290 lines
9.9 KiB
Typst
290 lines
9.9 KiB
Typst
#import "@preview/cetz:0.3.3"
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#set text(font: ("Noto Sans CJK SC", "Noto Serif CJK SC"), lang: "zh")
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#set page(paper: "a4", margin: (x: 2cm, y: 2cm))
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#set heading(numbering: "1.")
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#let answer(body) = block(
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fill: luma(240),
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stroke: (left: 2pt + blue),
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inset: 10pt,
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radius: 2pt,
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width: 100%,
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body
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)
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#align(center)[
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#text(size: 18pt, weight: "bold")[第13-14章 概率与贝叶斯网络 习题解答]
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#v(1em)
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// *生成时间:* #datetime.today().display()
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]
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= 题目 1
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*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 $P(a and c)$。
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#answer[
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根据乘法公式 $P(x and y) = P(x)P(y|x)$：
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$ P(a and b) &= P(a) dot P(b | a) = 0.3 times 0.2 = 0.06 \
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P(a and c) &= P(a) dot P(c | a) = 0.3 times 0.5 = 0.15 $
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因为 $0.06 < 0.15$，所以 *$P(a and b) < P(a and c)$*。
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*答案：B*
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]
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= 题目 2
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*题意：* $P(a or b)=0.7$, $P(a)=0.4$, $P(b)=0.5$，求 $P(a and b)$。
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#answer[
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根据概率加法公式（容斥原理）：
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$ P(a or b) = P(a) + P(b) - P(a and b) $
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代入已知数值：
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$ 0.7 = 0.4 + 0.5 - P(a and b) \
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P(a and b) = 0.9 - 0.7 = 0.2 $
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*答案：0.2*
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]
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= 题目 3
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*数据：* 全联合概率分布表如下（Cavity, Toothache, Catch）：
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#table(
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columns: (auto, auto, auto, auto, auto),
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inset: 8pt,
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align: center,
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[], [T, Catch], [T, $not$ Catch], [$not$ T, Catch], [$not$ T, $not$ Catch],
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[Cavity], [0.108], [0.012], [0.072], [0.008],
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[$not$ Cavity], [0.016], [0.064], [0.144], [0.576]
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)
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#answer[
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*1. 计算 $P("Toothache" or not "Cavity")$*
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利用补集思想：$P(A) = 1 - P(not A)$。
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事件的补集是 $not "Toothache" and "Cavity"$。
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对应表格中的项为：(Cavity, $not$ T, Catch) 和 (Cavity, $not$ T, $not$ Catch)。
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$ P(not T and "Cav") = 0.072 + 0.008 = 0.080 $
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$ P(T or not "Cav") = 1 - 0.080 = 0.92 $
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*2. 计算 $P(not "toothache" or "catch")$*
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补集是 $"toothache" and not "catch"$。
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对应表格列为 (T, $not$ Catch)，即第 2 列。
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$ P(T and not "Catch") = 0.012 ("Cav") + 0.064 (not "Cav") = 0.076 $
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$ P(not T or "Catch") = 1 - 0.076 = 0.924 $
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*3. 计算 $P("Toothache" or not "Cavity" | not "catch")$ 和 $P(not "catch" | "Toothache" or not "Cavity")$*
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令 $A = "Toothache" or not "Cavity"$, $B = not "catch"$。
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- *计算 $P(B)$*: Sum of all $not$ Catch columns (Col 2, Col 4).
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$P(B) = (0.012 + 0.064) + (0.008 + 0.576) = 0.076 + 0.584 = 0.66$
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- *计算 $P(A and B)$*: 在 $not$ Catch 列中满足 $T or not "Cav"$ 的项。
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- (Cav, T, $not$ C): 0.012 (满足 T)
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- ($not$ Cav, T, $not$ C): 0.064 (满足 T)
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- ($not$ Cav, $not$ T, $not$ C): 0.576 (满足 $not$ Cav)
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- (Cav, $not$ T, $not$ C): 0.008 (不满足，既无T也无$not$Cav)
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Sum $= 0.012 + 0.064 + 0.576 = 0.652$
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结果 1: $P(A | B) = 0.652 / 0.66 approx 0.9879$
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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)
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*4. 判断 Cavity 与 Toothache 是否独立*
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$ P("Cav") &= 0.108+0.012+0.072+0.008 = 0.2 \
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P("Toothache") &= 0.108+0.012+0.016+0.064 = 0.2 \
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P("Cav" and "T") &= 0.108 + 0.012 = 0.12 $
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检验：$P("Cav") times P("T") = 0.2 times 0.2 = 0.04$。
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因为 $0.12 != 0.04$，所以两者 *不独立*。
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]
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= 题目 4
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*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 $Z$？
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#answer[
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*结论：不能。* 条件独立性不具备传递性。
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*反例证明：*
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假设 $Z$ 是一个公平的硬币投掷（0 或 1）。
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令 $X = Z$（$X$ 完全依赖于 $Z$）。
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令 $W = Z$（$W$ 完全依赖于 $Z$）。
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令 $Y$ 是一个与 $X, W, Z$ 都完全独立的随机变量（例如掷骰子）。
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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。
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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。
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3. *检查 $X, W$ 关于 $Z$ 的独立性*：
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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：
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令 $X$ 和 $W$ 为同一变量的两个副本。显然它们是强相关的。
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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。
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]
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= 题目 5
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*题意：* 罕见病检测。$P(D)=0.001$。
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检测A：$P(+|D)=0.95, P(+|not D)=0.05$。
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检测B：$P(+|D)=0.90, P(+|not D)=0.10$。
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#answer[
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*a. 单次检测 A 为阳性，求 $P(D|A+)$*
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使用贝叶斯公式：
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$ P(D|A+) = (P(A+|D)P(D)) / P(A+) $
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$ P(A+) &= P(A+|D)P(D) + P(A+|not D)P(not D) \
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&= 0.95 times 0.001 + 0.05 times 0.999 \
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&= 0.00095 + 0.04995 = 0.0509 $
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$ P(D|A+) &= 0.00095 / 0.0509 approx 0.0187 (1.87%) $
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*b. 两次检测 A 和 B 均为阳性，求 $P(D|A+, B+)$*
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由于检测独立（Naive Bayes 假设）：
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$ P(A+, B+ | D) = 0.95 times 0.90 = 0.855 $
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$ P(A+, B+ | not D) = 0.05 times 0.10 = 0.005 $
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$ P(D | A+, B+) &= (P(A+, B+ | D)P(D)) / (P(A+, B+ | D)P(D) + P(A+, B+ | not D)P(not D)) \
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&= (0.855 times 0.001) / (0.855 times 0.001 + 0.005 times 0.999) \
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&= 0.000855 / (0.000855 + 0.004995) \
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&= 0.000855 / 0.00585 approx 0.1462 (14.62%) $
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*c. 解释*
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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。
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]
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= 题目 6
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*场景：* 骑车(B) 取决于 天气(W) 和 熬夜(S)。
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$P(W="Sun")=0.7, P(S="Yes")=0.4$。
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CPT 已知。
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#answer[
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*a. 贝叶斯网络与 CPT*
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结构：$W -> B <- S$ (V-structure)
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#cetz.canvas({
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import cetz.draw: *
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let r = 0.8
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// Nodes
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circle((0, 0), radius: r, name: "B")
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content("B", "B (骑车)")
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circle((-2, 3), radius: r, name: "W")
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content("W", "W (天气)")
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circle((2, 3), radius: r, name: "S")
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content("S", "S (熬夜)")
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// Edges
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line("W", "B", mark: (end: ">"))
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line("S", "B", mark: (end: ">"))
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})
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*条件概率表 (CPT) for B:*
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#table(
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columns: 3,
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align: center,
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[W (天气)], [S (熬夜)], [P(B=是 | W, S)],
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[晴 (Sun)], [否 (No)], [0.9],
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[晴 (Sun)], [是 (Yes)], [0.6],
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[雨 (Rain)], [否 (No)], [0.2],
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[雨 (Rain)], [是 (Yes)], [0.1]
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)
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*b. 推理：已知骑车(B=Yes)，求天气概率*
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我们需要计算 $P(W | B="Yes")$。根据贝叶斯法则：$P(W|B) = P(B|W)P(W) / P(B)$。
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首先计算边缘概率 $P(B="Yes")$。
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$ P(B=y) = sum_(w, s) P(B=y|w,s)P(w)P(s) $
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因 W 和 S 独立，联合概率直接相乘。
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1. $W="Sun", S="No"$: $0.9 times 0.7 times 0.6 = 0.378$
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2. $W="Sun", S="Yes"$: $0.6 times 0.7 times 0.4 = 0.168$
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3. $W="Rain", S="No"$: $0.2 times 0.3 times 0.6 = 0.036$
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4. $W="Rain", S="Yes"$: $0.1 times 0.3 times 0.4 = 0.012$
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$ P(B="Yes") = 0.378 + 0.168 + 0.036 + 0.012 = 0.594 $
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计算 $P(W="Sun" | B="Yes")$:
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分子为 W=Sun 的所有情况之和（项1 + 项2）：
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$ P(B=y, W="Sun") = 0.378 + 0.168 = 0.546 $
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$ P(W="Sun" | B="Yes") = 0.546 / 0.594 approx 0.919 $
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同理，$P(W="Rain" | B="Yes") = (0.036 + 0.012) / 0.594 approx 0.081$。
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*结论：这天最可能是晴天。*
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]
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= 题目 7
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*智能健康助手：* 变量 S(吸烟), G(基因), L(肺癌), C(咳嗽), X(胸片)。
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关系：$S->L, G->L, S->C, L->C, L->X$。
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#answer[
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*问题 1：贝叶斯网络图*
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#align(center)[
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#cetz.canvas({
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import cetz.draw: *
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// Manually positioning nodes for clarity
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let r = 0.5
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circle((0, 0), radius: r, name: "L")
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content("L", "L")
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circle((-2, 2), radius: r, name: "S")
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content("S", "S")
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circle((2, 2), radius: r, name: "G")
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content("G", "G")
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circle((-1, -2), radius: r, name: "C")
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content("C", "C")
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circle((1, -2), radius: r, name: "X")
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content("X", "X")
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line("S", "L", mark: (end: ">"))
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line("G", "L", mark: (end: ">"))
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line("L", "X", mark: (end: ">"))
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line("L", "C", mark: (end: ">"))
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// S -> C 直接边
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line("S", "C", mark: (end: ">"))
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})
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]
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*问题 2：概率推理 $P(L="Yes" | C="No", X="Yes")$*
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令 $alpha$ 为归一化常数。我们需要计算 $P(L=y, C=n, X=y)$ 和 $P(L=n, C=n, X=y)$。
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公式分解：$P(S,G,L,C,X) = P(S)P(G)P(L|S,G)P(C|L,S)P(X|L)$。
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求和消除 S, G：
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$ P(L, C, X) = P(X|L) sum_S sum_G P(C|L,S) P(L|S,G) P(S) P(G) $
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*Case 1: L = Yes* (且 $C=n, X=y$)
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因子 $P(X=y|L=y) = 0.9$。
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内部求和 $Sigma_(L=y)$:
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- $S=y, G=h$: $P(C=n|L=y,S=y)P(L=y|S=y,G=h)P(S=y)P(G=h) = 0.2 times 0.6 times 0.3 times 0.2 = 0.0072$
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- $S=y, G=l$: $0.2 times 0.3 times 0.3 times 0.8 = 0.0144$
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- $S=n, G=h$: $P(C=n|L=y,S=n)P(L=y|S=n,G=h)P(S=n)P(G=h) = 0.4 times 0.4 times 0.7 times 0.2 = 0.0224$
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- $S=n, G=l$: $0.4 times 0.1 times 0.7 times 0.8 = 0.0224$
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$Sigma_(L=y) = 0.0072 + 0.0144 + 0.0224 + 0.0224 = 0.0664$
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$P(L=y, C=n, X=y) = 0.9 times 0.0664 = 0.05976$
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*Case 2: L = No* (且 $C=n, X=y$)
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因子 $P(X=y|L=n) = 0.1$。
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内部求和 $Sigma_(L=n)$:
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- $S=y, G=h$: $P(C=n|L=n,S=y)P(L=n|S=y,G=h)P(S=y)P(G=h) = 0.7 times 0.4 times 0.3 times 0.2 = 0.0168$
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- $S=y, G=l$: $0.7 times 0.7 times 0.3 times 0.8 = 0.1176$
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- $S=n, G=h$: $P(C=n|L=n,S=n)P(L=n|S=n,G=h)P(S=n)P(G=h) = 0.9 times 0.6 times 0.7 times 0.2 = 0.0756$
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- $S=n, G=l$: $0.9 times 0.9 times 0.7 times 0.8 = 0.4536$
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$Sigma_(L=n) = 0.0168 + 0.1176 + 0.0756 + 0.4536 = 0.6636$
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$P(L=n, C=n, X=y) = 0.1 times 0.6636 = 0.06636$
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*最终归一化：*
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$ P(L="Yes" | ...) &= 0.05976 / (0.05976 + 0.06636) \
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&= 0.05976 / 0.12612 approx 0.4738 $
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*答案：47.38%*
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]
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