进行一种实验,各实验相互独立,若连续10次实验均失败,则进行该实验成功的可能性有多大?

 如题,这个问题可以用概率论的知识解答嘛,还是说这个题没有答案

或者说换种问法

袋子里有球若干,每个球除了颜色不同其他都一样,每次取出后放回。
现已尝试取10次,发现都是黑球,问这个袋子里不全是黑球的可能性有多少?

或者再换种问法

抛一枚硬币(我们不知道这枚硬币是不是错币),连续抛10次都是反面,问抛一次正面的可能性有多大?
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作者:Min.L
链接:https://www.zhihu.com/question/434921903/answer/1629038226
来源:知乎
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。

接着

的结论往后写一写。极大似然估计显然是不合适的。在这种极端值问题上,(在事先没有任何信息的情况下)用无信息先验的贝叶斯分析可能稍微靠谱一些。

https://www.zhihu.com/equation?tex=f%28%5Cmathbf%7Bx%7D%7Cp%29%5Cpropto+%281-p%29%5E%7B10%7D 是似然。加上一个 https://www.zhihu.com/equation?tex=p%5Csim+%5Cmathrm%7BUniform%7D%280%2C1%29 的先验;即 https://www.zhihu.com/equation?tex=%5Cpi%28p%29%3DI%280%3Cp%3C1%29 ,其中 https://www.zhihu.com/equation?tex=I%28%5Ccdot%29 是示性函数。这个先验被称为“无信息”,是因为它“没有偏袒 https://www.zhihu.com/equation?tex=p 的任何可能取值”。

由贝叶斯公式,后验: https://www.zhihu.com/equation?tex=%5Cpi%28p%7C%5Cmathbf%7Bx%7D%29%5Cpropto+f%28%5Cmathbf%7Bx%7D%7Cp%29%5Cpi%28p%29%5Cpropto%281-p%29%5E%7B10%7DI%280%3Cp%3C1%29

可以看出, https://www.zhihu.com/equation?tex=p%7C%5Cmathbf%7Bx%7D%5Csim+%5Cmathrm%7BBeta%7D%281%2C11%29 。有了后验分布,那么只要给定损失函数,就能找到一个合理的估计量 https://www.zhihu.com/equation?tex=%5Chat%7Bp%7D

如果啥损失函数也不知道,那么就默认使用后验期望作为估计量。 https://www.zhihu.com/equation?tex=%5Chat%7Bp%7D%3D%5Cfrac%7B1%7D%7B1%2B11%7D%3D%5Cfrac%7B1%7D%7B12%7D%5Capprox+0.0833

如果做了n次实验,总成功次数为y,那么使用均匀先验,得到的贝叶斯估计量为

https://www.zhihu.com/equation?tex=%5Chat%7Bp%7D%3D%5Cfrac%7By%2B1%7D%7Bn%2B2%7D 。当y=n时,也不会出现估计值为1的尴尬问题。


如果有先验信息,比如通过以往操作同类型实验的经验,模糊地认为成功率 https://www.zhihu.com/equation?tex=p 是不大于0.5的,那么可以用 https://www.zhihu.com/equation?tex=%5Cpi%28p%29%3D2I%280%3Cp%3C0.5%29 的均匀先验。10次实验成功0次,后验密度

https://www.zhihu.com/equation?tex=%5Cpi%28p%7C%5Cmathbf%7Bx%7D%29%3D%5Cfrac%7B1%7D%7BC%7D%5Ccdot+%281-p%29%5E%7B10%7D+I%280%3Cp%3C0.5%29 ,其中

https://www.zhihu.com/equation?tex=C%3D%5Cint_%7B0%7D%5E%7B0.5%7D%281-p%29%5E%7B10%7Ddp%3D%5Cfrac%7B1%7D%7B11%7D-%5Cfrac%7B1%7D%7B11%7D%280.5%29%5E%7B11%7D%5Capprox+0.09086

于是通过数值积分,可以算出贝叶斯估计值 0.08311

https://pic2.zhimg.com/80/v2-d0a58ff6760cdaf6d589bf1c5f83ce3f_720w.jpg?source=1940ef5c

这显然是估计问题;上来就“假设检验”的看得我脑壳疼。针对补充的提问进行一下更新:

抛一枚硬币,连续抛10次都是反面,问这枚硬币不是错币(两面都是反面)的可能性有多大?

还是记正面概率为 https://www.zhihu.com/equation?tex=p ,那么这个问题是在问 https://www.zhihu.com/equation?tex=%5Cmathbb%7BP%7D%28p%3E0%7C%E5%85%A8%E5%8F%8D%E9%9D%A2%29 的值,因此这里就必须将 https://www.zhihu.com/equation?tex=p 看成随机变量,而非固定参数。因为对不确定性的合理度量是概率,而定义概率需要随机变量。这就是一个显然的贝叶斯估计问题。记 https://www.zhihu.com/equation?tex=p 的参数空间 https://www.zhihu.com/equation?tex=%5CTheta%3D%5B0%2C1%29 【而不再是之前的 https://www.zhihu.com/equation?tex=%280%2C1%29 ;因为在之前的情况下,默认“实验成功率为0”是不可能事件;而现在由小球等具体情况可以知道, https://www.zhihu.com/equation?tex=p%3D0 是可能事件】。接下来需要确定先验 https://www.zhihu.com/equation?tex=%5Cpi%28p%29

首先需要考虑一个问题:如果对 https://www.zhihu.com/equation?tex=p 施加Beta或者均匀先验,后验还是连续的;因此单点概率为零。为了使问题有意义,有两种方法:

  1. 取定某个 https://www.zhihu.com/equation?tex=%5Cvarepsilon%3E0 ,使得对于所有 https://www.zhihu.com/equation?tex=p%5Cin%5B0%2C%5Cvarepsilon%29 ,都当作是等价零。
  2. 取定某个常数 https://www.zhihu.com/equation?tex=%5Cpi_0%5Cin%280%2C1%29 表示 https://www.zhihu.com/equation?tex=%5Cmathbb%7BP%7D%28p%3D0%29 ,再给定 https://www.zhihu.com/equation?tex=%280%2C1%29 上一个无信息先验密度 https://www.zhihu.com/equation?tex=%5Cpi_1 。从而先验为混合分布。当 https://www.zhihu.com/equation?tex=p%3D0 时, https://www.zhihu.com/equation?tex=X_1%2C...%2CX_%7B10%7D 退化为常数零,即 https://www.zhihu.com/equation?tex=%5Cmathbb%7BP%7D%28X_1%3DX_2%3D%5Ccdots%3DX_%7Bn%7D%3D0%7Cp%3D0%29%3D1

第二种比较常用。还是记 https://www.zhihu.com/equation?tex=Y%3D%5Csum_%7Bi%3D1%7D%5E%7Bn%7DX_i%5Csim%5Cmathrm%7BBinomial%7D%28n%2Cp%29 。按评论区老哥指出的,无信息先验 https://www.zhihu.com/equation?tex=%5Cpi_1 用Jeffreys的 https://www.zhihu.com/equation?tex=%5Cmathrm%7BBeta%7D%280.5%2C0.5%29 比较合适。当全反面时 https://www.zhihu.com/equation?tex=Y%3D0 。先求 https://www.zhihu.com/equation?tex=Y 的边缘(奇奇怪怪的斯蒂尔杰斯积分)

https://www.zhihu.com/equation?tex=%5Cbegin%7Balign%2A%7D+m%280%29%26%3D+%5Cint_%7B%5B0%2C1%29%7Df%280%7Cp%29dF%5E%7B%5Cpi%7D%28p%29%3D%5Cint_%7B%5C%7Bp%3D0%5C%7D%7Df%280%7Cp%29dF%5E%7B%5Cpi%7D%28p%29%2B%5Cint_%7B%5C%7Bp%3E0%5C%7D%7Df%280%7Cp%29dF%5E%7B%5Cpi%7D%28p%29%5C%5C+%26%3D+%5Cpi_0%2B%281-%5Cpi_0%29%5Cint_%7B%5C%7Bp%3E0%5C%7D%7Df%280%7Cp%29%5Cpi_1%28p%29dp%5C%5C+%26%3D+%5Cpi_0%2B%281-%5Cpi_0%29%5Cint_%7B0%7D%5E1%281-p%29%5En%5Cfrac%7B1%7D%7B%5C%7B%5CGamma%280.5%29%5C%7D%5E2%7Dp%5E%7B-0.5%7D%281-p%29%5E%7B-0.5%7Ddp%5C%5C+%26%3D%5Cpi_0%2B%281-%5Cpi_0%29%5Cfrac%7B%5CGamma%28n%2B0.5%29%7D%7B%5CGamma%28n%2B1%29%5CGamma%280.5%29%7D+%5Cend%7Balign%2A%7D


由贝叶斯公式得

https://www.zhihu.com/equation?tex=%5Cbegin%7Balign%2A%7D+%5Cmathbb%7BP%7D%28p%3D0%7CY%3D0%29%26%3D%5Cfrac%7B%5Cmathbb%7BP%7D%28Y%3D0%7Cp%3D0%29%5Cmathbb%7BP%7D%28p%3D0%29%7D%7Bm%280%29%7D%3D%5Cleft%281%2B%5Cfrac%7B1-%5Cpi_0%7D%7B%5Cpi_0%7D%5Cfrac%7B%5CGamma%28n%2B0.5%29%7D%7B%5CGamma%28n%2B1%29%5CGamma%280.5%29%7D%5Cright%29%5E%7B-1%7D+%5Cend%7Balign%2A%7D

不妨取 https://www.zhihu.com/equation?tex=%5Cpi_0%3D0.5 ,那么当 https://www.zhihu.com/equation?tex=n%3D10 时概率为0.8501976。

https://pic3.zhimg.com/80/v2-b1fde7af09921e4642cc29a7d3caa4e2_720w.jpg?source=1940ef5c 条件:n=10,硬币全反面

如果先验得认为“非错币”的可能性为“错币”的9倍,即 https://www.zhihu.com/equation?tex=%5Cpi_0%3D0.1 ,则需要9000多次反面才能使得后验概率达到0.95

https://pic2.zhimg.com/80/v2-9769165027b44018a6f71a9da93cb29c_720w.jpg?source=1940ef5c

惊闻“区间估计比点估计更优”的说法。回到原实验成功率问题,假定 https://www.zhihu.com/equation?tex=p%5Cin%280%2C1%29 。用Jeffreys先验可以求得10次失败的后验分布为 https://www.zhihu.com/equation?tex=%5Cmathrm%7BBeta%7D%280.5%2C10.5%29 ,点估计为后验期望 https://www.zhihu.com/equation?tex=0.5%2F11%3D0.04545455 ,最大后验密度(Highest Posterior Density) 95%可信区间为 https://www.zhihu.com/equation?tex=%280%2C0.1707731%29 【比(0,0.25)窄多了】

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