@@@贝叶斯后验概率-用新的信息来调整认知2023.12.5

import numpy as np

# 模拟扔骰子100次
np.random.seed(0)  # 设置随机种子以保证结果的可复现性
dice_rolls = np.random.randint(1, 7, 100)  # 生成100次1到6的随机数，代表扔骰子100次的结果

import pandas as pd
from scipy.stats import beta

# 设定两个不同的先验概率
# 先验 A: P(6) = 1/2，表示为 Beta 分布的参数 (α=2, β=2)
# 先验 B: P(6) = 1/6，表示为 Beta 分布的参数 (α=1, β=5)
priors = {
    "A": {"alpha": 2, "beta": 2},
    "B": {"alpha": 1, "beta": 5}
}

# 初始化一个 DataFrame 来存储后验概率的计算结果
posterior_results = pd.DataFrame(columns=["Throw", "Posterior_A", "Posterior_B"])

# 对于每次投掷，更新后验概率
for i in range(1, 101):
    six_count = (dice_rolls[:i] == 6).sum()
    non_six_count = i - six_count

    for prior_label, prior_params in priors.items():
        # 更新后验概率，Beta(alpha + successes, beta + failures)
        alpha_post = prior_params["alpha"] + six_count
        beta_post = prior_params["beta"] + non_six_count
        posterior_mean = alpha_post / (alpha_post + beta_post)

        posterior_results.loc[i, f"Posterior_{prior_label}"] = posterior_mean

posterior_results.tail(10)  # 显示最后10次投掷的后验概率结果

import matplotlib.pyplot as plt
import sympy as sp

# 定义符号
alpha_A, beta_A, alpha_B, beta_B, successes, failures = sp.symbols('alpha_A beta_A alpha_B beta_B successes failures')

# 定义贝塔分布的均值公式
posterior_mean = lambda alpha, beta: alpha / (alpha + beta)

# 定义先验参数
prior_params_A = {alpha_A: 2, beta_A: 2}  # 假设 A
prior_params_B = {alpha_B: 1, beta_B: 5}  # 假设 B

# 定义成功和失败的次数
successes_failures = {successes: 1, failures: 4}  # 1次成功（得到6点），4次失败（未得到6点）

# 计算后验概率的均值
posterior_A = posterior_mean(alpha_A + successes, beta_A + failures).subs(prior_params_A).subs(successes_failures)
posterior_B = posterior_mean(alpha_B + successes, beta_B + failures).subs(prior_params_B).subs(successes_failures)

# 准备绘制手写公式
plt.figure(figsize=(12, 8))
plt.text(0.1, 0.6, f"Posterior (A): {sp.latex(posterior_A)}", fontsize=12, ha='left')
plt.text(0.1, 0.3, f"Posterior (B): {sp.latex(posterior_B)}", fontsize=12, ha='left')
plt.axis('off')
plt.title('Handwritten Bayes Calculation')
plt.show()