\sum }_{i=1}^{n}P(E|F_i)P(F_i)$

Bayes’ formula:

$P(F_i|E)=\frac{P(E|F_i)P(F_i)}{{\sum }_{j=1}^{n}P(E|F_j)P(F_j)}$

Example: breast cancer screening（乳腺癌筛查）— prevalence $P(A)=0.01$, sensitivity $P(B|A)=0.85$, false-positive rate $P(B|A^c)=0.10$.

• $P(A|B)=\frac{0.85\times 0.01}{0.85\times 0.01+0.10\times 0.99}\approx 0.08$;
• A positive mammogram means “only” 8% chance of cancer — the low prevalence（先验）dominates;
• Key idea: posterior = prior × likelihood, then normalize（后验 $\propto$ 先验 × 似然）; test accuracy alone is not enough.

3.4 Independence（独立性）

Definition

$E$ and $F$ are independent if $P(E\cap F)=P(E)P(F)$, equivalently $P(E|F)=P(E)$ and $P(F|E)=P(F)$.

If $E,F$ are independent, then $E$ and $F^c$ are also independent.

Three events $E,F,G$ are independent only if the triple product AND all three pairwise products hold.

Example: two dice（掷两枚骰子）— $E=\{$sum is $8\}$, $F=\{$first die is $4\}$.

• $P(E)P(F)=\frac{5}{36}\times \frac{1}{6}=\frac{5}{126}$, but $P(E\cap F)=P(\{4,4\})=\frac{1}{36}$;
• Not independent.
• Key idea: independent ≠ mutually exclusive（互斥且都有正概率时不可能独立：$P(E\cap F)=0\ne P(E)P(F)>0$）.

3.5 Conditional Independence（条件独立性）

Definition

Given $E$ (and separately given $E^c$), events $T_1,\dots ,T_n$ are conditionally independent if for any subset $i_1,\dots ,i_r$:

$P(T_{i_1}\cap \cdots \cap T_{i_r}|E)={\prod }_{j=1}^{r}P(T_{i_j}|E)$

Example: sequential HIV tests（HIV 确诊复查）— sensitivity $P(T+|HIV+)=0.99$, specificity $P(T-|HIV-)=0.99$, prevalence $P(HIV+)=0.001$.

• After one positive: $P(HIV+|T_1^{+})=\frac{0.99\times 0.001}{0.99\times 0.001+0.01\times 0.999}\approx 0.0902$;
• After two positives (conditional independence assumed): $P(HIV+|T_1^{+}\cap T_2^{+})=\frac{0.99^2\times 0.001}{0.99^2\times 0.001+0.01^2\times 0.999}\approx 0.9075$;
• Key idea: one positive test is not conclusive, two independent positives raise 9% → 91% — this is the math behind “positive results need confirmation”; sequential updating lets $P(A|I_k)$ be computed from $P(A|I_{k-1})$（逐次更新）.