STA 256: Statistics and Probability I

Transcription

1 Al Nosedal. University of Toronto. Fall 2014

2

3 My momma always said: Life was like a box of chocolates. You never know what you re gonna get. Forrest Gump.

4 Experiment, outcome, sample space, and sample point When you toss a coin. It comes up heads or tails. Those are the only possibilities we allow. Tossing the coin is called an experiment. The results, namely H (heads) and T (tails) are called outcomes. There are only two outcomes here and none other. This set of outcomes namely, {H, T} is called a sample space. Each of the outcomes H and T is called a sample point.

5 Sample space Now suppose that you toss a coin twice in succession. Then there are four possible outcomes, HH, HT, TH, TT. These are the sample points in the sample space {HH, HT, TH, TT}. We will denote the sample space by S.

6 Event An event is a set of sample points. In the example of tossing a coin twice in succession, the event, the first toss results in heads, is the set {HH, HT}. Let A and B be events. By A B (read, A is a subset of B ) we mean that every point that is in A is also in B. If A B and B A, then A and B have to consist of the same points. In that case we write A = B.

7 Union of events If A and B are events, then, A or B, is also an event denoted by A B. For example, let A be: the first toss results in heads, and B be: the second toss results in tails, is the set {HH, HT, TH, TT}. Thus A B consists of all points that are either in A or in B.

8 A B B A

9 Intersection of events The event A and B, is denoted by A B. For example, if A and B are the events defined above, then A B is the event: the first toss results in heads and the second toss results in tails, which is the set {HT}. Thus A B consists of all points that belong to both A and B.

10 A B B A

11 Empty set The empty set, denoted by, contains no points. It is the impossible event. For instance, if A is the event that the first toss results in heads and C is the event that the first toss results in tails, then A C is the event that the first toss results in heads and the first toss results in tails. It is the empty set. If A C = we say that the events A and C are mutually exclusive. That means there are no sample points that are common to A and C. The events A and C in the last paragraph are mutually exclusive.

12 Complement The event that A does not occur is called the complement of A and is denoted by A c. It consists of all points that are not in A. If A is the event that the first toss results in heads, then A c is the event that the first toss does not result in heads. It is the set: {TH, TT}. Note that A A c = S and A A c =.

13 A c A

14 Example A survey is made of a population to find out how many of them own a home, how many own a car and how many are married. Let H, C and M stand respectively for the events, owning a home, owning a car and being married. What do the following symbols represent? 1. H M c. 2. (H M) c. 3. H c M c. 4. (H M) C.

15 Example A person who is 40 years old now will die before reaching age 120. The person could die at any time during the 80 years. So, if the outcome (sample point) is time of death, the sample space is the interval (0, 80). Determine the intervals that correspond to the following events: 1. The person lives past the age of 65 and dies before turning The person survives to age The person lives past the age of 65 but dies before turning 90 or the person lives beyond the age of 85.

16 Solution Let us denote the time of death (in years from now) by T. 1. This event corresponds to 25 < T < 50 or the interval (25, 50). 2. This corresponds to T > 30 or the interval (30, 80). 3. This corresponds to 25 < T < 50 or T > 45, that is (25, 50) (45, 80) = (25, 80).

17 Example The lifetime of a computer is less than 5 years. The lifetime of a printer is less than 8 years. Each outcome is a pair of numbers corresponding to the moments of breakdown of these machines. What is the sample space? Determine the set that corresponds to each of the following events. 1. The printer survives 5 years. 2. The computer outlives the printer. 3. The printer outlives the computer, but breaks down within a year after the computer breaks down.

18 Solution See Figures 1, 2, and 3. If x stands for the time that the computer breaks down and y for the time that the printer breaks down, then each sample point is described by an ordered pair of numbers (x, y). x can be any number between 0 and 5 and y can be any number between 0 and 8. So the sample space consists of all points (x, y) inside the rectangle 0 < x < 5, 0 < y < 8.

19 Solution (cont.) 1. The event that the printer survives 5 years corresponds to y > 5. This is the shaded region in the first figure. 2. The event that the computer outlives the printer corresponds to the set of all points x > y, that is, all the points below the line x = y. It is the shaded region shown in the second figure. 3. The printer outlives the computer means y > x. It breaks down within a year after the computer breaks down means that y < x + 1. So the event is represented by the region that lies between the lines x = y and y = x + 1, which is shown in the last figure.

20 Figure 1 (y > 5) y x

21 Figure 2 (x > y) y x

22 Figure 3 y x

23 Example An auto insurance company classifies each policyholder as i) young or old; ii) male or female; and iii) married or single. of these policyholders, 30% are young, 46% are male, and 70% are married. The policyholders can also be classified as 13.2% young males, 30.1% married males, and 14% young married persons. Finally, 6% of the policyholders are young married males. What proportion of the company s policyholders are young, female, and single?

24 Solution You can solve this with an elementary argument. Imagine that there are people. (You may use any number you want just turns out to be convenient for this calculation). We are given that out of these 10, 000 people 30% or 3000 are young. Of these, 1320 are young males. So the number of young females is = We are also given that there are 1400 young married people and 600 are young, married males. So the number of young married females is = 800. Since the total number of young females is 1680, the number of single young females is = 880. The proportion is 880/10000.

25 By now it must be clear that the population we considered in this problem can be thought of as a sample space, the set of people with a certain characteristic corresponds to an event and we assign a proportion to each event. The proportion that we assign to each event in a sample space is called the probability of that event. For example if 70% of a population own a car, we say that if we choose a person at random form that population, the probability that that person owns a car is We now give a formal definition of probability.

26 Definition Let S be the sample space. For each event A there is a number P(A), called the probability of the event A, with the following properties: 1. 0 P(A) P(S) = 1 and 3. If A 1, A 2,... are mutually exclusive, then P(A 1 A 2...) = P(A 1 ) + P(A 2 ) +... There can be infinitely many A i s.

27 Example For the data in the last Example, calculate the probability that a randomly chosen person is young or male or married.

28 Solution Let us denote young by A, male by B and married by C. Then P(A) = 0.3, P(B) = 0.46, P(C) = 0.7, P(A B) = 0.132, P(B C) = 0.301, P(C A) = 0.14 and P(A B C) = The proportion that is young or male or married is P(A B C) = =

29 Problem Verify the following: 1. (A B) c = A c B c. 2. (A B) c = A c B c.

30 Solution (1) 1. A B consists of all points that are common to both A and B. A point is in (A B) c means it is not in both A and B. That means it is at least not in one of the sets, A, B. That is so if and only if it is in A c or in B c. Therefore (A B) c = A c B c

31 Solution (2) 2. Applying result 1 to A c and B c (A c B c ) c = (A c ) c (B c ) c = A B Taking complements on both sides of last equation A c B c = (A B) c.

32 Problem Over a given day, The probability that the price of Stock A will go up 0.3. The probability that the price of Stock B will not go up is 0.6. The probability that the price neither stock will go up is 0.5. Calculate the probability that the prices of both stocks will go up.

33 Solution Let A denote the event that the price of stock A goes up and B that the price of stock B goes up. P(A) = 0.3, P(B) = 1 P(B c ) = = 0.4; P(A B) = 1 P(A c B c ) = = 0.5. Since P(A B) = P(A) + P(B) P(A B), 0.5 = P(A B) and P(A B) = 0.2.

34 Problem The probability that a visit to a primary care physician s (PCP) office results in neither lab work nor referral to a specialist is 35%. Of those coming to a PCP s office, 30% are referred to specialists and 40% require lab work. Determine the probability that a visit to a PCP s office results in both lab work and referral to a specialist.

35 Solution Let L stand for lab work and S for referral to a specialist. We are given that P(L c S c ) = 0.35, P(S) = 0.3 and P(L) = 0.4. We are asked to find P(L S). First note that L c S c = (L S) c and therefore P(L S) = P ((L c S c ) c ) = 1 P(L c S c ) = = Second, use the fact that P(L S) = P(L) + P(S) P(L S) which gives 0.65 = P(L S) and P(L S) = 0.05.

36 Problem You are given P(A B) = 0.7 and P(A B c ) = 0.9. Determine P(A).

37 Solution P(A B) = P(A) + P(B) P(A B) and P(A B c ) = P(A) + P(B c ) P(A B c ). Adding the two equations and noting that P(A B) + P(A B c ) = P(A), and that P(B) + P(B c ) = 1, we get = 2P(A) + 1 P(A) = P(A) + 1 and therefore P(A) = 0.6.

38 Problem Among a large group of patients recovering from shoulder injuries, it is found that 22% visit both a physical therapist an a chiropractor, whereas 12% visit neither of these. The probability that a patient visits a chiropractor exceeds by 0.14 the probability that a patient visits a physical therapist. Determine the probability that a randomly chosen member of this group visits a physical therapist.

39 Solution Let T stand for visit to a physical therapist and C for visit to a chiropractor. We are given that P(T C) = 0.22, P ((T C) c ) = 0.12 and P(C) = P(T ). We need P(T ). Since P ((T C) c ) = 0.12, P(T C) = = Recall that P(T C) = P(T ) + P(C) P(T C). Hence 0.88 = P(T ) + (P(T ) ) = 2P(T ) 0.08 P(T ) = = 0.48

40 Problem A survey of a group s viewing habits over the last year revealed the following information: 28% watched gymnastics 29% watched baseball 19% watched soccer 14% watched gymnastics and baseball 12% watched baseball and soccer 10% watched gymnastics and soccer 8% watched all three sports Calculate the percentage of the group that watched none of the three sports during the last year.

41 Solution Since G c B c S c = (G B S) c, P(G c B c S c ) = 1 P(G B S) = 1 P(G) P(B) P(S) + P(G B) + P(B S) + P(G S) P(G B S) = = 0.52

42 Conditional Probability This is one of the most important concepts in Probability. To illustrate it, let us look at an example. A blood test indicates the presence of a particular disease 95% of the time when the disease is actually present. The same test indicates the presence of the disease 0.5% of the time when the disease is not present. One percent of the population actually has the disease. Calculate the probability that a person has the disease given that the test indicates the presence of the disease.

43 Definition Let A and B be events. If P(A) 0, we define the conditional probability of the event B given A as P(B A) = P(B A) P(A)

44 Example You are given the following table for a loss: Amount of Loss Probability Given that the loss amount is positive, calculate the probability that it is more than 1.

45 Solution Let A be the event that the loss amount is positive, and B the event that the loss amount exceeds 1. Then A B is clearly equal to B because the loss will necessarily be positive if it exceeds 1 (i.e., B A). The probability that the claim amount is positive (event A) is = 0.6 and the probability that the claim amount is greater than 1 (event B) is = 0.3. The conditional probability is P(B A) = P(A B) P(A) = P(B) P(A) = = 0.5

46 Example An insurance office receives claims that are of three types, Life, Medical and Automobile. On the average 10% of the claims are Life claims, 40% are Medical claims and the rest are Automobile claims. A randomly chosen claim is not a Medical claim. Given that, what is the conditional probability that it is a Life claim?

47 Solution Let A be the event that the claim is a Life claim and B the event that it is not a Medical claim. P(A) = 0.1 and P(B) = = 0.6. Then the event A B is that it is a Life claim and not a Medical claim. Since a Life claim is not a Medical claim, A B is the same as the event that the claim is a life claim. Therefore P(A B) = P(A B) P(B) = P(A) P(B) = = 1 6

48 Independence Two events A and B are called independent if one has no effect on the other. That means that whether A is given or not is irrelevant to P(B), i. e., P(B A) = P(B). It follows from the definition of conditional independence that P(A B) = P(A)P(B)

49 Example In a certain population, 60% own a car, 30% own a house and 20% own a house and a car. Determine whether or not the events that a person owns a car and that a person owns a house are independent.

50 Solution Let A be the event that the person owns a car and B the event that the person owns a house. P(A) = 0.6, P(B) = 0.3 and P(A B) = = P(A)P(B). Hence the two events are not independent.

51 Example A dental insurance covers three procedures: orthodontics, fillings and extractions. During the lifetime of the policy, the probability that the policyholder needs: orthodontic work is 1/2 orthodontic work or a filling is 2/3 orthodontic work or an extraction is 3/4 a filling and an extraction 1/8. The event that the policyholder needs orthodontic work is independent of the need for either a filling or an extraction. Calculate the probability that the policyholder will need a filling or an extraction during the lifetime of the policy.

52 Solution Let O, F, and E denote the events of orthodontics, filling and extraction respectively. Then P(O) = 1/2, P(O F ) = 2/3, P(O E) = 3/4 and P(F E) = 1/8. We need P(F E). Recall that P(F E) = P(F ) + P(E) P(F E) = P(F ) + P(E) 1/8 So we need to find P(F ) and P(E).

53 Solution (cont.) P(O F ) = P(O)+P(F ) P(O F ) = P(O)+P(F ) P(O)P(F ) (because of independence) P(O F ) = 1/2 + P(F ) (1/2)P(F ) = 1/2 + (1/2)P(F ) = 2/3 Hence P(F ) = 1/3 P(O E) = P(O) + P(E) P(O)P(E) (because of independence) P(O E) = 1/2 + (1/2)P(E) = 3/4. Hence P(E) = 1/2, and P(F E) = (1/3) + (1/2) (1/8) = 17/24.

54 Example Workplace accidents are categorized as minor, moderate and severe. The probability that a given accident is minor is 0.5, that it is moderate is 0.4, and that it is severe is 0.1. Two accidents occur independently in one month. Calculate the probability that neither accident is severe and at most one is moderate.

55 Solution Let us denote by L 1, M 1 and S 1 respectively the events that the first accident is minor (light), moderate and severe. Similarly for the second accident denote with 2 as a subscript. Since neither should be S and at most one should be M, the only possibilities are L 1 L 2 or L 1 M 2 or M 1 L 2. Note that these possibilities are mutually exclusive. Since the two accidents are independent, the required probability is P(L 1 )P(L 2 ) + P(L 1 )P(M 2 ) + P(M 1 )P(L 2 ) = (0.5)(0.5) + (0.5)(0.4) + (0.4)(0.5) = 0.65

56 Example Suppose that 80 percent of used car buyers are good credit risks. Suppose, further, that the probability is 0.7 that an individual who is a good credit risk has a credit card, but that this probability is only 0.4 for a bad credit risk. Calculate the probability a. a randomly selected car buyer has a credit card. b. a randomly selected car buyer who has a credit card is a good risk. c. a randomly selected car buyer who does not have a credit card is a good risk.

57 Solution a. A = selecting a good credit risk. B = selecting an individual with a credit card. P(B) = P(A)P(B A) + P(A c )P(B A c ) P(B) = (0.8)(0.7) + (0.2)(0.4) = 0.64 b. P(A B) = P(A B) P(B) c. P(A B c ) = P(A Bc ) P(B c ) = P(A)P(B A) P(B) = (0.8)(0.7) 0.64 = 7 8 = P(A)P(Bc A) P(B c ) = (0.8)(0.3) 0.36 = 2 3

58 Example A local bank reviewed its credit card policy with the intention of recalling some of its credit cards. In the past approximately 5% of cardholders defaulted, leaving the bank unable to collect the outstanding balance. Hence, management established a prior probability of 0.05 that any particular cardholder will default. The bank also found that the probability of missing a monthly payment is 0.20% for customers who do not default. Of course, the probability of missing a monthly payment for those who default is 1.

59 Example (cont.) a. Given that a customer missed one or more monthly payments, compute the posterior probability that the customer will default. b. The bank would like to recall its card if the probability that a customer will default is greater than Should the bank recall its card if the customer misses a monthly payment? Why or why not?

60 Solution D = Default, D c = customer doesn t default, M = missed payment. a. P(D M) = P(D M) P(D M) P(M) = P(D M)+P(D c M). P(D) = 0.05 P(D c ) = 0.95 P(M D c ) = 0.20 P(M D) = 1 (0.05)(1) P(D M) = (0.05)(1)+(0.95)(0.20) = b. Yes, the bank should recall its card if the customer misses a monthly payment.

61 Law of Total Probability Suppose that A 1, A 2,..., A k are Mutually exclusive events and A 1 A 2... A k = S, the whole sample space. Let B be an event. Then P(B) = P(B A 1 ) + P(B A 2 ) P(B A k ).

62 Law of Total Probability (cont.) There is another way that we can write this. Since P(B A i ) = P(B A i )P(A i ), P(B) = P(B A 1 )P(A 1 ) + P(B A 2 )P(A 2 ) P(B A k )P(A k ). The above result is known as the Law of Total Probability.

63 Bayes Theorem P(A 1 B) = P(A 2 B) = where A1 A 2 = S and A 1 A 2 =. P(A 1 )P(B A 1 ) P(A 1 )P(B A 1 ) + P(A 2 )P(B A 2 ) P(A 2 )P(B A 2 ) P(A 1 )P(B A 1 ) + P(A 2 )P(B A 2 )

64 Bayes Theorem Quite generally, suppose that A 1, A 2,..., A k are mutually exclusive events and their union is the whole sample space, i. e., k j=1 P(A j) = 1, and B is an event with P(B) > 0. Then P(A i B) = A i B P(B) = P(B A i )P(A i ) k j=1 P(B A j)p(a j ). This is also known as Bayes formula.

65 Example An insurance company classifies drivers as High Risk, Standard and Preferred. 10% of the drivers in a population are High Risk, 60% are Standard and the rest are preferred. The probability of accident during a period is 0.3 for a High Risk driver, 0.2 for a Standard driver and 0.1 for a preferred driver. 1. Given that a person chosen at random had an accident during this period, find the probability that the person is Standard. 2. Given that a person chosen at random has not had an accident during this period, find the probability that the person is High Risk.

66 Solution Let H, S, P and A stand for High-Risk, Standard, Preferred and Accident and W stand for H, S or P. X Pr(W ) Pr(A W ) Pr(A W ) = Pr(A W )Pr(W ) H S P Total 1

67 Solution (1) 1. From Bayes Formula, taking the figures from the table, Pr(S A) = Pr(A S)Pr(S) Pr(A S)Pr(S) + Pr(A H)Pr(H) + Pr(A P)Pr(P) Pr(S A) = = 2 3

68 Solution (2) 2. For the second part, note that Pr(A c W ) = 1 Pr(A W ). This is because if a fraction of p amongst W get into an accident, then the fraction 1 p of W does not get into an accident. You can draw another table or observe that Pr(H A c ) = Pr(Ac H)Pr(H) Pr(A c ) = (1 0.3)(0.1) = = 7 82.

69 Problem Ten percent of a company s life insurance policyholders are smokers. The rest are nonsmokers. For each nonsmoker, the probability of dying during the year is For each smoker, the probability of dying during the year is Given that a policyholder has died, what is the probability that the policyholder was a smoker?

70 Solution Let S = smoker, N = Nonsmoker, D = Dying, and Ω = sample space. Note that S N = Ω and S N =. We also have that P(S) = 0.10, P(N) = 0.90, P(D N) = 0.01, and P(D S) = P(S D) = P(S D P(D) = P(D S)P(S) P(D S)P(S) + P(D N)P(N) P(S D) = (0.05)(0.10) (0.05)(0.10) + (0.01)(0.90) = 5 14

71 Problem An actuary studying the insurance preferences of automobile owners makes the following conclusions: An automobile owner is twice as likely to purchase collision coverage as disability coverage. The event that an automobile owner purchases collision coverage is independent of the event that he or she purchases disability coverage. The probability that an automobile owner purchases both collision and disability coverages is What is the probability that an automobile owner purchases neither collision nor disability coverage?

72 Solution Let C stand for purchase of collision coverage and D for purchasing disability coverage. Then we are given that P(C) = 2P(D), C and D are independent and P(C D) = P(C)P(D) = 0.15 (because of independence). This implies that 2P 2 (D) = 0.15, i. e. P 2 (D) = (or P(D) = 0.075) and P(C) = We need P(C c D c ).

73 Solution (cont.) P(C c D c ) = P((C D) c ) = 1 P(C D) = 1 P(C) P(D) + P(C D) = =

74 Problem The probability that a randomly chosen male has a circulation problem is Males who have a circulation problem are twice as likely to be smokers as those who do not have a circulation problem. What is the conditional probability that a male has a circulation problem, given that he is a smoker?

75 Solution Let C stand for circulation problem and S for smoker. We are given that P(C) = 0.25, and that P(S C) = 2P(S C c ). We need to find P(C S). P(C S) = P(C S) = P(S C)P(C) P(S C)P(C) + P(S C c )P(C c ) P(S C)(0.25) P(S C)(0.25) + (1/2)P(S C)(0.75) = 0.4.

76 Problem An actuary studied the likelihood that different types of drivers would be involved in at least one collision during any one-year period. The results of the study are presented below. Type of Percentage of Probability of at driver all drivers least one collision Teen 8% 0.15 Young adult 16% 0.08 Midlife 45% 0.04 Senior 31% 0.05 Total 100% Given that a driver has been involved in at least one collision in the past year, what is the probability that the driver is a young adult driver?

77 Solution Y = Young adult and C = collision. We have that P(C Y )P(Y ) = (0.16)(0.08), P(C T )P(T ) = 0.08)(0.15), P(C M)P(M) = (0.45)(0.04) and P(C S)P(S) = (0.31)(0.05). P(Y C) = (0.16)(0.08) (0.16)(0.08) + (0.08)(0.15) + (0.45)(0.04) + (0.31)(0.05) P(Y C) =

78 Problem Upon arrival at a hospital s emergency room, patients are categorized according to their condition as critical, serious, or stable. In the past year: 10% of the emergency room patients were critical; 30% of the emergency room patients were serious; the rest of the emergency room patients were stable; 40% of the critical patients died; 10% of the serious patients died; 1% of the stable patients died; Given that a patient survived, what is the probability that the patient was categorized as serious upon arrival? (0.2922)

79 Random variables Imagine the following simple situation. A loss is covered by an insurance policy. The probability that a claim is made on this policy is 0.45 and the probability that a claim is not made is There are two points in the sample space, namely a claim is made and a claim is not made. We have assigned a probability of 0.45 with the first outcome and a probability of 0.55 with the second. You can readily see that this situation is mathematically identical to the following one and many others like it. You have a biased coin. If you toss this coin the chances of heads is There are two outcomes. You have assigned a probability of 0.45 to heads and 0.55 to tails.

80 Random variables (cont.) It is rather clumsy to have to write out in words what the outcome is. Instead we can unify all such situations with exactly two outcomes by assigning the number 0 for one of the outcomes and the number 1 for the other. We can use a symbol X and say that X = 0 will correspond to the outcome that a claim is not made (toss results in tails) and X = 1 will correspond to the outcome that a claim is made (toss results in heads). Then we write P(X = 1) = 0.45 P(X = 0) = 0.55 (1) The entity X is called a random variable. A random variable associates a unique number with every outcome.

81 Example If a die is thrown and the number on the face up is observed, then the outcomes are the integers 1, 2, 3, 4, 5 and 6. Then we can define a random variable, X, such that its value is the outcome. If we assign the same probability for each outcome, then P(X = k) = 1/6; k = 1, 2, 3, 4, 5, 6 (2)

82 Probability Function Observe that in each of the cases that lead to equations 1 and 2, we defined a random variable and a probability for each value that the random variable assumes. That is, we have defined a function that associates with each value of X a unique number, that is a probability. Such a function is called the Probability Function of the random variable. You can also see that in each case the sum of all the probabilities equals 1.

83 Example A random loss, X, has the following probability function: x P(X = x) (You can verify that the probabilities add up to 1.)

84 Example (cont.) 1. Calculate the probability that loss is at least Given that the loss exceeds 85, calculate the conditional probability that it is at most Given that the loss is at most 400, calculate the conditional probability that it is at most 100.

85 Solution Since there are no losses between 85 and 100. P(X 85) = P(X = 100) + P(X = 200) + P(X = 400) +P(X = 1000) = = 0.8.

86 Solution P(X 400 X > 85) = = P(85<X 400) P(X >85) P(100 X 400) P(X 100) = = 7 8

87 Solution P(X 100 X 400) = P(X 100) P(X 400) = = 5 9

88 Example Using the previous example, find an expression for the function F(x) = P(X x).

89 Solution If x < 0, then F (x) = P(X x) = 0 because X does not assume negative values. If 0 x < 100, then there is exactly one loss to the amount of 0 in the interval (, x]. Therefore, F (x) = P(X x) = P(X = 0) = 0.2. If 100 x < 200, then there are exactly two losses in the interval (, x], one in the amount of 0 and the other 100. Therefore, F (x) = P(X x) = P(X = 0) + P(X = 100) = 0.5.

90 Solution (cont.) Similarly, if 200 x < 400, then F (x) = P(X x) = P(X = 0)+P(X = 100)+P(X = 200) = 0.8. If 400 x < 1000, then F (x) = P(X x) = P(X = 0) + P(X = 100) + P(X = 200) + P(X = 400) = 0.9. If 1000 x <, then all the losses are in the interval (, x] and so the sum of their probabilities has to be 1. F (x) = P(X x) = P(X = 0) + P(X = 100) + P(X = 200) + P(X = 400) + P(X = 1000) = 1.

91 Solution (cont.) F (x) = 0 < x < x < x < x < x < x <

92 Cumulative Distribution Function (CDF) The function F(x) = P(X x) is called the Cumulative Distribution Function (CDF) of the random variable X.

93 Continuous Random Variable A random variable X is called a continuous random variable if there is a nonnegative function f(x) such that F(x) = P(X x) = x f(y)dy More generally, for any set of real numbers A, P(X is in A) = f(x)dx In that case f is called the Probability Density Function (PDF) or the Density Function of X. A

94 Example The PDF of the random variable, X is f(x) = cx 2 for 0 < x < 1 and 0 elsewhere (c is constant). 1. Calculate c. 2. Find the CDF. 3. Calculate P(0.25 < X < 0.5 X > 0.25).

95 Solution 1 1. Since the PDF should integrate to 1, Hence, c = cx 2 dx = c 3 = 1

96 Solution 2 2. F (x) = x 0 x < 0 f (y)dy = x 3 0 x < 1 1 x 1

97 Solution 3 3. P(0.25 < X < 0.5 x > 0.25) = P(0.25 < X < 0.5) P(X > 0.25) = F (0.5) F (0.25) 1 F (0.25) P(0.25 < X < 0.5 x > 0.25) = (0.5)3 (0.25) 3 1 (0.25) 3 = 1 9

98 Example Let T denote the future lifetime (in years) of a newborn child. Suppose that the PDF of T is given by f(t) = 0 for t 0 and f(t) = a exp ct, t > 0, where a and c are constants. 1. Determine a in terms of c. 2. Find the CDF of T. 3. Given that the newborn survives to age y, determine the conditional probability that the person will live z more years.

99 Solution Let us determine a. We do that by using the fact that the PDF should integrate to 1. This requirement also implies that c > 0. Otherwise the integral will not exist. Therefore a = c. 0 a exp cx dx = a c = 1.

100 Solution It follows that and F (t) = P(T t) = c t 0 exp cu du = 1 exp ct P(T > t) = 1 F (t) = exp ct.

101 Solution The event that the newborn has survived to age y is the same as T > y and the event that the person will survive z more years is the same as T > y + z. P(T > y + z T > y) = P(T > y + z) P(T > y) = 1 F (y + z) 1 F (y) P(T > y + z T > y) = exp c(y+z) exp cy = exp cz.

102 Mixed distributions Example. A loss X has the PDF f(x) = exp x, x > 0. An insurance will pay the entire loss up to a maximum of 1. Determine the distribution of the payment.

103 Solution Let us denote by Y the payment. Then Y = X if X 1 and Y = 1 if X > 1. Therefore { P(X y) 0 < y 1 P(Y y) = 1 y > 1.

104 Solution Note that since the maximum that the insurance will pay is 1, it is certain that the payment is less than or equal to 1. Hence P(Y y) = 1 if y > 1. Since P(X y) = y 0 exp x dx = 1 exp y, we get the CDF of Y to be { 1 exp y 0 < y 1 F (y) = 1 y > 1.

105 Solution Notice that lim F (y) = lim y 1 y 1 (1 exp y ) = 1 exp 1 lim F (y) = 1. y 1 + Thus the CDF is discontinuous at 1. The jump in the value of F at 1 is 1 (1 exp 1 ) = exp 1. This is due to the fact that there is a point mass at 1.

106 Solution Since the maximum payment is 1, Y = 1 if X > 1 and P(Y = 1) = P(X > 1) = 1 exp x dx = exp 1. Thus the jump in the value of F at 1 precisely equals P(Y = 1). The distribution of Y is called mixed. It has a continuous part and a discrete part.

107 Solution We can write the PDF as a continuous part (by differentiating the CDF) and a discrete part as a point mass at 1. f (y) = F (y) = exp y, 0 < y < 1; P(Y = 1) = exp 1.

108 Example Let the CDF of X be: F (x) = 0 x < 0, x x < 1 3+x 16 1 x < x < 5 1 x 5. Find P(X = 1), P(X = 2), P(X = 4) and P(X = 5).

109 Solution Check the limits: lim y 1 F (x) = 1 2 /4 = 1/4 lim y 1 + F (x) = (3 + 1)/16 = 1/4 lim y 2 F (x) = (3 + 2)/16 = 5/16 lim y 2 + F (x) = 1/2 lim y 4 F (x) = lim y 4 F (x) = 1/2 lim y 5 F (x) = 1/2 lim y 5 + F (x) = 1

110 Solution There are discontinuities only at the points 2 and 5, with respective jumps 3/16 and 1/2. Therefore P(X = 1) = 0; P(X = 2) = 3/16; P(X = 4) = 0; P(X = 5) = 1/2.

111 Percentiles and mode If X is a continuous random variable and 0 < p < 1, then the 100p-th percentile of the distribution of X is the number x p such that P(X x p ) = p. The 50-th percentile is called the median.

112 Example If the PDF of X is f(x) = c exp cx, x > 0, find the 100p-th percentile.

113 Solution We need x p such that or P(X x p ) = F (x p ) = xp 0 c exp cx dx = 1 exp cxp = p x p = (1/c)ln(1 p).

114 Functions of a random variable Let X be a random variable with a given distribution and let Y = u(x) be another random variable defined by the given function u. How does one find the distribution of Y? The general way of doing this is by noting that F Y (y) = P(Y y) = P(u(X) y) = P(A), where A is the set of all points x such that u(x) y. The last can be, in principle, found from the distribution of X.

115 Example The PDF of X is f(x) = exp x, x > 0. Let Y = X. Find the PDF of Y.

116 Solution F Y (y) = P(Y y) = P( X y) = P(X y 2 ) = y 2 0 exp x dx = 1 exp y 2. f Y (y) = F Y (y) = [1 exp y 2 ] = 2y exp y 2.

117 Example f(x) = 1/4, 2 < x < 2. If Y = X 2, find the PDF of Y.

118 Solution P(Y y) = P(X 2 y). Now you have to be careful. Note that Y 0 since it is a square. As X takes values from 2 to 2, Y will take values between 0 and 4. X 2 y implies that X y or y X y. Therefore F Y (y) = P( y < X < y) = y y (1/4)dx = (1/2) y, 0 < y < 4. Differentiating, we get f Y (y) = (1/4)y 1/2 ; 0 < y < 4.

119 Example This year s loss has a PDF f(x) = 0.5 exp 0.5x, x > 0. Because of inflation, next year the loss will increase by 10%. Find the PDF of the next year s loss.

120 Solution If X is the loss this year and Y the loss next year then Y = u(x ) = 1.1X. Then F Y (y) = P(Y y) = P(1.1X y) = P(X y 1.1 ) = 0.5 y/1.1 0 exp 0.5x dx = 1 exp 0.5y/1.1. Therefore the PDF is ( ) f Y (y) = [1 exp 0.5y/1.1 ] = exp ( 1.1)y, y > 0 1.1

121 Example A random loss, X, has the PDF f (x) = exp x, x > 0. an insurance will pay the excess of a deductible 1. That means if the loss is less than or equal to 1, the insurance will pay nothing. If the loss exceeds 1, the insurance will pay the excess of X over 1. That is, if Y is the payment, then { 0 if X 1 Y = X 1 if X > 1. Find the Probability Function of Y.

122 Solution Note that there is a point mass at 0. For there will be no payment if X 1. P(Y = 0) = P(X 1) = Then for y > 0, 1 0 exp x dx = 1 exp 1. F Y (y) = P(Y y) = P(X 1 y) = P(X y + 1) = y+1 0 exp x dx = 1 exp y 1.

123 Solution Thus the PDF of Y is written as f Y (y) = F Y (y) = exp y 1, y > 0, P(Y = 0) = 1 exp 1.

124 Example A random loss, X, has the PDF f (x) = x2 9, 0 < x < 3 and 0 otherwise. An insurance will pay the entire loss up to a maximum of 2. Find the Probability Function of the payment.

125 Solution If Y is the payment, then Y is the smallest of X and 2. That is { X 0 X 2 Y = 2 X > 2.

126 Solution Again note that there is a point mass at 2. P(Y = 2) = P(X > 2) = x 2 dx = If y < 2, then Y = X and F Y (y) = P(Y y) = y 0 x 2 dx = y Differentiating we get f Y (y) = y 2 9, 0 < y < 2 P(Y = 2) =

127 Problem The PDF of a loss is f (x) = exp x, x > 0. An insurance will pay the entire loss up to a maximum of 2. Determine the PDF of the payment.

128 Solution Let Y be the payment. Then Y = X if X 2. So if y < 2, then P(Y y) = P(X y) = y 0 exp x dx = 1 exp y. If y 2, then P(Y y) = 1 since the maximum value that Y can assume is 2.

129 Solution F Y (y) = f Y (y) = exp y, 0 < y < 2, { 1 exp y 0 y < 2 1 y 2. P(Y = 2) = P(X > 2) = 2 exp x dx = exp 2

130 Problem The PDF of X is f (x) = c exp c(x 1) if x > 1 and 0 otherwise. If Y = 1/X, find the PDF of Y.

131 Solution When x = 1, y = 1 and as x, y 0. Therefore for 0 < y < 1, P(Y y) = P( 1 X y) = P(X 1 y ) = c 1/y exp c(x 1) dx = exp c(1/y 1). Therefore the pdf is f Y (y) = [exp c(1/y 1) ] = ( ) c y 2 exp c(1/y 1), 0 < y < 1.

132 Problem An insurance policy reimburses dental expense, X, up to a maximum benefit of 250. The probability density function for X is f (x) = c exp 0.004x, x 0 and 0 if x < 0. Calculate the median benefit for this policy.

133 Solution The CDF of X is F (x) = c x 0 exp 0.004t dt = ( c ) (1 exp 0.004x ) Since F ( ) = 1, c = and F (x) = 1 exp 0.004x. If payment is Y, then F Y (y) = 1 exp 0.004y if y < 250 and 1 if y 250. If the median m is less than 250, then it is given by 1 exp 0.004m = 1/2, orm = 250ln(2) 173

134 Problem The lifetime of a machine part has a continuous distribution on the interval (0, 40) with probability density function f (x) proportional 1 to. Calculate the probability that the lifetime of the (10+x) 2 machine part is less than 6.

135 Solution First find the constant. c 40 0 (10 + x) 2 dx = c ( ) = 1. This gives c = Hence F (6) = (10 + x) 2 dx = (12.5) ( ) =

136 Definition Let X be a random variable with Probability Function f (x) and u(x ) a function of X. We define the Expected value or Expectation of u(x ) as E[u(X )] = x=x i u(x)f (x) if X is discrete, E[u(X )] = E[u(X )] = u(x)f (x)dx A if X is continuous, u(x)f (x)dx + x=x i u(x)f (x) if the distribution is mixed, where A is the set of points where X is continuous and x i s the points where X is discrete.

137 Example Let the probability function of X be: f ( 1) = 1/4; f (0) = 1/4; f (1) = 1/4; f (2) = 1/8; f (3) = 1/8. Calculate E(X ) and E(exp tx ), where t is a constant.

138 Example Let the PDF of X be f (x) = c exp cx (x > 0, c a positive constant). Calculate E(X ) and E(exp tx ).

139 Example The CDF of X is 0 if x < 0 F (x) = x/3 if 0 x < 2 1 x 2. Calculate E(X ) and E(X 2 ).

140 Moments, mean and variance of a random variable The n-th moment of a random variable X is E(X n ). The first moment, E(X ), is called the mean of the expected value of X. The n-th central moment of X is defined as E({X E(X )} n ). The second central moment of X is called the variance of X and is denoted by Var(X ). Var(X ) = E({X E(X )} 2 ). Note that, since the variance is the expected value of a square, it cannot be negative.

141 Standard deviation and coefficient of variation We define the standard deviation (SD) and the coefficient of variation (CV) as SD(X ) = Var(X ) CV (X ) = SD(X ), E(X ) 0. E(X )

142 Example The Probability Function of X is P(X = c) = P(X = c) = 1/4, c > 2; P(X = 2) = 1/2. Calculate the coefficient of variation of X.

143 Linearity of the expected value If a and b are constants, then E[au(X ) + bv(x )] = ae[u(x )] + be[v(x )]. In particular, E[aX + b] = ae[x ] + b.

144 An alternative expression for the variance Var(X ) = E({X E(X )} 2 ) = E[X 2 2XE(X ) + {E(X )} 2 ] = E(X 2 ) 2E(X )E(X ) + {E(X )} 2 = E(X 2 ) {E(X )} 2. The standard notation is µ for the mean, σ 2 for the variance and σ for the standard deviation.

145 Example This year company will have fixed expenses of 100. The rest of the expenses is a random variable with PDF f (x) = 1 300, 0 < x < 300. Next year the fixed expenses will increase by 20% whereas the rest of the expenses will increase by 10%. Calculate the following: 1. The expected value of the total expenses this year, 2. The variance of the total expenses this year, 3. The variance of the total expenses next year.

146 The Moment Generating Function The Moment Generating Function or MGF for short, of a random variable X is defined by M(t) = E[exp Xt ] The motivation for this will be made clear shortly. As we mentioned for the PDF and CDF, if we want to emphasize that this is the MGF of X, we will write M X (t).

147 Example Find the MGFs of the following random variables: 1. P(X = 0) = q; P(X = 1) = 1 q. 2. P(X = n) = pq n, where n = 0, 1, 2,... and p = 1 q. 3. f (x) = 1 b a for a < x < b and 0 otherwise. 4. f (x) = c exp cx, x > 0.

148 Example The MGF of a random variable X is Calculate Var(X ). M(t) = expt t.

149 Problem Suppose that X is a continuous random variable that assumes only nonnegative values. That means the PDF f (x) is 0 for x < 0. Let us think of X as a loss. Suppose an insurance company will pay the entire loss up to a maximum of m. If we denote the payment by Y, then { X if X m Y = m if X > m. In other words Y = min(x, m), meaning Y is the smaller of X and m. One sometimes denotes min(x, m) by X m. Show that E(Y ) = m 0 [1 F (x)]dx.

150 Problem An insurance company s monthly claims are modeled by a continuous, positive random variable X, whose probability density function is proportional to (1 + x) 4, where 0 < x <. Determine the company s expected monthly claims. (1/2)

151 Problem A probability distribution of the claim sizes for an auto insurance policy is given in the table below: Claim size Probability What percentage of the claims are within one standard deviation of the mean claim size? (45%)

152 Problem An actuary determines that the claim size for a certain class of accidents is a random variable, X, with moment generating function M X (t) = (1 2500t) 4. Determine the standard deviation of the claim size for this class of accidents. (5000).

153 Problem A manufacturer s annual losses follow a distribution with density function f (x) = (2.5)(0.6) 2.5 x 3.5, x > 0.6 and 0 otherwise. To cover its losses, the manufacturer purchases an insurance policy with an annual deductible of 2. What is the mean of the manufacturer s annual losses not paid by the insurance policy? (0.934)

154 Problem An insurance company will pay the entire loss insured up to a maximum of 20. The CDF of the loss is: ( ) 10 3 F (x) = x Calculate the expected value of the payment. (40/9)

155 Joint distribution of discrete variables There are situations where one might be interested in more that one random variable. For example, an automobile insurance policy may cover collision and liability. The loss on collision and the loss on liability are random variables.

156 Let us consider two discrete random variables, say X and Y. These ideas can be readily generalized for more than two random variables. We can assign probabilities P(X = x i, Y = y j ) to all possible values x i and y j that X and Y can assume. Since these are probabilities, they have to have the following properties: 1. 0 p(x i, y j ) 1. j p(x i, y j ) = i 3. P[(X, Y ) D] = (x i,y j ) D P(X = xi, Y = y j ) = (x i,y j ) D p(xi, y j ) where D is a discrete set of points in the plane.

157 Joint Probability Function The function p(x i, y j ) is called the Joint Probability Function of the random variables X and Y. We can also define a Joint Cumulative Distribution Function by F (x, y) = P(X x, Y y)

158 Marginal Probability Functions Suppose for each fixed x i we sum the joint PF over all the possible values of Y. Then we get the PF of X. That is P(X = x i ) = j P(X = x i, Y = y j ) Similarly P(Y = y j ) = i P(X = x i, Y = y j ). These functions P(X = x i ) and P(Y = y j ) are called the Marginal Probability Functions of X and Y respectively. The term marginal refers to the fact they are the entries in the margins of the table as illustrated by the following example.

159 Example The joint PF of X and Y is given by: p(1,1) = 0.1 p(1,2) = 0.2 p(1,3) = 0.1 p(2,1) = 0.04 p(2,2) = 0.06 p(2,3) = 0.1 p(3,1) = 0.05 p(3,2) = 0.1 p(3,3) = 0.25 Calculate the marginal PFs of X and Y.

160 Joint density of continuous random variables Once again we will consider two random variables X and Y but which are now continuous. Analogous to the definition of probability for a single random variable, we can define a probability density function for these variables. So if we denote the density function (PDF) by f (x, y), then P((X, Y ) D) = f (x, y)da and D f (x, y)dx dy = 1

161 Joint Probability Density Function The function f (x, y) is called the Joint Probability Density Function of the pair of random variables (X, Y ). P(X x, Y y) = F (x, y) is called the joint CDF of the pair of random variables and It follows that F (x, y) = x y f (s, t)dt ds 2 F (x, y) = f (x, y) x y

162 Marginal densities The marginal densities f X and f Y are defined in a manner similar to the marginal PFs in the discrete case. These are PDFs. f X (x) = f Y (y) = f (x, y)dy f (x, y)dx

163 Example If f (x, y) = c exp x 2y, x > 0, y > 0 and 0 otherwise, calculate 1. c. 2. The marginal densities of X and Y. 3. P(1 < X < 2).

164 Example For the random variables in our last example, calculate P(X < Y ).

165 Example f (x, y) = 1/4 if 0 < x < 2 and 0 < y < 2. What is P(X + Y < 1)?

166 Example The joint PDF of X and Y is f (x, y) = cx, 0 < y < x < 2 and 0 elsewhere. Find 1. c. 2. The marginal densities of X and Y. 3. P(X < 2Y ).

167 Conditional distributions Let us first consider the discrete case. Let p(x, y) = P(X = x, Y = y) be the joint PF of the random variables, X and Y. Recall that the conditional probability of the occurrence of event A given that B has occurred is P(A B) = P(A B) P(B) If A is the event that X = x and B is the event that Y = y then P(A) = P(X = x) = p X (x), the marginal PF of X, P(B) = p Y (y), the marginal PF of Y and P(A B) = p(x, y), the joint PF of X and Y.

168 Conditional distributions (discrete case) We can then define a conditional probability function for the probability that X = x given Y = y by p X Y (x y) = P(X = x Y = y) = P(X = x, Y = y) P(Y = y) = p(x, y) p Y (y) Similarly p Y X (y x) = P(Y = y X = x) = P(X = x, Y = y) P(X = x) = p(x, y) p X (x)

169 Conditional densities (continuous case) In the continuous case we extend the same concept and define conditional densities or conditional PDFs by f X Y (x y) = f Y X (y x) = f (x, y) f Y (y) f (x, y) f X (x)

170 Example The joint PF of X and Y is given by: p(1,1) = 0.1 p(1,2) = 0.2 p(1,3) = 0.1 p(2,1) = 0.04 p(2,2) = 0.06 p(2,3) = 0.1 p(3,1) = 0.05 p(3,2) = 0.1 p(3,3) = 0.25 Find the conditional PF p X Y (x 1).

171 Example If f (x, y) = 2 exp x 2y, x > 0, y > 0 and 0 otherwise. Find the conditional densities, f X Y (x y) and f Y X (y x).

172 Example The joint PDF of X and Y is f (x, y) = 3 8x, 0 < y < x < 2 and 0 elsewhere. Calculate the conditional PDFs.

173 Independent random variables In one of our examples, and so f (x, y) = 2 exp x 2y, f X (x) = exp x, f Y (y) = 2 exp 2y, f (x, y) = f X (x)f Y (y). If the joint PF is the product of the marginal PFs we say that the random variables are independent.

174 Definition The random variables X 1, X 2,..., X n are said to be mutually independent if and only if in the discrete case and in the continuous case. p(x 1, x 2,..., x n ) = p X1 (x 1 )p X2 (x 2 )...p Xn (x n ) f (x 1, x 2,..., x n ) = f X1 (x 1 )f X2 (x 2 )...f Xn (x n )

175 Bivariate case If X and Y are independent, then the conditional density of X given Y = y is f X Y (x y) = f (x, y) f Y (y) = f X (x)f Y (y) = f X (x). f Y (y)

176 Example If the joint PDF of X and Y is f (x, y) = 1, 0 < x < 4, 0 < y < 2, and 0 elsewhere. 8 Determine whether X and Y are independent.

177 Functions of Random Variables If X and Y are random variables, then Z = g(x, Y ) is also a random variable. For example X and Y may be random losses and an insurance may pay the amount X + Y. In that case g(x, Y ) = X + Y. We will often be interested in the distribution of Z.

178 Example Let X and Y be random variables with the joint density function f (x, y) = exp (x+y) with 0 < x < and 0 < y <. An insurance policy is written to cover the loss X + Y. Find the probability that the loss is less than 1.

179 Example For the random variables X and Y of the previous example, let Z = max(x, Y ) and W = min(x, Y ). That is, Z is the larger of X and Y and W is the smaller of X and Y. Find the PDFs of Z and W.