Special Discrete Distributions Bernoulli Distribution A Bernoulli - PowerPoint PPT Presentation

Special Discrete Distributions

Bernoulli Distribution ◮ A Bernoulli trial is an experiment with two possible outcomes. A random variable X has a Bernoulli( p ) if  1 with probability p  X = 0 with probability 1 − p  X = 1 often termed as “success” and X = 0 often termed as “failure”. ◮ Example 1: toss a coin, head=“success” and tail=“failure”. ◮ Example 2: incidence of a disease, not infected=“success” and infected=“failure”.

Binomial experiments 1. The experiment consists of n repeated Bernoulli trials - each trial has only two possible outcomes labeled as success and failure; 2. The trials are independent - the outcome of any trial has no effect on the probability of the others; 3. The probability of success in each trial is constant which we denote by p .

Binomial distribution Let Y be the total number of successes in n trails, i.e., Y = X 1 + X 2 + · · · + X n , where X i ∼ Bernoulli ( p ) , then Y is said to have Binomial distribution with parameters n and p . Denote it as Y ∼ Binomial ( n , p ) . The probability mass function of Y is � n � p k ( 1 − p ) n − k for k = 0 , 1 , · · · , n . b ( k ; n , p ) = k

Example 1 Suppose a husband and a wife each is of genotype aA , where the gene a is recessive while A is dominant. Children who are of genotype aa have six toes, whereas those of genotype aA and AA are normal. If the couple has 6 children, what is the probability that two of them will have 6 toes?

Expectation and variance ◮ We have shown that E ( Y ) = np and Var ( Y ) = np ( 1 − p ) .

Discrete uniform distribution ◮ A random variable X has a discrete uniform on { 1 , · · · , N } if P ( X = k ) = 1 k = 1 , 2 , · · · , N , N where N is a specified integer. ◮ The mean and variance of X E ( X ) = N + 1 Var ( X ) = ( N + 1 )( N − 1 ) . 2 12

Poisson distribution Poisson distribution is often used to model the number of occurrence in a given interval such as: the number of customers arriving in a bank in a given time interval; the number of failures of a machine in a given period; the number of typos on a given page. The basic assumption is that, for a small time interval, the probability of an arrival is proportional to the length of the waiting time.

Definition of Poisson distribution A random variable X taking values in the non-negative integers has a Poisson( λ ) distribution if P ( X = k ) = e − λ λ k k = 0 , 1 , · · · k ! The expectation and variance of poisson distribution are E ( X ) = λ and Var ( X ) = λ.

Example Suppose that the number of typographical errors on a single page of this book has Poisson distribution with parameter λ = 1 2 . Calculate the probability that there is at least one error on this page.

The approximation by Binomial distribution If X ∼ Poisson ( λ ) and Y ∼ Binomial ( n , p ) , f X ( k ) = e − λ λ k k ! � n � p k ( 1 − p ) n − k . f Y ( k ) = k If p is small and n is large such that np → λ as n → ∞ , f Y ( k ) → f X ( k ) for every k .

The Geometric and Negative Binomial Distributions In a sequence of independent Bernoulli(p) trials, let the random variable X denote the trial at which r − th success occurs, where r is a fixed integer. The probability mass function for X is � k − 1 � p r ( 1 − p ) k − r f X ( k ) = for k = r , r + 1 , · · · r − 1 1. If r = 1 , X is said to be Geometric distributed with parameter p . 2. For general r , we say X has Negative Binomial distribution with parameters r and p . Denoted as NB(r,p).

Another way to define Negative Binomial ◮ Let Y = the number of failures before r -th success . Then Y = X − r . The PMF of Y is � r + k − 1 � � − r � p r ( 1 − p ) k = ( − 1 ) k p r ( 1 − p ) k f Y ( k ) = r − 1 k for k = 0 , 1 , · · · where � − r � = ( − r )( − r − 1 ) · · · ( − r − k + 1 ) . k k ! ◮ We have used � r + k − 1 � � r + k − 1 � � − r � = ( − 1 ) k = . r − 1 k k

The Banach match problem A pipe-smoking mathematician carries two match-boxes, one in his left hand and the other one in right-hand pocket. Initially, each box contains N matches. Each time the mathematician requires a match, he is equally likely to take it from either box. At the moment he first discovers one of the boxes to be empty, what is the probability that there are exactly k ( k = 0 , 1 , · · · , N ) matches in the other box?

Expectation and variance of NB( r , p ) Let D i be the additional number of trials for obtaining i -th success after i − 1 successes. Then D i are independent distributed as Geometric( p ) and X = D 1 + · · · + D r . It follows that r E ( D i ) = r 1 � E ( X ) = p i = 1 r Var ( D i ) = r 1 − p � Var ( X ) = . p 2 i = 1

Memoryless property of Geometric distribution If X ∼ Geometric ( p ) and any integers s > t , P ( X > s | X > t ) = P ( X > s − t ) = P ( X > s − t | X > 0 ) . The probability of obtaining s − t failures after having observed t failures is the same as the probability getting s − t failures at the first beginning. The geometric distribution “forgot” what has occurred. The probability depends on the length of the run, not on its location.