Binomial distribution

In probability theory and statistics, the binomial distribution with parameters and is the discrete probability distribution of the number of successes in a sequence of independent experiments, each asking a yes–no question, and each with its own Boolean-valued outcome: ''success'' (with probability ) or ''failure'' (with probability ). 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., , the binomial distribution is a Bernoulli distribution. The binomial distribution is the basis for the binomial test of statistical significance.

The binomial distribution is frequently used to model the number of successes in a sample of size drawn with replacement from a population of size . 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 much larger than , the binomial distribution remains a good approximation, and is widely used.

Definitions

Probability mass function

If the random variable follows the binomial distribution with parameters and , we write . The probability of getting exactly successes in independent Bernoulli trials (with the same rate ) is given by the probability mass function:
: $f(k,n,p) = \Pr(X = k) = \binomp^k(1-p)^$
for , where
: $\binom =\frac$
is the binomial coefficient. The formula can be understood as follows: is the probability of obtaining the sequence of independent Bernoulli trials in which trials are "successes" and the remaining trials result in "failure". Since the trials are independent with probabilities remaining constant between them, any sequence of trials with successes (and failures) has the same probability of being achieved (regardless of positions of successes within the sequence). There are $\binom$ such sequences, since the binomial coefficient $\binom$ counts the number of ways to choose the positions of the successes among the trials. The binomial distribution is concerned with the probability of obtaining ''any'' of these sequences, meaning the probability of obtaining one of them () must be added $\binom$ times, hence $\Pr(X = k) = \binom p^k (1-p)^$.

In creating reference tables for binomial distribution probability, usually, the table is filled in up to values. This is because for , the probability can be calculated by its complement as
: $f(k,n,p)=f(n-k,n,1-p).$

Looking at the expression as a function of , there is a value that maximizes it. This value can be found by calculating
: $\frac=\frac$
and comparing it to 1. There is always an integer that satisfies
: $(n+1)p-1 \leq M < (n+1)p.$

is monotone increasing for and monotone decreasing for , with the exception of the case where is an integer. In this case, there are two values for which is maximal: and . 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.

Equivalently, . Taking the floor function, we obtain .

Example

Suppose a biased coin comes up heads with probability 0.3 when tossed. The probability of seeing exactly 4 heads in 6 tosses is
: $f(4,6,0.3) = \binom0.3^4 (1-0.3)^= 0.059535.$

Cumulative distribution function

The cumulative distribution function can be expressed as:
: $F(k;n,p) = \Pr(X \le k) = \sum_^ p^i(1-p)^,$
where $\lfloor k\rfloor$ is the "floor" under , i.e. the greatest integer less than or equal to .

It can also be represented in terms of the regularized incomplete beta function, as follows:
: $\begin F(k;n,p) & = \Pr(X \le k) \\ &= I_(n-k, k+1) \\ & = (n-k) \int_0^ t^ (1-t)^k \, dt , \end$
which is equivalent to the cumulative distribution functions of the beta distribution and of the -distribution:
: $F(k;n,p) = F_\left(x=1-p;\alpha=n-k,\beta=k+1\right)$
: $F(k;n,p) = F_\left(x=\frac\frac;d_1=2(n-k),d_2=2(k+1)\right).$

Some closed-form bounds for the cumulative distribution function are given below.

Properties

Expected value and variance

If , that is, is a binomially distributed random variable, being the total number of experiments and ''p'' the probability of each experiment yielding a successful result, then the expected value of is:
: \operatorname = np.

This follows from the linearity of the expected value along with the fact that is the sum of identical Bernoulli random variables, each with expected value . In other words, if $X_1, \ldots, X_n$ are identical (and independent) Bernoulli random variables with parameter , then and
: \operatorname = \operatorname _1 + \cdots + X_n= \operatorname _1+ \cdots + \operatorname _n= p + \cdots + p = np.

The variance is:
: $\operatorname(X) = npq = np(1 - p).$

This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.

Higher moments

The first 6 central moments, defined as \mu _=\operatorname \left X-\operatorname [X^\right">.html" ;"title="X-\operatorname [X">X-\operatorname [X^\right, are given by
: $\begin \mu_1 &= 0, \\ \mu_2 &= np(1-p),\\ \mu_3 &= np(1-p)(1-2p),\\ \mu_4 &= np(1-p)(1+(3n-6)p(1-p)),\\ \mu_5 &= np(1-p)(1-2p)(1+(10n-12)p(1-p)),\\ \mu_6 &= np(1-p)(1-30p(1-p)(1-4p(1-p))+5np(1-p)(5-26p(1-p))+15n^2 p^2 (1-p)^2). \end$

The non-central moments satisfy
: \begin
\operatorname &= np, \\
\operatorname [X^2] &= np(1-p)+n^2p^2,
\end
and in general



:
\operatorname ^c= \sum_^c \left\ n^ p^k,

where $\textstyle \left\$ are the Stirling numbers of the second kind, and $n^ = n(n-1)\cdots(n-k+1)$ is the $k$th falling power of $n$.
A simple bound
follows by bounding the Binomial moments via the higher Poisson moments:
:
\operatorname ^c\le
\left(\frac\right)^c \le (np)^c \exp\left(\frac\right).

This shows that if $c=O(\sqrt)$, then \operatorname ^c/math> is at most a constant factor away from \operatorname c

Mode

Usually the mode of a binomial distribution is equal to $\lfloor (n+1)p\rfloor$, where $\lfloor\cdot\rfloor$ is the floor function. However, when is an integer and is neither 0 nor 1, then the distribution has two modes: and . When is equal to 0 or 1, the mode will be 0 and correspondingly. These cases can be summarized as follows:
: $\text = \begin \lfloor (n+1)\,p\rfloor & \text(n+1)p\text, \\ (n+1)\,p\ \text\ (n+1)\,p - 1 &\text(n+1)p\in\, \\ n & \text(n+1)p = n + 1. \end$

Proof: Let
: $f(k)=\binom nk p^k q^.$

For $p=0$ only $f(0)$ has a nonzero value with $f(0)=1$. For $p=1$ we find $f(n)=1$ and $f(k)=0$ for $k\neq n$. This proves that the mode is 0 for $p=0$ and $n$ for $p=1$.

Let $0 < p < 1$. We find
:$\frac = \frac$.

From this follows
: $\begin k > (n+1)p-1 \Rightarrow f(k+1) < f(k) \\ k = (n+1)p-1 \Rightarrow f(k+1) = f(k) \\ k < (n+1)p-1 \Rightarrow f(k+1) > f(k) \end$

So when $(n+1)p-1$ is an integer, then $(n+1)p-1$ and $(n+1)p$ is a mode. In the case that $(n+1)p-1\notin \Z$, then only $\lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor$ is a mode.

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 is an integer, then the mean, median, and mode coincide and equal .
* Any median must lie within the interval $\lfloor np \rfloor\leq m \leq \lceil np \rceil$.
* A median cannot lie too far away from the mean:$, m-np, \leq \min\$ .

* The median is unique and equal to when (except for the case when and is odd).
* When is a rational number (with the exception of \ and odd) the median is unique.

* When $p= \frac$ and is odd, any number in the interval $\frac \bigl(n-1\bigr)\leq m \leq \frac \bigl(n+1\bigr)$ is a median of the binomial distribution. If $p= \frac$ and is even, then $m= \frac$ is the unique median.

Tail bounds

For , upper bounds can be derived for the lower tail of the cumulative distribution function $F(k;n,p) = \Pr(X \le k)$, the probability that there are at most successes. Since $\Pr(X \ge k) = F(n-k;n,1-p)$, these bounds can also be seen as bounds for the upper tail of the cumulative distribution function for .

Hoeffding's inequality yields the simple bound
: $F(k;n,p) \leq \exp\left(-2 n\left(p-\frac\right)^2\right), \!$
which is however not very tight. In particular, for , we have that (for fixed , with ), but Hoeffding's bound evaluates to a positive constant.

A sharper bound can be obtained from the Chernoff bound:
: $F(k;n,p) \leq \exp\left(-nD\left(\frac\parallel p\right)\right)$
where is the relative entropy (or Kullback-Leibler divergence) between an -coin and a -coin (i.e. between the and distribution):
: $D(a\parallel p)=(a)\ln\frac+(1-a)\ln\frac. \!$

Asymptotically, this bound is reasonably tight; see for details.

One can also obtain ''lower'' bounds on the tail , known as anti-concentration bounds. By approximating the binomial coefficient with Stirling's formula it can be shown that
: $F(k;n,p) \geq \frac \exp\left(-nD\left(\frac\parallel p\right)\right),$
which implies the simpler but looser bound
: $F(k;n,p) \geq \frac1 \exp\left(-nD\left(\frac\parallel p\right)\right).$

For and for even , it is possible to make the denominator constant:
: $F(k;n,\tfrac) \geq \frac \exp\left(- 16n \left(\frac -\frac\right)^2\right). \!$

Statistical inference

Estimation of parameters

When is known, the parameter can be estimated using the proportion of successes:
: $\widehat = \frac.$
This estimator is found using maximum likelihood estimator and also the method of moments. This estimator is unbiased and uniformly with minimum variance, proven using Lehmann–Scheffé theorem, since it is based on a minimal sufficient and complete statistic (i.e.: ). It is also consistent both in probability and in MSE. This statistic is asymptotically normal thanks to the central limit theorem, because it is the same as taking the mean over Bernoulli samples. It has a variance of $var(\widehat) = \frac$, a property which is used in various ways, such as in Wald's confidence intervals.

A closed form Bayes estimator for also exists when using the Beta distribution as a conjugate prior distribution. When using a general $\operatorname(\alpha, \beta)$ as a prior, the posterior mean estimator is:
: $\widehat_b = \frac.$
The Bayes estimator is asymptotically efficient and as the sample size approaches infinity (), it approaches the MLE solution. The Bayes estimator is biased (how much depends on the priors), admissible and 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, $\operatorname(\alpha=1, \beta=1) = U(0,1)$, the posterior mean estimator becomes:
:$\widehat_b = \frac.$
(A 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 $\operatorname(\alpha=\frac, \beta=\frac)$, which leads to the estimator:
: $\widehat_ = \frac.$

When estimating with very rare events and a small (e.g.: if ), then using the standard estimator leads to $\widehat = 0,$ which sometimes is unrealistic and undesirable. In such cases there are various alternative estimators. One way is to use the Bayes estimator $\widehat_b$, leading to:
: $\widehat_b = \frac.$
Another method is to use the upper bound of the confidence interval obtained using the rule of three:
: $\widehat_ = \frac.$

Confidence intervals for the parameter p

Even for quite large values of ''n'', the actual distribution of the mean is significantly nonnormal. 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''₁ is the number of successes out of ''n'', the total number of trials
* $\widehat = \frac$ is the proportion of successes
* $z$ is the $1 - \tfrac\alpha$ quantile of a standard normal distribution (i.e., probit) corresponding to the target error rate $\alpha$. For example, for a 95% confidence level the error $\alpha$ = 0.05, so $1 - \tfrac\alpha$ = 0.975 and $z$ = 1.96.

Wald method

: $\widehat \pm z \sqrt .$

A continuity correction of may be added.

Agresti–Coull method

: $\tilde \pm z \sqrt$

Here the estimate of is modified to
: $\tilde= \frac$

This method works well for and . See here for $n\leq 10$. For use the Wilson (score) method below.

Arcsine method

: $\sin^2 \left(\arcsin \left(\sqrt\right) \pm \frac \right).$

Wilson (score) method

The notation in the formula below differs from the previous formulas in two respects:
* Firstly, has a slightly different interpretation in the formula below: it has its ordinary meaning of 'the th quantile of the standard normal distribution', rather than being a shorthand for 'the th quantile'.
* Secondly, this formula does not use a plus-minus to define the two bounds. Instead, one may use $z = z_$ to get the lower bound, or use $z = z_$ to get the upper bound. For example: for a 95% confidence level the error $\alpha$ = 0.05, so one gets the lower bound by using $z = z_ = z_ = - 1.96$, and one gets the upper bound by using $z = z_ = z_ = 1.96$.
: $\frac$

Comparison

The so-called "exact" ( Clopper–Pearson) method is the most conservative. (''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.

Related distributions

Sums of binomials

If and are independent binomial variables with the same probability , then is again a binomial variable; its distribution is :
: \begin
\operatorname P(Z=k) &= \sum_^k\left binomi p^i (1-p)^\rightleft binom p^ (1-p)^\right\
&= \binomk p^k (1-p)^
\end

A Binomial distributed random variable can be considered as the sum of Bernoulli distributed random variables. So the sum of two Binomial distributed random variables and is equivalent to the sum of Bernoulli distributed random variables, which means . This can also be proven directly using the addition rule.

However, if and do not have the same probability , then the variance of the sum will be smaller than the variance of a binomial variable distributed as .

Poisson binomial distribution

The binomial distribution is a special case of the Poisson binomial distribution, which is the distribution of a sum of independent non-identical Bernoulli trials .

Ratio of two binomial distributions

This result was first derived by Katz and coauthors in 1978.

Let and be independent. Let .

Then log(''T'') is approximately normally distributed with mean log(''p''₁/''p''₂) and variance .

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_X'' and taking the balls that hit and throwing them to another basket ''U_Y''. If ''p'' is the probability to hit ''U_X'' then ''X'' ~ B(''n'', ''p'') is the number of balls that hit ''U_X''. If ''q'' is the probability to hit ''U_Y'' then the number of balls that hit ''U_Y'' is ''Y'' ~ B(''X'', ''q'') and therefore ''Y'' ~ B(''n'', ''pq'').

Since $X \sim B(n, p)$ and $Y \sim B(X, q)$, by the law of total probability,
: \begin
\Pr = m&= \sum_^ \Pr = m \mid X = k\Pr = k\\ pt &= \sum_^n \binom \binom p^k q^m (1-p)^ (1-q)^
\end
Since $\tbinom \tbinom = \tbinom \tbinom,$ the equation above can be expressed as
: \Pr = m= \sum_^ \binom \binom p^k q^m (1-p)^ (1-q)^
Factoring $p^k = p^m p^$ and pulling all the terms that don't depend on $k$ out of the sum now yields
: \begin
\Pr = m&= \binom p^m q^m \left( \sum_^n \binom p^ (1-p)^ (1-q)^ \right) \\ pt &= \binom (pq)^m \left( \sum_^n \binom \left(p(1-q)\right)^ (1-p)^ \right)
\end
After substituting $i = k - m$ in the expression above, we get
: \Pr = m= \binom (pq)^m \left( \sum_^ \binom (p - pq)^i (1-p)^ \right)
Notice that the sum (in the parentheses) above equals $(p - pq + 1 - p)^$ by the binomial theorem. Substituting this in finally yields
: \begin
\Pr =m&= \binom (pq)^m (p - pq + 1 - p)^\\ pt &= \binom (pq)^m (1-pq)^
\end
and thus $Y \sim B(n, pq)$ as desired.

Bernoulli distribution

The Bernoulli distribution is a special case of the binomial distribution, where . Symbolically, has the same meaning as . Conversely, any binomial distribution, , is the distribution of the sum of independent Bernoulli trials, , each with the same probability .

Normal approximation

If is large enough, then the skew of the distribution is not too great. In this case a reasonable approximation to is given by the normal distribution
: $\mathcal(np,\,np(1-p)),$
and this basic approximation can be improved in a simple way by using a suitable continuity correction.
The basic approximation generally improves as increases (at least 20) and is better when is not near to 0 or 1. Various rules of thumb may be used to decide whether is large enough, and is far enough from the extremes of zero or one:
* One rule is that for the normal approximation is adequate if the absolute value of the skewness is strictly less than 0.3; that is, if
*: $\frac=\frac1\left, \sqrt-\sqrt\,\<0.3.$
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
*: $\mu\pm3\sigma=np\pm3\sqrt\in(0,n).$
: This 3-standard-deviation rule is equivalent to the following conditions, which also imply the first rule above.
:: $n>9 \left(\frac \right)\quad\text\quad n>9\left(\frac\right).$

The rule $np\pm3\sqrt\in(0,n)$ is totally equivalent to request that
: np-3\sqrt>0\quad\text\quad np+3\sqrt
Moving terms around yields:
: $np>3\sqrt\quad\text\quad n(1-p)>3\sqrt.$
Since 0, we can apply the square power and divide by the respective factors $np^2$ and $n(1-p)^2$, to obtain the desired conditions:
: $n>9 \left(\fracp\right) \quad\text\quad n>9 \left(\frac\right).$
Notice that these conditions automatically imply that $n>9$. On the other hand, apply again the square root and divide by 3,
: $\frac3>\sqrt>0 \quad \text \quad \frac3 > \sqrt>0.$
Subtracting the second set of inequalities from the first one yields:
: $\frac3>\sqrt-\sqrt>-\frac3;$
and so, the desired first rule is satisfied,
: $\left, \sqrt-\sqrt\,\<\frac3.$

* Another commonly used rule is that both values and must be greater than 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.

Assume that both values $np$ and $n(1-p)$ are greater than 9. Since $0< p<1$, we easily have that
: $np\geq9>9(1-p)\quad\text\quad n(1-p)\geq9>9p.$
We only have to divide now by the respective factors $p$ and $1-p$, to deduce the alternative form of the 3-standard-deviation rule:
: $n>9 \left(\fracp\right) \quad\text\quad n>9 \left(\frac\right).$

The following is an example of applying a continuity correction. Suppose one wishes to calculate for a binomial random variable . If has a distribution given by the normal approximation, then is approximated by . 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 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 is a sum of independent, identically distributed Bernoulli variables with parameter . This fact is the basis of a hypothesis test, a "proportion z-test", for the value of using , the sample proportion and estimator of , in a common test statistic.

For example, suppose one randomly samples 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
: $\sigma = \sqrt$

Poisson approximation

The binomial distribution converges towards the Poisson distribution as the number of trials goes to infinity while the product converges to a finite limit. Therefore, the Poisson distribution with parameter can be used as an approximation to of the binomial distribution if is sufficiently large and is sufficiently small. According to rules of thumb, this approximation is good if and such that , or if and such that , or if and .NIST/SEMATECH
"6.3.3.1. Counts Control Charts"
''e-Handbook of Statistical Methods.''

Concerning the accuracy of Poisson approximation, see Novak, ch. 4, and references therein.

Limiting distributions

* '' Poisson limit theorem'': As approaches and approaches 0 with the product held fixed, the distribution approaches the Poisson distribution with expected value .
* '' de Moivre–Laplace theorem'': As approaches while remains fixed, the distribution of
*: $\frac$
: approaches the normal distribution with expected value 0 and variance 1. This result is sometimes loosely stated by saying that the distribution of is 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 PMF of successes given independent events each with a probability of success.
Mathematically, when and , the beta distribution and the binomial distribution are related by a factor of :
: $\operatorname(p;\alpha;\beta) = (n+1)B(k;n;p)$

Beta distributions also provide a family of prior probability distributions for binomial distributions in Bayesian inference:
: $P(p;\alpha,\beta) = \frac.$
Given a uniform prior, the posterior distribution for the probability of success given independent events with observed successes is a beta distribution.

Computational methods

Random number generation

Methods for random number generation where the marginal distribution is a binomial distribution are well-established.
One way to generate random variates 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.

History

This distribution was derived by Jacob Bernoulli. He considered the case where where is the probability of success and and are positive integers. Blaise Pascal had earlier considered the case where , tabulating the corresponding binomial coefficients in what is now recognized as Pascal's triangle.

See also

* Logistic regression
* Multinomial distribution
* Negative binomial distribution
* Beta-binomial distribution
* Binomial measure, an example of a multifractal measure.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''
* Statistical mechanics
* Piling-up lemma, the resulting probability when XOR-ing independent Boolean variables

References

Further reading

*
*

External links

* Interactive graphic
Univariate Distribution Relationships

Binomial distribution formula calculator
* Difference of two binomial variables
X-Y
o
, X-Y,

Querying the binomial probability distribution in WolframAlpha
* Confidence (credible) intervals for binomial probability, p

available a
causaScientia.org

{{DEFAULTSORT:Binomial Distribution
Discrete distributions
Factorial and binomial topics
Conjugate prior distributions
Exponential family distributions