Bernoulli Binomial Negative Binomial Poisson Hypergeometric Topic 9 Examples of Mass Functions and Densities Discrete Random Variables 1 / 12 Bernoulli Binomial Negative Binomial Poisson Hypergeometric Outline Bernoulli Binomial Negative Binomial Poisson Hypergeometric 2 / 12 Bernoulli Binomial Negative Binomial Poisson Hypergeometric Introduction Write fX (xjθ) = PθfX = xg for the mass function of the given family of discrete random variables depending on the parameter θ. We will • use the expression Family(θ) as shorthand for this family • followed by the R command family and • the state space S. 3 / 12 Bernoulli Binomial Negative Binomial Poisson Hypergeometric Bernoulli Random Variables Ber(p) on S = f0; 1g 0 with probability (1 − p); f (xjp) = = px (1 − p)1−x : X 1 with probability p; This is the simplest random variable, taking on only two values, namely,0 and1. Think of it as the outcome of a Bernoulli trial, i.e., a single toss of an unfair coin that turns up heads with probability p. 4 / 12 Bernoulli Binomial Negative Binomial Poisson Hypergeometric Binomial Random Variables Bin(n; p) (R command binom) on S = f0; 1;:::; ng n f (xjp) = px (1 − p)n−x : X x The binomial distribution arises from computing the probability of x successes in n Bernoulli trials. Considered in this way, the family Ber(p) is also Bin(1; p). Exercise. Enter in R plot(c(0:12),dbinom(c(0:12),12,p,),type="h") for p=0.25,0.5,0.75. Describe what you see. 5 / 12 Bernoulli Binomial Negative Binomial Poisson Hypergeometric Negative Binomial Random Variables Negbin(n; p) (R command nbinom) on S = N n + x − 1 f (xjp) = pn(1 − p)x : X x This random variable is the number of failed Bernoulli trials before the n-th success. To find the mass function, • For the outcome fX = xg, the n-th success must occur on the n + x-th trial. So, • we must have n − 1 successes and x failures in first n + x − 1 Bernoulli trials • followed by success on the last trial. 6 / 12 Bernoulli Binomial Negative Binomial Poisson Hypergeometric Negative Binomial Random Variables The first n + x − 1 trials and the last trial are independent and so the probabilities multiply. PpfX = xg = Ppfn − 1 successes in n + x − 1 trials; success in the n − x-th trialg = Ppfn − 1 successes in n + x − 1 trialsgPpfsuccess in the n − x-th trialg n + x − 1 n + x − 1 = pn−1(1 − p)x · p = pn(1 − p)x n − 1 x The first factor is computed from the binomial distribution, the second from the Bernoulli distribution. Note the use of the identity m m = k m − k in giving the final formula. 7 / 12 Bernoulli Binomial Negative Binomial Poisson Hypergeometric Poisson Random Variables Pois(λ) (R command pois) on S = N λx f (xjλ) = e−λ: X x! The Poisson distribution approximates of the binomial distribution for n large, p small, but the product λ = np is moderate. Examples. • bacterial colonies • n is the number of bacteria and p is the probability of a mutation, • recombination events during meiosis • n is the number of nucleotides on a chromosome and p is the probability of a recombination event occurring at a particular nucleotide. The product λ is the mean number of events. 8 / 12 Bernoulli Binomial Negative Binomial Poisson Hypergeometric Poisson Random Variables The approximation is based on the limit λn lim 1 − = e−λ: n!1 n We compute binomial probabilities, replace p by λ/n and take a limit as n ! 1. In this computation, we use the fact that for a fixed value of x, (n) λ−x x ! 1 and 1 − ! 1 as n ! 1 nx n 9 / 12 Bernoulli Binomial Negative Binomial Poisson Hypergeometric Poisson Random Variables n 0 n λ n −λ PfX = 0g = 0 p (1 − p) = 1 − n ≈ e n 1 n−1 λ λ n−1 −λ PfX = 1g = 1 p (1 − p) = n n 1 − n ≈ λe n 2 n−2 n(n−1) λ 2 λ n−2 n(n−1) λ2 λ n−2 λ2 −λ PfX = 2g = 2 p (1 − p) = 2 n 1 − n = n2 2 1 − n ≈ 2 e . x x n x n−x (n)x λ x λ n−x (n)x λ λ n−x λ −λ PfX = xg = x p (1 − p) = x! n 1 − n = nx x! 1 − n ≈ x! e : Exercise. Explain why 1 1 X X λx f (x) = e−λ = 1: X x! x=0 x=0 10 / 12 Bernoulli Binomial Negative Binomial Poisson Hypergeometric Hypergeometric Random Variables Hyper(m; n; k) (R command hyper) on S = fmaxf0; k − ng;:::; minfm; kgg m n x k−x fX (xjm; n; k) = m+n k • Begin with an urn holding m white balls and n black balls. • Remove k at random and • let the random variable X denote the number of white balls. 11 / 12 Bernoulli Binomial Negative Binomial Poisson Hypergeometric Hypergeometric Random Variables We consider equally likely outcomes to determine fX (xjm; n; k) = Pm;n;k fX = xg. • The total number of possible outcomes, #(Ω), namely, the number of ways to choose k balls out of an urn containing m + n balls. m + n : k This will be the denominator for the probability. • For the numerator, the outcomes that result in x white balls from the total of m, we must also choose k − x black balls from the total of n. By the multiplication property, the number of ways#( Ax ) to accomplish this is product. m n : x k − x Finally, fX (xjm; n; k) = #(Ax )=#(Ω). 12 / 12.