1 Intro to Probability Zhi Wei 2 Outline Basic concepts in probability theory Random variable and...

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Intro to Probability

Zhi Wei

Outline Basic concepts in probability theory Random variable and probability

distribution Bayes’ rule

Introduction

Probability is the study of randomness and uncertainty.

In the early days, probability was associated with games of chance (gambling).

Simple Games Involving Probability

Game: A fair die is rolled. If the result is 2, 3, or 4, you win $1; if it is 5, you win $2; but if it is 1 or 6, you lose $3.

Should you play this game?

Random Experiment

a random experiment is a process whose outcome is uncertain.

Examples: Tossing a coin once or several times Picking a card or cards from a deck Measuring temperature of patients ...

Sample SpaceThe sample space is the set of all possible outcomes.

Simple EventsThe individual outcomes are called simple events.

EventAn event is any subset of the whole sample space

Events & Sample Spaces

Example

Experiment: Toss a coin 3 times.

Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

Examples of events include A = {at least two heads}

B = {exactly two tails.}

= {HHH, HHT,HTH, THH}

= {HTT, THT,TTH}

Basic Concepts (from Set Theory)

A B, the union of two events A and B, is the event consisting of all outcomes that are either in A or in B or in both events.

A B (AB), the intersection of two events A and B, is the event consisting of all outcomes that are in both events.

Ac, the complement of an event A, is the set of all outcomes in that are not in A.

A-B, the set difference, is the event consisting of all outcomes that in A but not in B

When two events A and B have no outcomes in common, they are said to be mutually exclusive, or disjoint, events.

Example

Experiment: toss a coin 10 times and the number of heads is observed.

Let A = { 0, 2, 4, 6, 8, 10}.

B = { 1, 3, 5, 7, 9}, C = {0, 1, 2, 3, 4, 5}.

A B= {0, 1, …, 10} = .

A B contains no outcomes. So A and B are mutually exclusive.

Cc = {6, 7, 8, 9, 10}, A C = {0, 2, 4}, A-C={6,8,10}

Rules Commutative Laws:

A B = B A, A B = B A Associative Laws:

(A B) C = A (B C ) (A B) C = A (B C) .

Distributive Laws: (A B) C = (A C) (B C) (A B) C = (A C) (B C)

DeMorgan’s Laws:

. ,1111

Venn Diagram

A BA∩BA BA∩B

Probability

A Probability is a number assigned to each subset (events) of a sample space .

Probability distributions satisfy the following rules:

Axioms of Probability

For any event A, 0 P(A) 1.

P() =1.

If A1, A2, … An is a partition of A, then

P(A) = P(A1) + P(A2) + ...+ P(An)

(A1, A2, … An is called a partition of A if A1 A2 … An = A and A1, A2, … An are mutually exclusive.)

Properties of Probability

For any event A, P(Ac) = 1 - P(A).

P(A-B)=P(A) – P(A B) If A B, then P(A - B) = P(A) – P(B)

For any two events A and B, P(A B) = P(A) + P(B) - P(A B). For three events, A, B, and C, P(ABC) = P(A) + P(B) + P(C) - P(AB) - P(AC) - P(BC) + P(AB C).

Example

In a certain population, 10% of the people are rich, 5% are famous, and 3% are both rich and famous. A person is randomly selected from this population. What is the chance that the person is

not rich? rich but not famous? either rich or famous?

Joint Probability For events A and B, joint probability

Pr(AB) stands for the probability that both events happen.

Example: A={HT}, B={HT, TH}, what is the joint probability Pr(AB)?

Independence

Two events A and B are independent in casePr(AB) = Pr(A)Pr(B)

Independence

Example 1: Drug test

Women Men

Success 200 1800

Failure 1800 200

A = {A patient is a Women}

B = {Drug fails}

Are A and B independent?

Independence

Example 1: Drug test

Example 2: toss a coin 3 times, Let = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.A={having both T and H}, B={at most one T}Are A and B independent? How about toss 2 times?

Women Men

Success 200 1800

Failure 1800 200

A = {A patient is a Women}

B = {Drug fails}

Are A and B independent?

Independence If A and B independent, then A and Bc, B and Ac, Ac and Bc

independent

Consider the experiment of tossing a coin twice Example I:

A = {HT, HH}, B = {HT} Will event A independent from event B?

Example II: A = {HT}, B = {TH} Is event A independent from event B?

Disjoint Independence

If A is independent from B, B is independent from C, will A be independent from C?

If A and B are events with Pr(A) > 0, the conditional probability of B given A is

Conditioning

Pr( )Pr( | )

Example: Drug test

Conditioning

Pr( )Pr( | )

Women Men

Success 200 1800

Failure 1800 200

A = {Patient is a Women}

B = {Drug fails}

Pr(B|A) = ?

Pr(A|B) = ?

Example: Drug test

Given A is independent from B, what is the relationship between Pr(A|B) and Pr(A)?

Conditioning

Pr( )Pr( | )

Women Men

Success 200 1800

Failure 1800 200

A = {Patient is a Women}

B = {Drug fails}

Pr(B|A) = ?

Pr(A|B) = ?

Which Drug is Better ?

Simpson’s Paradox: View I

Drug I Drug II

Success 219 1010

Failure 1801 1190

A = {Using Drug I}

B = {Using Drug II}

C = {Drug succeeds}

Pr(C|A) ~ 10%

Pr(C|B) ~ 50%

Drug II is better than Drug I

Simpson’s Paradox: View II

Female Patient

A = {Using Drug I}

B = {Using Drug II}

C = {Drug succeeds}

Pr(C|A) ~ 10%

Pr(C|B) ~ 5%

Female Patient

A = {Using Drug I}

B = {Using Drug II}

C = {Drug succeeds}

Pr(C|A) ~ 10%

Pr(C|B) ~ 5%

Male Patient

A = {Using Drug I}

B = {Using Drug II}

C = {Drug succeeds}

Pr(C|A) ~ 100%

Pr(C|B) ~ 50%

Female Patient

A = {Using Drug I}

B = {Using Drug II}

C = {Drug succeeds}

Pr(C|A) ~ 10%

Pr(C|B) ~ 5%

Male Patient

A = {Using Drug I}

B = {Using Drug II}

C = {Drug succeeds}

Pr(C|A) ~ 100%

Pr(C|B) ~ 50%

Drug I is better than Drug II

Conditional Independence

Event A and B are conditionally independent given C in case

Pr(AB|C)=Pr(A|C)Pr(B|C) A set of events {Ai} is conditionally independent

given C in case

Pr( ... | ) Pr( | )n

A A C A C

Conditional Independence (cont’d) Example: There are three events: A, B, C

Pr(A) = Pr(B) = Pr(C) = 1/5 Pr(AC) = Pr(BC) = 1/25, Pr(AB) = 1/10 Pr(ABC) = 1/125 Whether A, B are independent? Whether A, B are conditionally independent

given C? A and B are independent A and B are

conditionally independent

Random Variable and Distribution

A random variable X is a numerical outcome of a random experiment

The distribution of a random variable is the collection of possible outcomes along with their probabilities: Categorical case: Numerical case:

Pr( ) ( )X x p x

Pr( ) ( )b

aa X b p x dx

Random Variables Distributions Cumulative Probability Distribution (CDF):

• Probability Density Function (PDF):Probability Density Function (PDF):

Random Variable: Example Let S be the set of all sequences of three rolls of

a die. Let X be the sum of the number of dots on the three rolls.

What are the possible values for X? Pr(X = 5) = ?, Pr(X = 10) = ?

Expectation value Division of the stakes problem

Henry and Tony play a game. They toss a fair coin, if get a Head, Henry wins; Tail, Tony wins. They contribute equally to a prize pot of $100, and agree in advance that the first player who has won 3 rounds will collect the entire prize. However, the game is interrupted for some reason after 3 rounds. They got 2 H and 1 T. How should they divide the pot fairly?

a) It seems unfair to divide the pot equally Since Henry has won 2 out of 3 rounds. Then, how about Henry gets 2/3 of $100?

b) Other thoughts?

X 0 100

P 0.25 0.75

X is what Henry will win if the game not interrupted

Expectation Definition: the expectation of a random

variable is , discrete case

, continuous case Properties

Summation: For any n≥1, and any constants k1,…,kn

Product: If X1, X2, …, Xn are independent

[ ] Pr( )x

E X x X x [ ] ( )E X xp x dx

[ ] ( )n n

i ii i

E X E X

[ ] ( )n n

i i i ii i

E k X k E X

Expectation: Example Let S be the set of all sequence of three rolls of a

die. Let X be the sum of the number of dots on the three rolls.

What is E(X)?

Let S be the set of all sequence of three rolls of a die. Let X be the product of the number of dots on the three rolls.

What is E(X)?

Variance Definition: the variance of a random variable X

is the expectation of (X-E[x])2 :

Properties For any constant C, Var(CX)=C2Var(X) If X1, X2, …, Xn are independent

( ) (( [ ]) )

( [ ] 2 [ ])

( ) [ ] 2 ( [ ])

[ ] [ ]

Var X E X E X

E X E X XE X

E X E X E XE X

E X E X

1 2 1 2( ... ) ( ) ( ) ... ( )n nVar X X X Var X Var X Var X

Bernoulli Distribution

The outcome of an experiment can either be success (i.e., 1) and failure (i.e., 0).

Pr(X=1) = p, Pr(X=0) = 1-p, or

E[X] = p, Var(X) = p(1-p) Using sample() to generate Bernoulli samples> n = 10; p = 1/4; > sample(0:1, size=n, replace=TRUE, prob=c(1-p, p))[1] 0 1 0 0 0 1 0 0 0 1

1( ) (1 )x xp x p p

Binomial Distribution n draws of a Bernoulli distribution

Xi~Bernoulli(p), X=i=1n Xi, X~Bin(p, n)

Random variable X stands for the number of times that experiments are successful.

n = the number of trials x = the number of successes p = the probability of success

E[X] = np, Var(X) = np(1-p)

otherwise0

,...,2,1,0)1()()Pr(

xpxXxnx

the binomial distribution in R

dbinom(x, size, prob) Try 7 times, equally

likely succeed or fail> dbinom(3,7,0.5)[1] 0.2734375

>barplot(dbinom(0:7,7,0.5),names.arg=0:7)

0 1 2 3 4 5 6 7

what if p ≠ 0.5? > barplot(dbinom(0:7,7,0.1),names.arg=0:7)

0 1 2 3 4 5 6 7

Which distribution has greater variance?

0 1 2 3 4 5 6 7

p = 0.5 p = 0.1

var = n*p*(1-p) = 7*0.1*0.9=7*0.09var = n*p*(1-p) = 7*0.5*0.5 = 7*0.25

briefly comparing an experiment to a distribution

nExpr = 1000tosses = 7; y=rep(0,nExpr);for (i in 1:nExpr) {

x = sample(c("H","T"), tosses, replace = T)y[i] = sum(x=="H")

}hist(y,breaks=-0.5:7.5)lines(0:7,dbinom(0:7,7,0.5)*n

Expr) points(0:7,dbinom(0:7,7,0.5)

*nExpr)

theoreticaldistribution

result of 1000 trials

Histogram of y

0 2 4 6

Cumulative distribution

P(X=x)

0 1 2 3 4 5 6 7

P(X≤x)

> barplot(dbinom(0:7,7,0.5),names.arg=0:7) > barplot(pbinom(0:7,7,0.5),names.arg=0:7)

cumulative distribution

0 1 2 3 4 5 6 7

tive d

P(X=x) P(X≤x)

example: surfers on a website Your site has a lot of visitors 45% of whom

are female You’ve created a new section on

gardening Out of the first 100 visitors, 55 are female. What is the probability that this many or

more of the visitors are female? P(X≥55) = 1 – P(X≤54) = 1-

pbinom(54,100,0.45)

Another way to calculate cumulative probabilities ?pbinom P(X≤x) = pbinom(x, size, prob, lower.tail = T) P(X>x) = pbinom(x, size, prob, lower.tail = F)

> 1-pbinom(54,100,0.45)[1] 0.02839342

> pbinom(54,100,0.45,lower.tail=F)[1] 0.02839342

Female surfers visiting a section of a website

0 6 13 21 29 37 45 53 61 69 77 85 93

what is the area under the curve?

Cumulative distribution

0 6 13 21 29 37 45 53 61 69 77 85 93

> 1-pbinom(54,100,0.45)[1] 0.02839342

Plots of Binomial Distribution

Another discrete distribution: hypergeometric

Randomly draw n elements without replacement from a set of N elements, r of which are S’s (successes) and (N-r) of which are F’s (failures)

hypergeometric random variable x is the number of S’s in the draw of n elements

hypergeometric example fortune cookies there are N = 20 fortune cookies r = 18 have a fortune, N-r = 2 are empty What is the probability that out of n = 5 cookies,

s=5 have a fortune (that is we don’t notice that some cookies are empty)

> dhyper(5, 18, 2, 5) [1] 0.5526316

So there is a greater than 50% chance that we won’t notice.

Gene Set Enrichment Analysis

hypergeometric and binomial When the population N is (very) big, whether one

samples with or without replacement is pretty much the same

100 cookies, 10 of which are empty

1 2 3 4 5

binomial

hypergeometric

number of full cookies out of 5

code aside> x = 1:5> y1 = dhyper(1:5,90,10,5)> y2 = dbinom(1:5,5,0.9)> tmp = as.matrix(t(cbind(y1,y2)))> barplot(tmp,beside=T,names.arg=x)

hypergeometric probability

binomial probability

Poisson distribution # of events in a given interval

e.g. number of light bulbs burning out in a building in a year # of people arriving in a queue per minute

= mean # of events in a given interval

E[X] = , Var(X) = !

Example: Poisson distribution You got a box of 1,000 widgets. The manufacturer says that the failure

rate is 5 per box on average. Your box contains 10 defective widgets.

What are the odds?> ppois(9,5,lower.tail=F)[1] 0.03182806 Less than 3%, maybe the manufacturer is

not quite honest. Or the distribution is not Poisson?

Poisson approximation to binomial If n is large (e.g. > 100) and n*p is moderate (p

should be small) (e.g. < 10), the Poisson is a good approximation to the binomial with = n*p

0 1 2 3 4 5 6 7 8 9 11 13 15

binomialPoisson

Plots of Poisson Distribution

Normal (Gaussian) Distribution Normal distribution (aka “bell curve”) fits many biological data well

e.g. height, weight serves as an approximation to binomial,

hypergeometric, Poisson because of the Central Limit Theorem

Well studied

Normal (Gaussian) Distribution X~N(,)

E[X]= , Var(X)= 2

If X1~N(1,1), X2~N(2,2), and X1, X2 are independent X= X1+ X2 ?

X= X1- X2 ?

1 ( )( ) exp

1 ( )Pr( ) ( ) exp

xa X b p x dx dx

sampling from a normal distributionx <- rnorm(1000)h <- hist(x, plot=F)ylim <-

range(0,h$density,dnorm(0))

hist(x,freq=F,ylim=ylim)curve(dnorm(x),add=T)

Histogram of x

-4 -2 0 2 4

Normal Approximation based on Central Limit Theorem

Central Limit Theorem If xi ~i.i.d with (μ, σ2) and when n is large, then

(x1+…+xn)/n ~ N(μ, σ2/n)

Or (x1+…+xn) ~ N(nμ, nσ2)

Example A population is evenly divided on an issue (p=0.5). For a random

sample of size 1000, what is the probability of having ≥550 in favor of it?

n=1000, xi~Bernoulli (p=0.5), i.e. E(xi)=p; V(xi)=p(1-p)

(x1+…+xn) ~ Binomial(n=1000, p=0.5)

Pr((x1+…+xn)>=550) =1-pbinom(550,1000,0.5)

Normal Approximation:(x1+…+xn) ~ N(np, np(1-p))=N(500, 250)

Pr((x1+…+xn)>=550) =1-pnorm(550, mean=500, sd=sqrt(250))63

d, p, q, and r functions in R In R, a set of functions have been implemented

for each of almost all known distributions. r<distname>(n,<parameters>)

Possible distributions: binom, pois, hyper, norm, beta, chisq, f, gamma, t, unif, etc

You find other characteristics of distributions as well d<dist>(x,<parameters>): density at x p<dist>(x,<parameters>): cumulative distribution

function to x q<dist>(p,<parameters>): inverse cdf

Example: Uniform Distribution The uniform distr. On [a,b] has two

parameter. The family name is unif. In R, the parameters are named min and max

> dunif(x=1, min=0, max=3)[1] 0.3333333> punif(q=2, min=0, max=3)[1] 0.6666667> qunif(p=0.5, min=0, max=3)[1] 1.5> runif(n=5, min=0, max=3)[1] 2.7866852 1.9627136 0.9594195 2.6293273 1.6277597

Lab Exercise Using R for Introductory StatisticsPage 39: 2.4, 2.8-2.10Page 54: 2.16, 2.23, 2.35, 2.26Page 66: 2.30, 2.32, 2.34-2.36, 2.39, 2.41,

2.42, 2.43-2.46

Given two events A and B and suppose that Pr(A) > 0. Then

Example:

Bayes’ Rule

Pr(W|R) R R

W 0.7 0.4

W 0.3 0.6

R: It is a rainy day

W: The grass is wet

Pr(R|W) = ?

Pr(R) = 0.8

)Pr()|Pr(

Bayes’ Rule

W 0.7 0.4

W 0.3 0.6

R: It rains

W: The grass is wet

Information

Pr(W|R)

Inference

Pr(R|W)

Pr( | ) Pr( )Pr( | )

E H HH E

Bayes’ Rule

W 0.7 0.4

W 0.3 0.6

R: The weather rains

W: The grass is wet

Hypothesis H Evidence EInformation: Pr(E|H)

Inference: Pr(H|E) PriorLikelihoodPosterior

Summation (Integration) out tip Suppose that B1, B2, … Bk form a partition of :

B1 B2 … Bk = and B1, B2, … Bk are mutually exclusive

Suppose that Pr(Bi) > 0 and Pr(A) > 0. Then

Pr( | ) Pr( )Pr( | )

Pr( | ) Pr( )

Pr( ) Pr( | )

A B BB A

Pr( | ) Pr( )Pr( | )

Pr( | ) Pr( )

Pr( ) Pr( | )

A B BB A

Pr( | ) Pr( )Pr( | )

Pr( | ) Pr( )

Pr( ) Pr( | )

A B BB A

Key: Joint distribution!

Application of Bayes’ Rule

R It rains

W The grass is wet

U People bring umbrella

Pr(W|R) R R

W 0.7 0.4

W 0.3 0.6

Pr(U|R) R R

U 0.9 0.2

U 0.1 0.8

Pr(U|W) = ?

Pr(R) = 0.8

A More Complicated Example

R It rains

W The grass is wet

Pr(W|R) R R

W 0.7 0.4

W 0.3 0.6

Pr(U|R) R R

U 0.9 0.2

U 0.1 0.8

Pr(U|W) = ?

Pr(R) = 0.8

A More Complicated Example

R It rains

W The grass is wet

Pr(W|R) R R

W 0.7 0.4

W 0.3 0.6

Pr(U|R) R R

U 0.9 0.2

U 0.1 0.8

Pr(U|W) = ?

Pr(R) = 0.8

Acknowledgments Peter N. Belhumeur: for some of the slides adapted or

modified from his lecture slides at Columbia University Rong Jin: for some of the slides adapted or modified from

his lecture slides at Michigan State University Jeff Solka: for some of the slides adapted or modified from

his lecture slides at George Mason University Brian Healy: for some of the slides adapted or modified

from his lecture slides at Harvard University

1 Intro to Probability Zhi Wei 2 Outline Basic concepts in probability theory Random variable and...

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