Two characteristic properties of a random experiment:

Transcription

1 1 Lecture Probability as Relative Frequency A single coin flipping experiment. Observe a coin flipped many times. What probability we can prescribe to the event that a head turns up? We expect the coin to be fair, we expect a head turn up approximately half of the times the coin if flipped. Probability of a head = for large number of experiments. #of times a head turns up # of coin flips 1 2, Probability of an event = Relative Frequency of the event. Two characteristic properties of a random experiment: 1. We can not predict the outcome of experiment. 2. We can estimate relative frequencies of the outcomes A box contains n balls; m are red, n m are black. A ball is drawn at random. What is the probability that the ball is red? The ball is black? The ball is either red or black? Two cons are flipped. Write a model of this experiment. Find the probability that there is at least on head on two coins. 1.2 Mathematical Model of Probability Outcomes of a random experiment are observed. The set of all possible outcomes is called Sample space: S. Outcomes are labeled by small letters a,b,c... In this notation S = {a,b,c...} An event is a subset of S. Events are denoted by capital letters A,B,C,D,E... A set containing no elements is call a null set: /0. An event A is a subset of event B is every element of A is also an element of B : A B, If x A then x B. 1

2 1. Operation on events. 2. Venn Diagrams. 3. Properties of set Operations. De Morgan s Laws. 4. Proving the Properties of set operations. 5. Union and intersection of several sets. 6. Disjoint sets. 7. Property that for any two sets A,B : A = (AB) (AB), with two disjoint sets AB and AB. Problem Prove the De Morgan s Law n i=1 A i = n i=1 A i. 2

3 2 Lecture Axioms of Probability The probability is a function from the set of events into the interval [0,1], that verifies the following properties: 1. P(A) 0, for any event A. 2. P(S) = 1; 3. For any pairwise disjoint events A 1,A 2,A 3,...,A n : Properties of the probabilities. Problem Show that P(/0) = 0. P(A 1... A n ) = P(A 1 ) P(A n ). Solution Since S /0 = S, and S and /0 are disjoint (why?) we can use axiom 3 to get which implies that P(/0) = 0. 1 = P(S) = P(S) + P(/0) = 1 + P(/0), Problem Show that for any event A, P(A) = 1 P(A), P(A) = 1 P(A). Solution Using the set identity S = A A, and axiom 3 we get 1 = P(S) = P(A) + P(A), from which the formulas follow. Problem Show that for any event A, 0 P(A) 1. 3

4 Solution The first inequality P(A) 0 is the statement of axiom 1. From the last problem we get P(A) = 1 P(A) 1. Problem Show that Hint: Use the fact that P(A B) = P(A) + P(B) P(AB). A = AB AB, B = BA AB, A B = AB BA AB. Solution Using the hint and axiom 3 we get P(A) = P(AB) + P(AB) P(B) = P(BA) + P(AB) P(A B) = P(AB) + P(BA) + P(AB). Subtracting the sum of the first two equations from the last we get the formula. Problem Show that P(A B C) = P(A) + P(B) + P(C) P(AB) P(AC) P(BC) + P(ABC). Hint: Use the formula from the previous problem. Solution Introduce set D = A B. Using the previous formula P(D C) = P(D) + P(C) P(DC) = P(A B) + P(C) P((A B)C) = P(A) + P(B) P(AB) + P(C) P(AC BC) = P(A) + P(B) P(AB) + P(C) (P(AC) + P(BC) P(ABC)) = P(A) + P(B) + P(C) P(AB) P(AC) P(BC) + P(ABC). 4

5 2.2 Probability Space The sample space S, the set of events F, and the probability P(A) that verifies the axioms of probability is called a Probability Space: Examples of Probability Spaces. (S,F,P). 1. The classical probability: S = {a 1,...,a n } and all outcomes are equally likely. The probability to get an outcome P(a i ) = 1 n, i = 1..n. If an event A contains k outcomes, then P(A) = k n. 2. The area function. S is a square (or any other figure) of area 1. Events are the subsets of S. P(A) = Area(A). Show that P(A) is a valid probability function. 3. Empirical Probability. An experiment with n outcomes {a 1,...,a n } is repeated N times. Define P(a i ) = # of times a i occurs. N 4. Belief Probability. Betting on sports. The odds are computed as a fraction of all money people bet on a particular team to all money against the team. The corresponding probabilities are defined are ratios of all money people bet on a team to the total bets. An example is given in the table below. 5

6 Team Odds Probability of winning Team 1 3:17 Team 2 1:1 Team 3 7: Note that all probabilities sum up to 1. The odds let you quickly compute your return if a team wins. For example if you bet $N for team 1 and it wins, you return (the money you get in the excess of your bet) is $ 17 3 N. Note, that according to the table people rate team 1 lower than other teams. 3 Lecture Problems involving proportions 1. In a group of 10,000 people, 1,000 have home insurance; 2,000 have health insurance; 300 have both home and health insurance. Find probability that a randomly selected person has at least one type of insurance. Solve the problem in two ways: counting the number of people in the event in question and using the formula for the probability of union of two events. Answer: P = In a group of 100 students, 50 take Math, 60 take Chem, and 10 do not take Math. and do not take Chem. Find the probability that a random student takes Math and Chem. Answer: P =.2. 6

7 3. In a small town 25% of population have Life insurance; 10% have home insurance; 20% have health insurance; 5% have home and life insurance; 10% have health and life insurance; 3% have home and health insurance; 1% have all 3 types of insurance. Find the probability that a randmly selected person has at least one type of insurance. Solve the problem in two ways: using Venn diagram and the formula for the union of three events. Answer: P = A group of people is polled on 3 questions. 55% said Yes on Question 3; 32.5% said Yes exactly twice; 10% said Yes on all three questions; 12.5% said Yes on Question 1 and Question 2. Find the probability that a randomly selected person answered No, No, Yes on Question 1, Question 2 and Question 3. Hint: Draw the Venn diagram and denote the unknown probabilities by letters. Set up and solve the system of equations for unknown probabilities. Answer: P = Classical Probability Recall the definition of the classical probability from Lecture 2: we have an experiment with finitely many outcomes, that are equally likely. For an event A, P(A) = # of outcomes in A # of outcomes in S. 1. A well balanced die is rolled 3 times. Find: (a) the probability that the same number appears 3 times; (b) the probability that the last two numbers are the same; (c) the probability that the same number appears at least twice. 7

8 Answers: 6/216 = 1/36; 36/216 = 1/6; 1/6. The number of ways to select a combination k objects from n objects equals ( ) n n! = k k!(n k)!. When we select a combination we don t return the selected object to the original set, and we don t pay attention to the order in which objects are selected. 2. Lottery Problem. In the Texas powerball lottery a combination of 5 numbers in the range chosen at random. The lottery pays $1,000,000 to everyone who correctly guesses 5 numbers numbers (no repetitions allowed). It pays out $7 to anyone who guess 3 numbers. Find the probability of the corresponding events. ( ) 69 1 P( 5 numbers correct) = ( 5 64 ) P( exactly 3 numbers match) = 3)( 2 ) ( 69 5 Find the probability that at least three numbers match. ( 5 66 ) P( at least 3 numbers correct) = 3)( 2 ) The lottery ticket costs $2. Is it worth playing this game repeatedly? We ll get back to this question when we introduce the expectation of a random variable. 3. Committee problem. A committee of 5 people is selected at random from a pool of 15 women and 10 men. Find the probability that the committee contains all men. Find the probability that 2 men and 3 women are selected. Sample Space: all combinations of 5 people out of 25. #S = ( 25 5 ). ( 69 5 Let A be the event that only men are selected. #A = ( 10) 5. ( ) ( ) P(A) = /. 5 5 Let A be the event that 3 women and 2 men are selected. #A = ( 15 3 ( 25 P(A) = ( 15 3 )( 10 2 ) / 5 ). )( ) 8

9 Problem Find a flaw in the solution to the following problem. A box contains 5 black and 5 white balls. Two balls are drawn at random. Lets count the probability that at least two balls are white. The total number of combinations is ( 10) 2. There are ) ways to select one white ball. Since the remaining ball can be of any color, ( 5 1 there are ( 9 1) ways to select the remaining ball. So, The correct answer: P(at least one white) = P(at least one white) = ( 5 )( 9 1 ( 1) 11 ). 2 ( 5 )( 5 ) ( ( 2) 11 ) Card problem. From a standard deck of cards (52=13*4) 6 cards are chosen at random. Find the probability that there is only one ace in a hand. Find the probability that there are exactly two aces in a hand. #S = ( 52 6 ). Let A the event that a hand of six cards has one ace. A = {all combinations of 6 cards with one ace}. #A = ( 4)( 48 ) 1 5. P(A) = ( 4 1 )( 48 5 ) / ( 52 6 ). Let A the event that a hand of six cards has two aces. #A = ( 4)( 48 ) 2 4. )( ( P(A) = ( ) / 6 ). 5. The key problem. In a set of 10 keys only one opens a door. You choose 7 keys at random. What is the probability that the right key is among selected. S = { combinations of 7 keys out of 20}. #S = ( 10) 7. Let A be the event that the right key is among 7 selected. A = {all combinations of 7 keys from 10 that include the right key}. #A = ( 9 6). P(A) = ( ) 9 / 6 ( 10 7 ) = 7/10. 9

10 6. Sock problem. A drawer contains 10 pairs of socks of different colors. You select 8 socks at random. Find the probability that you don t have a matching pair. #S = ( 20) 8. Let A be the set of all combinations of socks of different colors. ( ) 10 #A = 2 8 ; 8 first select 8 different colors out of 10. Each color can be represented by 1 out of two socks of that color. ( ) ( ) P(A) = 2 8 / Another Sock problem A drawer contains 20 pairs of socks of two colors. What is the minimum number of socks you have to draw to get at least one complete pair with probability Selecting objects with replacement.3 contracts are assigned at random to 4 firms (a firm might get multiple contracts). Find the probability that all contracts go to a single firm. S = {(a 1,a 2,a 3 ), a i { f irm1, f irm2, f irm3, f irm4}}, assignment of firms to contracts 1 3. #S = 4 3. P(A) = ( ) 4 /4 3 = 1/16. 1 (This is also a slot machine problem: 3 objects are selected from 4 with replacement. You win when all objects are the same.) When choosing a combination of k objects, we do not pay attention to the order in which the objects a chosen. That is, we can select a combination in many different ways. Sometimes it is important to now what objects we select first, second and so on. The number of ways to select k objects from a set of n objects when objects are distinguished by the order in which they are selected equals ( ) n k! = n! k (n k)!. 9. Birthday problem. Find the probability that at least two people in a group of r people have birthdays on the same day. Assume that every year has 365 days. Answer: P = 1 365! 1. For r = 80, P.99. (365 r)! 365r 10

11 Yet another way to make a selection is to select a combination from the set of n objects, when duplicates are allowed: you can think of this as selecting an object, remembering what you ve selected, returning the object back to the set, and repeating the process k 1 times. This is called selecting with the replacement. Problem (this requires some thinking) Show that the number of combinations when selecting with replacement equals: ( ) n + k 1. n A Variation of a lottery problem. In a certain lottery 8 numbers are selected from the set of numbers Duplicates are allowed. Find the probability to win the lottery. The probability is one over the total number of combinations. In this case: ( ) 47 P = 1/ Lecture Conditional Probability A survey of 10,000 people in a small town shows the following information Men Women Total Smokers 1, ,500 Non-smokers 3,000 5,500 8,500 Total 4,000 6,000 10,000 A person is chosen at random. Let S be the event that the person is a smoker, N non-smoker, M male. P(S) = 3/20, P(N) = 17/20, P(M) = 2/5. P(MS) = 1/10, P(M S) = 9/20. Find the probability that the person is a smoker if we know that the person is male? P(S M) = 1/4. Find the probability that the person is female if the person is non-smoker? P(W N) = 11/17. The last two examples are conditional probabilities. In general, P(A B) = # of outcomes in AB # of outcomes in B, 11

12 or P(A B) = P(AB) P(B). Sometimes it is convenient to use the conditional probability to determine P(AB) : P(AB) = P(A B)P(B). Problem What is more likely for a man to be smoker or for a woman to be a smoker? Hint Compare P(S M) and P(S W). Problem If we pick a smoker, is this person more likely to be a man or woman? Hint Compare P(M S) and P(W S). Problem A die is rolled. If the die shows an even number, what is the probability of getting 4? What is the probability of getting an even number, if the die shows 4. Solution 1/3, 1 Problem Show that for a fixed B F, P(A B), as a function of A, is a probability function on S. Problem You have 3 boxes. Box X has 10 light bulbs of which 4 are defective, box Y has 6 light bulbs of which 1 is defective, box Z has 8 light bulbs of which 3 are defective. A box is chosen at random and then a bulb is randomly selected from this box. Find the probability that the bulb is not defective. If the bulb is not-defective, find the probability it came from box Z. Let N,D be the events that the bulb is not defective and defective. Let X,Y,Z be the events that box X,Y,Z is selected. Draw the tree diagram and label the branches with correct probabilities. P(N) = P(N(X Y Z)) = P(NX NY NZ) = P(NX) + P(NY ) + P(NZ). P(NX) = 1/3 3/5 = 1/5, P(NY ) = 5/6 1/3 = 5/18, P(NZ) = 5/8 1/3 = 5/24. P(N) = 1/3 + 5/6 + 5/24. The second probability P(Z N) = P(ZN) P(N) =

13 Problem How to ask an uncomfortable question. You want to poll people on a question that they may feel uncomfortable to given an answer. To deal with that the statisticians came up with the following solution. You tell the person that you want to poll to flip a coin, but don t show it to you. If the coin shows a head the person should answer Yes; if the coin shows a tail the person should provide a correct answer. From polling a large number of people you compute p e the proportion of people who said Yes in the experiment. Find the proportion of people would actually answered Yes. Draw the tree diagram. Two events A,B are independent if If events are independent p e = p + (1 p)1/2, p = 2p e 1. P(AB) = P(A)P(B). P(A B) = P(A), P(B A) = P(B). Problem A box contains 5 black and 6 white balls. Two balls are drawn with replacement. Let A be the event that first ball is black, and B the event that second ball is white. Compute P(A), P(B), P(AB), P(B A), P(A B). Solution P(A) = 5/11, P(B) = 6/11, P(AB) = 30/121, P(B A) = P(B), P(A B) = P(A). Problem Repeat the problem for the selection without replacement. P(A) = 5/11, P(B) = 6/11, as before, but P(AB) = 3/10. In this situation 5 Lecture The Law of Total Probability. Let H i, i = 1..k, be pairwise disjoint events such that k i=1 H i = S. The Law of Total Probability states that for every event A, Problem Proof this formula. P(A) = k i=1 P(A H i )P(H i ). 13

14 Problem An insurance company divides new customers into high and low risk groups. From statistical study the company finds that 1 out of 3 people in the high risk group will have an accident in 1 year period, and 1 out of 10 in low-risk group will have an accident in a 1-year period. If the high-risk group is 30% of all potential customers, what is the probability that a new policyholder will have an accident within one year? Solution P(A) = Bayes Theorem. Let H i, i = 1..k, be pairwise disjoint events such that For every event A, with P(A) > 0, This is called Bayes formula. k i=1 H i = S. P(H j A) = P(A H j)p(h j ) k i=1 P(A H i)p(h i ). Problem In the conditions of the previous problem, suppose a new policyholder has an accident in the first year. What is the probability that he is in high-risk group? Solution P(H 1 A) =.59 Problem The problem above estimates the chance for a customer to be in high-risk group. Now the Law of Total Probability can be used again to re-evaluate the chance that a customer will have an accident in one year. Compute this probability. Problem A laboratory blood test is 95% effective in detecting a certain disease. The test in false positive for 1 out of 100 healthy person. If.5% of the population has the disease, what is the probability a person has the disease, given that the result is positive? Solution P =.323. If H,D,T are the events that a random person is healthy, ill, and the test is positive then, 1 P(D T ) =. P(T H)P(H) 1 + P(T D)P(D) 14

15 The relatively low probability is explained by the fact that the disease is rare: there is P(D) = probability that the person has the disease, to start with. In this situation, the positive blood test is more likely to be false positive. 6 Lecture Random Variables Let X(x) be a function from set A to set B. The set {X = b,} for b B, denotes the set of points a A such that X(a) = b. The set {X < b,} for b B, denotes the set of points a A such that X(a) < b. In general, the set {b 1 < X < b 2,} for b 1,b 2 B, denotes the set of points a A such that b 1 < X(a) < b 2. For function X(x) = (x 2) 2 from R to R determine the sets {X = 3}, {X < 2}, {X 100} and {1 X < 4}. For function X(x) from set A = {a,b,c,d,e} to set B = { 3,0,1,2} given by the table x X(x) a 2 b -3 c 2 d 0 e 1 determine the sets {X = 2}, {X < 2}, {X 100} and {1 X < 3}. Consider a game: two coins are tossed. You earn $1 for each head; You lose $3 for two tails. 15

16 Let X be the amount you get (win/lose) after one round of this game. Describe the statistical properties of X. Sample space Table of values of X : S = {HT, T H, HH, T T }. P(HT ) =... = P(T T ) = 1/4. Outcome X(a) HH 2 HT 1 TH 1 TT -3 Events: {X = 3} = {T T }, {X = 1} = {HT, T H}, {X = 2} = {HH}. Probabilities: P(X = 3) = 1/4, P(X = 1) = 1/2, P(X = 2) = 1/4. Find the probability of the event {X > 0}. {X > 0} = {T H, HT, HH}, P(X > 0) = 3/4. Definition 1. A function from the set of outcomes into real numbers is called a random variable. If a random variable takes at most countably many values, then it is called discrete random variable. Definition 2 (Probability Mass Function). The probability mass function of a discrete random variable X is a function from set of values of X into real numbers such that p X (x) = P(X = x), for any value x of X. 16

17 Typically p X is given by the table of its values. For the example above, the table of the probability mass function: x p X 1/4 1/2 1/4 p X can also be represented by a Bar Graph. A box contains 10 bulbs of which 3 are defective. 4 bulbs are selected at random without replacement. Let X be the number of defective bulbs in the sample. Write down the table of the probability mass function p X. Properties of p X : For any value x of X : 0 p X (x) 1, if x 1,x 2,... is the list of values of X, then p X (x i ) = 1. i Definition 3. The Cumulative Distribution Function, (or CDF), is defined for all values x as F X (x) = P(X x). Find the CDF for X from the previous problem. Properties of CDFs. for any x, 0 F X (x) 1; lim F X(x) = 0; x lim F X(x) = 1; x + F X (x) is continuous from the right: F X (x) = lim y x+ F X(y). For a function F(x) = 0 x < x < /2 x < x < x

18 verify that F is a valid CDF. Find PMF for F. Find the probability P(X < 3), if the random variable X has F as its CDF. Find P(X 0.5). 6.2 Notion of Expectation For the game described at the begging of Lecture 7, estimate Averaged Accumulated Wealth after N rounds of the game. After N turns of the game we expect to lose $3 in N/4 games; win $1 in N/2 games; win $2 in N/4 games. Total accumulated wealth: ( 3)N/4 + N/4 + 2N/4 = N/4. The Averaged Accumulated Wealth per game equals 1/4. This number is called the expectation of X : E[X] = x i P(X = x i ) = x i p X (x i ). i i 1. Key problem. A man has N keys on a key chain. One key unlocks the door. A man randomly selects a key and tries it. If the key doesn t open the door he puts it aside and select another key from the set of the remaining keys. Let X be the number of tries needed to open the door. Find PMF of X and the expectation E[X]. Answer: E[X] = (N + 1)/2. Hint: P(X = k) = 1/N. This is statistically equivalent to the probability to put the correct key to position k, among N positions. It can also be computed using the conditional probability. 18

19 2. Solve the problem under the condition that the man always returns a key back to the chain. Answer: E[X] = k=1 (k/n)(1 1/N)k 1 = N. Hint: use independence of outcomes of successive tries. Properties of the Expectation If X a is constant, E[X] = E[a] = a; E[aX + b] = ae[x] + b; Problem Let X be a random variable with PMF: E[g(X)] = g(x i )p X (x i ). i Problem Find E[X 2 ]. x p X 1/4 1/8 1/2 1/8 Problem Find the probability mass function of X Variance Compare two games. Game 1: Win $1 for each head; Lose $3 for getting two tails. Game 2: 19

20 Win $3 for getting two heads; Win $1 for one of each; Lose $4 for getting two tails. Denote by X 1,X 2 the amount you win(lose) when playing Game 1 and Game 2, respectively. Compare E[X 1 ], E[X 2 ] and the Bar graphs of p X1 and p X2. 20

21 7 Review problems for Test 1 1. For any sets A,B show that AB A A B. 2. Show that if A B then B A. 3. Show that A = AB AB. 4. Show that A b = B BA. 5. Show that ( n i=1 A i)b = n i=1 (A ib). 6. List axioms of the Probability. 7. Show that P(A) = 1 P(A), for any event A. 8. Show that if A B, then P(A) P(B). 9. Prove that for any sets A,B : 10. Prove that for any sets A,B : P(A B) = P(A) + P(B) P(AB). P(AB) = P(A) P(AB) students are polled to see whether or not they have studied French or German. The poll shows that 25 studied French, 20 studied German, and 5 studied both. Find the probability that a randomly selected student: (a) Studied only French; (b) Did not study German; (c) Studied French or German; (d) Studied neither language. Write each events above using the set notation F for the set of student who studied French, G the set of students who studied German and set operations. 12. Review problems involving three kinds of sets. 13. Review the following problems: Committee problem, Card problem, Key problem, Birthday problem. 21

22 14. In a community, 36% of families own a dog; 22% of the families that own a dog, also own a can; 30% of the families own a cat. Find the probability that a randomly selected family owns a dog and a cat. Find the probability that a family that owns a cat also owns a dog. Let X be a random variable. Definition 4. The variance of X is defined as V (X) = E[(X E[X]) 2 ], and standard deviation σ = E[(X E[X]) 2 ]. Problem Compute V (X 1 ) and V (X 2 ) from the coin tossing experiment. Problem Grade distribution in class. Problem Let X be a discrete random variable with mean µ = E[X]. Suppose that the variance V (X) = 0. What is the prob. mass function of X? Properties of Variance V (ax + b) = a 2 V (X); V (X) = E[X 2 ] (E[X]) 2. Tchebyshev s Inequality 22

23 Theorem 1. Let X be a discrete random variable with mean µ and standard deviation σ. For any k > 0, P( X µ < kσ) 1 1 k 2, or, equivalently P( X µ kσ) 1 k 2. Proof. Let X be a random variable with PMF: Then, x a 1 a 2... a n p X p 1 p 2... p n σ 2 = E[(X µ) 2 ] = (a i µ) 2 p i i = i: a i µ <kσ (a i µ) 2 p i + i: a i µ kσ (a i µ) 2 p i. It follows that But Thus we showed that k 2 σ 2 i: a i µ kσ p i σ 2. p i = P( X µ kσ). i: a i µ kσ P( X µ kσ) 1 k 2. Problem A factory produces on average 120 items per week, with standard deviation σ = 10. Estimate the probability for the weekly production to be between 100 and 140 items. Find the shortest interval certain to contain at leas 90% of the weekly production level. Answer: P = 3/4. Interval [ , ]. 8 Lecture Binomial B(n, p). Special Discrete Distributions 23

24 2. Geometric Geom(p). 3. Poisson Pois(λ). 4. Negative Binomial NB(r, p). For each of the distribution we will be interested in the description of random experiments leading to these distributions, and the statistical properties: the mean E[X], the variance V (X). 8.1 Binomial Distribution Definition 5. n Bernoulli Trials: an experiment with only two outcomes {S, F} is independently repeated n times. We set p = P(S), 1 p = P(F). Definition 6. Let X be a random variable that counts the number of successes, S, in n Bernoulli Trials. X is called the Binomial random variable B(n, p). Sample Space: all sequences of n letters S or F. S = {(SSSSFFFSFSFSFFF)} P(SSSSFFFSFSFSFFF) = p 7 (1 p) 8. Let X be the number of S s in a sequence. ( ) n P(X = k) = p k (1 p) n k, k k = 0,1,..,n. Problem Verify that P(X = k) is a valid probability mass functions. Problem The probability that a student answers each question correctly on a test is 1/4 and is independent of questions. If a test has 10 problems what is the probability that the student answers at least 3 correctly? Problem An automotive parts manufacturer produced transmission for a certain vehicle. There is 2% chance that any given transmission is defective. Determine that more than 2 transmissions in a box of 100 are defective. Problem Auto insurance is analyzing the claim frequency on a block of 250 policies. Historical data suggest that 10% of the policyholders will submit at least one claim in the coming year. What is the probability that more than 12% of the policyholders will submit at least one claim in the coming year. 24

25 Problem Show that E[X] = np. Problem A current price of a particular stock is $10. On any particular day the stock price either increases by $2, with the probability 60%, or decreases by $3, with the probability 40%. Determine the expected value of the stock price in 4 days from now. Problem Show that V (X) = np(1 p). 8.2 Geometric Distribution Definition 7. In a repeated Bernoulli Trials, let X counts the number of F s before the first S. X is called the geometric random variable, Geom(p). Let X be geometric random variable. P(X = 0) = p, P(X = 1) = (1 p)p,...,p(x = k) = (1 p) k p, k = 0,1,2,... Problem Verify that P(X = k) is a valid probability mass functions. Problem Show that E[X] = 1 p p, V (X) = 1 p p 2. Problem The probability to win a jackpot on a slot machine is 1/16. Find the probability that 4 or more plays are needed to win a jackpot. 8.3 Poisson Distribution An insurance company estimates that there are 150 claims are filed on average per year. What is the probability that a company gets exactly k claims? There can be many answers to it. It might be that every year the company gets exactly 150 claims. In this case P(150 claims) = 1. Or it might get 100 claims one year and 200 claims the following year, in which case P(100 claims) = 1/2, P(200 claims) = 1/2, P(150 claims) = 0. Other arrangements are possible, leading to different answers. There is not enough information to answer the question. We will build a reasonable model based on the following assumptions. 25

26 1. Let us divide one year into small intervals I k, k = 1..n, so small that we re almost certain that no more than one claim can be submitted during each interval I k. 2. Let us assume that the event that a claim is filed during interval I j is independent from the event that a claim is filed during interval I k, for any j k. 3. We estimate from the conditions of the problem that the probability that a claim is filed during interval I k is p = 150/n. Under these assumptions we re dealing with n Bernoulli Trials with success probability p. Thus, the number of claims, X is B(n, p). P(X = k) = ( ) n p k (1 p) n k. k Take the limit n in the above expression and show that it converges to P(X = k) = 150k e 150, k = 0,1,2,... k! Definition 8. We say that X is a Poisson random variable with parameter 150, Pois(150). Problem Verify that P(X = k) is a valid probability mass functions. Problem Let X be Pois(λ). Compute E[X], and V (X) : E[X] = λ, V (X) = λ. Problem Calls to a telephone hot line service are made randomly and independently at the expected rate of 2 per minute. Find the probability that the service recieves fewer than 5 calls in the next minute. P(X < 5) = k e 2 = k! Problem A book of 500 pages contains 300 typos that are distributed randomly. Find the probability that a given page contains: a)exactly two misprints; b) 2 or more misprints. Problem The Poisson Distribution Pois(λ) is used as an approximation to Binomial B(n, p) when n is large, p is small, and np = λ is moderate. Solve the following problem using the Binomial and Poisson distributions. Compare the answers. 26

27 8.4 Negative Binomial Distribution Let r 1. Suppose we repeat Bernoulli trials until we get success S r times. Let X denote the number of failures before r th success. For example, an outcome with r = 3 could be P(FFFSSFFFS) = (1 p) 6 p 3 To find P(X = k) we have to count the number of combination of r 1 letters S among k + r 1 letters F and S. Thus, ( ) k + r 1 P(X = k) = (1 p) k p r, k = 0,1,2,3... r 1 Definition 9. X is called negative binomial random variable, NB(r, p). Problem Probability to hit the target is 60%. The target is destroyed after 3 hits. Find the probability that it takes more than 6 shots to destroy the target. Let X be the number of times we miss, before target is destroyed. X is NB(3,.6). We need the probability that it takes 7 or more shots. With 3 hits, we need the number of missed shots to be 4 or more. 9 Lecture 9 Test I P(X 4) = k=4 ( k ) (.4) k (.6) Lecture Gambler Ruin Problem Let X be a payoff if a coin toss game such that P(X = 1) = P(X = 1) = 1/2. Suppose you start with x dollars and play the game n times. Let X 1,X 2,...,X n be payoffs in each of the game. Your accumulated wealth after n games is S n = x + X 1 + X X n. Each S n is a discrete random variable. The sequence S n, n = 1,2,..., is called a Random Walk. If you compute you expected accumulated wealth after n turns: E[S n ] = x, 27

28 as you would expect not to gain anything from this game. Suppose that the game for you ends once you have no money left and cannot borrow. This is expressed by the condition S n = 0, to which we refer as to go bankrupt. Lets show that P( to go bankrupt ) = 1. For simplicity, lets assume that x is an integer. The probability in question is a function of x only, and we denote it q x = P( to go bankrupt if you start with $x ). We going to look at how q x depends on x. First of all, q 0 = 1. If you have no money, you re bankrupt. Using the law of total probability we can write: q x = 1 2 q x q x+1, x 1. This is a system of difference equations. We solve by trying a formula q x = Ax + B, for some A,B. It follows from the initial condition that B = 1. Then, q x = Ax + 1. But if A 0, then for large x, q x is either negative, or greater than 1, which is not possible for a probability. So, A = 0, and q x = 1, for all x. Suppose you have an option to stop or continue playing anytime you want, but you have to stop if you go bankrupt: S n = 0. Consider the following strategy: set a goal of N dollars. You decided to stop the game when you reach the goal, provided, of course that didn t go bankrupt before. Find the probability that you reach the goal before going bankrupt. Set p x as the probability to reach the goal before going bankrupt when we start with $x dollars. Use the Law of Total Probability to show that p x verifies a system of difference equations: 1 p x = p x p 1 x 1 2, 1 x N 1 p 0 = 0, p N = 1. Solve the system of equations assuming p x = Ax + B, for some A,B. Answer: p x = x/n. Let Y be your total winnings (the return) in the above game. Find E[Y ]. Answer: E[Y ] = (N x)x/n x(1 x/n) = 0. 28

29 Find the probability q x to go bankrupt before reaching the goal, if you start with $x dollars. Answer: q x = 1 x/n. What happens if N +? 10.2 Continuous Random Variables Let X be a random variable, i.e., the function from the sample space S into R. The cumulative distribution function (CDF) is defined as F X (x) = P(X x). Definition 10. X is a continuous random variable if F X (x) is a differentiable function. The derivative f (x) = df X dx is called the probability density function (PDF) of X. Properties of continuous random variables 1. PDF f (x) is a non-negative function f (x)dx = For any a b, P(a < X b) = F X (b) F X (a) = b a f (x)dx 4. For any x, P(X = x) = For any a b, P(a < X < b) = P(a X b) = P(a X < b) = b a f (x)dx. Definition 11. A random variable X which is neither discrete or continuous, is called a mixed random variable. 29

30 10.3 Expectation and Variance 1. The expectation of X is defined as an integral + E[X] = x f (x)dx. 2. The variance and the standard deviations are defined as V (X) = + (x E[X]) 2 f (x)dx = σ = V (X). 3. Tschebyshev s inequality: for any k > 0, + P( X µ > k) σ 2 k Properties of the Expectation and Variance: x 2 f (x)dx (E[X]) 2, E[aX + b] = ae[x] + b, V (ax + b) = a 2 V (X). Problem Prove both formulas. Hint: if PDF of X is f (x) what is the PDF of Y = ax + b? Find the PDF of X and its mean E[X] Uniform Distribution X is a uniform random variable on interval [a, b] if its PDF, f (x) takes a constant value on the interval [a,b] and is zero otherwise: { 1 a x b, f (x) = b a 0 x < a or x > b. Problem Find the mean E[x], and the variance V (X) of a uniform random variable. Answer E[X] = b a (b a)2 2, V (X) = 12. Problem Let X be the time (in minutes) a bus arrives at the bus stop. Suppose that X is a uniform random variable over the interval from 1:00pm to 2:00pm. Find the probability that the bus will arrive between 1:30 and 1:45pm. Now, you arrive at the bus stop at 1:15pm and you find out that the has not yet passed. Find the probability the bus arrives in the next 20 minutes. Hint: For the second question you need to find P(15 < X 35 X > 15). 30

31 10.5 Exponential Random Variable Let X be a Poisson random variable Pois(λ), which counts a number of occurrences of some event in a unit of time, starting at time zero. Let T be the time of the first event occurred. For time t > 0, what is the probability that P(T > t)? One way to find this probability is the following. Let Y counts the same events in the interval [0,t]. Y is Poisson Pois(λt). This can be shown using a binomial approximation of the Poisson distribution. Then Why? Then, since The CDF of T : P(T > t) = P(Y = 0). P(Y = 0) = e λt, P(T > t) = e λt. F T (t) = 1 e λt, t > 0, and zero otherwise. The PDF of T : { λe λt t > 0, (10.1) {?} f (t) = 0 t 0. Definition 12. T with PDF (??) is called an exponential random variable, Exp(1/λ). The textbook uses a parameter θ = 1 λ for the formula for f (t). θ has a meaning of an average waiting time, since We can also write the PDF of T : f (t) = E[T ] = θ. { 1θ e t θ t > 0, 0 t 0. Problem Show that the time between i th and (i+1) th events is exponential with PDF given above Properties of Exponential Distribution Problem Find the mean E[X], and the variance V (X) of an exponential random variable. Answer: E[X] = 1/λ, V (X) = 1/λ 2. Problem Exponential distribution is a unique distribution that has the property of being memoryless. For any t > s 0, Verify this property. P(T t + s T > s) = P(T t). 31

32 Problem This property can be equivalently expressed as P(T > t + s T > s) = P(T > t). Problem Suppose that a number of miles that a car can run before its battery wears out is exponential random variable with an average value of 10,000 miles. Find the probability that a car still operational after the first 5,000 miles. Find the probability that the battery will work after 25,000 miles given that didn t fail for the first 20,000 miles. Answer: Both probabilities are the same and equal The next problem shows a connection between Poisson and Uniform distributions. Problem Let N be Poisson Pois(λ) random variable that counts the number of customers in a store in the first hour after the store opens. Let X denote the time of arrival of the first customer. Show that conditioned on the event {N = 1}, P(0 < X x N = 1} = x, x (0,1). This means that conditioned on fact that there is only one customer in the first hour, the time of her arrival is uniform random variable. Hint: think of a Poisson random variable as a binomial random variable B(n,λ/n), when n is large, and use the formula P(0 < X x N = 1} = 10.6 Normal Distribution P({0 < X x}{n = 1}). P(N = 1) X is called a normal random variable with mean µ and standard deviation σ, if its PDF, 1 (x µ)2 f (x) = e 2σ 2. 2πσ 2 Notation: X N(µ,σ). Problem If X is N(µ,σ) then Y = ax + b is N(aµ + b, a σ) Standard Normal random variables Z N(0,1) is called standard normal distribution. If X is N(µ,σ) then Z = X µ σ is a standard normal random variable. 32

33 It is common to find the values of the probability P(0 Z z) = z 0 1 2π e z2 /2 dz = Φ(z), to be given in the Table of the Standard Normal Variable, or programmed in a calculator. Problem Let X be a normal random var. N(10,2). Using the table values of Φ(z) determine P(8 X 13). Answer: P = Problem (Prob from the textbook). A machine produces steel shafts where the diameter has a normal distribution with mean and standard deviation 0.01 inch. Quality control requires diameters fit into the interval 1.00 ± 0.02 inches. What percentage of output will fail the quality control? Answer: 7.3% Three Percentiles The following rules are often use estimate the probabilities of a normal random variable X : P(µ σ X µ + σ) = 68%, P(µ 2σ X µ + 2σ) = 95%, P(µ 3σ X µ + 3σ) = 99.7% Normal Approximation of Binomial Let X be a binomial B(n, p). E[X] = np, σ = np(1 p). The random variable X np np(1 p) is a random variable with zero mean and standard deviation 1. Let Z be a standard normal variable. Theorem 2 (De Moivre-Laplace theorem). For any a b, as n +, ( ) P a X np b P(a Z b). np(1 p) Approximation is good when np(1 p)

34 Problem (From [3]) An unknown fraction p of a population are smokers. A random sampling with replacement of size n is taken. How large n should be so that the error in estimating p is less than 0.5% Let X be the number of smokers in the sample. X is binomial B(n, p) random variable. The estimated number of smokers is We d like to have X n X/n p with high probability. For that, we will select a confidence level, say 0.95, and require that P( X/n p 0.005) Rewrite probability as P( X/n p 0.005) = P( 0.005n X np 0.005n) ( = P n X np n ). p(1 p) np(1 p) p(1 p) By the De-Moivre-Laplace theorem the last probability is well approximated by ( ) ( P n Z n n = 2Φ ), p(1 p) p(1 p) p(1 p) where Z is a standard normal variable. From the table we find that n p(1 p) 1.96, and n 392 p(1 p). But for any p [0, 1], p(1 p) 1/4. Thus, it is enough to take n 392/2 = 196, n Further examples of random variables Problem Suppose a man, standing at the point (0, 1), shoots an arrow at the target that located at (0,0) by randomly choosing an angle from [ π/2,π/2]. Let X denote the distance along the x-axis from the origin to the point where the arrow lands. Find the PDF of this random variable. Compute E[X]. A random variable with such PDF is called Cauchy Random Variable. A random variable that has either infinite expectation or infinity variance is sometimes referred to as a heavy tail random variable (or distribution). 34

35 Solution Let Θ be the angle the archer selects. Θ is uniform random variable Uni([ π/2, π/2]. The distance from the target X = 1 tanθ. Let x > 0. P(X x) = P( tanθ 1/x) = 2P(tanΘ 1/x) It follows that Observe now that E[X] = 2 π = 2P(Θ arctan(1/x)) = f (x) = d dx P(X x) = 2 π x 1 + x 2 dx = 1 + π z π/2 1 π dθ = 2 ( π ) π 2 arctan(1/x). arctan(1/x) x 2. dz = ln(1 + z) π + 0 = +. Problem (Compound distribution) Suppose you are given a choice to play Game 1 or Game 2. The payoff in Game 1 is a random variable X 1 with CDF F 1 (x), and the payoff in Game 2 is a random variable X 2 with CDF F 2 (x). After comparing X 1 and X 2 you have decided that you have a slight preference to play Game 1 than Game 2, but you also like to play Game 2 from time to time. Let p (0, 1), with p > 1/2, be the weight that describes you preference of Game 1 over Game 2. In deciding which game to play, you use the following setup: flip an unfair coin that lands heads with probability p. If coin lands heads, play Game1 and get payoff X 1. ; if tails play Game 2 and get payoff X 2. Find the CDF of the payoff in this compound game. Answer F(x) = pf 1 (x) + (1 p)f 2 (x). 11 Multivariate Random Distributions 11.1 Discrete bi-variate random variables Let X,Y be two discrete random variables with values of X : and values of Y : {x 1,...,x n } {y 1,...,y m }. Definition 13. The joint probability mass function of X and Y, is a function p(x,y) : p(x i,y j ) = P({X = x i }{Y = y j }), i = 1..n, j = 1..m. 35

36 The PMF of (X,Y ) is non-negative and n m i=1 j=1 p(x i,y j ) = 1. Marginal distributions. The PMF of X, p X (x) can be determined from p(x,y) by p X (x i ) = p(x i,y j ), j and likewise, p Y (y) : p Y (y j ) = p(x i,y j ). i p X, p Y are called marginal probability mass functions. The information about the joint PMF can be describe by a contingency table: x 1... x n p Y (y) y 1 p(x 1,y 1 )... p(x n,y 1 ) p Y (y 1 ) y m p(x 1,y m )... p(x n,y m ) p Y (y m ) p X (x) p X (x 1 )... p X (x n ) 1 Problem You have 3 boxes. Box 1 has 10 light bulbs of which 4 are defective, box 2 has 6 light bulbs of which 1 is defective, box 3 has 8 light bulbs of which 3 are defective. A box is chosen at random and then a bulb is randomly selected from this box. Let X be the box selected, and Y be the random variable { 1 the bulb is defective Y = 0 the bulb is non-defective Find joint probability mass function for (X,Y ). Problem Find the marginal probability mass functions. Definition 14. Random variables X,Y are called independent if for all (x,y), p(x,y) = p X (x)p Y (y) Continuous bi-variate random variables Definition 15. We say that a pair of random variables X,Y are continuous bi-variate random variables if there is a function f (x,y) such that P(a < X b, c < Y d) = b d a c f (x,y)dydx, for any a,b,c,d. The function f (x,y) is called the joint probability density function. 36

37 Properties of joint PDF 1. f (x,y) is a non-negative function and + f (x,y)dydx = For any set A R 2, P((X,Y ) A) = A f (x,y)dydx Marginal PDFs We can obtain PDF of each variable separately from the following argument: b ( + ) P(a < X b) = P(a < X b, < Y < + ) = f (x,y)dy dx. a So, In a similar way f X (x) = f Y (y) = + + f (x,y)dy. f (x,y)dx. f X (x), f Y (y) are called marginal PDFs of f (x,y). Definition 16. Random variables X,Y are called independent if f (x,y) = f X (x) f Y (y), (x,y). Problems (From [5]) A joint probability density function of a random variables X,Y is given by { 2(1 x) 0 x 1, 0 y 1 f (x,y) = 0 otherwise Find P(X < 0.5, 0.4 < Y < 0.7). Answer: Find P(X < Y, Y < 0.5). 37

38 (From [2]) Suppose that a point is chosen at random insider a circle of radius R centered at the origin. Let X,Y be the coordinates of the point. The join PDF is given as { c x f (x,y) = 2 + y 2 < R 2 0 otherwise for some constant c. 1. Find c. 2. Find the marginal PDF of X and Y. 3. Let Z be the distance from (X,Y ) to the origin. Find the PDF of Z. 4. Find E[Z]. Note, that the expected value is not R/2, but R/ 2, this is the radius that divides the area of the circle in half. For a pair of random variables X,Y with joint PDF { e (x+y) 0 < x < +, 0 < y < + f (x,y) = 0 otherwise find the PDF of the function Z = X/Y. Answer: f Z (z) = 1/(z + 1) 2, 0 < z < Some special multivariate distributions 1. Bivariate uniform distribution: let A be a set in positive area in R 2. If (X,Y ) are the coordinates of a point we choose at random inside A, then the joint PDF equals: { 1/ A (x,y) A f (x,y) = 0 otherwise Problem Show that if A is a square [a,b] [c,d], then (X,Y ) are independent random variables. 2. Bivariate normal distribution. (X,Y ) is said to be bivariate normal variables if for some numbers (µ 1, µ 2 ) and (a,b,c) such that a > 0, (ab c 2 ) > 0, f (x,y) = 1 2π (ab c 2 ) e a(x µ 1)2 /2 c(x µ 1 )(y µ 2 ) b(y µ 2 )2 /2, 3. Multinomial distribution. This is a multivariate discrete distribution, that generalizes binomial distribution. Suppose we deal with n independent repetitions of an experiment that has k outcomes with probabilities p 1, p 2,..., p k, p k = 1. i Let X 1 count the number of times outcome 1 occurred, X 2 counts the number of times 2 occurred, and so on. Note that always, X X k = n. 38

39 Then P(X 1 = n 1,X 2 = n 2,...,X k = n k ) = n! n 1!n 2!...n k! pn 1 1 pn pn k k. This is called the multinomial distribution Expectation, Variance, Covariance and Correlation 1. The expectations E[X],E[Y ] are given by the formulas E[X] = x f X (x)dx = x f (x,y)dxdy, E[Y ] = y f Y (y)dy = y f (x,y)dxdy, V (X) = E[(X E[X]) 2 ], V (Y ) = E[(Y E[Y ]) 2 ]. 2. The covariance of X,Y is defined as cov(x,y ) = E[(X E[X])(Y E[Y ])] = E[XY ] E[X]E[Y ]. 3. The correlation of X and Y, is defined as ρ(x,y ) = cov(x,y ) V (X)V (Y ). 4. Show that for all random variables X,Y. 1 ρ(x,y ) 1, 5. If X,Y are independent random variables, then E[XY ] = E[X]E[Y ], cov(x,y ) = Note, however, that cov(x,y ) = 0 doesn t always imply that X,Y are independent random variables. Consider an example: X is uniform on [ 1, 1] and Y = X 2. Show that cov(x,y ) = 0, but X and Y are not independent. Problem Compute correlation ρ for random variables (X,Y ), with PDF { 1/2 0 < x < y, 0 < y < 2 f (x,y) = 0 otherwise 39

40 Answer: E[X] = 2/3, E[Y ] = 4/3, E[XY ] = 1, cov(x,y ) = 1/6, V (X) = 2/9, V (Y ) = 2/9, ρ = 1/2. Further properties of expectations. Let X i, i = 1..n, be random variables of any kind. Let S n = X X n. Then, and Var(S n ) = i E[S n ] = E[X i ], i Var(X i ) + cov(x i,x j ). i< j An important special case: when X i and X j are independent for all i j, then Var(S n ) = Var(X i ). i Problem Use the last formula to compute the variance of the a binomial random variable X B(n, p) The Weak Law of Large Numbers and The Central Limit Theorem Consider a sequence X 1,X 2,X 3,..., of pair-wise independent random variables with the same PDFs. Denote µ = E[X 1 ], σ = V (X 1 ). Think of X i as repeated measurements of some physical parameter (say electric current). We d like to study the statistics of the average Lets compute E[S N ] and V (S N ) : S N = X X N. N E[S N ] = µ, V (S N ) = σ 2 N. Notice that V (S N ) decreases to zero as N +, and E[S N ] is independent of N. Thus, the effect of averaging produces a more accurate estimate of µ as the number of measurement increases. To describe this effect mathematically, we use Tschebyshev s inequality for S N : Thus, for any ε > 0, P( S N µ > ε) V (S N) ε 2 = σ 2 ε 2 N. limp( S N µ > ε) = 0. This is the Weak Law of Large Numbers. 40

41 Central Limit Theorem. There is even more that one can say about S N. Lets normalize S N so that is has zero mean and unit standard deviation: Z N = S N µ σ X X N N = σ. N Then, the Central Limit Theorem states that for any a,b, where Z is N(0,1) random variable. limp(a < Z N < b) = P(a < Z < B), To state it slightly differently: the average of the sum of large number of pairwise independent, identically distributed random variables approximately equals to X X N N µ + σ N Z, i.e. equals to the mean µ with the error that has normal distribution Bi-variate Normal Distribution Problem Determine E[X], E[Y ],Var(X),Var(Y ), cov(x,y ) for bi-variate normal random variable. Problem Show bi-variate normal random variables (X,Y ) are independent iff cov(x,y ) = 0. Problem Show that there is a linear transformation that makes (X,Y ) independent. Let (X,Y ) be the deviation from the center of a target in a shooting experiment. Let f (x,y) be the joint PDF of (X,Y ). We will postulate that 1. (X,Y ) are independent and identically distributed, i.e.: for some PDF f 0, f (x,y) = f 0 (x) f 0 (y), 2. and the probability density g(r,θ) of the random variables (R,Θ) the polar coordinates of (X,Y ), does not depend on θ. We will prove that under these two conditions for some σ. Show first that f (x,y) = 1 x 2 +y 2 2πσ 2 e 2σ 2, g(r,θ) = r f (r cosθ,r sinθ). 41

42 The second assumptions implies that for some function p(r), f (r cosθ,r sinθ) = p(r)/r, or, f (x,y) = p( x 2 + y 2 )/ x 2 + y 2 h(x 2 + y 2 ). The first assumption implies then, that f 0 (x) f 0 (y) = h(x 2 + y 2 ), for some function h of a single argument. Take y = 0, and assume f 0 (0) 0. Then f 0 (x) = h(x 2 )/ f 0 (0). Call c = f 0 (0). Then, h(x 2 )h(y 2 ) = c 2 h(x 2 + y 2 ), for all x,y. The only function that verifies that equation is h(x 2 ) = c 2 e αx2, for any α. Thus we showed that f (x,y) = c 2 e α(x2 +y 2). Since f (x,y) is a PDF, the integral over all of R 2 must be equal to 1. This implies that for some σ, f is as claimed Conditional Distribution Problem Let Y be a randomly chosen number from the list {1,...,n}. Let X be the outcome of the flip of biased coin with the probability of Tails, p = 1/Y. Let p(x,y) be the joint PMF of (X,Y ). For a fixed value of y, the function p(x, y) P(Y = y) is a probability mass function of X. Verify that. Let (X,Y ) be a pair of discrete random variables with values {x 1,...,x n }, and {y 1,...,y m } and the joint PMF p(x,y). Definition 17. The conditional distribution of X given Y = y k, with P(Y = y k ) > 0, is the probability mass function p (X Y =yk )(x i ) = P(X = x i Y = y k ), i = 1...n. 42