可信度的估计

• 二项分布中的\(p\) 服从Beta分布 $ {\rm beta}(\alpha, \beta)$, 密度函数 \(\frac1{B(\alpha, \beta)} x^{\alpha-1} (1-x)^{\beta -1}\)
• 均值 \(\frac \alpha {\alpha + \beta}\)
• 方差 \(\frac {\alpha \beta} {(\alpha+\beta)^2 (\alpha+ \beta + 1) } ​\)

from scipy.stats import betadef confidence(n_bad, n_good, tol=2):''' 返回估计的坏率p, 以及在tol倍标准差下的可信度'''a, b = n_bad+1, n_good+1p = a / (a+b)v = beta.std(a, b)up, low =  min(1, p + v*tol), max(0, p - v*tol)d = beta.cdf(up, a,b) -  beta.cdf(low, a,b)return p, v, dtest_set = [(500, 20000, 2), (1000, 200000, 2), (2000, 200000, 2), (5000, 200000, 2),(500,  100000, 2), (1000, 100000, 2), (2000, 100000, 2), (5000, 100000, 2), (2000, 10000, 2),
]print("  bad;  total; 均值p;    标准差v;     均值的相对误差e;  置信度")
for (n_bad, n_good, tol)  in  test_set:p,v,d = confidence(n_bad, n_good, tol)ss = ('{:5d};{:7d}; p={p:0.4f}; v={v:0.6f}; e={e:0.3f}; '  + '均值在[p - {t}v, p + {t}v]的概率 {d:2.2f}%').format(n_bad, n_bad+n_good, p=p,v=v, c=v/p, d =d*100,t=tol, e=tol*v/p)print(ss)

  bad;  total; 均值p;    标准差v;     均值的相对误差e;  置信度500;  20500; p=0.0244; v=0.001078; e=0.088; 均值在[p - 2v, p + 2v]的概率 95.46%1000; 201000; p=0.0050; v=0.000157; e=0.063; 均值在[p - 2v, p + 2v]的概率 95.46%2000; 202000; p=0.0099; v=0.000220; e=0.044; 均值在[p - 2v, p + 2v]的概率 95.45%5000; 205000; p=0.0244; v=0.000341; e=0.028; 均值在[p - 2v, p + 2v]的概率 95.45%500; 100500; p=0.0050; v=0.000222; e=0.089; 均值在[p - 2v, p + 2v]的概率 95.46%1000; 101000; p=0.0099; v=0.000312; e=0.063; 均值在[p - 2v, p + 2v]的概率 95.46%2000; 102000; p=0.0196; v=0.000434; e=0.044; 均值在[p - 2v, p + 2v]的概率 95.45%5000; 105000; p=0.0476; v=0.000657; e=0.028; 均值在[p - 2v, p + 2v]的概率 95.45%2000;  12000; p=0.1667; v=0.003402; e=0.041; 均值在[p - 2v, p + 2v]的概率 95.45%

结论: 坏样本大于2000以上, 在95%置信度下, 坏率的相对误差<5%