2. 1 Intro to Probability Zhi Wei

3. 2 Outline  Basic concepts in probability theory  Random variable and probability distribution  Bayes ’ rule

4. 3 Introduction  Probability is the study of randomness and uncertainty.  In the early days, probability was associated with games of chance (gambling).

5. 4 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?

6. 5 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 ...

7. 6 Sample Space The sample space is the set of all possible outcomes. Simple Events The individual outcomes are called simple events. Event An event is any subset of the whole sample space Events & Sample Spaces

8. 7 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}

9. 8 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.  A c, 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.

10. 9 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.  C c = {6, 7, 8, 9, 10}, A  C = {0, 2, 4}, A-C={6,8,10}

11. 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:

12. 11 Venn Diagram  A B A∩B A B

13. 12 Probability  A Probability is a number assigned to each subset (events) of a sample space .  Probability distributions satisfy the following rules:

14. 13 Axioms of Probability  For any event A, 0  P(A)  1.  P() =1.  If A 1, A 2, … A n is a partition of A, then P(A) = P(A 1 ) + P(A 2 ) +...+ P(A n ) (A 1, A 2, … A n is called a partition of A if A 1  A 2  …  A n = A and A 1, A 2, … A n are mutually exclusive.)

15. 14 Properties of Probability  For any event A, P(A c ) = 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(ABC) = P(A) + P(B) + P(C) - P(AB) - P(AC) - P(BC) + P(AB C).

16. 15 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?

17. 16 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)?

18. 17 Independence  Two events A and B are independent in case Pr(AB) = Pr(A)Pr(B)

19. 18 Independence  Two events A and B are independent in case Pr(AB) = Pr(A)Pr(B)  Example 1: Drug test WomenMen Success2001800 Failure1800200 A = {A patient is a Women} B = {Drug fails} Are A and B independent?

20. 19 Independence  Two events A and B are independent in case Pr(AB) = Pr(A)Pr(B)  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? WomenMen Success2001800 Failure1800200 A = {A patient is a Women} B = {Drug fails} Are A and B independent?

21. 20 Independence  If A and B independent, then A and B c, B and A c, A c and B c 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?

22. 21  If A and B are events with Pr(A) > 0, the conditional probability of B given A is Conditioning

23. 22  If A and B are events with Pr(A) > 0, the conditional probability of B given A is  Example: Drug test Conditioning WomenMen Success2001800 Failure1800200 A = {Patient is a Women} B = {Drug fails} Pr(B|A) = ? Pr(A|B) = ?

24. 23  If A and B are events with Pr(A) > 0, the conditional probability of B given A is  Example: Drug test  Given A is independent from B, what is the relationship between Pr(A|B) and Pr(A)? Conditioning WomenMen Success2001800 Failure1800200 A = {Patient is a Women} B = {Drug fails} Pr(B|A) = ? Pr(A|B) = ?

25. 24 Which Drug is Better ?

26. 25 Simpson ’ s Paradox: View I Drug IDrug II Success2191010 Failure18011190 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

27. 26 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%

28. 27 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% Male Patient A = {Using Drug I} B = {Using Drug II} C = {Drug succeeds} Pr(C|A) ~ 100% Pr(C|B) ~ 50%

29. 28 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% 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

30. 29 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 {A i } is conditionally independent given C in case

31. 30 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

32. 31 Outline  Basic concepts in probability theory  Random variable and probability distribution  Bayes ’ rule

33. 32 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:

34. 33 Random Variables Distributions  Cumulative Probability Distribution (CDF): Probability Density Function (PDF): Probability Density Function (PDF):

35. 34 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) = ?

36. 35 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? X0100 P0.250.75 X is what Henry will win if the game not interrupted

37. 36 Expectation  Definition: the expectation of a random variable is, discrete case, continuous case  Properties Summation: For any n≥1, and any constants k 1, …,k n Product: If X 1, X 2, …, X n are independent

38. 37 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)?

39. 38 Variance  Definition: the variance of a random variable X is the expectation of (X-E[x]) 2 :  Properties For any constant C, Var(CX)=C 2 Var(X) If X 1, X 2, …, X n are independent

40. 39 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

41. 40 Binomial Distribution  n draws of a Bernoulli distribution X i ~Bernoulli(p), X= i=1 n X i, 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)

42. 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)

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

44. Which distribution has greater variance? p = 0.5 p = 0.1 var = n*p*(1-p) = 7*0.1*0.9=7*0.09 var = n*p*(1-p) = 7*0.5*0.5 = 7*0.25

45. briefly comparing an experiment to a distribution nExpr = 1000 tosses = 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)* nExpr) points(0:7,dbinom(0:7,7,0.5 )*nExpr) theoretical distribution result of 1000 trials Histogram of y y Frequency 0246 0 50 100 150 200 250 300

46. Cumulative distribution P(X=x)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)

47. cumulative distribution 01234567 probability distribution 0.0 0.2 0.4 0.6 0.8 1.0 01234567 cumulative distribution 0.0 0.2 0.4 0.6 0.8 1.0 P(X=x)P(X≤x)

48. 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)

49. 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

50. Female surfers visiting a section of a website what is the area under the curve?

51. Cumulative distribution <3 % > 1-pbinom(54,100,0.45) [1] 0.02839342

52. 51 Plots of Binomial Distribution

53. 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

54. 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

55. 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 binomial hypergeometric number of full cookies out of 5

56. 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

57. 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) =

58. 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?

59. 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 0123456789111315 binomial Poisson 0.00 0.05 0.10 0.15

60. 59 Plots of Poisson Distribution

61. 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

62. 61 Normal (Gaussian) Distribution  X~N(,)  E[X]= , Var(X)=  2  If X 1 ~N( 1, 1 ), X 2 ~N( 2, 2 ), and X 1, X 2 are independent X= X 1 + X 2 ? X= X 1 - X 2 ?

63. sampling from a normal distribution x <- rnorm(1000) h <- hist(x, plot=F) ylim <- range(0,h$density,dnor m(0)) hist(x,freq=F,ylim=ylim) curve(dnorm(x),add=T)

64. Normal Approximation based on Central Limit Theorem  Central Limit Theorem If x i ~i.i.d with (μ, σ 2 ) and when n is large, then (x 1 +…+x n )/n ~ N(μ, σ 2 /n) Or (x 1 +…+x n ) ~ N(nμ, nσ 2 )  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, x i ~Bernoulli (p=0.5), i.e. E(x i )=p; V(x i )=p(1-p) (x 1 +…+x n ) ~ Binomial(n=1000, p=0.5) Pr((x 1 +…+x n )>=550) =1-pbinom(550,1000,0.5) Normal Approximation: (x 1 +…+x n ) ~ N(np, np(1-p))=N(500, 250) Pr((x 1 +…+x n )>=550) =1-pnorm(550, mean=500, sd=sqrt(250)) 63

65. 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 (n, ) Possible distributions: binom, pois, hyper, norm, beta, chisq, f, gamma, t, unif, etc  You find other characteristics of distributions as well d (x, ): density at x p (x, ): cumulative distribution function to x q (p, ): inverse cdf 64

66. 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 65

67. Lab Exercise  Using R for Introductory Statistics Page 39: 2.4, 2.8-2.10 Page 54: 2.16, 2.23, 2.35, 2.26 Page 66: 2.30, 2.32, 2.34-2.36, 2.39, 2.41, 2.42, 2.43-2.46 66

68. 67 Outline  Basic concepts in probability theory  Random variable and probability distribution  Bayes ’ rule

69. 68  Given two events A and B and suppose that Pr(A) > 0. Then  Example: Bayes ’ Rule Pr(W|R)R RR W0.70.4 WW 0.30.6 R: It is a rainy day W: The grass is wet Pr(R|W) = ? Pr(R) = 0.8

70. 69 Bayes ’ Rule R RR W0.70.4 WW 0.30.6 R: It rains W: The grass is wet RW Information Pr(W|R) Inference Pr(R|W)

71. 70 Bayes ’ Rule R RR W0.70.4 WW 0.30.6 R: The weather rains W: The grass is wet Hypothesis H Evidence E Information: Pr(E|H) Inference: Pr(H|E) PriorLikelihoodPosterior

72. 71 Summation (Integration) out tip  Suppose that B 1, B 2, … B k form a partition of : B 1  B 2  …  B k =  and B 1, B 2, … B k are mutually exclusive Suppose that Pr(B i ) > 0 and Pr(A) > 0. Then

73. 72 Summation (Integration) out tip  Suppose that B 1, B 2, … B k form a partition of : B 1  B 2  …  B k =  and B 1, B 2, … B k are mutually exclusive Suppose that Pr(B i ) > 0 and Pr(A) > 0. Then

74. 73 Summation (Integration) out tip  Suppose that B 1, B 2, … B k form a partition of : B 1  B 2  …  B k =  and B 1, B 2, … B k are mutually exclusive Suppose that Pr(B i ) > 0 and Pr(A) > 0. Then Key: Joint distribution!

75. 74 Application of Bayes’ Rule RIt rains WThe grass is wet UPeople bring umbrella Pr(UW|R)=Pr(U|R)Pr(W|R) Pr(UW| R)=Pr(U| R)Pr(W| R) R WU Pr(W|R)R RR W0.70.4 WW 0.30.6 Pr(U|R)R RR U0.90.2 UU 0.10.8 Pr(U|W) = ? Pr(R) = 0.8

76. 75 A More Complicated Example RIt rains WThe grass is wet UPeople bring umbrella Pr(UW|R)=Pr(U|R)Pr(W|R) Pr(UW| R)=Pr(U| R)Pr(W| R) R WU Pr(W|R)R RR W0.70.4 WW 0.30.6 Pr(U|R)R RR U0.90.2 UU 0.10.8 Pr(U|W) = ? Pr(R) = 0.8

77. 76 A More Complicated Example RIt rains WThe grass is wet UPeople bring umbrella Pr(UW|R)=Pr(U|R)Pr(W|R) Pr(UW| R)=Pr(U| R)Pr(W| R) R WU Pr(W|R)R RR W0.70.4 WW 0.30.6 Pr(U|R)R RR U0.90.2 UU 0.10.8 Pr(U|W) = ? Pr(R) = 0.8

78. 77 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