1. Geometric Poisson Negative Binomial Gamma
Chapter 3b
Geometric
Poisson
Negative Binomial
Gamma

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

3. 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

4. Bernoulli distribution
Expectation:
Variance:

5. Binomial distribution
The binomial distribution is just n independent Bernoullis
added up.
It is the number of “successes” in n trials.
If Z1, Z2, …, Zn are Bernoulli, then X is binomial:

6. Binomial distribution
The binomial distribution is just n independent Bernoullis
added up. Testing for defects “with replacement.”
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

7. 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) = (1-p)3
We want P(X=1):
Can happen three ways: 100, 010, 001
p(1-p)(1-p)+(1-p)p(1-p)+(1-p)(1-p)p = 3p(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 = 3p2(1-p)
We want P(X=3):
Can happen one way: 111
ppp = p3

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

9. Binomial distribution
Typical shape of binomial:
Symmetric

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

11. 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?

12. Binomial distribution
Less than 5% or
1 in 20.
So, it’s unlikely
that his success
rate is 0.4.
Note:

13. 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:

14. 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.

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

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

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

18. 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.

19. 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.

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

21. Poisson distribution
Shape
Low 
Med 
High 

22. 2.4 – The Poisson Distribution
Definition 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

23. Example 2.4.1
Suppose a 1000 ft2 lawn contains 3000 dandelions. Find the probability that a randomly chosen 1 ft2 section of lawn contains exactly five dandelions.
“Occurrence” = dandelion
“Interval” = 1 ft2 section of lawn
X = number of dandelions in a 1 ft2 section of lawn
Assume X is Poisson

24. Example 2.4.1

25. Negative Binomial Probability Distribution

26. 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

27. 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

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

29. 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-1Cr-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

30. 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)/p2
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.

31. 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

32. 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 Suppose the company plans to drill a series of wells looking for three successful wells.
a) What is the probability that the third success will be achieved with the 8th well drilled?   
b) What is the probability that the third success will be achieved with the 20th well drilled?
P(x) = x-1Cr-1 pr(1 – p)x-r
p = 0.3
P(8) = 8-1C3-1 p3(1 – p)8-3
= 7C2 (0.3)3(0.7)5
= (21)(0.027)( ) =
P(20) = C3-1 p3(1 – p)20-3
= 19C2 (0.3)3(0.7)17
= (171)(0.027)( ) =

33. 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 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 its third productive well.
Mean μx = r/p = 3 / 0.3 = 10
Standard Deviation σx = r(1-p)/p² = 3(0.7)/(0.3)² =

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

35. 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

36. The gamma distribution
The gamma distribution arises naturally as the distribution of a series of iid random exponential variables
The gamma distribution has expectation a/b and variance a/b2
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)
a = b = 0.5
a = b = 2
a = b = 10

37. 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
The expectation is a/(a + b)
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
a = b = 0.5
a = b = 10
a = b = 2