可信度的估计
- 二项分布中的(p) 服从Beta分布 $ { m beta}(alpha, eta)$, 密度函数 (frac1{B(alpha, eta)} x^{alpha-1} (1-x)^{eta -1})
- 均值 (frac alpha {alpha + eta})
- 方差 (frac {alpha eta} {(alpha+eta)^2 (alpha+ eta + 1) } )
from scipy.stats import beta
def 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+1
p = 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, d
test_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%