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第864行：
第864行： − + − + 第872行：
第872行： − + 第963行：
第963行： − − \sqrt{} − − <math> \end{align}</math> 第974行：
第970行： − + 第983行：
第979行： − − <math>\begin{align}</math> − + 第1,017行：
第1,011行： − + 第1,023行：
第1,017行： − 因此Y \sim B(n, pq)为所需值。+ 第1,058行：
第1,052行： − + − − :<math>\operatorname P(Z=k) &= \sum_{i=0}^k\left[\binom{n}i p^i (1-p)^{n-i}\right]\left[\binom{m}{k-i} p^{k-i} (1-p)^{m-k+i}\right]\\</math> 第1,066行：
第1,058行： − <math> &= \binom{n+m}k p^k (1-p)^{n+m-k}</math>+ 第1,153行：
第1,145行： − + − + − &= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}</math>+ 第1,178行：
第1,170行： − :<math>\Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]</math>+ − + − &= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right) − − \end{align}</math>
无编辑摘要
<font color="#ff8000">Wald 法</font>
<font color="#ff8000">Wald 法</font>
: <math> \widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .</math>
:<math> \widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .</math>
<math>\frac{}\widehat{p\,} + \frac{z^2}{2n} + z</math>
<math>\frac{p}{z^2}{2n}\widehat{p\,} + \frac{z^2}{2n} + z</math>
: A [[continuity correction]] of 0.5/''n'' may be added. {{clarify|date=July 2012}};
: A [[continuity correction]] of 0.5/''n'' may be added. {{clarify|date=July 2012}};
可以添加一个0.5/''n''连续调整。（2012年7月更新）
可以添加一个0.5/''n''连续调整。（2012年7月更新）
<math> \sqrt{\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} </math>
<math> \sqrt{\frac{p}{n}\widehat{p\,}(1 - \widehat{p\,}){n} </math>
==== Agresti–Coull method ====
==== Agresti–Coull method ====
<math> &= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}</math>
<math> &= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}</math>
<math>\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} </math>
<math>\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} </math>
由于<math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}</math>，上述方程可表示为
由于<math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}</math>，上述方程可表示为
<math> \frac{z^2}{4 n^2}</math>
<math>\frac{z^2}{4 n^2}</math>
<math> \Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} </math>
<math> \Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} </math>
}{
}{
<math> 1 + \frac{z^2}{n}</math>
<math> 1 + \frac{z^2}{n}</math>
<math> \Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]}</math><ref>{{cite book
<math>\Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]}</math><ref>{{cite book
<math> &= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)</math>
<math> &= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)</math>
}}</ref>
}}</ref>
<math> &= \binom{n}{m} (pq)^m (1-pq)^{n-m}</math>
<math> &= \binom{n}{m} (pq)^m (1-pq)^{n-m}</math>
==== Comparison ====
==== Comparison ====
and thus Y \sim B(n, pq) as desired.
and thus Y \sim B(n, pq) as desired.
因此<math>Y \sim B(n, pq)</math>为所需值。
The exact ([[Binomial proportion confidence interval#Clopper–Pearson interval|Clopper–Pearson]]) method is the most conservative.<ref name="Brown2001" />
The exact ([[Binomial proportion confidence interval#Clopper–Pearson interval|Clopper–Pearson]]) method is the most conservative.<ref name="Brown2001" />
<math>\operatorname P(Z=k) &= \sum_{i=0}^k\left[\binom{n}i p^i (1-p)^{n-i}\right]\left[\binom{m}{k-i} p^{k-i} (1-p)^{m-k+i}\right]\\</math>
Binomial [[probability mass function and normal probability density function approximation for n = 6 and p = 0.5]]
Binomial [[probability mass function and normal probability density function approximation for n = 6 and p = 0.5]]
二项式n = 6 and p = 0.5的概率质量函数和正态概率密度函数近似
二项式n = 6 and p = 0.5的概率质量函数和正态概率密度函数近似
<math> &= \binom{n+m}k p^k (1-p)^{n+m-k}</math>
:<math>\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]</math>
:<math>\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]</math>
<math>P(p;\alpha,\beta) = \frac{p^{\alpha-1}(1-p)^{\beta-1}}{\mathrm{B}(\alpha,\beta)}.</math>
<math>P(p;\alpha,\beta) =\frac{p^{\alpha-1}(1-p)^{\beta-1}}{\mathrm{B}(\alpha,\beta)}.</math>
<math>P (p; alpha，beta) = frac { p ^ { alpha-1}(1-p) ^ { beta-1}{ mathrm { b }(alpha，beta)}}.
<math>P (p; alpha，beta) = frac { p ^ { alpha-1}(1-p) ^ { beta-1}{ mathrm { b }(alpha，beta)}}.</math>
<math> &= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}</math></math>
Given a uniform prior, the posterior distribution for the probability of success given independent events with observed successes is a beta distribution.
Given a uniform prior, the posterior distribution for the probability of success given independent events with observed successes is a beta distribution.
<font color="#ff8000">边缘分布 marginal distribution </font>是二项分布较完善的随机数产生方法。
<font color="#ff8000">边缘分布 marginal distribution </font>是二项分布较完善的随机数产生方法。
<math>\Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]</math>
One way to generate random samples from a binomial distribution is to use an inversion algorithm. To do so, one must calculate the probability that for all values from through . (These probabilities should sum to a value close to one, in order to encompass the entire sample space.) Then by using a pseudorandom number generator to generate samples uniformly between 0 and 1, one can transform the calculated samples into discrete numbers by using the probabilities calculated in the first step.
One way to generate random samples from a binomial distribution is to use an inversion algorithm. To do so, one must calculate the probability that for all values from through . (These probabilities should sum to a value close to one, in order to encompass the entire sample space.) Then by using a pseudorandom number generator to generate samples uniformly between 0 and 1, one can transform the calculated samples into discrete numbers by using the probabilities calculated in the first step.
一种从二项分布中产生随机样本的方法是使用<font color="#ff8000">反演算法 inversion algorithm </font>。要做到这一点，我们必须计算从到的所有值的概率。(为了包含整个样本空间，这些概率的和应该接近于1。)然后，通过使用伪随机数生成器来生成介于0和1之间的样本，可以使用在第一步计算出的概率将计算出的样本转换成离散数。
一种从二项分布中产生随机样本的方法是使用<font color="#ff8000">反演算法 inversion algorithm </font>。要做到这一点，我们必须计算从到的所有值的概率。(为了包含整个样本空间，这些概率的和应该接近于1。)然后，通过使用伪随机数生成器来生成介于0和1之间的样本，可以使用在第一步计算出的概率将计算出的样本转换成离散数。
:<math>&= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)</math>
After substituting <math> i = k - m </math> in the expression above, we get
After substituting <math> i = k - m </math> in the expression above, we get