Binomial Probability Distribution Family

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A Binomial Probability Distribution Family, $B(n,p)$, is a finite support discrete probability distribution family for a binomial random variable.

References

2015

• (Wikipedia, 2015) ⇒ http://en.wikipedia.org/wiki/binomial_distribution Retrieved:2015-6-15.
• (Wikipedia, 2015) ⇒ http://en.wikipedia.org/wiki/binomial_distribution#Specification Retrieved:2015-6-15.
• In general, if the random variable X follows the binomial distribution with parameters n and p, we write X ~ B(np). The probability of getting exactly k successes in n trials is given by the probability mass function: : $f(k;n,p) = \Pr(X = k) = {n\choose k}p^k(1-p)^{n-k}$ for k = 0, 1, 2, ..., n, where : ${n\choose k}=\frac{n!}{k!(n-k)!}$ is the binomial coefficient, hence the name of the distribution. The formula can be understood as follows: we want exactly k successes (pk) and n − k failures (1 − p)n − k. However, the k successes can occur anywhere among the n trials, and there are ${n\choose k}$ 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 : $f(k,n,p)=f(n-k,n,1-p).$ Looking at the expression ƒ(knp) as a function of k, there is a k value that maximizes it. This k value can be found by calculating : $\frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)}$ and comparing it to 1. There is always an integer M that satisfies :$(n+1)p-1 \leq M \lt (n+1)p.$

ƒ(knp) 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 ƒ 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.

Recurrence relation $\left\{p (n-k) \text{Prob}(k)+(k+1) (p-1) \text{Prob}(k+1)=0,\text{Prob}(0)=( 1-p)^n\right\}$