SanJoséStateUniversity Math161A:AppliedProbability&Statistics...Specialdiscretedistributions(part1)...

San José State University

Math 161A: Applied Probability & Statistics

Special discrete distributions (part 1)

Prof. Guangliang Chen

This lecture: Sections 3.4, 3.5

• Bernoulli

• Binomial

• HyperGeometric

Next lecture: Sections 3.5, 3.6

• Geometric

• Negative Binomial

• Poisson


Our treatment plan for each distribution

• Examples

• Definition (including pmf)

• Expected value

• Variance

• Other useful properties (if any)

Prof. Guangliang Chen | Mathematics & Statistics, San José State University 3/26


Bernoulli distribution

Consider the following experiments:

Example 0.1 (Toss a fair coin). X = 1 (heads) and 0 (tails).

Example 0.2 (Randomly select a ball from an urn that has 10 red and 20green balls). Let Y = 1 (if the selected ball is red) and 0 (otherwise).

Example 0.3 (Randomly select an individual from a population 40% ofwhich have certain characteristic). Let Z = 1 (if the selected individualhas the characteristic) and 0 (otherwise).



These experiments all share the following traits:

• There is only one trial;

• It has only two outcomes, “success” or “failure”;

• The probability of getting a success is some number p;

• X is a indicator variable: X = 1 (success) or 0 (failure)

We say that such a random variable has a Bernoulli distribution withparameter p, and denote it as X ∼ Bernoulli(p).

Such an experiment is called a Bernoulli trial.



Example 0.4 (Toss a fair coin). X = 1 (heads) and 0 (tails).

Answer: X ∼ Bernoulli(12)

Example 0.5 (Randomly select a ball from an urn that has 10 red and 20green balls). Let Y = 1 (if the selected ball is red) and 0 (otherwise).

Answer: Y ∼ Bernoulli(13)

Example 0.6 (Randomly select an individual from a population 40% ofwhich have certain characteristic). Let Z = 1 (if the selected individualhas the characteristic) and 0 (otherwise).

Answer: Z ∼ Bernoulli(0.4)



Clearly, if a discrete random variable X follows a Bernoulli distributionwith parameter p, then its pmf has the following form (and vice versa):

fX(x) = px(1− p)1−x, x = 0, 1

(and fX(x) = 0 for all other x)

x 0 1P (X = x) 1− p p 0 1

b

b

1− p

p |



Theorem 0.1. Let X ∼ Bernoulli(p). Then

E(X) = pVar(X) = p(1− p)

Proof. We have already obtained these results previously.



Binomial

The following experiments are identical in nature:

• (Toss a fair coin 10 times): X = #heads

• (Answer 10 multiple-choiced questions by random guessing): X=#correctly answered questions

• (Select with replacement 10 balls at random from an urn containing30 red and 20 blue balls): X= # red balls obtained



We make the following abstraction:

• There are n repeated trials in the experiment

• Each trial has only two outcomes, “success” and “failure”

• The probability p of getting successes is fixed

• The n Bernoulli trials are independent

• X denotes the total number of successes



In short, X counts the total number of successes in n independent Bernoullitrials with fixed probability of success p.

1 2 n

In the above scenario, we say that X follows a binomial distributionwith parameters n, p, and write X ∼ B(n, p)



Example.

• (Toss a fair coin 10 times) X = #heads B(10, 12)

• (Answer 10 multiple-choiced questions by random guessing) X=#correctly answered questions B(10, 14)

• (Draw with replacement 10 balls from an urn containing 30 red and20 blue balls at random) X= # red balls obtained B(10, 0.6)

Question. In the last example, if balls are drawn without replacement, isX still a binomial random variable? Why?



Theorem 0.2. The pmf of X ∼ B(n, p) is

fX(x) =(

n

x

)px(1− p)n−x, x = 0, 1, . . . , n.

1 2 n

H HH

How to understand this result:

• (nx): # ways of having x successes in n trials• px: probability of having exactly x successes

• (1− p)n−x: probability of having exactly n− x failures



Example 0.7. What is the probability of getting 0 heads in 10 independentflips of a fair coin? Exactly 1 head? At least two heads?



Example 0.8 (Answer 10 multiple-choiced questions by random guessing).Let X= #correctly answered questions. Find P (X = x) for x = 0, 2, 9.



0 2 4 6 8 10

x

0

0.05

0.1

0.15

0.2

0.25

0.3

f(x)

B(n,p) pmfs with fixed n=10 and different values of p

p=0.5p=0.25p=0.6



Theorem 0.3. Let X ∼ B(n, p). Then

E(X) = np,Var(X) = np(1− p)

Proof. We have already proved the two results previously.



Two “no meal” jokes

Q: What is the binomial distribution?

A: A free lunch program (buy no meal).

Q: What do you call a mathematician’s parrot that has not been fed?

A: A polynomial (poly no meal)



Hypergeometric

Recall the example in which we draw 10 balls at random from an urncontaining 30 red and 20 blue balls, and let X= # red balls.

We obtained that X ∼ B(10, 0.6) if the experiment is performedwith replacement:

fX(x) =(

10x

)0.6x(1− 0.6)10−x, x = 0, 1, . . . , 10

We also mentioned that X is not binomial if it is without replacement.



In fact, in the latter case (without replacement), X has the following pmf:

fX(x) =(30

x

)( 2010−x

)(5010) , x = 0, 1, . . . , 10

where

• (30x ): #ways of choosing x red balls out of 30• ( 20n−x): #ways of choosing 10− x blue balls out of 20• (5010): #ways of choosing 10 balls out of 50 in total (ignoring color)

This is an example of a hypergeometric distribution with parametersN = 50, r = 30, n = 10.



Def 0.1 (X ∼ HyperGeom(N, r, n)).We say that any random variable Xwith a pmf of the following form

fX(x) =(r

x

)(N−rn−x

)(Nn

) , for x in rangehas a Hypergeometric distributionwith parameters N, r, n).

(r red)

N balls in total

(N − r blue)

select n at random

No replacement

X= #red balls

obtained

Remark. Typically, Range(X) = {0, 1, . . . , n}. In the cases when there arefew red/blue balls, Range(X) ( {0, 1, . . . , n}. For example, if r = 5, N =25 and n = 10, then Range(X) = {0, 1, . . . , 5} ⊂ {0, 1, . . . , 10}.



Example 0.9. Draw, without replacement, n voters at random fromthe whole pool of N that are registered, r of which support certaincandidate. Let X = #selected supporters of the candidate. Then X ∼HyperGeom(N, r, n).

r individuals

Population of size NSample n

X = # individuals

from subpopulation



The hypergeometric pmf has a complicated formula, but it can be wellapproximated by the binomial pmf in the setting of large N and large r.

Theorem 0.4. When N, r are both large (relative to n), then

HyperGeom(N, r, n) ≈ B(n, p = rN

).

Remark. In the preceding polling example, both r and N are typicallylarge (e.g., in the order of millions) and n is at most a thousand. Theabove theorem implies that, the number of selected voters that support aparticular candicate is approximately binomial: X approx∼ B(n, p = rN ).



500 red

499 blue

498 red

500 blue

499 red

500 blue

500 red

500 blue

499 red

499 blue

499 red

499 blue

500 red

498 blue

5001000

5001000

499999

500999

500999

499999

497 red

500 blue

498998

500 red

497 blue498998

b

b

b

b

b

b

b

b

b

b

b

b



Example 0.10. Select 20 balls at random from an urn containing 500red and 500 blue balls, and let X= number of red balls in the selection.Find both the exact value of P (X = 10) and its binomial approximation(Answer: .1780, .1762).

0 2 4 6 8 10 12 14 16 18 200

0.05

0.1

0.15

0.2

HG(1000,500,20)B(20,0.5)



Theorem 0.5. Let X ∼ HyperGeom(N, r, n) and p = rN . Then

E(X) = nrN

= np

Var(X) = np(1− p)(

N − nN − 1

)

Remark. Compare with B(n, p = rN ) corresponding to with replacement:

• The expected values are identical, but

• the variance of the hypergeometric distribution is smaller.


SanJoséStateUniversity Math161A:AppliedProbability&Statistics...Specialdiscretedistributions(part1)...

Documents

Transcript of SanJoséStateUniversity Math161A:AppliedProbability&Statistics...Specialdiscretedistributions(part1)...