updated ch13-14 report
This commit is contained in:
6702
bayesian/main.pdf
Normal file
6702
bayesian/main.pdf
Normal file
File diff suppressed because one or more lines are too long
289
bayesian/main.typ
Normal file
289
bayesian/main.typ
Normal file
@ -0,0 +1,289 @@
|
||||
#import "@preview/cetz:0.3.3"
|
||||
|
||||
#set text(font: ("Noto Sans CJK SC", "Noto Serif CJK SC"), lang: "zh")
|
||||
#set page(paper: "a4", margin: (x: 2cm, y: 2cm))
|
||||
#set heading(numbering: "1.")
|
||||
|
||||
#let answer(body) = block(
|
||||
fill: luma(240),
|
||||
stroke: (left: 2pt + blue),
|
||||
inset: 10pt,
|
||||
radius: 2pt,
|
||||
width: 100%,
|
||||
body
|
||||
)
|
||||
|
||||
#align(center)[
|
||||
#text(size: 18pt, weight: "bold")[第13-14章 概率与贝叶斯网络 习题解答]
|
||||
|
||||
#v(1em)
|
||||
// *生成时间:* #datetime.today().display()
|
||||
]
|
||||
|
||||
= 题目 1
|
||||
*题意：* 已知 $P(a) = 0.3$, $P(b | a) = 0.2$, $P(c | a) = 0.5$。比较 $P(a and b)$ 与 $P(a and c)$。
|
||||
|
||||
#answer[
|
||||
根据乘法公式 $P(x and y) = P(x)P(y|x)$：
|
||||
$ P(a and b) &= P(a) dot P(b | a) = 0.3 times 0.2 = 0.06 \
|
||||
P(a and c) &= P(a) dot P(c | a) = 0.3 times 0.5 = 0.15 $
|
||||
因为 $0.06 < 0.15$，所以 *$P(a and b) < P(a and c)$*。
|
||||
|
||||
*答案：B*
|
||||
]
|
||||
|
||||
= 题目 2
|
||||
*题意：* $P(a or b)=0.7$, $P(a)=0.4$, $P(b)=0.5$，求 $P(a and b)$。
|
||||
|
||||
#answer[
|
||||
根据概率加法公式（容斥原理）：
|
||||
$ P(a or b) = P(a) + P(b) - P(a and b) $
|
||||
代入已知数值：
|
||||
$ 0.7 = 0.4 + 0.5 - P(a and b) \
|
||||
P(a and b) = 0.9 - 0.7 = 0.2 $
|
||||
|
||||
*答案：0.2*
|
||||
]
|
||||
|
||||
= 题目 3
|
||||
*数据：* 全联合概率分布表如下（Cavity, Toothache, Catch）：
|
||||
|
||||
#table(
|
||||
columns: (auto, auto, auto, auto, auto),
|
||||
inset: 8pt,
|
||||
align: center,
|
||||
[], [T, Catch], [T, $not$ Catch], [$not$ T, Catch], [$not$ T, $not$ Catch],
|
||||
[Cavity], [0.108], [0.012], [0.072], [0.008],
|
||||
[$not$ Cavity], [0.016], [0.064], [0.144], [0.576]
|
||||
)
|
||||
|
||||
#answer[
|
||||
*1. 计算 $P("Toothache" or not "Cavity")$*
|
||||
|
||||
利用补集思想：$P(A) = 1 - P(not A)$。
|
||||
事件的补集是 $not "Toothache" and "Cavity"$。
|
||||
对应表格中的项为：(Cavity, $not$ T, Catch) 和 (Cavity, $not$ T, $not$ Catch)。
|
||||
$ P(not T and "Cav") = 0.072 + 0.008 = 0.080 $
|
||||
$ P(T or not "Cav") = 1 - 0.080 = 0.92 $
|
||||
|
||||
*2. 计算 $P(not "toothache" or "catch")$*
|
||||
|
||||
补集是 $"toothache" and not "catch"$。
|
||||
对应表格列为 (T, $not$ Catch)，即第 2 列。
|
||||
$ P(T and not "Catch") = 0.012 ("Cav") + 0.064 (not "Cav") = 0.076 $
|
||||
$ P(not T or "Catch") = 1 - 0.076 = 0.924 $
|
||||
|
||||
*3. 计算 $P("Toothache" or not "Cavity" | not "catch")$ 和 $P(not "catch" | "Toothache" or not "Cavity")$*
|
||||
|
||||
令 $A = "Toothache" or not "Cavity"$, $B = not "catch"$。
|
||||
|
||||
- *计算 $P(B)$*: Sum of all $not$ Catch columns (Col 2, Col 4).
|
||||
$P(B) = (0.012 + 0.064) + (0.008 + 0.576) = 0.076 + 0.584 = 0.66$
|
||||
- *计算 $P(A and B)$*: 在 $not$ Catch 列中满足 $T or not "Cav"$ 的项。
|
||||
- (Cav, T, $not$ C): 0.012 (满足 T)
|
||||
- ($not$ Cav, T, $not$ C): 0.064 (满足 T)
|
||||
- ($not$ Cav, $not$ T, $not$ C): 0.576 (满足 $not$ Cav)
|
||||
- (Cav, $not$ T, $not$ C): 0.008 (不满足，既无T也无$not$Cav)
|
||||
Sum $= 0.012 + 0.064 + 0.576 = 0.652$
|
||||
|
||||
结果 1: $P(A | B) = 0.652 / 0.66 approx 0.9879$
|
||||
结果 2: $P(B | A) = P(A and B) / P(A) = 0.652 / 0.92 approx 0.7087$ (其中 P(A) 来自第1小问)
|
||||
|
||||
*4. 判断 Cavity 与 Toothache 是否独立*
|
||||
|
||||
$ P("Cav") &= 0.108+0.012+0.072+0.008 = 0.2 \
|
||||
P("Toothache") &= 0.108+0.012+0.016+0.064 = 0.2 \
|
||||
P("Cav" and "T") &= 0.108 + 0.012 = 0.12 $
|
||||
检验：$P("Cav") times P("T") = 0.2 times 0.2 = 0.04$。
|
||||
因为 $0.12 != 0.04$，所以两者 *不独立*。
|
||||
]
|
||||
|
||||
= 题目 4
|
||||
*题意：* 设 $X, Y$ 条件独立于 $Z$（即 $X perp Y | Z$），$Y, W$ 条件独立于 $Z$（即 $Y perp W | Z$）。问 $X, W$ 是否条件独立于 $Z$？
|
||||
|
||||
#answer[
|
||||
*结论：不能。* 条件独立性不具备传递性。
|
||||
|
||||
*反例证明：*
|
||||
假设 $Z$ 是一个公平的硬币投掷（0 或 1）。
|
||||
令 $X = Z$（$X$ 完全依赖于 $Z$）。
|
||||
令 $W = Z$（$W$ 完全依赖于 $Z$）。
|
||||
令 $Y$ 是一个与 $X, W, Z$ 都完全独立的随机变量（例如掷骰子）。
|
||||
|
||||
1. *检查 $X, Y$ 关于 $Z$ 的独立性*：给定 $Z$， $X$ 变为常数。常数与任何变量独立，故成立。
|
||||
2. *检查 $Y, W$ 关于 $Z$ 的独立性*：同理，给定 $Z$，$W$ 变为常数，故成立。
|
||||
3. *检查 $X, W$ 关于 $Z$ 的独立性*：
|
||||
给定 $Z$，$X$ 和 $W$ 的值完全确定且相同（例如若 $Z=1$，则 $X=1, W=1$）。虽然在 $Z$ 固定的情况下它们的方差为0（技术上可视作独立），但如果我们考虑一种因果结构：
|
||||
令 $X$ 和 $W$ 为同一变量的两个副本。显然它们是强相关的。
|
||||
|
||||
更直观的例子：$X$ 和 $W$ 互为因果或由共同隐变量控制，而它们都与 $Y$ 独立。仅仅知道它们分别与 $Y$ 独立，无法切断 $X$ 和 $W$ 之间的联系。
|
||||
]
|
||||
|
||||
= 题目 5
|
||||
*题意：* 罕见病检测。$P(D)=0.001$。
|
||||
检测A：$P(+|D)=0.95, P(+|not D)=0.05$。
|
||||
检测B：$P(+|D)=0.90, P(+|not D)=0.10$。
|
||||
|
||||
#answer[
|
||||
*a. 单次检测 A 为阳性，求 $P(D|A+)$*
|
||||
|
||||
使用贝叶斯公式：
|
||||
$ P(D|A+) = (P(A+|D)P(D)) / P(A+) $
|
||||
$ P(A+) &= P(A+|D)P(D) + P(A+|not D)P(not D) \
|
||||
&= 0.95 times 0.001 + 0.05 times 0.999 \
|
||||
&= 0.00095 + 0.04995 = 0.0509 $
|
||||
$ P(D|A+) &= 0.00095 / 0.0509 approx 0.0187 (1.87%) $
|
||||
|
||||
*b. 两次检测 A 和 B 均为阳性，求 $P(D|A+, B+)$*
|
||||
|
||||
由于检测独立（Naive Bayes 假设）：
|
||||
$ P(A+, B+ | D) = 0.95 times 0.90 = 0.855 $
|
||||
$ P(A+, B+ | not D) = 0.05 times 0.10 = 0.005 $
|
||||
|
||||
$ P(D | A+, B+) &= (P(A+, B+ | D)P(D)) / (P(A+, B+ | D)P(D) + P(A+, B+ | not D)P(not D)) \
|
||||
&= (0.855 times 0.001) / (0.855 times 0.001 + 0.005 times 0.999) \
|
||||
&= 0.000855 / (0.000855 + 0.004995) \
|
||||
&= 0.000855 / 0.00585 approx 0.1462 (14.62%) $
|
||||
|
||||
*c. 解释*
|
||||
|
||||
联合检测通过引入第二次独立测试，极大地降低了*假阳性率*（从 0.05 降至 $0.05 times 0.10 = 0.005$）。虽然真阳性率也略有下降，但分母中占主导地位的假阳性项（由 $P(not D)$ 权重放大）被大幅削减，从而显著提升了后验概率。
|
||||
]
|
||||
|
||||
= 题目 6
|
||||
*场景：* 骑车(B) 取决于 天气(W) 和 熬夜(S)。
|
||||
$P(W="Sun")=0.7, P(S="Yes")=0.4$。
|
||||
CPT 已知。
|
||||
|
||||
#answer[
|
||||
*a. 贝叶斯网络与 CPT*
|
||||
|
||||
结构：$W -> B <- S$ (V-structure)
|
||||
|
||||
#cetz.canvas({
|
||||
import cetz.draw: *
|
||||
|
||||
let r = 0.8
|
||||
// Nodes
|
||||
circle((0, 0), radius: r, name: "B")
|
||||
content("B", "B (骑车)")
|
||||
|
||||
circle((-2, 3), radius: r, name: "W")
|
||||
content("W", "W (天气)")
|
||||
|
||||
circle((2, 3), radius: r, name: "S")
|
||||
content("S", "S (熬夜)")
|
||||
|
||||
// Edges
|
||||
line("W", "B", mark: (end: ">"))
|
||||
line("S", "B", mark: (end: ">"))
|
||||
})
|
||||
|
||||
*条件概率表 (CPT) for B:*
|
||||
#table(
|
||||
columns: 3,
|
||||
align: center,
|
||||
[W (天气)], [S (熬夜)], [P(B=是 | W, S)],
|
||||
[晴 (Sun)], [否 (No)], [0.9],
|
||||
[晴 (Sun)], [是 (Yes)], [0.6],
|
||||
[雨 (Rain)], [否 (No)], [0.2],
|
||||
[雨 (Rain)], [是 (Yes)], [0.1]
|
||||
)
|
||||
|
||||
*b. 推理：已知骑车(B=Yes)，求天气概率*
|
||||
|
||||
我们需要计算 $P(W | B="Yes")$。根据贝叶斯法则：$P(W|B) = P(B|W)P(W) / P(B)$。
|
||||
|
||||
首先计算边缘概率 $P(B="Yes")$。
|
||||
$ P(B=y) = sum_(w, s) P(B=y|w,s)P(w)P(s) $
|
||||
因 W 和 S 独立，联合概率直接相乘。
|
||||
|
||||
1. $W="Sun", S="No"$: $0.9 times 0.7 times 0.6 = 0.378$
|
||||
2. $W="Sun", S="Yes"$: $0.6 times 0.7 times 0.4 = 0.168$
|
||||
3. $W="Rain", S="No"$: $0.2 times 0.3 times 0.6 = 0.036$
|
||||
4. $W="Rain", S="Yes"$: $0.1 times 0.3 times 0.4 = 0.012$
|
||||
|
||||
$ P(B="Yes") = 0.378 + 0.168 + 0.036 + 0.012 = 0.594 $
|
||||
|
||||
计算 $P(W="Sun" | B="Yes")$:
|
||||
分子为 W=Sun 的所有情况之和（项1 + 项2）：
|
||||
$ P(B=y, W="Sun") = 0.378 + 0.168 = 0.546 $
|
||||
$ P(W="Sun" | B="Yes") = 0.546 / 0.594 approx 0.919 $
|
||||
|
||||
同理，$P(W="Rain" | B="Yes") = (0.036 + 0.012) / 0.594 approx 0.081$。
|
||||
|
||||
*结论：这天最可能是晴天。*
|
||||
]
|
||||
|
||||
= 题目 7
|
||||
*智能健康助手：* 变量 S(吸烟), G(基因), L(肺癌), C(咳嗽), X(胸片)。
|
||||
关系：$S->L, G->L, S->C, L->C, L->X$。
|
||||
|
||||
#answer[
|
||||
*问题 1：贝叶斯网络图*
|
||||
|
||||
#align(center)[
|
||||
#cetz.canvas({
|
||||
import cetz.draw: *
|
||||
// Manually positioning nodes for clarity
|
||||
let r = 0.5
|
||||
circle((0, 0), radius: r, name: "L")
|
||||
content("L", "L")
|
||||
|
||||
circle((-2, 2), radius: r, name: "S")
|
||||
content("S", "S")
|
||||
|
||||
circle((2, 2), radius: r, name: "G")
|
||||
content("G", "G")
|
||||
|
||||
circle((-1, -2), radius: r, name: "C")
|
||||
content("C", "C")
|
||||
|
||||
circle((1, -2), radius: r, name: "X")
|
||||
content("X", "X")
|
||||
|
||||
line("S", "L", mark: (end: ">"))
|
||||
line("G", "L", mark: (end: ">"))
|
||||
line("L", "X", mark: (end: ">"))
|
||||
line("L", "C", mark: (end: ">"))
|
||||
|
||||
// S -> C 直接边
|
||||
line("S", "C", mark: (end: ">"))
|
||||
})
|
||||
]
|
||||
|
||||
*问题 2：概率推理 $P(L="Yes" | C="No", X="Yes")$*
|
||||
|
||||
令 $alpha$ 为归一化常数。我们需要计算 $P(L=y, C=n, X=y)$ 和 $P(L=n, C=n, X=y)$。
|
||||
公式分解：$P(S,G,L,C,X) = P(S)P(G)P(L|S,G)P(C|L,S)P(X|L)$。
|
||||
求和消除 S, G：
|
||||
$ P(L, C, X) = P(X|L) sum_S sum_G P(C|L,S) P(L|S,G) P(S) P(G) $
|
||||
|
||||
*Case 1: L = Yes* (且 $C=n, X=y$)
|
||||
因子 $P(X=y|L=y) = 0.9$。
|
||||
内部求和 $Sigma_(L=y)$:
|
||||
- $S=y, G=h$: $P(C=n|L=y,S=y)P(L=y|S=y,G=h)P(S=y)P(G=h) = 0.2 times 0.6 times 0.3 times 0.2 = 0.0072$
|
||||
- $S=y, G=l$: $0.2 times 0.3 times 0.3 times 0.8 = 0.0144$
|
||||
- $S=n, G=h$: $P(C=n|L=y,S=n)P(L=y|S=n,G=h)P(S=n)P(G=h) = 0.4 times 0.4 times 0.7 times 0.2 = 0.0224$
|
||||
- $S=n, G=l$: $0.4 times 0.1 times 0.7 times 0.8 = 0.0224$
|
||||
|
||||
$Sigma_(L=y) = 0.0072 + 0.0144 + 0.0224 + 0.0224 = 0.0664$
|
||||
$P(L=y, C=n, X=y) = 0.9 times 0.0664 = 0.05976$
|
||||
|
||||
*Case 2: L = No* (且 $C=n, X=y$)
|
||||
因子 $P(X=y|L=n) = 0.1$。
|
||||
内部求和 $Sigma_(L=n)$:
|
||||
- $S=y, G=h$: $P(C=n|L=n,S=y)P(L=n|S=y,G=h)P(S=y)P(G=h) = 0.7 times 0.4 times 0.3 times 0.2 = 0.0168$
|
||||
- $S=y, G=l$: $0.7 times 0.7 times 0.3 times 0.8 = 0.1176$
|
||||
- $S=n, G=h$: $P(C=n|L=n,S=n)P(L=n|S=n,G=h)P(S=n)P(G=h) = 0.9 times 0.6 times 0.7 times 0.2 = 0.0756$
|
||||
- $S=n, G=l$: $0.9 times 0.9 times 0.7 times 0.8 = 0.4536$
|
||||
|
||||
$Sigma_(L=n) = 0.0168 + 0.1176 + 0.0756 + 0.4536 = 0.6636$
|
||||
$P(L=n, C=n, X=y) = 0.1 times 0.6636 = 0.06636$
|
||||
|
||||
*最终归一化：*
|
||||
$ P(L="Yes" | ...) &= 0.05976 / (0.05976 + 0.06636) \
|
||||
&= 0.05976 / 0.12612 approx 0.4738 $
|
||||
|
||||
*答案：47.38%*
|
||||
]
|
||||
Reference in New Issue
Block a user