Multinomial Geometric Hypergeometric Poisson
Math 141 Lecture 4: Distributions Related To The Binomial Distribution
Albyn Jones1
1Library 304 [email protected] www.people.reed.edu/∼jones/courses/141
Albyn Jones Math 141 Multinomial Geometric Hypergeometric Poisson Outline
Review The Multinomial Distribution The Geometric and Negative Binomial Distributions The Hypergeometric Distribution The Poisson Distribution
Albyn Jones Math 141 Multinomial Geometric Hypergeometric Poisson Examples of Different Experiments
Binomial: Count the number of Heads in a fixed number of tosses. Geometric: Count the number of Tails before the first Head. Negative Binomial: Count the number of Tails before before the k-th Head. Hypergeometric: Count the number of red cards dealt in a poker hand. Poisson A model for rare events.
Albyn Jones Math 141 Multinomial Geometric Hypergeometric Poisson Review: The Binomial
Dichotomous Trials: Each trial results in either a ‘Success’, S, or a ‘Failure’ F. Independence: Successive trials are independent; knowing we got S on one trial does not help us predict the outcome of any other trial. Constant probability: Each trial has the same probability P(S) = p, and P(F) = 1 − p. X counts the number of S’s. n the number of trials is fixed.
Albyn Jones Math 141 Multinomial Geometric Hypergeometric Poisson X ∼ Binomial(n, p) Probabilities
Let p = P(S) and q = 1 − p = P(F), then n (X = k) = pk qn−k P k
And in R, the density function is given by
P(X = k) = dbinom(k, n, p)
Albyn Jones Math 141 Multinomial Geometric Hypergeometric Poisson The Multinomial
Polychotomous Trials: Each trial results in one of a fixed set of possible outcomes {E1, E2,... EN }. Example: die rolls, Ω = {1, 2, 3, 4, 5, 6}. Independence: Successive trials are independent; knowing the outcome of one trial does not help us predict the outcome of any other trial. Constant probability: Each trial has the same probability for each possibility.
X1, X2,... XN count the number of events of each type. n the number of trials is fixed.
Albyn Jones Math 141 Multinomial Geometric Hypergeometric Poisson Multinomial Probabilities
Good news! The typical computation involves collapsing categories to create a binomial.
Example: Roll a fair die 20 times. For each roll, there are six possible outcomes, so we have a Multinomial Distribution. What is the probability of rolling three 6’s in 20 rolls? Let X be the number of 6’s in 20 rolls. What is the distribution of X?
X ∼ Binomial(20, 1/6)
Albyn Jones Math 141 Multinomial Geometric Hypergeometric Poisson But Since You Asked...
For n independent Multinomial(p1, p2,..., pN ) trials, where the probability of observing category i is pi . Let Xi be the count of events observed in category i, where
N N X X Xi = n and pi = 1 i=1 i=1
n! k1 k2 kN P(X1 = k1,..., XN = kN ) = p1 p2 ... pN k1!k2! ... kN ! with R functions: rmultinom, dmultinom.
Albyn Jones Math 141 Multinomial Geometric Hypergeometric Poisson Example: Election Polls
Suppose we ask registered Republicans for their preference: Bachman, Gingrich, Perry, Romney, or ‘none of the above’ (Ron Paul is invisible :-). The Poll We ask 1000 randomly chosen Republicans for their preference, and let {9, 290, 108, 395, 198} be the counts for those 5 options, in order. The Population Proportions Suppose that the actual probabilities are (in order) {.01,.3,.1,.4,.19}. The Probability: dmultinom(c(9, 290, 108, 395 , 198),1000, c(.01, .3, .1, .4, .19)) [1] 2.572792e-06 That looks small, but even the most likely outcome has small probability (about 5 × 10−6)!
Albyn Jones Math 141 Multinomial Geometric Hypergeometric Poisson The Geometric Distribution
Dichotomous Trials: Each trial results in either a ‘Success’, S, or a ‘Failure’ F. Independence: Successive trials are independent; knowing we got S on one trial does not help us predict the outcome of any other trial. Constant probability: Each trial has the same probability P(S) = p, and P(F) = 1 − p = q. X counts the number of F’s before the first S. Question: What is the probability we see k failures before the first success?
k Pr(X = k) = P(F1, F2, F3,..., Fk , S) = p · q
Albyn Jones Math 141 Multinomial Geometric Hypergeometric Poisson Example
Roll a fair die repeatedly until getting the first 6. What are p and q here? What is the probability it comes on the 6th roll, that is we have 5 non-sixes before the first 6?
55 1 (X = 5) = ≈ .067 P 6 6
Albyn Jones Math 141 Multinomial Geometric Hypergeometric Poisson The Negative Binomial Distribution
Like the Geometric: Dichotomous outcomes {F, S}, independent trials, constant probability. X counts the number of F’s before the rth S. Question: What is the probability we see k failures before the rth success? Hint The last trial must be an S, so we see k failures and (r − 1) successes (in any order), followed by a success.
r − 1 + k r − 1 + k Pr(X = k) = ·pr−1 ·qk ·p = ·qk ·pr k k
Albyn Jones Math 141 Multinomial Geometric Hypergeometric Poisson Example
Roll a fair die repeatedly until getting the third 6. What are p and q here? What is the probability it comes on the 10th roll, that is we have 7 non-sixes before the third 6?
9 57 13 (X = 7) = ≈ .0465 P 7 6 6
Albyn Jones Math 141 Multinomial Geometric Hypergeometric Poisson Connections
The number of Failures observed before getting the rth Success is clearly the sum of The number of Failures observed before the 1st Success The number of Failures observed between the 1st and 2nd Successes The number of Failures observed between the 2nd and 3rd Successes and so on. Theorem The sum of r independent Geometric(p) RV’s has a NegativeBinomial(r, p) distribution.
Albyn Jones Math 141 Multinomial Geometric Hypergeometric Poisson The Hypergeometric Distribution
Sampling from a finite population of two categories A and B without replacement. Non-Independence, Non-constant Probability: Successive trials are dependent; every trial changes the sample space and probabilities for the subsequent trials. X counts the number of A’s. n the number of trials is fixed.
Albyn Jones Math 141 Multinomial Geometric Hypergeometric Poisson Hypergeometric Probabilities
Suppose we have a bag with A alabaster and B black marbles, well mixed. We extract n marbles. Let X be the number of alabaster marbles drawn. For 0 ≤ k ≤ min(A, n), the probability of drawing k alabaster and n − k black marbles is given by