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{{short description|Probability distribution}}
{{Redirect|Binomial model|the binomial model in options pricing|Binomial options pricing model}}
<!-- EDITORS! Please see [[Wikipedia:WikiProject Probability#Standards]] for a discussion of standards used for probability distribution articles such as this one. -->
{{Probability distribution
| name = Binomial distribution
| type = mass
| pdf_image = [[File:Binomial distribution pmf.svg|300px|Probability mass function for the binomial distribution]]
| cdf_image = [[File:Binomial distribution cdf.svg|300px|Cumulative distribution function for the binomial distribution]]
| parameters = <math>n \in \{0, 1, 2, \ldots\}</math> – number of trials<br /><math>p \in [0,1]</math> – success probability for each trial<br /><math>q = 1 - p</math>
| cdf = <math>I_q(n - \lfloor k \rfloor, 1 + \lfloor k \rfloor)</math> (the [[Beta_function#Incomplete_beta_function|regularized incomplete beta function]])
| kurtosis = <math>\frac{1-6pq}{npq}</math>
| entropy = <math>\frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)</math><br /> in [[Shannon (unit)|shannon]]s. For [[nat (unit)|nats]], use the natural log in the log.
| mgf = <math>(q + pe^t)^n</math>
| char = <math>(q + pe^{it})^n</math>
| pgf = <math>G(z) = [q + pz]^n</math>
| fisher = <math> g_n(p) = \frac{n}{pq} </math><br />(for fixed <math>n</math>)
}}
{{Probability fundamentals}}
[[File:Pascal's triangle; binomial distribution.svg|thumb|280px|Binomial distribution for {{math|p {{=}} 0.5}}<br />with {{mvar|n}} and {{mvar|k}} as in [[Pascal's triangle]]<br /><br />The probability that a ball in a [[Bean machine|Galton box]] with 8 layers ({{math|''n'' {{=}} 8}}) ends up in the central bin ({{math|''k'' {{=}} 4}}) is {{math|70/256}}.]]
In [[probability theory]] and [[statistics]], the '''binomial distribution''' with parameters {{mvar|n}} and {{mvar|p}} is the [[discrete probability distribution]] of the number of successes in a sequence of {{mvar|n}} [[statistical independence|independent]] [[experiment (probability theory)|experiment]]s, each asking a [[yes–no question]], and each with its own [[Boolean-valued function|Boolean]]-valued [[outcome (probability)|outcome]]: ''success'' (with probability {{mvar|p}}) or ''failure'' (with probability {{math|''q'' {{=}} 1 − ''p''}}). A single success/failure experiment is also called a [[Bernoulli trial]] or Bernoulli experiment, and a sequence of outcomes is called a [[Bernoulli process]]; for a single trial, i.e., {{math|''n'' {{=}} 1}}, the binomial distribution is a [[Bernoulli distribution]]. The binomial distribution is the basis for the [[binomial test]] of [[statistical significance]].<ref>{{Cite book|last=Westland|first=J. Christopher|title=Audit Analytics: Data Science for the Accounting Profession|publisher=Springer|year=2020|isbn=978-3-030-49091-1|___location=Chicago, IL, USA|pages=53}}</ref>
The binomial distribution is frequently used to model the number of successes in a sample of size {{mvar|n}} drawn [[with replacement]] from a population of size {{mvar|N}}. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a [[hypergeometric distribution]], not a binomial one. However, for {{mvar|N}} much larger than {{mvar|n}}, the binomial distribution remains a good approximation, and is widely used.
==
=== Probability mass function ===
<math display="block">f(k,n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}</math>
for {{math|''k'' {{=}} 0, 1, 2, ..., ''n''}}, where
<math display="block">\binom{n}{k} =\frac{n!}{k!(n-k)!}</math>
is the [[binomial coefficient]]. The formula can be understood as follows: {{math|''p''{{sup|''k''}} ''q''{{sup|''n''−''k''}}}} is the probability of obtaining the sequence of {{mvar|n}} independent Bernoulli trials in which {{mvar|k}} trials are "successes" and the remaining {{math|''n'' − ''k''}} trials result in "failure". Since the trials are independent with probabilities remaining constant between them, any sequence of {{mvar|n}} trials with {{mvar|k}} successes (and {{math|''n'' − ''k''}} failures) has the same probability of being achieved (regardless of positions of successes within the sequence). There are <math display="inline">\binom{n}{k}</math> such sequences, since the binomial coefficient <math display="inline">\binom{n}{k}</math> counts the number of ways to choose the positions of the {{mvar|k}} successes among the {{mvar|n}} trials. The binomial distribution is concerned with the probability of obtaining ''any'' of these sequences, meaning the probability of obtaining one of them ({{math|''p''{{sup|''k''}} ''q''{{sup|''n''−''k''}}}}) must be added <math display="inline">\binom{n}{k}</math> times, hence <math display="inline">\Pr(X = k) = \binom{n}{k} p^k (1-p)^{n-k}</math>.
In creating reference tables for binomial distribution probability, usually, the table is filled in up to {{math|''n''/2}} values. This is because for {{math|''k'' > ''n''/2}}, the probability can be calculated by its complement as
<math display="block">f(k,n,p)=f(n-k,n,1-p). </math>
Looking at the expression {{math|''f''(''k'', ''n'', ''p'')}} as a function of {{mvar|k}}, there is a {{mvar|k}} value that maximizes it. This {{mvar|k}} value can be found by calculating
<math display="block"> \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} </math>
and comparing it to 1. There is always an integer {{mvar|M}} that satisfies<ref>{{cite book |last=Feller |first=W. |title=An Introduction to Probability Theory and Its Applications |url=https://archive.org/details/introductiontopr01wfel |url-access=limited |year=1968 |publisher=Wiley |___location=New York |edition=Third |page=[https://archive.org/details/introductiontopr01wfel/page/n167 151] (theorem in section VI.3) }}</ref>
<math display="block">(n+1)p-1 \leq M < (n+1)p.</math>
{{math|''f''(''k'', ''n'', ''p'')}} is monotone increasing for {{math|''k'' < ''M''}} and monotone decreasing for {{math|''k'' > ''M''}}, with the exception of the case where {{math|(''n'' + 1)''p''}} is an integer. In this case, there are two values for which {{mvar|f}} is maximal: {{math|(''n'' + 1) ''p''}} and {{math|(''n'' + 1) ''p'' − 1}}. {{mvar|M}} is the ''most probable'' outcome (that is, the most likely, although this can still be unlikely overall) of the Bernoulli trials and is called the [[Mode (statistics)|mode]].
Equivalently, {{math|''M'' − ''p'' < ''np'' ≤ ''M'' + 1 − ''p''}}. Taking the [[Floor and ceiling functions|floor function]], we obtain {{math|''M'' {{=}} floor(''np'')}}.{{NoteTag|Except the trivial case {{math|''p'' {{=}} 0}}, which must be checked separately.}}
=== Example ===
Suppose a [[fair coin|biased coin]] comes up heads with probability 0.3 when tossed. The probability of seeing exactly 4 heads in 6 tosses is
<math display="block">f(4,6,0.3) = \binom{6}{4} 0.3^4 (1-0.3)^{6-4}= 0.059535.</math>
=== Cumulative distribution function ===
<math display="block">F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},</math>
where <math>\lfloor k\rfloor</math> is the "floor" under {{math|''k''}}, i.e. the [[floor and ceiling functions|greatest integer]] less than or equal to {{math|''k''}}.
It can also be represented in terms of the [[regularized incomplete beta function]], as follows:<ref>{{cite book |last=Wadsworth |first=G. P. |title=Introduction to Probability and Random Variables |year=1960 |publisher=McGraw-Hill |___location=New York |page=[https://archive.org/details/introductiontopr0000wads/page/52 52] |url=https://archive.org/details/introductiontopr0000wads |url-access=registration }}</ref>
<math display="block">\begin{align}
F(k;n,p) & = \Pr(X \le k) \\
&= I_{1-p}(n-k, k+1) \\
& = (n-k) {n \choose k} \int_0^{1-p} t^{n-k-1} (1-t)^k \, dt ,
\end{align}</math>
which is equivalent to the [[cumulative distribution function]]s of the [[beta distribution]] and of the [[F-distribution|{{mvar|F}}-distribution]]:<ref>{{cite journal |last=Jowett |first=G. H. |year=1963 |title=The Relationship Between the Binomial and F Distributions |journal=Journal of the Royal Statistical Society, Series D |volume=13 |issue=1 |pages=55–57 |doi=10.2307/2986663 |jstor=2986663 }}</ref>
<math display="block">F(k;n,p) = F_{\text{beta-distribution}}\left(x=1-p;\alpha=n-k,\beta=k+1\right)</math>
<math display="block">F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).</math>
Some closed-form bounds for the cumulative distribution function are given [[#Tail bounds|below]].
== Properties ==
=== Expected value and variance ===
If {{math|''X'' ~ ''B''(''n'', ''p'')}}, that is, {{math|''X''}} is a binomially distributed random variable, {{mvar|n}} being the total number of experiments and ''p'' the probability of each experiment yielding a successful result, then the [[expected value]] of {{math|''X''}} is:<ref>See [https://proofwiki.org/wiki/Expectation_of_Binomial_Distribution Proof Wiki]</ref>
<math display="block"> \operatorname{E}[X] = np.</math>
This follows from the linearity of the expected value along with the fact that {{mvar|X}} is the sum of {{mvar|n}} identical Bernoulli random variables, each with expected value {{mvar|p}}. In other words, if <math>X_1, \ldots, X_n</math> are identical (and independent) Bernoulli random variables with parameter {{mvar|p}}, then {{math|1=''X'' = ''X''<sub>1</sub> + ... + ''X''<sub>''n''</sub>}} and
<math display="block">\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.</math>
The [[variance]] is:
<math display="block"> \operatorname{Var}(X) = npq = np(1 - p).</math>
This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.
=== Higher moments ===
<!-- Please stop changing the equation \mu_1 = 0, it is correct. The first central moment is not the mean. -->
The first 6 [[central moment]]s, defined as <math> \mu _{c}=\operatorname {E} \left[(X-\operatorname {E} [X])^{c}\right] </math>, are given by
<math display="block">\begin{align}
\mu_1 &= 0, \\
\mu_2 &= np \left(1-p\right), \\
\mu_3 &= np \left(1-p\right) \left(1-2p\right), \\
\mu_4 &= np \left(1-p\right) \left[1 + \left(3n-6\right) p \left(1-p\right)\right],\\
\mu_5 &= np \left(1-p\right) \left(1-2p\right) \left[1 + \left(10n-12\right) p \left(1-p\right)\right],\\
\mu_6 &= np \left(1-p\right) \left[1 - 30p\left(1-p\right)\left[1-4p(1-p)\right] + 5np \left(1-p\right)\left[5 - 26p\left(1-p\right)\right] + 15n^2 p^2 \left(1-p\right)^2\right].
\end{align}</math>
The non-central moments satisfy
<math display="block">\begin{align}
\operatorname {E}[X] &= np, \\
\operatorname {E}[X^2] &= np(1-p)+n^2p^2,
\end{align}</math>
and in general<ref name="Andreas2008">
{{citation
|last1=Knoblauch |first1=Andreas
|title=Closed-Form Expressions for the Moments of the Binomial Probability Distribution
|year=2008
|journal=SIAM Journal on Applied Mathematics
|url=https://www.jstor.org/stable/40233780
|volume=69
|issue=1
|pages=197–204
|doi=10.1137/070700024
|jstor=40233780
|url-access=subscription
}}</ref><ref name="Nguyen2021">
{{citation
|last1=Nguyen |first1=Duy
|title=A probabilistic approach to the moments of binomial random variables and application
|year=2021
|journal=The American Statistician
|url=https://www.tandfonline.com/doi/abs/10.1080/00031305.2019.1679257?journalCode=utas20
|volume=75
|issue=1
|pages=101–103
|doi=10.1080/00031305.2019.1679257
|s2cid=209923008
|url-access=subscription
}}</ref>
<math display="block">
\operatorname {E}[X^c] = \sum_{k=0}^c \left\{ {c \atop k} \right\} n^{\underline{k}} p^k,
</math>
where <math display="inline"> \left\{{c\atop k}\right\}</math> are the [[Stirling numbers of the second kind]], and <math>n^{\underline{k}} = n(n-1)\cdots(n-k+1)</math> is the <math>k</math>-th [[Falling and rising factorials|falling power]] of <math>n</math>.
A simple bound
<ref>{{Citation |last1=D. Ahle |first1=Thomas |title=Sharp and Simple Bounds for the raw Moments of the Binomial and Poisson Distributions
|year=2022
|volume=182
|doi=10.1016/j.spl.2021.109306
|journal=Statistics & Probability Letters
|page=109306 |arxiv=2103.17027 }}</ref> follows by bounding the Binomial moments via the [[Poisson distribution#Higher moments|higher Poisson moments]]:
<math display="block">
\operatorname {E}[X^c] \le
\left[\frac{c}{\ln\left(1+\frac{c}{np}\right)}\right]^c \le (np)^c \exp\left(\frac{c^2}{2np}\right).
</math>
This shows that if <math>c=O(\sqrt{np})</math>, then <math>\operatorname {E}[X^c]</math> is at most a constant factor away from <math>\operatorname {E}[X]^c</math>
=== Mode ===
Usually the [[mode (statistics)|mode]] of a binomial {{math|''B''(''n'', ''p'')}} distribution is equal to <math>\lfloor (n+1)p\rfloor</math>, where <math>\lfloor\cdot\rfloor</math> is the [[floor function]]. However, when {{math|(''n'' + 1)''p''}} is an integer and {{math|''p''}} is neither 0 nor 1, then the distribution has two modes: {{math|(''n'' + 1)''p''}} and {{math|(''n'' + 1)''p'' − 1}}. When {{math|''p''}} is equal to 0 or 1, the mode will be 0 and {{math|''n''}} correspondingly. These cases can be summarized as follows:
<math display="block">\text{mode} =
\begin{cases}
\lfloor (n+1)\,p\rfloor & \text{if }(n+1)p\text{ is 0 or a noninteger}, \\
(n+1)\,p\ \text{ and }\ (n+1)\,p - 1 &\text{if }(n+1)p\in\{1,\dots,n\}, \\
n & \text{if }(n+1)p = n + 1.
\end{cases}</math>
'''Proof:''' Let
<math display="block">f(k)=\binom nk p^k q^{n-k}.</math>
For <math>p=0</math> only <math>f(0)</math> has a nonzero value with <math>f(0)=1</math>. For <math>p=1</math> we find <math>f(n)=1</math> and <math>f(k)=0</math> for <math>k\neq n</math>. This proves that the mode is 0 for <math>p=0</math> and <math>n</math> for <math>p=1</math>.
<math display="block">\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}.</math>
From this follows
<math display="block">\begin{align}
k > (n+1)p-1 \Rightarrow f(k+1) < f(k) \\
k = (n+1)p-1 \Rightarrow f(k+1) = f(k) \\
k < (n+1)p-1 \Rightarrow f(k+1) > f(k)
\end{align}</math>
So when <math>(n+1)p-1</math> is an integer, then <math>(n+1)p-1</math> and <math>(n+1)p</math> is a mode. In the case that <math>(n+1)p-1\notin \Z</math>, then only <math>\lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor</math> is a mode.<ref>See also {{cite web |first=André |last=Nicolas |title=Finding mode in Binomial distribution |work=[[Stack Exchange]] |date=January 7, 2019 |url=https://math.stackexchange.com/q/117940 }}</ref>
=== Median ===
In general, there is no single formula to find the [[median]] for a binomial distribution, and it may even be non-unique. However, several special results have been established:
* If {{math|''np''}} is an integer, then the mean, median, and mode coincide and equal {{math|''np''}}.<ref>{{cite journal|last=Neumann|first=P.|year=1966|title=Über den Median der Binomial- and Poissonverteilung|journal=Wissenschaftliche Zeitschrift der Technischen Universität Dresden|volume=19|pages=29–33|language=de}}</ref><ref>Lord, Nick. (July 2010). "Binomial averages when the mean is an integer", [[The Mathematical Gazette]] 94, 331-332.</ref>
* Any median {{math|''m''}} must lie within the interval <math>\lfloor np \rfloor\leq m \leq \lceil np \rceil</math>.<ref name="KaasBuhrman">{{cite journal|first1=R.|last1=Kaas|first2=J.M.|last2=Buhrman|title=Mean, Median and Mode in Binomial Distributions|journal=Statistica Neerlandica|year=1980|volume=34|issue=1|pages=13–18|doi=10.1111/j.1467-9574.1980.tb00681.x}}</ref>
* A median {{math|''m''}} cannot lie too far away from the mean:<math>|m-np|\leq \min\{{\ln2}, \max\{p,1-p\}\}</math>.<ref name="Hamza">
{{cite journal
| last1 = Hamza | first1 = K.
| doi = 10.1016/0167-7152(94)00090-U
| title = The smallest uniform upper bound on the distance between the mean and the median of the binomial and Poisson distributions
| journal = Statistics & Probability Letters
| volume = 23
| pages = 21–25
| year = 1995
}}</ref>
* The median is unique and equal to {{math|1=''m'' = [[Rounding|round]](''np'')}} when {{math|1={{abs|''m'' − ''np''}} ≤ min{{brace|''p'', 1 − ''p''}}}} (except for the case when {{math|1=''p'' = 1/2}} and {{math|''n''}} is odd).<ref name="KaasBuhrman"/>
* When {{math|''p''}} is a rational number (with the exception of {{math|1=''p'' = 1/2}}\ and {{math|''n''}} odd) the median is unique.<ref name="Nowakowski">
{{cite journal
| last1 = Nowakowski | first1 = Sz.
| doi = 10.37418/amsj.10.4.9
| issn=1857-8365
| title = Uniqueness of a Median of a Binomial Distribution with Rational Probability
| journal = Advances in Mathematics: Scientific Journal
| volume = 10
| issue = 4
| pages = 1951–1958
| year = 2021
| arxiv = 2004.03280
| s2cid = 215238991
}}</ref>
* When <math display="inline">p = \tfrac{1}{2} </math> and {{math|''n''}} is odd, any number {{math|''m''}} in the interval <math display="inline"> \frac{1}{2} \left(n-1\right) \leq m \leq \frac{1}{2} \left(n+1\right)</math> is a median of the binomial distribution. If <math display="inline">p = \tfrac{1}{2} </math> and {{math|''n''}} is even, then <math display="inline">m = \tfrac{n}{2} </math> is the unique median.
=== Tail bounds ===
For {{math|''k'' ≤ ''np''}}, upper bounds can be derived for the lower tail of the cumulative distribution function <math>F(k;n,p) = \Pr(X \le k)</math>, the probability that there are at most {{math|''k''}} successes. Since <math>\Pr(X \ge k) = F(n-k;n,1-p) </math>, these bounds can also be seen as bounds for the upper tail of the cumulative distribution function for {{math|''k'' ≥ ''np''}}.
[[Hoeffding's inequality]] yields the simple bound
<math display="block"> F(k;n,p) \leq \exp\left(-2 n\left(p-\frac{k}{n}\right)^2\right), \!</math>
which is however not very tight. In particular, for {{math|1=''p'' = 1}}, we have that {{math|1=''F''(''k''; ''n'', ''p'') = 0}} (for fixed {{math|''k''}}, {{math|''n''}} with {{math|''k'' < ''n''}}), but Hoeffding's bound evaluates to a positive constant.
A sharper bound can be obtained from the [[Chernoff bound]]:<ref name="ag">{{cite journal |first1=R. |last1=Arratia |first2=L. |last2=Gordon |title=Tutorial on large deviations for the binomial distribution |journal=Bulletin of Mathematical Biology |volume=51 |issue=1 |year=1989 |pages=125–131 |doi=10.1007/BF02458840 |pmid=2706397 |s2cid=189884382 }}</ref>
<math display="block"> F(k;n,p) \leq \exp\left(-n D{\left(\frac{k}{n}\parallel p\right)}\right) </math>
where {{math|''D''(''a'' ∥ ''p'')}} is the [[Kullback–Leibler divergence|relative entropy (or Kullback-Leibler divergence)]] between an {{math|''a''}}-coin and a {{math|''p''}}-coin (i.e. between the {{math|Bernoulli(''a'')}} and {{math|Bernoulli(''p'')}} distribution):
<math display="block"> D(a\parallel p)=(a)\ln\frac{a}{p}+(1-a)\ln\frac{1-a}{1-p}. \!</math>
Asymptotically, this bound is reasonably tight; see <ref name="ag"/> for details.
One can also obtain ''lower'' bounds on the tail {{math|''F''(''k''; ''n'', ''p'')}}, known as anti-concentration bounds. By approximating the binomial coefficient with [[Stirling's approximation|Stirling's formula]] it can be shown that<ref>{{cite book |author1=Robert B. Ash |title=Information Theory |url=https://archive.org/details/informationtheor00ashr |url-access=limited |date=1990 |publisher=Dover Publications |page=[https://archive.org/details/informationtheor00ashr/page/n81 115]|isbn=9780486665214 }}</ref>
<math display="block"> F(k;n,p) \geq \frac{1}{\sqrt{8n\tfrac{k}{n}(1-\tfrac{k}{n})}} \exp\left(-n D{\left(\frac{k}{n}\parallel p\right)}\right),</math>
which implies the simpler but looser bound
<math display="block"> F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).</math>
For {{math|1=''p'' = 1/2}} and {{math|''k'' ≥ 3''n''/8}} for even {{math|''n''}}, it is possible to make the denominator constant:<ref>{{cite web |last1=Matoušek |first1=J. |last2=Vondrak |first2=J. |title=The Probabilistic Method |work=lecture notes |url=https://www.cs.cmu.edu/afs/cs.cmu.edu/academic/class/15859-f09/www/handouts/matousek-vondrak-prob-ln.pdf |archive-url=https://ghostarchive.org/archive/20221009/https://www.cs.cmu.edu/afs/cs.cmu.edu/academic/class/15859-f09/www/handouts/matousek-vondrak-prob-ln.pdf |archive-date=2022-10-09 |url-status=live }}</ref>
<math display="block"> F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!</math>
== Statistical inference ==
=== Estimation of parameters ===
{{see also|Beta distribution#Bayesian inference}}
When {{math|''n''}} is known, the parameter {{math|''p''}} can be estimated using the proportion of successes:
<math display="block"> \widehat{p} = \frac{x}{n}.</math>
This estimator is found using [[maximum likelihood estimator]] and also the [[method of moments (statistics)|method of moments]]. This estimator is [[Bias of an estimator|unbiased]] and uniformly with [[Minimum-variance unbiased estimator|minimum variance]], proven using [[Lehmann–Scheffé theorem]], since it is based on a [[minimal sufficient]] and [[Completeness (statistics)|complete]] statistic (i.e.: {{math|''x''}}). It is also [[Consistent estimator|consistent]] both in probability and in [[Mean squared error|MSE]]. This statistic is [[Asymptotic distribution|asymptotically]] [[normal distribution|normal]] thanks to the [[central limit theorem]], because it is the same as taking the [[arithmetic mean|mean]] over Bernoulli samples. It has a variance of <math> \operatorname{Var}(\hat{p}) = \frac{p(1-p)}{n}</math>, a property which is used in various ways, such as in [[Binomial_proportion_confidence_interval#Wald_interval|Wald's confidence intervals]].
A closed form [[Bayes estimator]] for {{math|''p''}} also exists when using the [[Beta distribution]] as a [[Conjugate prior|conjugate]] [[prior distribution]]. When using a general <math>\operatorname{Beta}(\alpha, \beta)</math> as a prior, the [[Bayes estimator#Posterior mean|posterior mean]] estimator is:
<math display="block"> \widehat{p}_b = \frac{x+\alpha}{n+\alpha+\beta}.</math>
The Bayes estimator is [[Asymptotic efficiency (Bayes)|asymptotically efficient]] and as the sample size approaches infinity ({{math|''n'' → ∞}}), it approaches the [[Maximum likelihood estimation|MLE]] solution.<ref>{{Cite journal |last=Wilcox |first=Rand R. |date=1979 |title=Estimating the Parameters of the Beta-Binomial Distribution |url=http://journals.sagepub.com/doi/10.1177/001316447903900302 |journal=Educational and Psychological Measurement |language=en |volume=39 |issue=3 |pages=527–535 |doi=10.1177/001316447903900302 |s2cid=121331083 |issn=0013-1644|url-access=subscription }}</ref> The Bayes estimator is [[Bias of an estimator|biased]] (how much depends on the priors), [[Bayes estimator#Admissibility|admissible]] and [[Consistent estimator|consistent]] in probability. Using the Bayesian estimator with the Beta distribution can be used with [[Thompson sampling]].
For the special case of using the [[standard uniform distribution]] as a [[non-informative prior]], <math>\operatorname{Beta}(\alpha{=}1,\, \beta{=}1) = U(0,1)</math>, the posterior mean estimator becomes:
<math display="block"> \widehat{p}_b = \frac{x+1}{n+2}.</math>
(A [[Bayes estimator#Posterior mode|posterior mode]] should just lead to the standard estimator.) This method is called the [[rule of succession]], which was introduced in the 18th century by [[Pierre-Simon Laplace]].
When relying on [[Jeffreys prior]], the prior is <math display="inline">\operatorname{Beta}(\alpha{=}\tfrac{1}{2}, \, \beta{=}\tfrac{1}{2})</math>,<ref>Marko Lalovic (https://stats.stackexchange.com/users/105848/marko-lalovic), Jeffreys prior for binomial likelihood, URL (version: 2019-03-04): https://stats.stackexchange.com/q/275608</ref> which leads to the estimator:
<math display="block"> \widehat{p}_{Jeffreys} = \frac{x+\frac{1}{2}}{n+1}.</math>
When estimating {{math|''p''}} with very rare events and a small {{math|''n''}} (e.g.: if {{math|1=''x'' = 0}}), then using the standard estimator leads to <math> \widehat{p} = 0,</math> which sometimes is unrealistic and undesirable. In such cases there are various alternative estimators.<ref>{{cite journal |last=Razzaghi |first=Mehdi |title=On the estimation of binomial success probability with zero occurrence in sample |journal=Journal of Modern Applied Statistical Methods |volume=1 |issue=2 |year=2002 |pages=326–332 |doi=10.22237/jmasm/1036110000 |doi-access=free }}</ref> One way is to use the Bayes estimator <math> \widehat{p}_b</math>, leading to:
<math display="block"> \widehat{p}_b = \frac{1}{n+2}.</math>
Another method is to use the upper bound of the [[confidence interval]] obtained using the [[Rule of three (statistics)|rule of three]]:
<math display="block"> \widehat{p}_{\text{rule of 3}} = \frac{3}{n}.</math>
=== Confidence intervals for the parameter p ===
{{Main|Binomial proportion confidence interval}}
{{see also|Z-test#Comparing the Proportions of Two Binomials}}
Even for quite large values of ''n'', the actual distribution of the mean is significantly nonnormal.<ref name=Brown2001>{{Citation |first1=Lawrence D. |last1=Brown |first2=T. Tony |last2=Cai |first3=Anirban |last3=DasGupta |year=2001 |title = Interval Estimation for a Binomial Proportion |url=http://www-stat.wharton.upenn.edu/~tcai/paper/html/Binomial-StatSci.html |journal=Statistical Science |volume=16 |issue=2 |pages=101–133 |access-date = 2015-01-05 |doi=10.1214/ss/1009213286|citeseerx=10.1.1.323.7752 }}</ref> Because of this problem several methods to estimate confidence intervals have been proposed.
In the equations for confidence intervals below, the variables have the following meaning:
* ''n''<sub>1</sub> is the number of successes out of ''n'', the total number of trials
* <math> \widehat{p\,} = \frac{n_1}{n}</math> is the proportion of successes
* <math>z</math> is the <math>1 - \tfrac{1}{2}\alpha</math> [[quantile]] of a [[standard normal distribution]] (i.e., [[probit]]) corresponding to the target error rate <math>\alpha</math>. For example, for a 95% [[confidence level]] the error <math>\alpha</math> = 0.05, so <math>1 - \tfrac{1}{2}\alpha</math> = 0.975 and <math>z</math> = 1.96.
==== Wald method ====
{{Main|Binomial proportion confidence interval#Wald interval}}
<math display="block"> \widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .</math>
A [[continuity correction]] of {{math|0.5/''n''}} may be added.{{clarify|date=July 2012}}
==== Agresti–Coull method ====
{{Main|Binomial proportion confidence interval#Agresti–Coull interval}}
<ref name=Agresti1988>{{Citation |last1=Agresti |first1=Alan |last2=Coull |first2=Brent A. |date=May 1998 |title=Approximate is better than 'exact' for interval estimation of binomial proportions |url = http://www.stat.ufl.edu/~aa/articles/agresti_coull_1998.pdf |journal=The American Statistician |volume=52 |issue=2 |pages=119–126 |access-date=2015-01-05 |doi=10.2307/2685469 |jstor=2685469 }}</ref>
<math display="block"> \tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } }</math>
Here the estimate of {{math|''p''}} is modified to
<math display="block"> \tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } </math>
This method works well for {{math|''n'' > 10}} and {{math|''n''<sub>1</sub> ≠ 0, ''n''}}.<ref>{{cite web|last1=Gulotta|first1=Joseph|title=Agresti-Coull Interval Method|url=https://pellucid.atlassian.net/wiki/spaces/PEL/pages/25722894/Agresti-Coull+Interval+Method#:~:text=The%20Agresti%2DCoull%20Interval%20Method,%2C%20or%20per%20100%2C000%2C%20etc|website=pellucid.atlassian.net|access-date=18 May 2021}}</ref> See here for <math>n\leq 10</math>.<ref>{{cite web|title=Confidence intervals|url=https://www.itl.nist.gov/div898/handbook/prc/section2/prc241.htm|website=itl.nist.gov|access-date=18 May 2021}}</ref> For {{math|1=''n''<sub>1</sub> = 0, ''n''}} use the Wilson (score) method below.
==== Arcsine method ====
{{Main|Binomial proportion confidence interval#Arcsine transformation}}
<ref name="Pires00">{{cite book |last=Pires |first=M. A. |chapter-url=https://www.math.tecnico.ulisboa.pt/~apires/PDFs/AP_COMPSTAT02.pdf |archive-url=https://ghostarchive.org/archive/20221009/https://www.math.tecnico.ulisboa.pt/~apires/PDFs/AP_COMPSTAT02.pdf |archive-date=2022-10-09 |url-status=live |chapter=Confidence intervals for a binomial proportion: comparison of methods and software evaluation |editor-last=Klinke |editor-first=S. |editor2-last=Ahrend |editor2-first=P. |editor3-last=Richter |editor3-first=L. |title=Proceedings of the Conference CompStat 2002 |others=Short Communications and Posters |year=2002 }}</ref>
<math display="block">\sin^2 \left(\arcsin \left(\sqrt{\hat{p}}\right) \pm \frac{z}{2\sqrt{n}} \right).</math>
==== Wilson (score) method ====
{{Main|Binomial proportion confidence interval#Wilson score interval}}
The notation in the formula below differs from the previous formulas in two respects:<ref name="Wilson1927">{{Citation |last = Wilson |first=Edwin B. |date = June 1927 |title = Probable inference, the law of succession, and statistical inference |url = http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |journal = Journal of the American Statistical Association |volume=22 |issue=158 |pages=209–212 |access-date= 2015-01-05 |doi = 10.2307/2276774 |url-status=dead |archive-url = https://web.archive.org/web/20150113082307/http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |archive-date = 2015-01-13 |jstor = 2276774 }}</ref>
* Firstly, {{math|''z''<sub>''x''</sub>}} has a slightly different interpretation in the formula below: it has its ordinary meaning of 'the {{math|''x''}}th quantile of the standard normal distribution', rather than being a shorthand for 'the {{math|(1 − ''x'')}}th quantile'.
* Secondly, this formula does not use a plus-minus to define the two bounds. Instead, one may use <math>z = z_{\alpha / 2}</math> to get the lower bound, or use <math>z = z_{1 - \alpha/2}</math> to get the upper bound. For example: for a 95% confidence level the error <math>\alpha</math> = 0.05, so one gets the lower bound by using <math>z = z_{\alpha/2} = z_{0.025} = - 1.96</math>, and one gets the upper bound by using <math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>.
<math display="block">\frac{
\hat{p} + \frac{z^2}{2n} + z
\sqrt{
\frac{\hat{p} \left(1 - \hat{p}\right)}{n} +
\frac{z^2}{4 n^2}
}
}{
1 + \frac{z^2}{n}
}</math><ref>
{{cite book
| chapter = Confidence intervals
| chapter-url = http://www.itl.nist.gov/div898/handbook/prc/section2/prc241.htm
| title = Engineering Statistics Handbook
| publisher = NIST/Sematech
| year = 2012
| access-date = 2017-07-23
}}</ref>
==== Comparison ====
The so-called "exact" ([[Binomial proportion confidence interval#Clopper–Pearson interval|Clopper–Pearson]]) method is the most conservative.<ref name="Brown2001" /> (''Exact'' does not mean perfectly accurate; rather, it indicates that the estimates will not be less conservative than the true value.)
The Wald method, although commonly recommended in textbooks, is the most biased.{{clarify|reason=what sense of bias is this|date=July 2012}}
== Related distributions ==
=== Sums of binomials ===
If {{math|''X'' ~ B(''n'', ''p'')}} and {{math|''Y'' ~ B(''m'', ''p'')}} are independent binomial variables with the same probability {{math|''p''}}, then {{math|''X'' + ''Y''}} is again a binomial variable; its distribution is {{math|1=''Z'' = ''X'' + ''Y'' ~ B(''n'' + ''m'', ''p'')}}:<ref>{{cite book |last1=Dekking |first1=F.M. |last2=Kraaikamp |first2=C. |last3=Lopohaa |first3=H.P. |last4=Meester |first4=L.E. |title=A Modern Introduction of Probability and Statistics |date=2005 |publisher=Springer-Verlag London |isbn=978-1-84628-168-6 |edition=1 |url=https://www.springer.com/gp/book/9781852338961}}</ref>
<math display="block">\begin{align}
\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]\\
&= \binom{n+m}k p^k (1-p)^{n+m-k}
\end{align}</math>
A Binomial distributed random variable {{math|''X'' ~ B(''n'', ''p'')}} can be considered as the sum of {{math|''n''}} Bernoulli distributed random variables. So the sum of two Binomial distributed random variables {{math|''X'' ~ B(''n'', ''p'')}} and {{math|''Y'' ~ B(''m'', ''p'')}} is equivalent to the sum of {{math|''n'' + ''m''}} Bernoulli distributed random variables, which means {{math|1=''Z'' = ''X'' + ''Y'' ~ B(''n'' + ''m'', ''p'')}}. This can also be proven directly using the addition rule.
However, if {{math|''X''}} and {{math|''Y''}} do not have the same probability {{math|''p''}}, then the variance of the sum will be [[Binomial sum variance inequality|smaller than the variance of a binomial variable]] distributed as {{math|B(''n'' + ''m'', {{overline|''p''}})}}.
=== Poisson binomial distribution ===
The binomial distribution is a special case of the [[Poisson binomial distribution]], which is the distribution of a sum of {{math|''n''}} independent non-identical [[Bernoulli trials]] {{math|B(''p''<sub>''i''</sub>)}}.<ref>
{{cite journal
| volume = 3
| issue = 2
| pages = 295–312
| last = Wang
| first = Y. H.
| title = On the number of successes in independent trials
| journal = Statistica Sinica
| year = 1993
| url = http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf
| url-status = dead
| archive-url = https://web.archive.org/web/20160303182353/http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf
| archive-date = 2016-03-03
}}
</ref>
=== Ratio of two binomial distributions ===
This result was first derived by Katz and coauthors in 1978.<ref name=Katz1978>{{cite journal |last1=Katz |first1=D. |display-authors=1 |first2=J. |last2=Baptista |first3=S. P. |last3=Azen |first4=M. C. |last4=Pike |year=1978 |title=Obtaining confidence intervals for the risk ratio in cohort studies |journal=Biometrics |volume=34 |issue=3 |pages=469–474 |doi=10.2307/2530610 |jstor=2530610 }}</ref>
Let {{nowrap|''X'' ~ B(''n'', ''p''<sub>1</sub>)}} and {{nowrap|''Y'' ~ B(''m'', ''p''<sub>2</sub>)}} be independent. Let {{nowrap|1=''T'' = (''X''/''n'') / (''Y''/''m'')}}.
Then log(''T'') is approximately normally distributed with mean log(''p''<sub>1</sub>/''p''<sub>2</sub>) and variance {{nowrap|((1/''p''<sub>1</sub>) − 1)/''n'' + ((1/''p''<sub>2</sub>) − 1)/''m''}}.
=== Conditional binomials ===
If ''X'' ~ B(''n'', ''p'') and ''Y'' | ''X'' ~ B(''X'', ''q'') (the conditional distribution of ''Y'', given ''X''), then ''Y'' is a simple binomial random variable with distribution ''Y'' ~ B(''n'', ''pq'').
For example, imagine throwing ''n'' balls to a basket ''U<sub>X</sub>'' and taking the balls that hit and throwing them to another basket ''U<sub>Y</sub>''. If ''p'' is the probability to hit ''U<sub>X</sub>'' then ''X'' ~ B(''n'', ''p'') is the number of balls that hit ''U<sub>X</sub>''. If ''q'' is the probability to hit ''U<sub>Y</sub>'' then the number of balls that hit ''U<sub>Y</sub>'' is ''Y'' ~ B(''X'', ''q'') and therefore ''Y'' ~ B(''n'', ''pq'').
{{hidden begin|style=width:60%|ta1=center|border=1px #aaa solid|title=[Proof]}}
Since <math> X \sim B(n, p) </math> and <math> Y \sim B(X, q) </math>, by the [[law of total probability]],
<math display="block">\begin{align}
\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]
&= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}
\end{align}</math>
Since <math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math> the equation above can be expressed as
<math display="block"> \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>
Factoring <math> p^k = p^m p^{k-m} </math> and pulling all the terms that don't depend on <math> k </math> out of the sum now yields
<math display="block">\begin{align}
\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]
&= \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>
After substituting <math> i = k - m </math> in the expression above, we get
<math display="block"> \Pr[Y = m] = \binom{n}{m} (pq)^m \left( \sum_{i=0}^{n-m} \binom{n-m}{i} (p - pq)^i (1-p)^{n-m - i} \right) </math>
Notice that the sum (in the parentheses) above equals <math> (p - pq + 1 - p)^{n-m} </math> by the [[binomial theorem]]. Substituting this in finally yields
<math display="block">\begin{align}
\Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]
&= \binom{n}{m} (pq)^m (1-pq)^{n-m}
\end{align}</math>
and thus <math> Y \sim B(n, pq) </math> as desired.
{{hidden end}}
=== Bernoulli distribution ===
The [[Bernoulli distribution]] is a special case of the binomial distribution, where {{math|1=''n'' = 1}}. Symbolically, {{math|''X'' ~ B(1, ''p'')}} has the same meaning as {{math|''X'' ~ Bernoulli(''p'')}}. Conversely, any binomial distribution, {{math|B(''n'', ''p'')}}, is the distribution of the sum of {{math|''n''}} independent [[Bernoulli trials]], {{math|Bernoulli(''p'')}}, each with the same probability {{math|''p''}}.<ref>{{cite web|last1=Taboga|first1=Marco|title=Lectures on Probability Theory and Mathematical Statistics|url=https://www.statlect.com/probability-distributions/binomial-distribution#hid3|website=statlect.com|access-date=18 December 2017}}</ref>
=== Normal approximation ===
{{see also|Binomial proportion confidence interval#Normal approximation interval}}
[[File:Binomial Distribution.svg|right|250px|thumb|Binomial [[probability mass function]] and normal [[probability density function]] approximation for {{math|1=''n'' = 6}} and {{math|1=''p'' = 0.5}}]]
If {{math|''n''}} is large enough, then the skew of the distribution is not too great. In this case a reasonable approximation to {{math|B(''n'', ''p'')}} is given by the [[normal distribution]]
<math display="block"> \mathcal{N}(np,\,np(1-p)),</math>
and this basic approximation can be improved in a simple way by using a suitable [[continuity correction]].
The basic approximation generally improves as {{math|''n''}} increases (at least 20) and is better when {{math|''p''}} is not near to 0 or 1.<ref name="bhh">{{cite book|title=Statistics for experimenters|url=https://archive.org/details/statisticsforexp00geor|url-access=registration|author=Box, Hunter and Hunter|publisher=Wiley|year=1978|page=[https://archive.org/details/statisticsforexp00geor/page/130 130]|isbn=9780471093152}}</ref> Various [[Rule of thumb|rules of thumb]] may be used to decide whether {{math|''n''}} is large enough, and {{math|''p''}} is far enough from the extremes of zero or one:
* One rule<ref name="bhh"/> is that for {{math|''n'' > 5}} the normal approximation is adequate if the absolute value of the skewness is strictly less than 0.3; that is, if <math display="block">\frac{|1-2p|}{\sqrt{np(1-p)}}=\frac1{\sqrt{n}}\left|\sqrt{\frac{1-p}p}-\sqrt{\frac{p}{1-p}}\,\right|<0.3.</math>
This can be made precise using the [[Berry–Esseen theorem]].
* A stronger rule states that the normal approximation is appropriate only if everything within 3 standard deviations of its mean is within the range of possible values; that is, only if <math display="block"> \mu \pm 3\sigma = n p \pm 3 \sqrt{np(1-p)}\in(0,n).</math>
: This 3-standard-deviation rule is equivalent to the following conditions, which also imply the first rule above. <math display="block">n>9 \left(\frac{1-p}{p} \right)\quad\text{and}\quad n>9\left(\frac{p}{1-p}\right).</math>
{{hidden begin|style=width:66%|ta1=center|border=1px #aaa solid|title=[Proof]}}
The rule <math> np\pm3\sqrt{np(1-p)}\in(0,n)</math> is totally equivalent to request that
<math display="block">np-3\sqrt{np(1-p)}>0\quad\text{and}\quad np+3\sqrt{np(1-p)}<n.</math>
Moving terms around yields:
<math display="block">np>3\sqrt{np(1-p)}\quad\text{and}\quad n(1-p)>3\sqrt{np(1-p)}.</math>
Since <math>0<p<1</math>, we can apply the square power and divide by the respective factors <math>np^2</math> and <math>n(1-p)^2</math>, to obtain the desired conditions:
<math display="block">n>9 \left(\frac{1-p}p\right) \quad\text{and}\quad n>9 \left(\frac{p}{1-p}\right).</math>
Notice that these conditions automatically imply that <math>n>9</math>. On the other hand, apply again the square root and divide by 3,
<math display="block">\frac{\sqrt{n}}3>\sqrt{\frac{1-p}p}>0 \quad \text{and} \quad \frac{\sqrt{n}}3 > \sqrt{\frac{p}{1-p}}>0.</math>
Subtracting the second set of inequalities from the first one yields:
<math display="block">\frac{\sqrt{n}}3>\sqrt{\frac{1-p}p}-\sqrt{\frac{p}{1-p}}>-\frac{\sqrt{n}}3;</math>
and so, the desired first rule is satisfied,
<math display="block">\left|\sqrt{\frac{1-p}p}-\sqrt{\frac{p}{1-p}}\,\right|<\frac{\sqrt{n}}3.</math>
{{hidden end}}
* Another commonly used rule is that both values {{math|''np''}} and {{math|''n''(1 − ''p'')}} must be greater than<ref>{{Cite book |last=Chen |first=Zac |title=H2 Mathematics Handbook |publisher=Educational Publishing House |year=2011 |isbn=9789814288484 |edition=1 |___location=Singapore |pages=350}}</ref><ref>{{Cite web |date=2023-05-29 |title=6.4: Normal Approximation to the Binomial Distribution - Statistics LibreTexts |url=https://stats.libretexts.org/Courses/Las_Positas_College/Math_40:_Statistics_and_Probability/06:_Continuous_Random_Variables_and_the_Normal_Distribution/6.04:_Normal_Approximation_to_the_Binomial_Distribution |access-date=2023-10-07 |archive-date=2023-05-29 |archive-url=https://web.archive.org/web/20230529211919/https://stats.libretexts.org/Courses/Las_Positas_College/Math_40:_Statistics_and_Probability/06:_Continuous_Random_Variables_and_the_Normal_Distribution/6.04:_Normal_Approximation_to_the_Binomial_Distribution |url-status=bot: unknown }}</ref> or equal to 5. However, the specific number varies from source to source, and depends on how good an approximation one wants. In particular, if one uses 9 instead of 5, the rule implies the results stated in the previous paragraphs.
{{hidden begin|style=width:66%|ta1=center|border=1px #aaa solid|title=[Proof]}}
Assume that both values <math>np</math> and <math>n(1-p)</math> are greater than 9. Since <math>0< p<1</math>, we easily have that
<math display="block">np\geq9>9(1-p)\quad\text{and}\quad n(1-p)\geq9>9p.</math>
We only have to divide now by the respective factors <math>p</math> and <math>1-p</math>, to deduce the alternative form of the 3-standard-deviation rule:
<math display="block">n>9 \left(\frac{1-p}p\right) \quad\text{and}\quad n>9 \left(\frac{p}{1-p}\right).</math>
{{hidden end}}
The following is an example of applying a [[continuity correction]]. Suppose one wishes to calculate {{math|Pr(''X'' ≤ 8)}} for a binomial random variable {{math|''X''}}. If {{math|''Y''}} has a distribution given by the normal approximation, then {{math|Pr(''X'' ≤ 8)}} is approximated by {{math|Pr(''Y'' ≤ 8.5)}}. The addition of 0.5 is the continuity correction; the uncorrected normal approximation gives considerably less accurate results.
This approximation, known as [[de Moivre–Laplace theorem]], is a huge time-saver when undertaking calculations by hand (exact calculations with large {{math|''n''}} are very onerous); historically, it was the first use of the normal distribution, introduced in [[Abraham de Moivre]]'s book ''[[The Doctrine of Chances]]'' in 1738. Nowadays, it can be seen as a consequence of the [[central limit theorem]] since {{math|B(''n'', ''p'')}} is a sum of {{math|''n''}} independent, identically distributed [[Bernoulli distribution|Bernoulli variables]] with parameter {{math|''p''}}. This fact is the basis of a [[hypothesis test]], a "proportion z-test", for the value of {{math|''p''}} using {{math|''x''/''n''}}, the sample proportion and estimator of {{math|''p''}}, in a [[common test statistics|common test statistic]].<ref>[[NIST]]/[[SEMATECH]], [http://www.itl.nist.gov/div898/handbook/prc/section2/prc24.htm "7.2.4. Does the proportion of defectives meet requirements?"] ''e-Handbook of Statistical Methods.''</ref>
For example, suppose one randomly samples {{math|''n''}} people out of a large population and ask them whether they agree with a certain statement. The proportion of people who agree will of course depend on the sample. If groups of ''n'' people were sampled repeatedly and truly randomly, the proportions would follow an approximate normal distribution with mean equal to the true proportion ''p'' of agreement in the population and with standard deviation
<math display="block">\sigma = \sqrt{\frac{p(1-p)}{n}}</math>
=== Poisson approximation ===
The binomial distribution converges towards the [[Poisson distribution]] as the number of trials goes to infinity while the product {{math|''np''}} converges to a finite limit. Therefore, the Poisson distribution with parameter {{math|1=''λ'' = ''np''}} can be used as an approximation to {{math|B(''n'', ''p'')}} of the binomial distribution if {{math|''n''}} is sufficiently large and {{math|''p''}} is sufficiently small. According to rules of thumb, this approximation is good if {{math|''n'' ≥ 20}} and {{math|''p'' ≤ 0.05}}<ref>{{cite news |date=2023-03-28 |title=12.4 – Approximating the Binomial Distribution {{!}} STAT 414 |newspaper=Pennstate: Statistics Online Courses |url=https://online.stat.psu.edu/stat414/lesson/12/12.4 |access-date=2023-10-08 |archive-date=2023-03-28 |archive-url=https://web.archive.org/web/20230328081322/https://online.stat.psu.edu/stat414/lesson/12/12.4 |url-status=bot: unknown }}</ref> such that {{math|''np'' ≤ 1}}, or if {{math|''n'' > 50}} and {{math|''p'' < 0.1}} such that {{math|''np'' < 5}},<ref>{{Cite book |last=Chen |first=Zac |title=H2 mathematics handbook |publisher=Educational publishing house |year=2011 |isbn=9789814288484 |edition=1 |___location=Singapore |pages=348}}</ref> or if {{math|''n'' ≥ 100}} and {{math|''np'' ≤ 10}}.<ref name="nist">[[NIST]]/[[SEMATECH]], [http://www.itl.nist.gov/div898/handbook/pmc/section3/pmc331.htm "6.3.3.1. Counts Control Charts"], ''e-Handbook of Statistical Methods.''</ref><ref>{{Cite web |date=2023-03-13 |title=The Connection Between the Poisson and Binomial Distributions |url=https://mathcenter.oxford.emory.edu/site/math117/connectingPoissonAndBinomial/ |access-date=2023-10-08 |archive-date=2023-03-13 |archive-url=https://web.archive.org/web/20230313085931/https://mathcenter.oxford.emory.edu/site/math117/connectingPoissonAndBinomial/ |url-status=bot: unknown }}</ref>
Concerning the accuracy of Poisson approximation, see Novak,<ref>Novak S.Y. (2011) Extreme value methods with applications to finance. London: CRC/ Chapman & Hall/Taylor & Francis. {{ISBN|9781-43983-5746}}.</ref> ch. 4, and references therein.
=== Limiting distributions ===
* ''[[Poisson limit theorem]]'': As {{math|''n''}} approaches {{math|∞}} and {{math|''p''}} approaches 0 with the product {{math|''np''}} held fixed, the {{math|Binomial(''n'', ''p'')}} distribution approaches the [[Poisson distribution]] with [[expected value]] {{math|1=''λ'' = ''np''}}.<ref name="nist"/>
* ''[[de Moivre–Laplace theorem]]'': As {{math|''n''}} approaches {{math|∞}} while {{math|''p''}} remains fixed, the distribution of <math display="block">\frac{X-np}{\sqrt{np(1-p)}}</math> approaches the [[normal distribution]] with expected value 0 and [[variance]] 1. This result is sometimes loosely stated by saying that the distribution of {{math|''X''}} is [[Asymptotic normality|asymptotically normal]] with expected value 0 and [[variance]] 1. This result is a specific case of the [[central limit theorem]].
=== Beta distribution ===
The binomial distribution and beta distribution are different views of the same model of repeated Bernoulli trials. The binomial distribution is the [[Probability mass function|PMF]] of {{mvar|k}} successes given {{mvar|n}} independent events each with a probability {{mvar|p}} of success.
Mathematically, when {{math|1=''α'' = ''k'' + 1}} and {{math|1=''β'' = ''n'' − ''k'' + 1}}, the beta distribution and the binomial distribution are related by{{clarification needed|date=March 2023|
reason=Is the left hand side referring to a probability density, and the right hand side to a probability mass function? Clearly a beta distributed random variable can not be a scalar multiple of a binomial random variable given that the former is continuous and the latter discrete. In any case, it would seem to be more correct to say that this relationship means that the PDF of one is related to the PMF of the other, rather than appearing to say that the _distributions_ (often interchangeable with their CDFs) are directly related to one another.
}} a factor of {{math|''n'' + 1}}:
<math display="block">\operatorname{Beta}(p;\alpha;\beta) = (n+1)B(k;n;p)</math>
[[Beta distribution]]s also provide a family of [[prior distribution|prior probability distribution]]s for binomial distributions in [[Bayesian inference]]:<ref name=MacKay>{{cite book| last=MacKay| first=David| title = Information Theory, Inference and Learning Algorithms|year=2003| publisher=Cambridge University Press; First Edition |isbn=978-0521642989}}</ref>
<math display="block">P(p;\alpha,\beta) = \frac{p^{\alpha-1}(1-p)^{\beta-1}}{\operatorname{Beta}(\alpha,\beta)}.</math>
Given a uniform prior, the posterior distribution for the probability of success {{mvar|p}} given {{mvar|n}} independent events with {{mvar|k}} observed successes is a beta distribution.<ref>{{Cite web|url=https://www.statlect.com/probability-distributions/beta-distribution|title = Beta distribution}}</ref>
== Computational methods ==
=== Random number generation ===
{{further|Pseudo-random number sampling}}
Methods for [[random number generation]] where the [[marginal distribution]] is a binomial distribution are well-established.<ref>Devroye, Luc (1986) ''Non-Uniform Random Variate Generation'', New York: Springer-Verlag. (See especially [http://luc.devroye.org/chapter_ten.pdf Chapter X, Discrete Univariate Distributions])</ref><ref>
{{cite journal
| pages = 216–222
| year = 1988
| doi = 10.1145/42372.42381
| last2 = Schmeiser| first1 = V.
| volume = 31| first2 = B. W.
| journal = Communications of the ACM
| title = Binomial random variate generation
| last1 = Kachitvichyanukul| issue = 2
| s2cid = 18698828
}}</ref>
One way to generate [[random variate]]s samples from a binomial distribution is to use an inversion algorithm. To do so, one must calculate the probability that {{math|1=Pr(''X'' = ''k'')}} for all values {{mvar|k}} from {{math|0}} through {{mvar|n}}. (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.
== History ==
This distribution was derived by [[Jacob Bernoulli]]. He considered the case where {{math|1=''p'' = ''r''/(''r'' + ''s'')}} where {{math|''p''}} is the probability of success and {{math|''r''}} and {{math|''s''}} are positive integers. [[Blaise Pascal]] had earlier considered the case where {{math|1=''p'' = 1/2}}, tabulating the corresponding binomial coefficients in what is now recognized as [[Pascal's triangle]].<ref name=":1">{{cite book|last=Katz|first=Victor|title=A History of Mathematics: An Introduction|publisher=Addison-Wesley|year=2009|isbn=978-0-321-38700-4|pages=491|chapter=14.3: Elementary Probability}}</ref>
== See also ==
{{Portal|Mathematics}}
* [[Logistic regression]]
* [[Multinomial distribution]]
* [[Negative binomial distribution]]
* [[Beta-binomial distribution]]
* Binomial measure, an example of a [[Multifractal system|multifractal]] [[measure (mathematics)|measure]].<ref>Mandelbrot, B. B., Fisher, A. J., & Calvet, L. E. (1997). A multifractal model of asset returns. ''3.2 The Binomial Measure is the Simplest Example of a Multifractal''</ref>
* [[Statistical mechanics]]
* [[Piling-up lemma]], the resulting probability when [[XOR]]-ing independent Boolean variables
== Notes ==
{{reflist|group=note}}
== References ==
{{reflist|colwidth=30em}}
== Further reading ==
* {{cite book |first=Werner Z. |last=Hirsch |title=Introduction to Modern Statistics |___location=New York |publisher=MacMillan |year=1957 |chapter=Binomial Distribution—Success or Failure, How Likely Are They? |pages=140–153 |chapter-url=https://books.google.com/books?id=KostAAAAIAAJ&pg=PA140 }}
* {{cite book |first1=John |last1=Neter |first2=William |last2=Wasserman |first3=G. A. |last3=Whitmore |title=Applied Statistics |___location=Boston |publisher=Allyn & Bacon |edition=Third |year=1988 |isbn=0-205-10328-6 |pages=185–192 }}
== External links ==
{{Commons category|Binomial distributions}}{{Wikifunctions|Z20094|}}
* Interactive graphic: [http://www.math.wm.edu/~leemis/chart/UDR/UDR.html Univariate Distribution Relationships]
* [http://www.fxsolver.com/browse/formulas/Binomial+distribution Binomial distribution formula calculator]
* Difference of two binomial variables: [https://math.stackexchange.com/q/1065487 X-Y] or [https://math.stackexchange.com/q/562119 |X-Y|]
* [http://www.wolframalpha.com/input/?i=Prob+x+%3E+19+if+x+is+binomial+with+n+%3D+36++and+p+%3D+.6 Querying the binomial probability distribution in WolframAlpha]
* Confidence (credible) intervals for binomial probability, p: [https://causascientia.org/math_stat/ProportionCI.html online calculator] available at [https://causascientia.org causaScientia.org]
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{{ProbDistributions|discrete-finite}}
{{DEFAULTSORT:Binomial Distribution}}
[[Category:Discrete distributions]]
[[Category:Factorial and binomial topics]]
[[Category:Conjugate prior distributions]]
[[Category:Exponential family distributions]]
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