Chapter 3 b Geometric Poisson Negative Binomial Gamma

• The Most Common • Discrete Probability Distributions • (some discussed before) 1) --- Bernoulli distribution 2) --- Binomial 3) --- Geometric 4) --- Poisson 2

Bernoulli distribution • The Bernoulli distribution is the “coin flip” distribution. • X is Bernoulli if its probability function is: X=1 is usually interpreted as a “success. ” E. g. : X=1 for heads in coin toss X=1 for male in survey X=1 for defective in a test of product X=1 for “made the sale” tracking performance 3

Bernoulli distribution • Expectation: • Variance: 4

Binomial distribution • The binomial distribution is just n independent Bernoullis • added up. • It is the number of “successes” in n trials. • If Z 1, Z 2, …, Zn are Bernoulli, then X is binomial: 5

Binomial distribution • The binomial distribution is just n independent Bernoullis • added up. Testing for defects “with replacement. ” – – 6 Have many light bulbs Pick one at random, test for defect, put it back If there are many light bulbs, do not have to replace

Binomial distribution • Let’s figure out a binomial r. v. ’s probability function. • Suppose we are looking at a binomial with n=3. • We want P(X=0): – Can happen one way: 000 – (1 -p)(1 -p) = (1 -p)3 • We want P(X=1): – Can happen three ways: 100, 010, 001 – p(1 -p)+(1 -p)p(1 -p)+(1 -p)p = 3 p(1 -p)2 • We want P(X=2): – Can happen three ways: 110, 011, 101 – pp(1 -p)+(1 -p)pp+p(1 -p)p = 3 p 2(1 -p) • We want P(X=3): – Can happen one way: 111 – ppp = p 3

Binomial distribution • So, binomial r. v. ’s probability function 8

Binomial distribution • Typical shape of binomial: – Symmetric 9

• Expectation: Variance: Aside: If V(X) = V(Y). And? But 10 Hmm…

Binomial distribution • A salesman claims that he closes a deal 40% of the time. • This month, he closed 1 out of 10 deals. • How likely is it that he did 1/10 or worse given his claim? 11

Binomial distribution Note: 12 Less than 5% or 1 in 20. So, it’s unlikely that his success rate is 0. 4.

Binomial and normal / Gaussian distribution The normal distribution is a good approximation to the binomial distribution. (“large” n, small skew. ) B(n, p) Prob. density function:

Geometric Distribution • A geometric distribution is usually interpreted as number of time periods until a failure occurs. • Imagine a sequence of coin flips, and the random variable X is the flip number on which the first tails occurs. • The probability of a head (a success) is p. 14

Geometric • Let’s find the probability function for the geometric distribution: etc. So, (x is a positive integer)

Geometric • Notice, there is no upper limit on how large X can be • Let’s check that these probabilities add to 1: 16 Geometric series

Geometric differentiate both sides w. r. t. p: See Rosen page 158, example 17. • Expectation: Variance: 17

Poisson distribution • The Poisson distribution is typical of random variables which represent counts. – Number of requests to a server in 1 hour. – Number of sick days in a year for an employee. 18

• The Poisson distribution is derived from the following underlying arrival time model: – The probability of an unit arriving is uniform through time. – Two items never arrive at exactly the same time. – Arrivals are independent --- the arrival of one unit does not make the next unit more or less likely to arrive quickly. 19

Poisson distribution • The probability function for the Poisson distribution with parameter is: • is like the arrival rate --- higher means more/faster arrivals 20

Poisson distribution • Shape Low 21 Med High

2. 4 – The Poisson Distribution Definition 2. 4. 1 A random variable X has a Poisson distribution if its p. m. f. is where λ > 0 is a constant. – Describes the number of “occurrences” over a randomly selected “interval” – The parameter λ is the “average” number of occurrences per interval

Example 2. 4. 1 Suppose a 1000 ft 2 lawn contains 3000 dandelions. Find the probability that a randomly chosen 1 ft 2 section of lawn contains exactly five dandelions. – “Occurrence” = dandelion – “Interval” = 1 ft 2 section of lawn – X = number of dandelions in a 1 ft 2 section of lawn – Assume X is Poisson

Example 2. 4. 1

Negative Binomial Probability Distribution

Objectives • Determine whether a probability experiment is a geometric, hypergeometric or negative binomial experiment • Compute probabilities of geometric, hypergeometric and negative binomial experiments • Compute the mean and standard deviation of a geometric, hypergeometric and negative binomial random variable • Construct geometric, hypergeometric and negative binomial probability histograms

Vocabulary • Trial – each repetition of an experiment • Success – one assigned result of a binomial experiment • Failure – the other result of a binomial experiment • PDF – probability distribution function • CDF – cumulative (probability) distribution function, computes probabilities less than or equal to a specified value

Criteria for a Negative Binomial Probability Experiment An experiment is said to be a negative binomial experiment provided: 1. Each repetition is called a trial. 2. For each trial there are two mutually exclusive (disjoint) outcomes: success or failure 3. The probability of success is the same for each trial of the experiment 4. The trials are independent 5. The trials are repeated until r successes are observed, where r is specified in advance.

Negative Binomial Notation When we studied the Binomial distribution, we were only interested in the probability for a success or a failure to happen. The negative binomial distribution addresses the number of trials necessary before the rth success. If the trials are repeated x times until the rth success, we will have had x – r failures. If p is the probability for a success and (1 – p) the probability for a failure, the probability for the rth success to occur at the xth trial will be P(x) = x-1 Cr-1 pr(1 – p)x-r x = r, r+1, r+2, … Where r number of successes is observed in x number of trials of a binomial experiment with success rate of p

Mean and Standard Deviation of a Negative Binomial RV A negative binomial experiment with probability of success p has Mean μx = r/p Standard Deviation σx = r(1 -p)/p 2 Where r number of successes is observed in x number of trials of a binomial experiment with success rate of p Note that the geometric distribution is a special case of the negative binomial distribution with k = 1.

Examples of Negative Binomial PDF • Number of cars arriving at a service station until the fourth one that needs brake work • Flipping a coin until the fourth tail is observed • Number of planes arriving at an airport until the second one that needs repairs • Number of house showings before an agent gets her third sale • Length of time (in days) until the second sale of a large computer system

Example 1 The drilling records for an oil company suggest that the probability the company will hit oil in productive quantities at a certain offshore location is 0. 3. Suppose the company plans to drill a series of wells looking for three successful wells. P(x) = x-1 Cr-1 pr(1 – p)x-r p = 0. 3 a) What is the probability that the third success will be achieved with the 8 th well drilled? P(8) = 8 -1 C 3 -1 p 3(1 – p)8 -3 = 7 C 2 (0. 3)3(0. 7)5 = (21)(0. 027)(0. 16807) = 0. 0953 b) What is the probability that the third success will be achieved with the 20 th well drilled? P(20) = 20 -1 C 3 -1 p 3(1 – p)20 -3 = 19 C 2 (0. 3)3(0. 7)17 = (171)(0. 027)(0. 00233) = 0. 0107

Example 1 cont The drilling records for an oil company suggest that the probability the company will hit oil in productive quantities at a certain offshore location is 0. 3. Suppose the company plans to drill a series of wells looking for three successful wells. c) Find the mean and standard deviation of the number of wells that must be drilled before the company hits third productive well. Mean μx = r/p = 3 / 0. 3 = 10 Standard Deviation σx = r(1 -p)/p² = 3(0. 7)/(0. 3)² = 4. 8305

Example 2 A standard, fair die is thrown until 3 aces occur. Let Y denote the number of throws. a)Find the mean of Y E(Y) = r/p = 3/(1/6) = 18 b)Find the variance of Y V(Y) = r(1 -p)/p² = 3(5/6)/(1/6)² = 90 c)Find the probability that at least 20 throws will needed P(at least 20) = P(Y≥ 20) = 1 – P(Y<20) = 1 – [P(3) + P(4) + … + P(18) + P(19)] P (Y≥ 20) = 0. 3643

Summary • Summary – Negative Binomial has 5 characteristics to be met – Looking for the rth success (becomes Geometric for r = 1) – Computer applet required for pdf and cdf – Not on the AP

The gamma distribution • The gamma distribution arises naturally as the distribution of a series of iid random exponential variables a = b = 0. 5 a=b=2 a = b = 10 • The gamma distribution has expectation a/b and variance a/b 2 • More generally, a need not be an integer (for example, the Chi-square distribution with one degree of freedom is a G(½, ½) distribution)

The beta distribution • The beta distribution models random variables that take the value [0, 1] • It arises naturally as the proportional ratio of two gamma distributed random variables a = b = 0. 5 a = b = 10 • The expectation is a/(a + b) a=b=2 • In Bayesian statistics, the beta distribution is the natural prior for binomial proportions (beta-binomial) – The Dirichlet distribution generalises the beta to more than 2 proportions