bayesian lab finished
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bayesian/bayesian_network.py
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bayesian/bayesian_network.py
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import random
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# CPTs
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# P(C)
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P_C = {True: 0.5, False: 0.5}
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# P(S | C)
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P_S_given_C = {
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True: {True: 0.10, False: 0.90}, # C=T
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False: {True: 0.50, False: 0.50} # C=F
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}
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# P(R | C)
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P_R_given_C = {
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True: {True: 0.80, False: 0.20}, # C=T
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False: {True: 0.20, False: 0.80} # C=F
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}
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# P(W | S, R)
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P_W_given_SR = {
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(True, True): {True: 0.99, False: 0.01},
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(True, False): {True: 0.90, False: 0.10},
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(False, True): {True: 0.90, False: 0.10},
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(False, False): {True: 0.00, False: 1.00}
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}
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def sample_variable(prob_true):
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"""Returns True with probability prob_true, else False"""
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return random.random() < prob_true
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def get_prior_sample():
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"""Generates a sample from the joint distribution P(C, S, R, W)"""
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c = sample_variable(P_C[True])
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s = sample_variable(P_S_given_C[c][True])
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r = sample_variable(P_R_given_C[c][True])
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w = sample_variable(P_W_given_SR[(s, r)][True])
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return {'C': c, 'S': s, 'R': r, 'W': w}
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def rejection_sampling(query_var, query_val, evidence, num_samples):
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"""
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Estimates P(query_var = query_val | evidence) using Rejection Sampling.
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evidence is a dict, e.g., {'S': True, 'W': True}
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"""
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consistent_count = 0
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query_match_count = 0
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for _ in range(num_samples):
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sample = get_prior_sample()
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# Check if sample is consistent with evidence
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is_consistent = True
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for var, val in evidence.items():
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if sample[var] != val:
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is_consistent = False
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break
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if is_consistent:
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consistent_count += 1
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if sample[query_var] == query_val:
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query_match_count += 1
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if consistent_count == 0:
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return 0.0 # Avoid division by zero
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return query_match_count / consistent_count
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def likelihood_weighting(query_var, query_val, evidence, num_samples):
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"""
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Estimates P(query_var = query_val | evidence) using Likelihood Weighting.
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evidence is a dict, e.g., {'S': True, 'W': True}
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"""
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weighted_counts = {True: 0.0, False: 0.0}
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for _ in range(num_samples):
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weight = 1.0
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sample = {}
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# C
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if 'C' in evidence:
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sample['C'] = evidence['C']
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weight *= P_C[sample['C']]
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else:
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sample['C'] = sample_variable(P_C[True])
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# S
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if 'S' in evidence:
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sample['S'] = evidence['S']
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weight *= P_S_given_C[sample['C']][sample['S']]
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else:
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sample['S'] = sample_variable(P_S_given_C[sample['C']][True])
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# R
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if 'R' in evidence:
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sample['R'] = evidence['R']
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weight *= P_R_given_C[sample['C']][sample['R']]
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else:
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sample['R'] = sample_variable(P_R_given_C[sample['C']][True])
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# W
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if 'W' in evidence:
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sample['W'] = evidence['W']
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weight *= P_W_given_SR[(sample['S'], sample['R'])][sample['W']]
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else:
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sample['W'] = sample_variable(P_W_given_SR[(sample['S'], sample['R'])][True])
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# Accumulate weight for the value of the query variable in this sample
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weighted_counts[sample[query_var]] += weight
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total_weight = sum(weighted_counts.values())
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if total_weight == 0:
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return 0.0
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return weighted_counts[query_val] / total_weight
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if __name__ == "__main__":
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evidence = {'S': True, 'W': True}
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query_var = 'R'
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query_val = True
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# Exact Calculation (calculated manually in thought process)
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# Numerator: 0.0891
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# Denominator: 0.2781
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exact_prob = 0.0891 / 0.2781
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print(f"Exact Probability P(R=T | S=T, W=T): {exact_prob:.6f}")
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print("-" * 60)
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print(f"{'Samples':<10} | {'Rejection Sampling':<20} | {'Likelihood Weighting':<20}")
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print("-" * 60)
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for N in [100, 1000, 10000, 100000, 1000000]:
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rs_result = rejection_sampling(query_var, query_val, evidence, N)
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lw_result = likelihood_weighting(query_var, query_val, evidence, N)
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print(f"{N:<10} | {rs_result:<20.6f} | {lw_result:<20.6f}")
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57
bayesian/generate_data.py
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bayesian/generate_data.py
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import pandas as pd
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import numpy as np
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# 定义真实的条件概率表 (Ground Truth CPTs)
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# 学生需要从生成的数据中“反向学习”出这些概率
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PROBS = {
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'IsSmoker': 0.2,
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'VisitAsia': 0.05,
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'HasLungCancer': { # Given IsSmoker
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(1,): 0.1,
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(0,): 0.01
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},
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'HasTuberculosis': { # Given VisitAsia
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(1,): 0.08,
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(0,): 0.001
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},
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'XRayPositive': { # Given HasLungCancer, HasTuberculosis
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(1, 1): 0.98,
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(1, 0): 0.95,
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(0, 1): 0.90,
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(0, 0): 0.05
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},
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'HasDyspnea': { # Given HasLungCancer
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(1,): 0.7,
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(0,): 0.2
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}
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}
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def generate_data(num_samples=1000):
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data = []
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for _ in range(num_samples):
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# Sample from parent nodes
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is_smoker = np.random.binomial(1, PROBS['IsSmoker'])
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visit_asia = np.random.binomial(1, PROBS['VisitAsia'])
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# Sample from children nodes based on parents
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p_lc = PROBS['HasLungCancer'][(is_smoker,)]
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has_lc = np.random.binomial(1, p_lc)
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p_tb = PROBS['HasTuberculosis'][(visit_asia,)]
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has_tb = np.random.binomial(1, p_tb)
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p_xray = PROBS['XRayPositive'][(has_lc, has_tb)]
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xray_pos = np.random.binomial(1, p_xray)
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p_dys = PROBS['HasDyspnea'][(has_lc,)]
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has_dys = np.random.binomial(1, p_dys)
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data.append([is_smoker, visit_asia, has_lc, has_tb, xray_pos, has_dys])
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df = pd.DataFrame(data, columns=['IsSmoker', 'VisitAsia', 'HasLungCancer',
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'HasTuberculosis', 'XRayPositive', 'HasDyspnea'])
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df.to_csv('data.csv', index=False)
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print(f"Generated data.csv with {num_samples} samples.")
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if __name__ == '__main__':
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generate_data(num_samples=2000)
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6702
bayesian/homework.pdf
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6702
bayesian/homework.pdf
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289
bayesian/homework.typ
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bayesian/homework.typ
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#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: ">"))
|
||||
})
|
||||
|
||||
*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%*
|
||||
]
|
||||
177
bayesian/labtemplate.typ
Executable file
177
bayesian/labtemplate.typ
Executable file
@ -0,0 +1,177 @@
|
||||
#let times = "Times LT Pro"
|
||||
#let times = "Times New Roman"
|
||||
#let song = (times, "Noto Serif CJK SC")
|
||||
#let hei = (times, "Noto Sans CJK SC")
|
||||
#let kai = (times, "Noto Serif CJK SC")
|
||||
#let xbsong = (times, "Noto Serif CJK SC")
|
||||
#let fsong = (times, "Noto Serif CJK SC")
|
||||
#let code = (times, "JetBrains Mono")
|
||||
#let nudtlabpaper(title: "",
|
||||
author: "",
|
||||
id: "",
|
||||
training_type:"",
|
||||
grade: "",
|
||||
major: "",
|
||||
department: "",
|
||||
advisor: "",
|
||||
jobtitle: "",
|
||||
lab: "",
|
||||
date: "",
|
||||
header_str: "",
|
||||
simple_cover: "",
|
||||
body) = {
|
||||
// Set the document's basic properties.
|
||||
set document(author: author, title: title)
|
||||
set page(
|
||||
|
||||
margin: (left: 30mm, right: 30mm, top: 30mm, bottom: 30mm),
|
||||
)
|
||||
|
||||
// Title row.
|
||||
v(158pt)
|
||||
align(center)[
|
||||
#block(text(weight: 700, size: 30pt, font: hei, tracking: 15pt, "人工智能实验报告"))
|
||||
]
|
||||
// align(center)[
|
||||
// #block(text(weight: 700, size: 30pt, font: song, tracking: 15pt, "本科实验报告"))
|
||||
// ]
|
||||
|
||||
v(83pt)
|
||||
pad(
|
||||
left: 1em,
|
||||
right: 1em,
|
||||
grid(
|
||||
columns: (80pt, 1fr),
|
||||
rows: (17pt, auto),
|
||||
text(weight: 700, size: 16pt, font: song, "实验名称："),
|
||||
align(center, text(weight: "regular", size: 16pt, font: song, title)),
|
||||
text(""),
|
||||
line(length: 100%)
|
||||
)
|
||||
// #block(text(weight: 700, 1.75em, title))
|
||||
// underline(text(weight: 700, size: 16pt, font: song, title))
|
||||
)
|
||||
|
||||
// Author information.
|
||||
|
||||
v(82.5pt)
|
||||
|
||||
if simple_cover {
|
||||
pad(
|
||||
left: 10%,
|
||||
right: 10%,
|
||||
grid(
|
||||
columns: (100pt, 1fr),
|
||||
rows: (15pt, 8pt, 15pt, 8pt, 15pt, 8pt),
|
||||
text(size: 14pt, font: song, "学员姓名："),
|
||||
align(center, text(size: 14pt, font: song, author)),
|
||||
text(""),
|
||||
line(length: 100%),
|
||||
text(size: 14pt, font: song, "学 号："),
|
||||
align(center, text(size: 14pt, font: times, id)),
|
||||
text(""),
|
||||
line(length: 100%),
|
||||
text(size: 14pt, font: song, "实验日期："),
|
||||
align(center, text(size: 14pt, font: song, date)),
|
||||
text(""),
|
||||
line(length: 100%),
|
||||
)
|
||||
)
|
||||
} else {
|
||||
grid(
|
||||
columns: (0.25fr, 0.25fr, 0.25fr, 0.25fr),
|
||||
rows: (15pt, 8pt, 15pt, 8pt, 15pt, 8pt, 15pt, 8pt, 15pt),
|
||||
text(size: 14pt, font: song, tracking: 10pt, "学员姓名"),
|
||||
align(center, text(size: 14pt, font: song, author)),
|
||||
text(size: 14pt, font: song, tracking: 54pt, "学号"),
|
||||
align(center, text(size: 14pt, font: times, id)),
|
||||
text(""),
|
||||
line(length: 100%),
|
||||
text(""),
|
||||
line(length: 100%),
|
||||
text(size: 14pt, font: song, tracking: 9pt, "培养类型"),
|
||||
align(center, text(size: 14pt, font: song, training_type)),
|
||||
text(size: 14pt, font: song, tracking: 54pt, "年级"),
|
||||
align(center, text(size: 14pt, font: times, grade)),
|
||||
text(""),
|
||||
line(length: 100%),
|
||||
text(""),
|
||||
line(length: 100%),
|
||||
text(size: 14pt, font: song, tracking: 54pt, "专业"),
|
||||
align(center, text(size: 14pt, font: song, major)),
|
||||
text(size: 14pt, font: song, tracking: 9pt, "所属学院"),
|
||||
align(center, text(size: 14pt, font: song, department)),
|
||||
text(""),
|
||||
line(length: 100%),
|
||||
text(""),
|
||||
line(length: 100%),
|
||||
text(size: 14pt, font: song, tracking: 9pt, "指导教员"),
|
||||
align(center, text(size: 14pt, font: song, advisor)),
|
||||
text(size: 14pt, font: song, tracking: 54pt, "职称"),
|
||||
align(center, text(size: 14pt, font: song, jobtitle)),
|
||||
text(""),
|
||||
line(length: 100%),
|
||||
text(""),
|
||||
line(length: 100%),
|
||||
text(size: 14pt, font: song, tracking: 20pt, "实验室"),
|
||||
align(center, text(size: 14pt, font: song, lab)),
|
||||
text(size: 14pt, font: song, tracking: 9pt, "实验时间"),
|
||||
align(center, text(size: 14pt, font: song, date)),
|
||||
text(""),
|
||||
line(length: 100%),
|
||||
text(""),
|
||||
line(length: 100%),
|
||||
)
|
||||
}
|
||||
|
||||
v(50.5pt)
|
||||
align(center, text(font: hei, size: 15pt, "国防科技大学教育训练部制"))
|
||||
|
||||
pagebreak()
|
||||
|
||||
set page(
|
||||
margin: (left: 30mm, right: 30mm, top: 30mm, bottom: 30mm),
|
||||
numbering: "i",
|
||||
number-align: center,
|
||||
)
|
||||
|
||||
v(14pt)
|
||||
align(center)[
|
||||
#block(text(font: hei, size: 14pt, "《本科实验报告》填写说明"))
|
||||
]
|
||||
|
||||
v(14pt)
|
||||
text("")
|
||||
par(first-line-indent: 2em, text(font: song, size: 12pt, "实验报告内容编排应符合以下要求："))
|
||||
|
||||
par(first-line-indent: 2em, text(font: fsong, size: 12pt, "（1）采用A4（21cm×29.7cm）白色复印纸，单面黑字。上下左右各侧的页边距均为3cm；缺省文档网格：字号为小4号，中文为宋体，英文和阿拉伯数字为Times New Roman，每页30行，每行36字；页脚距边界为2.5cm，页码置于页脚、居中，采用小5号阿拉伯数字从1开始连续编排，封面不编页码。"))
|
||||
|
||||
par(first-line-indent: 2em, text(font: fsong, size: 12pt, "（2）报告正文最多可设四级标题，字体均为黑体，第一级标题字号为4号，其余各级标题为小4号；标题序号第一级用“一、”、“二、”……，第二级用“（一）”、“（二）” ……，第三级用“1.”、“2.” ……，第四级用“（1）”、“（2）” ……，分别按序连续编排。"))
|
||||
|
||||
par(first-line-indent: 2em, text(font: fsong, size: 12pt, "（3）正文插图、表格中的文字字号均为5号。"))
|
||||
|
||||
pagebreak()
|
||||
|
||||
set page(
|
||||
margin: (left: 30mm, right: 30mm, top: 30mm, bottom: 30mm),
|
||||
numbering: "1",
|
||||
number-align: center,
|
||||
)
|
||||
|
||||
set heading(numbering: "1.1")
|
||||
// set text(font: hei, lang: "zh")
|
||||
|
||||
show heading: it => box(width: 100%)[
|
||||
#v(0.50em)
|
||||
#set text(font: hei)
|
||||
#counter(heading).display()
|
||||
// #h(0.5em)
|
||||
#it.body
|
||||
]
|
||||
// Main body.
|
||||
set par(justify: true)
|
||||
|
||||
body
|
||||
}
|
||||
|
||||
#let para(t) = par(first-line-indent: 2em, text(font: song, size: 10.5pt, t))
|
||||
8945
bayesian/main.pdf
8945
bayesian/main.pdf
File diff suppressed because one or more lines are too long
430
bayesian/main.typ
Normal file → Executable file
430
bayesian/main.typ
Normal file → Executable file
@ -1,289 +1,177 @@
|
||||
#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
|
||||
#import "labtemplate.typ": *
|
||||
#show: nudtlabpaper.with(
|
||||
author: "程景愉",
|
||||
id: "202302723005",
|
||||
title: "贝叶斯网络编程实验",
|
||||
training_type: "普通本科生",
|
||||
grade: "2023级",
|
||||
major: "网络工程",
|
||||
department: "计算机学院",
|
||||
advisor: "祝恩",
|
||||
jobtitle: "教授",
|
||||
lab: "306-707",
|
||||
date: "2025.12.31",
|
||||
header_str: "贝叶斯网络编程实验报告",
|
||||
simple_cover: true,
|
||||
)
|
||||
|
||||
#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()
|
||||
#set page(header: [
|
||||
#set par(spacing: 6pt)
|
||||
#align(center)[#text(size: 11pt)[人工智能实验报告]]
|
||||
#v(-0.3em)
|
||||
#line(length: 100%, stroke: (thickness: 1pt))
|
||||
],)
|
||||
|
||||
#show heading: it => if it.level == 1 [
|
||||
#v(1em)
|
||||
#set text(font: hei, size: 16pt)
|
||||
#it.body
|
||||
#v(0.5em)
|
||||
] else [
|
||||
#v(0.8em)
|
||||
#set text(font: "New Computer Modern", weight: "bold", style: "italic")
|
||||
#it.body
|
||||
#v(0.5em)
|
||||
]
|
||||
|
||||
= 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 $P(a and c)$。
|
||||
#set heading(numbering: none)
|
||||
|
||||
#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*
|
||||
|
||||
#outline(title: "目录",depth: 2, indent: 1em)
|
||||
|
||||
#set enum(indent: 0.5em,body-indent: 0.5em,)
|
||||
#pagebreak()
|
||||
|
||||
= 实验介绍
|
||||
#para[
|
||||
本次实验旨在实现并比较两种贝叶斯网络近似推理算法：拒绝采样（Rejection Sampling）和似然加权（Likelihood Weighting）。
|
||||
实验使用的贝叶斯网络模型包含四个布尔变量：Cloudy (C), Sprinkler (S), Rain (R), WetGrass (W)。
|
||||
网络结构为 $C \to S, C \to R, S \to W, R \to W$。
|
||||
实验目标是根据给定的条件概率表（CPTs），在已知证据 $S=T$ 和 $W=T$ 的情况下，估计 $R=T$ 的后验概率 $P(R=T | S=T, W=T)$。
|
||||
]
|
||||
|
||||
= 题目 2
|
||||
*题意：* $P(a or b)=0.7$, $P(a)=0.4$, $P(b)=0.5$，求 $P(a and b)$。
|
||||
= 实验内容
|
||||
#para[
|
||||
实验主要包含以下内容：
|
||||
]
|
||||
+ 根据题目描述的贝叶斯网络结构和条件概率表，构建数据结构表示该网络。
|
||||
+ 实现拒绝采样算法：从先验分布中采样，拒绝与证据不一致的样本，利用保留样本中查询变量的频率作为后验概率的估计。
|
||||
+ 实现似然加权算法：固定证据变量的值，对非证据变量进行采样，并根据证据变量的条件概率计算样本权重，利用加权频率估计后验概率。
|
||||
+ 比较两种算法在不同采样数量（N=100, 1000, ...）下的估计结果与真实值的偏差，分析算法收敛性。
|
||||
|
||||
#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*
|
||||
= 实验情况
|
||||
+ 使用 Python 语言实现算法。
|
||||
+ 代码需清晰易读，包含必要的注释。
|
||||
+ 计算并输出 $P(R=T | S=T, W=T)$ 的精确值以便对比。
|
||||
+ 撰写实验报告，展示核心代码和实验结果。
|
||||
|
||||
= 实验步骤与实现
|
||||
|
||||
== 网络定义与概率表
|
||||
首先定义网络的条件概率表（CPTs）。
|
||||
```python
|
||||
# CPTs
|
||||
P_C = {True: 0.5, False: 0.5}
|
||||
|
||||
# P(S | C)
|
||||
P_S_given_C = {
|
||||
True: {True: 0.10, False: 0.90}, # C=T
|
||||
False: {True: 0.50, False: 0.50} # C=F
|
||||
}
|
||||
|
||||
# P(R | C)
|
||||
P_R_given_C = {
|
||||
True: {True: 0.80, False: 0.20}, # C=T
|
||||
False: {True: 0.20, False: 0.80} # C=F
|
||||
}
|
||||
|
||||
# P(W | S, R)
|
||||
P_W_given_SR = {
|
||||
(True, True): {True: 0.99, False: 0.01},
|
||||
(True, False): {True: 0.90, False: 0.10},
|
||||
(False, True): {True: 0.90, False: 0.10},
|
||||
(False, False): {True: 0.00, False: 1.00}
|
||||
}
|
||||
```
|
||||
|
||||
== 拒绝采样 (Rejection Sampling)
|
||||
拒绝采样通过生成先验分布的样本，并丢弃那些与证据不符的样本来工作。
|
||||
```python
|
||||
def rejection_sampling(query_var, query_val, evidence, num_samples):
|
||||
consistent_count = 0
|
||||
query_match_count = 0
|
||||
|
||||
for _ in range(num_samples):
|
||||
sample = get_prior_sample() # 从联合分布采样
|
||||
|
||||
# 检查样本是否与证据一致
|
||||
is_consistent = True
|
||||
for var, val in evidence.items():
|
||||
if sample[var] != val:
|
||||
is_consistent = False
|
||||
break
|
||||
|
||||
if is_consistent:
|
||||
consistent_count += 1
|
||||
if sample[query_var] == query_val:
|
||||
query_match_count += 1
|
||||
|
||||
if consistent_count == 0: return 0.0
|
||||
return query_match_count / consistent_count
|
||||
```
|
||||
|
||||
== 似然加权 (Likelihood Weighting)
|
||||
似然加权强制证据变量取观测值，并将样本权重设为证据变量在其父节点给定情况下的似然概率乘积。
|
||||
```python
|
||||
def likelihood_weighting(query_var, query_val, evidence, num_samples):
|
||||
weighted_counts = {True: 0.0, False: 0.0}
|
||||
|
||||
for _ in range(num_samples):
|
||||
weight = 1.0
|
||||
sample = {}
|
||||
|
||||
# 以 C, S, R, W 的拓扑顺序采样/赋值
|
||||
# 若变量在证据中，则赋值为证据值并更新权重
|
||||
# 若变量不在证据中，则根据父节点采样
|
||||
# (代码细节略，参见附件源程序)
|
||||
# ...
|
||||
|
||||
# 累加权重
|
||||
weighted_counts[sample[query_var]] += weight
|
||||
|
||||
total_weight = sum(weighted_counts.values())
|
||||
if total_weight == 0: return 0.0
|
||||
return weighted_counts[query_val] / total_weight
|
||||
```
|
||||
|
||||
= 实验结果
|
||||
#para[
|
||||
首先通过精确计算（枚举联合概率分布）得到理论真实值：
|
||||
$P(R=T | S=T, W=T) approx 0.320388$
|
||||
]
|
||||
|
||||
= 题目 3
|
||||
*数据：* 全联合概率分布表如下（Cavity, Toothache, Catch）：
|
||||
#para[
|
||||
下表展示了在不同采样数量 $N$ 下，两种算法的估计结果：
|
||||
]
|
||||
|
||||
#table(
|
||||
columns: (auto, auto, auto, auto, auto),
|
||||
inset: 8pt,
|
||||
columns: (auto, auto, auto),
|
||||
inset: 10pt,
|
||||
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]
|
||||
[*样本数量 N*], [*拒绝采样结果*], [*似然加权结果*],
|
||||
[100], [0.480000], [0.377902],
|
||||
[1000], [0.353741], [0.311148],
|
||||
[10000], [0.320734], [0.317552],
|
||||
[100000], [0.318791], [0.321722],
|
||||
[1000000], [0.319916], [0.319471],
|
||||
)
|
||||
|
||||
#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 $
|
||||
= 实验总结
|
||||
|
||||
*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 $
|
||||
+ *收敛性*：从实验结果可以看出，随着采样数量 $N$ 的增加，拒绝采样和似然加权算法的结果都逐渐收敛于真实值 $0.320388$。
|
||||
|
||||
*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)
|
||||
+ *算法比较*：
|
||||
- *拒绝采样*：实现简单，直接利用先验分布采样。但在证据概率较低时（本例中证据发生的概率约为 0.278，尚可），会拒绝大量样本，导致计算资源浪费。在样本量较小（如 N=100）时波动较大。
|
||||
- *似然加权*：利用了证据信息，所有生成的样本都是有效的（权重不为0），因此在相同采样数量下，通常能提供比拒绝采样更稳定的估计，尤其是当证据发生概率很低时优势更明显。
|
||||
|
||||
*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%*
|
||||
]
|
||||
+ *结论*：两种算法均能有效进行近似推理。对于本实验中的简单网络和非极端概率证据，两者在 $N=10000$ 以上时均能给出较好的估计结果。
|
||||
|
||||
34
bayesian/task.txt
Normal file
34
bayesian/task.txt
Normal file
@ -0,0 +1,34 @@
|
||||
作业二：实现贝叶斯网络近似推理算法
|
||||
（一）问题描述
|
||||
实现两种基本的近似推理算法——拒绝采样和似然加权，并在一个简单的
|
||||
贝叶斯网络（PPT 示例）上进行应用和比较。
|
||||
网络结构图:
|
||||
C -> S
|
||||
C -> R
|
||||
S -> W
|
||||
R -> W
|
||||
变量解释:
|
||||
C(Cloudy): 是否多云 (Boolean)
|
||||
S(Sprinkler): 洒水器是否开启 (Boolean)
|
||||
R(Rain): 是否下雨 (Boolean)
|
||||
W(WetGrass): 草坪是否湿润 (Boolean)
|
||||
条件概率表 (CPTs):
|
||||
P(C=T) = 0.5
|
||||
P(S | C):
|
||||
P(S=T | C=T) = 0.10
|
||||
P(S=T | C=F) = 0.50
|
||||
P(R | C):
|
||||
P(R=T | C=T) = 0.80
|
||||
P(R=T | C=F) = 0.20
|
||||
P(W | S, R):
|
||||
P(W=T | S=T, R=T) = 0.99
|
||||
P(W=T | S=T, R=F) = 0.90
|
||||
P(W=T | S=F, R=T) = 0.90
|
||||
P(W=T | S=F, R=F) = 0.00
|
||||
|
||||
(二) 任务要求
|
||||
实现拒绝采样算法和似然加权算法，算法需要能够估计条件概率 P(X | e)，
|
||||
其中 X 是查询变量，e 是证据。使用你实现的两种算法估计 P(R | S=T,
|
||||
W=T)。
|
||||
|
||||
要求撰写实验报告，提供源程序。
|
||||
BIN
bayesian/实验题（贝叶斯网络）.pdf
Normal file
BIN
bayesian/实验题（贝叶斯网络）.pdf
Normal file
Binary file not shown.
Reference in New Issue
Block a user