Probability mass function


Cumulative distribution function


Notation  B(n, p) 

Parameters  n ? N_{0}  number of trials p ? [0,1]  success probability in each trial 
Support  k ? { 0, ..., n }  number of successes 
pmf  
CDF  
Mean  
Median  or 
Mode  or 
Variance  
Skewness  
Ex. kurtosis  
Entropy  in shannons. For nats, use the natural log in the log. 
MGF  
CF  
PGF  
Fisher information 
(for fixed ) 
In probability theory and statistics, the binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent experiments, each asking a yesno question, and each with its own booleanvalued outcome: a random variable containing single bit of information: success/yes/true/one (with probability p) or failure/no/false/zero (with probability 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., n = 1, the binomial distribution is a Bernoulli distribution. The binomial distribution is the basis for the popular binomial test of statistical significance.
The binomial distribution is frequently used to model the number of successes in a sample of size n drawn with replacement from a population of size 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 N much larger than n, the binomial distribution remains a good approximation, and is widely used.
In general, if the random variable X follows the binomial distribution with parameters n ? N and p ? [0,1], we write X ~ B(n, p). The probability of getting exactly k successes in n trials is given by the probability mass function:
for k = 0, 1, 2, ..., n, where
is the binomial coefficient, hence the name of the distribution. The formula can be understood as follows. k successes occur with probability p^{k} and n  k failures occur with probability (1  p)^{n  k}. However, the k successes can occur anywhere among the n trials, and there are different ways of distributing k successes in a sequence of n trials.
In creating reference tables for binomial distribution probability, usually the table is filled in up to n/2 values. This is because for k > n/2, the probability can be calculated by its complement as
Looking at the expression ?(k, n, p) as a function of k, there is a k value that maximizes it. This k value can be found by calculating
and comparing it to 1. There is always an integer M that satisfies
?(k, n, p) is monotone increasing for k < M and monotone decreasing for k > M, with the exception of the case where (n + 1)p is an integer. In this case, there are two values for which ? is maximal: (n + 1)p and (n + 1)p  1. M is the most probable (most likely) outcome of the Bernoulli trials and is called the mode. Note that the probability of it occurring can be fairly small.
The cumulative distribution function can be expressed as:
where is the "floor" under k, i.e. the greatest integer less than or equal to k.
It can also be represented in terms of the regularized incomplete beta function, as follows:^{[1]}
Some closedform bounds for the cumulative distribution function are given below.
Suppose a biased coin comes up heads with probability 0.3 when tossed. What is the probability of achieving 0, 1,..., 6 heads after six tosses?
If X ~ B(n, p), that is, X is a binomially distributed random variable, n being the total number of experiments and p the probability of each experiment yielding a successful result, then the expected value of X is:^{[3]}
For example, if n = 100, and p =1/4, then the average number of successful results will be 25.
Proof: We calculate the mean, ?, directly calculated from its definition
and the binomial theorem:
It is also possible to deduce the mean from the equation whereby all are Bernoulli distributed random variables with (=1 if the i experiment succeed and =0 otherwise). We get:
A more deductive proof using calculus is shown below. This relies on defining a new function which only replaces one , and then plugging back in at the end to recover the binomial distribution.
The variance is:
Proof: Let where all are independently Bernoulli distributed random variables. Since , we get:
Another proof, in the same vein as the section on the Mean, can be done using calculus. Keep in mind that all the functions integrated below are zero at zero, but the binomial distribution is: . If . then . By contrast, one can check that as defined below.
Now let's backtrack to . We have to solve the following for :
We can just use the operation: . This will undo all of the steps we have done.
Usually the mode of a binomial B(n, p) distribution is equal to , where is the floor function. However, when (n + 1)p is an integer and p is neither 0 nor 1, then the distribution has two modes: (n + 1)p and (n + 1)p  1. When p is equal to 0 or 1, the mode will be 0 and n correspondingly. These cases can be summarized as follows:
Proof: Let
For only has a nonzero value with . For we find and for . This proves that the mode is 0 for and for .
Let . We find
From this follows
So when is an integer, then and is a mode. In the case that , then only is a mode.^{[4]}
In general, there is no single formula to find the median for a binomial distribution, and it may even be nonunique. However several special results have been established:
If two binomially distributed random variables X and Y are observed together, estimating their covariance can be useful. The covariance is
In the case n = 1 (the case of Bernoulli trials) XY is nonzero only when both X and Y are one, and ?_{X} and ?_{Y} are equal to the two probabilities. Defining p_{B} as the probability of both happening at the same time, this gives
and for n independent pairwise trials
If X and Y are the same variable, this reduces to the variance formula given above.
If X ~ B(n, p) and Y ~ B(m, p) are independent binomial variables with the same probability p, then X + Y is again a binomial variable; its distribution is Z=X+Y ~ B(n+m, p):
However, if X and Y do not have the same probability p, then the variance of the sum will be smaller than the variance of a binomial variable distributed as
This result was first derived by Katz et al in 1978.^{[9]}
Let p_{1} and p_{2} be the probabilities of success in the binomial distributions B(X,n) and B(Y,m) respectively. Let T = (X/n)/(Y/m).
Then log(T) is approximately normally distributed with mean log(p_{1}/p_{2}) and variance ((1/p_{1})  1)/n + ((1/p_{2})  1)/m.
If X ~ B(n, p) and, conditional on X, Y ~ B(X, q), then Y is a simple binomial 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 and , by the law of total probability,
Since , the equation above can be expressed as
Factoring and pulling all the terms that don't depend on out of the sum now yields
After substituting in the expression above, we get
Notice that the sum (in the parentheses) above equals by the binomial theorem. Substituting this in finally yields
and thus as desired.
The Bernoulli distribution is a special case of the binomial distribution, where n = 1. Symbolically, X ~ B(1, p) has the same meaning as X ~ B(p). Conversely, any binomial distribution, B(n, p), is the distribution of the sum of n Bernoulli trials, B(p), each with the same probability p.^{[10]}
The binomial distribution is a special case of the Poisson binomial distribution, or general binomial distribution, which is the distribution of a sum of n independent nonidentical Bernoulli trials B(p_{i}).^{[11]}
If n is large enough, then the skew of the distribution is not too great. In this case a reasonable approximation to B(n, p) is given by the normal distribution
and this basic approximation can be improved in a simple way by using a suitable continuity correction. The basic approximation generally improves as n increases (at least 20) and is better when p is not near to 0 or 1.^{[12]} Various rules of thumb may be used to decide whether n is large enough, and p is far enough from the extremes of zero or one:
The rule is totally equivalent to request that
Moving terms around yields:
Since , we can apply the square power and divide by the respective factors and , to obtain the desired conditions:
Notice that these conditions automatically imply that . On the other hand, apply again the square root and divide by 3,
Subtracting the second set of inequalities from the first one yields:
and so, the desired first rule is satisfied,
Assume that both values and are greater than 9. Since , we easily have that
We only have to divide now by the respective factors and , to deduce the alternative form of the 3standarddeviation rule:
The following is an example of applying a continuity correction. Suppose one wishes to calculate Pr(X X. If Y has a distribution given by the normal approximation, then Pr(X Y
This approximation, known as de MoivreLaplace theorem, is a huge timesaver when undertaking calculations by hand (exact calculations with large 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 B(n, p) is a sum of n independent, identically distributed Bernoulli variables with parameter p. This fact is the basis of a hypothesis test, a "proportion ztest", for the value of p using x/n, the sample proportion and estimator of p, in a common test statistic.^{[13]}
For example, suppose one randomly samples 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
The binomial distribution converges towards the Poisson distribution as the number of trials goes to infinity while the product np remains fixed or at least p tends to zero. Therefore, the Poisson distribution with parameter ? = np can be used as an approximation to B(n, p) of the binomial distribution if n is sufficiently large and p is sufficiently small. According to two rules of thumb, this approximation is good if n >= 20 and p n >= 100 and np [14]
Concerning the accuracy of Poisson approximation, see Novak,^{[15]} ch. 4, and references therein.
Beta distributions provide a family of prior probability distributions for binomial distributions in Bayesian inference:^{[16]}
Even for quite large values of n, the actual distribution of the mean is significantly nonnormal.^{[17]} 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:
The notation in the formula below differs from the previous formula's in two respects:
^{[21]}
The exact (ClopperPearson) method is the most conservative.^{[17]}
The Wald method, although commonly recommended in textbooks, is the most biased.^{[clarification needed]}
Methods for random number generation where the marginal distribution is a binomial distribution are wellestablished.^{[22]}^{[23]}
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 P(X=k) for all values k from 0 through 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 U[0,1] into discrete numbers by using the probabilities calculated in step one.
For k np, upper bounds for the lower tail of the distribution function can be derived. Recall that , the probability that there are at most k successes.
Hoeffding's inequality yields the bound
and Chernoff's inequality can be used to derive the bound
Moreover, these bounds are reasonably tight when p = 1/2, since the following expression holds for all k >= 3n/8^{[24]}
However, the bounds do not work well for extreme values of p. In particular, as p 1, value F(k;n,p) goes to zero (for fixed k, n with k<n) while the upper bound above goes to a positive constant. In this case a better bound is given by ^{[25]}
where D(a  p) is the relative entropy between an acoin and a pcoin (i.e. between the Bernoulli(a) and Bernoulli(p) distribution):
Asymptotically, this bound is reasonably tight; see ^{[25]} for details. An equivalent formulation of the bound is
Both these bounds are derived directly from the Chernoff bound. It can also be shown that,
This is proved using the method of types (see for example chapter 12 of Elements of Information Theory by Cover and Thomas ^{[26]}).
We can also change the in the denominator to , by approximating the binomial coefficient with Stirling's formula.^{[27]}
This distribution was derived by James Bernoulli. He considered the case where p = r/(r+s) where p is the probability of success and r and s are positive integers. Blaise Pascal had earlier considered the case where p = 1/2.
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