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1 P U B L I C A T I O N S The Experts In Actuarial Career Advancement Product Previe For More Information: Support@ActexMadRivercom or call 1(800)

2 SECTION 1 - BASIC PROBABILITY CONCEPTS 37 SECTION 1 - BASIC PROBABILITY CONCEPTS PROBABILITY SPACES AND EVENTS Sample point and sample space: A sample point is the simple outcome of a random experiment The probability space (also called sample space) is the collection of all possible sample points related to a specified experiment When the experiment is performed, one of the sample points ill be the outcome The probability space is the "full set" of possible outcomes of the experiment Mutually exclusive outcomes: Outcomes are mutually exclusive if they cannot occur simultaneously They are also referred to as disjoint outcomes Exhaustive outcomes: Outcomes are exhaustive if they combine to be the entire probability space, or equivalently, if at least one of the outcomes must occur henever the experiment is performed Event: Any collection of sample points, or any subset of the probability space is referred to as an event We say that "event E has occurred" if the experimental outcome as one of the sample points in E Union of events Eand F : E F denotes the union of events Eand F, and consists of all sample points that are in either E or F 8 Union of events E" ß E ß ÞÞÞß E8 : E" E â E8 œ E3 denotes the union of 3œ" the events E" ß E ß ÞÞÞß E8, and consists of all sample points that are in at least one of the E3's This definition can be extended to the union of infinitely many events 8 Intersection of events E" ß E ß ÞÞÞß E8 : E" E â E8 œ E3 denotes the 3œ" intersection of the events E" ß E ß ÞÞÞß E8, and consists of all sample points that are simultaneously in all of the E3 's ( E F is also denoted E F or EF)

3 38 SECTION 1 - BASIC PROBABILITY CONCEPTS Mutually exclusive events E" ß E ß ÞÞÞß E8 : To events are mutually exclusive if they have no sample points in common, or equivalently, if they have empty intersection Events E" ß E ß ÞÞÞß E8 are mutually exclusive if E3 E4 œ g for all 3 Á 4, here g denotes the empty set ith no sample points Mutually exclusive events cannot occur simultaneously Exhaustive events F" ß F ß ÞÞÞß F8 : If F" F â F8 œ W, the entire probability space, then the events F ß F ß ÞÞÞß F " 8 are referred to as exhaustive events Complement of event E : The complement of event E consists of all sample points in - the probability space that are not in E The complement is denoted Eß µ Eß E or E and is equal to ÖB À B  E When the underlying random experiment is performed, to say that the complement of E has occurred is the same as saying that E has not occurred Subevent (or subset) E of event F : If event F contains all the sample points in event E, then E is a subevent of F, denoted E F The occurrence of event E implies that event F has occurred Partition of event E : Events G ß G ß ÞÞÞß G form a partition of event E if 8 E œ G 3œ" and the G 3 's are mutually exclusive " 8 3 DeMorgan's Las: (i) ÐE FÑ œe F, to say that E Fhas not occurred is to say that Ehas not occurred and F has not occurred ; this rule generalizes to any number of events; 8 8 E3 œ ÐE" E â E8Ñ œe" E â E8 œ E3 3œ" 3œ" (ii) ÐE FÑ œe F, to say that E Fhas not occurred is to say that either Ehas not occurred or F has not occurred (or both have not occurred) ; this rule generalizes to any 8 8 number of events, E3 œ ÐE E â E Ñ œe E â E œ E 3œ" 3œ" Indicator function for event E : The function " 8 " 8 3 M ÐBÑ œ E " if B E! if BÂE is the indicator function for event E, here B denotes a sample point MEÐBÑ is 1 if event E has occurred Basic set theory as revieed in Section 0 of these notes The Venn diagrams presented there apply in the sample space and event context presented here

4 SECTION 1 - BASIC PROBABILITY CONCEPTS 39 Example 1-1: Suppose that an "experiment" consists of tossing a six-faced die The probability space of outcomes consists of the set Ö"ß ß $ß %ß &ß ', each number being a sample point representing the number of spots that can turn up hen the die is tossed The outcomes " and (or more formally, Ö" and Ö ) are an example of mutually exclusive outcomes, since they cannot occur simultaneously on one toss of the die The collection of all the outcomes (sample points) 1 to 6 are exhaustive for the experiment of tossing a die since one of those outcomes must occur The collection Öß %ß ' represents the event of tossing an even number hen tossing a die We define the folloing events E œ Ö"ß ß $ œ "a number less than % is tossed", F œ Öß %ß ' œ "an even number is tossed", G œ Ö% œ "a 4 is tossed", H œ Ö œ "a 2 is tossed" Then (i) E F œ Ö"ß ß $ß %ß ', (ii) E F œ Ö, (iii) E and G are mutually exclusive since E G œ g (hen the die is tossed it is not possible for both E and G to occur), (iv) H F, (v) E œ Ö%ß &ß ' (complement of E), (vi) F œ Ö"ß $ß &, (vii) E F œ Ö"ß ß $ß %ß ', so that ÐE FÑ œ Ö& œ E F (this illustrates one of DeMorgan's Las)

5 40 SECTION 1 - BASIC PROBABILITY CONCEPTS Some rules concerning operations on events: (i) E ÐF F â F ÑœÐE F Ñ ÐE F Ñ â ÐE F Ñ " 8 " 8 E ÐF F â F ÑœÐE F Ñ ÐE F Ñ â ÐE F Ñ " 8 " 8 Eß F ß F ß ÞÞÞß F " 8 and for any events 8 (ii) If F" ß F ß ÞÞÞß F8 are exhaustive events F3 œ W, the entire probability space, 3œ" then for any event E, EœE ÐF F â F ÑœÐE F Ñ ÐE F Ñ â ÐE F Ñ " 8 " 8 If F ß F ß ÞÞÞß F " 8 are exhaustive and mutually exclusive events, then they form a partition of the probability space For example, the events F" œ Ö"ß, F œ Ö$ß % and F œ Ö&ß ' $ form a partition of the probability space for the outcomes of tossing a single die The general idea of a partition is illustrated in the diagram belo As a special case of a partition, if F is any event, then F and F form a partition of the probability space We then get the folloing identity for any to events E and F: E œ E ÐF F Ñ œ ÐE FÑ ÐE F Ñ ; note also that E F and E F form a partition of event E B 1 B 2 A B A B AB 1 AB 2 A B 3 AB 4 A B B B 3 4 (iii) For any event E, E E œ W, the entire probability space, and E E œ g (iv) E F œöbàb Eand BÂF is sometimes denoted EF, and consists of all sample points that are in event E but not in event F (v) If E Fthen E FœFand E FœE

6 SECTION 1 - BASIC PROBABILITY CONCEPTS 41 PROBABILITY Probability function for a discrete probability space: A discrete probability space (or sample space) is a set of a finite or countable infinite number of sample points TÒ+ 3Ó or : 3 denotes the probability that sample point (or outcome Ñ+ 3 occurs There is some underlying "random experiment" hose outcome ill be one of the 's in the probability space Each time the experiment is performed, + 3 one of the + 3's ill occur The probability function T must satisfy the folloing to conditions: (i)!ÿtò+óÿ" 3 for each + 3 in the sample space, and D (ii) TÒ+ " ÓTÒ+ Óâ œ TÒ+ 3Ó œ "(total probability for a probability space is alays 1) all 3 This definition applies to both finite and infinite probability spaces Tossing an ordinary die is an example of an experiment ith a finite probability space Ö"ß ß $ß %ß &ß ' An example of an experiment ith an infinite probability space is the tossing of a coin until the first head appears The toss number of the first head could be any positive integer, 1, or 2, or 3, The probability space for this experiment is the infinite set of positive integers Ö"ß ß $ß ÞÞÞ since the first head could occur on any toss starting ith the first toss The notion of discrete random variable covered later is closely related to the notion of discrete probability space and probability function Uniform probability function: If a probability space has a finite number of sample points, say 5 points, + " ß + ß ÞÞÞß + 5, then the probability function is said to be uniform if each sample point has the same probability of occurring ; T Ò+ 3Ó œ " 5 for each 3 œ "ß ß ÞÞÞß 5 Tossing a fair die ould be an example of this, ith 5œ' Probability of event E : An event E consists of a subset of sample points in the probability space In the case of a discrete probability space, the probability of E is T ÒEÓ œ D T Ò+ 3Ó, the sum of T Ò+ 3Ó over all sample points in event E + E 3 " Example 1-2: In tossing a "fair" die, it is assumed that each of the six faces has the same chance of ' " turning up If this is true, then the probability function TÐ4Ñœ ' for 4œ"ßß$ß%ß&ß' is a uniform probability function on the sample space Ö"ß ß $ß %ß &ß ' of The event "an even number is tossed" is E œ Öß %ß ', and has probability " " " " TÒEÓ œ ' ' ' œ 2

7 42 SECTION 1 - BASIC PROBABILITY CONCEPTS Continuous probability space: An experiment can result in an outcome hich can be any real number in some interval For example, imagine a simple game in hich a pointer is spun randomly and ends up pointing at some spot on a circle The angle from the vertical (measured clockise) is beteen! and 1 The probability space is the interval Ð!ß 1Ó, the set of possible angles that can occur We regard this as a continuous probability space In the case of a continuous probability space (an interval), e describe probability by assigning probability to subintervals rather than individual points If the spin is "fair", so that all points on the circle are equally likely to occur, then intuition suggests that the probability assigned to an interval ould be the fraction that the interval is of the full circle For instance, the probability that the pointer ends up beteen "3 O'clock" and "9 O'clock" (beteen 1Î or 90 and $ 1Î or 270 from the vertical) ould be 5, since that subinterval is one-half of the full circle The notion of a continuous random variable, covered later in this study guide, is related to a continuous probability space Some rules concerning probability: (i) TÒWÓœ" if W is the entire probability space (hen the underlying experiment is performed, some outcome must occur ith probability 1; for instance W œ Ö"ß ß $ß %ß &ß ' for the die toss example) (ii) TÒgÓœ! (the probability of no face turning up hen e toss a die is 0) (iii) If events E" ß E ß ÞÞÞß E8 are mutually exclusive (also called disjoint) then 8 8 TÒ E3 ÓœTÒE" E â E8 ÓœTÒE" ÓTÒE ÓâTÒE8Óœ TÒE3Ó 3œ" 3œ" This extends to infinitely many mutually exclusive events This rule is similar to the rule discussed in Section 0 of this study guide, here it as noted that the number of elements in the union of mutually disjoint sets is the sum of the numbers of elements in each set When e have mutually exclusive events and e add the event probabilities, there is no double counting (iv) For any event E,!ŸTÒEÓŸ"Þ (v) If E F then TÒEÓŸTÒFÓ

8 SECTION 1 - BASIC PROBABILITY CONCEPTS 43 (vi) For any events EF, and G, TÒE FÓœTÒEÓTÒFÓTÒE FÓ This relationship can be explained as follos We can formulate E Fas the union of three mutually exclusive events as follos: E F œ ÐE F Ñ ÐE FÑ ÐF E Ñ This is expressed in the folloing Venn diagram AB AB B A Since these are mutually exclusive events, it follos that TÒE FÓ œ TÒE F ÓTÒE FÓTÒF EÓ From the Venn diagram e see that E œ ÐE F Ñ ÐE FÑ, so that TÒEÓœTÒE F ÓTÒE FÓ, and e also see that TÒF EÓ œ TÒFÓTÒE FÓ It then follos that T ÒE FÓ œ ÐT ÒE F Ó T ÒE FÓÑ T ÒF E Ó œ T ÒEÓ T ÒFÓ T ÒE FÓ Þ We subtract T ÒE FÓ because T ÒEÓ T ÒFÓ counts T ÒE FÓ tice T ÒE FÓ is the probability that at least one of the to events Eß F occurs This as revieed in Section 0, here a similar rule as described for counting the number of elements in E F For the union of three sets e have TÒE F GÓ œ TÒEÓTÒFÓTÒGÓTÒE FÓTÒE GÓTÒF GÓTÒE F GÓ (vii) For any event E, TÒEÓœ"TÒEÓ (viii) For any events E and F, TÒEÓœTÒE FÓTÒE F Ó (this as mentioned in (vi), it is illustrated in the Venn diagram above) 8 (ix) For exhaustive events F" ß F ß ÞÞÞß F8, T Ò F3Ó œ " 3œ" If F" ß F ß ÞÞÞß F8 are exhaustive and mutually exclusive, they form a partition of the entire probability space, and for any event E, TÒEÓœTÒE F ÓTÒE F ÓâTÒE F Ó œtòe F Ó " 8 3 3œ" (x) If T is a uniform probability function on a probability space ith 5 points, and if event E consists of 7 of those points, then TÒEÓ œ 7 5 8

9 44 SECTION 1 - BASIC PROBABILITY CONCEPTS (xi) The ords "percentage" and "proportion" are used as alternatives to "probability" As an example, if e are told that the percentage or proportion of a group of people that are of a certain type is 20%, this is generally interpreted to mean that a randomly chosen person from the group has a 20% probability of being of that type This is the "long-run frequency" interpretation of probability As another example, suppose that e are tossing a fair die In the " long-run frequency interpretation of probability, to say that the probability of tossing a 1 is ' is the same as saying that if e repeatedly toss the die, the proportion of tosses that are 1's ill " approach ' 8 8 (xii) for any events E" ß E ß ÞÞÞß E8, T Ò E3Ó Ÿ T ÒE3Ó, ith equality holding if and only if the 3œ" events are mutually exclusive 3œ" Example 1-3: Suppose that TÒE FÓ œ Þß TÒEÓ œ Þ'ß and TÒFÓ œ Þ& Find T ÒE F Ó ß T ÒE F Ó, T ÒE FÓ and T ÒE FÓ Solution: Using probability rules e get the folloing TÒE F Ó œ TÒÐE FÑÓ œ "TÒE FÓ œ Þ)Þ T ÒE FÓ œ T ÒEÓ T ÒFÓ T ÒE FÓ œ Þ' Þ& Þ œ Þ* ptòe FÓœTÒÐE FÑÓœ"TÒE FÓœ"Þ*œÞ"Þ T ÒFÓ œ T ÒF EÓ T ÒF E Ó p T ÒE FÓ œ Þ& Þ œ Þ$ Þ T ÒE FÓ œ T ÒE Ó T ÒFÓ T ÒE FÓ œ Þ% Þ& Þ$ œ Þ' The folloing Venn diagrams illustrate the various combinations of intersections, unions and complements of the events E and F A B A B A B PA [ ] 6 PB [ ] 5 PA [ B] 2

10 SECTION 1 - BASIC PROBABILITY CONCEPTS 45 A B A B PA [ B] 4 PA [ B] 3 From the folloing Venn diagrams e see that TÒE FÓœTÒEÓTÒE FÓœÞ'ÞœÞ% and TÒE FÓœTÒFÓTÒE FÓœÞ&ÞœÞ$Þ A B PA [ ] PA [ B] PA [ B] A B PB [ ] PA [ B] PA [ B] The folloing Venn diagrams shos ho to find TÒE FÓ A B PA [ B] PA [ B] PA [ B] PA [ B]

11 46 SECTION 1 - BASIC PROBABILITY CONCEPTS The relationship TÒE FÓ œ TÒEÓTÒFÓTÒE FÓ is explained in the folloing Venn PA [ B] PA [ ] PB [ ] PA [ B] The components of the events and their probabilities are summarized in the folloing diagram ( AB) AB 1 AB AB A B We can represent a variety of events in Venn diagram form and find their probabilities from the component events described in the previous diagram For instance, the complement of E is the combined shaded region in the folloing Venn diagram, and the probability is this probability also from T ÒE Ó œ " T ÒEÓ œ " Þ' œ Þ%Þ TÒEÓœÞ$Þ"œÞ%Þ We can get A

12 SECTION 1 - BASIC PROBABILITY CONCEPTS 47 Another efficient ay of summarizing the probability information for events E and F is in the form of a table TÒEÓ œ Þ' (given) TÒEÓ T ÒFÓ œ Þ& (given) T ÒE FÓ œ Þ (given) T ÒE FÓ TÒF Ó TÒE F Ó TÒE F Ó Complementary event probabilities can be found from T ÒE Ó œ " T ÒEÓ œ Þ% T ÒF Ó œ " T ÒFÓ œ Þ& Also, across each ro or along each column, the "intersection probabilities" add up to the single event probability at the end or top: and T ÒFÓ œ T ÒE FÓ T ÒE FÓ p Þ& œ Þ T ÒE FÓ p T ÒE FÓ œ Þ$, T ÒEÓ œ T ÒE FÓ T ÒE F Ó p Þ' œ Þ T ÒE F Ó p T ÒE F Ó œ Þ%, and T ÒE Ó œ T ÒE FÓ T ÒE F Ó p Þ% œ Þ$ T ÒE F Ó p T ÒE F Ó œ Þ" T ÒF Ó œ T ÒE F Ó T ÒE F Ó p Þ& œ Þ% T ÒE F Ó p T ÒE F Ó œ Þ" or These calculations can be summarized in the next table TÒEÓ œ Þ' (given) Ê TÒEÓ œ "TÒEÓ œ Þ% Ë Ë T ÒFÓ œ Þ& É T ÒE FÓ œ Þ T ÒE FÓ œ Þ$ TÒF Ó œþ& É TÒE F Ó œþ% TÒE F Ó œþ" Example 1-4: A survey is made to determine the number of households having electric appliances in a certain city It is found that 75% have radios ( V), 65% have electric irons ( M), 55% have electric toasters ( X), 50% have ( MV), 40% have ( VX), 30% have ( MX), and 20% have all three Find the probability that a household has at least one of these appliances Solution: TÒV M XÓ œ TÒVÓTÒMÓTÒXÓ TÒV MÓTÒV XÓTÒM XÓTÒV M XÓ œþ(&þ'&þ&&þ&þ%þ$þœþ*&

13 48 SECTION 1 - BASIC PROBABILITY CONCEPTS The folloing diagram deconstructs the three events I 05 R T The entries in this diagram ere calculated from the "inside out" For instance, since TÐV MÑ œ Þ& (given), and since TÐV M X Ñ œ Þ$ TÐV M XÑ œ Þ, since (also given), it follos that Þ&œTÐV MÑœTÐV M XÑTÐV M X ÑœÞTÐV M X Ñ (this uses the rule TÐEÑ œ TÐE FÑTÐE F Ñ, here E œ V M and F œ X) This is illustrated in the folloing diagram I 05 R RIT RIT 10 T The value "05" that is inside the diagram for event Vrefers to TÐV M X Ñ (the proportion ho have radios but neither irons nor toasters) This can be found in the folloing ay First e find TÐV M Ñ: Þ(& œ TÐVÑ œ TÐV MÑTÐV M Ñ œ Þ&TÐV M Ñ p TÐV M Ñ œ Þ&Þ

14 SECTION 1 - BASIC PROBABILITY CONCEPTS 49 TÐV M Ñis the proportion ith radios but not irons; this is the "05" inside V combined ith the "2" in the loer triangle inside V X Then e find TÐV M XÑ: Þ%œTÐV XÑœTÐV M XÑTÐV M XÑ œþtðv M XÑ ptðv M XÑœÞÞ Finally e find TÐV M X Ñ: Þ&œTÐV MÑœTÐV M XÑTÐV M X Ñ œþtðv M X Ñ ptðv M X ÑœÞ!&Þ The other probabilities in the diagram can be found in a similar ay Notice that TÐV M XÑ is the sum of the probabilities of all the disjoint pieces inside the three events, TÐV M XÑœÞ!&Þ!&Þ!&Þ"ÞÞ$ÞœÞ*& We can also use the rule TÐV M XÑ œ TÐVÑTÐMÑTÐXÑTÐV MÑTÐV XÑTÐM XÑTÐV M XÑ œ Þ(& Þ'& Þ&& Þ& Þ%Þ $ Þ œ Þ*& Either ay, this implies that 5% of the households have none of the three appliances It is possible that information is given in terms of numbers of units in each category rather than proportion of probability of each category that as given in Example 1-4 Example 1-5: In a survey of 120 students, the folloing data as obtained 60 took English, 56 took Math, 42 took Chemistry, 34 took English and Math, 20 took Math and Chemistry, 16 took English and Chemistry, 6 took all three subjects Find the number of students ho took (i) none of the subjects, (ii) Math, but not English or Chemistry, (iii) English and Math but not Chemistry Solution: Since I Q has 34 and I Q G has 6, it follos that I Q G has 28 The other entries are calculated in the same ay (very much like the previous example) (i) The total number of students taking any of the three subjects is I Q G, and is "')'"!)"%"œ*% The remaining '(out of 120) students are not taking any of the three subjects (this could be described as the set I Q G ) (ii) Q I G has 8 students (iii) I Q G has 28 students

15 50 SECTION 1 - BASIC PROBABILITY CONCEPTS Example 1-5 continued The folloing diagram illustrates ho the numbers of students can be deconstructed E 16 M C

16 PROBLEM SET 1 51 PROBLEM SET 1 Basic Probability Concepts 1 A survey of 1000 people determines that 80% like alking and 60% like biking, and all like at least one of the to activities What is the probability that a randomly chosen person in this survey likes biking but not alking? A) 0 B) 1 C) 2 D) 3 E) 4 2 (SOA) Among a large group of patients recovering from shoulder injuries, it is found that 22% visit both a physical therapist and a chiropractor, hereas 12% visit neither of these The probability that a patient visits a chiropractor exceeds by 014 the probability that a patient visits a physical therapist Determine the probability that a randomly chosen member of this group visits a physical therapist A) 026 B) 038 C) 040 D) 048 E) (SOA) An insurer offers a health plan to the employees of a large company As part of this plan, the individual employees may choose exactly to of the supplementary coverages A, B, and C, or they may choose no supplementary coverage The proportions of the company s employees that choose coverages A, B, and C are ", ", and & % $ ", respectively Determine the probability that a randomly chosen employee ill choose no supplementary coverage A) 0 B) C) D) E) %( " *( ( "%% "%% *

17 52 PROBLEM SET 1 4 (SOA) An auto insurance company has 10,000 policyholders Each policyholder is classified as (i) young or old; (ii) male or female; and (iii) married or single Of these policyholders, 3000 are young, 4600 are male, and 7000 are married The policyholders can also be classified as 1320 young males, 3010 married males, and 1400 young married persons Finally, 600 of the policyholders are young married males Ho many of the company s policyholders are young, female, and single? A) 280 B) 423 C) 486 D) 880 E) (SOA) The probability that a visit to a primary care physician s (PCP) office results in neither lab ork nor referral to a specialist is 35% Of those coming to a PCP s office, 30% are referred to specialists and 40% require lab ork Determine the probability that a visit to a PCP s office results in both lab ork and referral to a specialist A) 005 B) 012 C) 018 D) 025 E) (SOA) You are given TÒE FÓ œ!þ( and TÒE F Ó œ!þ* Determine TÒEÓÞ A) 02 B) 03 C) 04 D) 06 E) 08 7 (SOA) A survey of a group s vieing habits over the last year revealed the folloing information: (i) 28% atched gymnastics (ii) 29% atched baseball (iii) 19% atched soccer (iv) 14% atched gymnastics and baseball (v) 12% atched baseball and soccer (vi) 10% atched gymnastics and soccer (vii) 8% atched all three sports Calculate the percentage of the group that atched none of the three sports during the last year A) 24 B) 36 C) 41 D) 52 E) 60

18 PROBLEM SET (SOA) Under an insurance policy, a maximum of five claims may be filed per year by a policyholder Let : 8 be the probability that a policyholder files 8 claims during a given year, here 8 œ!ß "ß ß $ß %ß & An actuary makes the folloing observations: i) : 8 : 8" for 8œ!ß"ßß$ß% ii) The difference beteen : 8and : 8" is the same for 8 œ!ß "ß ß $ß % iii) Exactly 40% of policyholders file feer than to claims during a given year Calculate the probability that a random policyholder ill file more than three claims during a given year A) 014 B) 016 C) 027 D) 029 E) (SOA) The probability that a member of a certain class of homeoners ith liability and property coverage ill file a liability claim is 004, and the probability that a member of this class ill file a property claim is 010 The probability that member of this class ill file a liability claim but not a property claim is 001 Calculate the probability that a randomly selected member of this class of homeoners ill not file a claim of either type A) 0850 B) 0860 C) 0864 D) 0870 E) (SOA) A mattress store sells only, king, queen and tin-size mattresses Sales records at the store indicate that one-fourth as many queen-size mattresses are sold as king and tin-size mattresses combined Records also indicate that three times as many king-size mattresses are sold as tin-size mattresses Calculate the probability that the next mattress sold is either king or queen-size A) 012 B) 013 C) 0080 D) 085 E) 095

19 54 PROBLEM SET 1 PROBLEM SET 1 SOLUTIONS 1 Let Eœ"like alking" and Fœ"like biking" We use the interpretation that "percentage" and "proportion" are taken to mean "probability" We are given TÐEÑ œ Þ)ß TÐFÑ œ Þ' and TÐE FÑ œ " From the diagram belo e can see that since TÐE FÑ œ TÐEÑTÐE FÑ p TÐE FÑ œ Þ E FœE ÐF EÑ but (and) not alking In a similar ay e get TÐE F Ñ œ Þ% A e have is the proportion of people ho like biking B 4 4 A B 2 An algebraic approach is the folloing Using the rule TÐE FÑ œ TÐEÑTÐFÑTÐE FÑ, e get " œ Þ) Þ' T ÐE FÑ p T ÐE FÑ œ Þ% Then, using the rule TÐFÑ œ TÐF EÑTÐF EÑ, e get TÐF EÑ œ Þ'Þ% œ Þ Anser: C 2 G - chiropractor visit ; X - therapist visit We are given TÐG XÑ œ Þ, TÐG X Ñ œ Þ", TÐGÑ œ TÐXÑÞ"% )) œ " T ÐG X Ñ œ T ÐG X Ñ œ T ÐGÑ T ÐX Ñ T ÐG X Ñ œ T ÐX Ñ Þ"% T ÐX Ñ Þ p T ÐX Ñ œ Þ%) Anser: D 3 Since someone ho chooses coverage must choose exactly to supplementary coverages, in order for someone to choose coverage A, they must choose either A-and-B or A-and-C Thus, the " proportion of % of individuals that choose A is TÒE FÓTÒE GÓ œ " % (here this refers to the probability that someone chosen at random in the company chooses coverage A) In a similar ay e get Then, ÐT ÒE FÓ T ÒE GÓÑ ÐT ÒF EÓ T ÒF GÓÑ ÐT ÒG EÓ T ÒG FÓÑ œðtòe FÓTÒE GÓTÒF GÓÑ œ " " & % $ " œ" It follos that TÒE FÓTÒE GÓTÒF GÓ œ "

20 PROBLEM SET continued This is the probability that a randomly chosen individual chooses some form of coverage, since if someone ho chooses coverage chooses exactly to of A,B, and C Therefore, the probability that a randomly chosen individual does not choose any coverage is the probability of the complementary event, hich is also " Anser: C 4 We identify the folloing subsets of the set of 10,000 policyholders: ]œ young, ith size 3000 (so that ] œold has size 7000), Qœmale, ith size 4600 (so that Q œfemale has size 5400), and Gœmarried, ith size 7000 (so that G œsingle has size 3000) We are also given that ] Qhas size 1320, Q Ghas size 3010, ] G has size 1400, and ] Q Ghas size 600 We ish to find the size of the subset ] Q G We use the folloing rules of set theory: (i) if to finite sets are disjoint (have no elements in common, also referred to as empty intersection), then the total number of elements in the union of the to sets is the sum of the numbers of elements in each of the sets; (ii) for any sets Eand F, EœÐE FÑ ÐE FÑ, and E F and E F are disjoint Applying rule (ii), e have ] œ Ð] QÑ Ð] Q Ñ Applying rule (i), it follos that the size of ] Q must be œ1680 We no apply rule (ii) to ] G to get ] GœÐ] G QÑ Ð] G QÑ Applyng rule (i), it follos that ] G Q has size œ800 No applying rule (ii) to ] Q e get ] Q œð] Q GÑ Ð] Q GÑ Applying rule (i), it follos that ] Q G has size œ880 Within the "Young" category, hich e are told is 3000, e can summarize the calculations in the folloing table This is a more straightforard solution Married Single 1400 (given) 1600 œ Male 600 (given) 720 œ (given) Female 800 œ œ œ Anser: D

21 56 PROBLEM SET 1 5 We identify events as follos: PÀ VÀ lab ork needed referral to a specialist needed We are given T ÒP V Ó œ Þ$& ß T ÒVÓ œ Þ$ ß T ÒPÓ œ Þ% It follos that T ÒP VÓ œ " T ÒP V Ó œ Þ'&, and then since T ÒP VÓ œ T ÒPÓ T ÒVÓ T ÒP VÓ, e get T ÒP VÓ œ Þ$ Þ% Þ'& œ Þ!& L R L R These calculations can be summarized in the folloing table P ß Þ% P ß Þ' given Þ' œ " Þ% Vß Þ$ P V P V given Þ!& œ Þ% Þ$& Þ& œ Þ$ Þ!& V, Þ( P V P V, Þ$& Þ(œ"Þ$ Þ$&œÞ(Þ$& given Anser: A 6 T ÒE FÓ œ T ÒEÓ T ÒFÓ T ÒE FÓ ß T ÒE F Ó œ T ÒEÓ T ÒF Ó T ÒE F Ó We use the relationship TÒEÓ œ TÒE FÓTÒE F Ó Then TÒE FÓTÒE F Ó œ TÒEÓTÒFÓTÒE FÓ TÒEÓTÒF ÓTÒE F Ó œtòeó"tòeóœtòeó" (since TÒFÓTÒFÓœ" ) Therefore, Þ(Þ*œTÒEÓ" so that TÒEÓœÞ'

22 PROBLEM SET continued An alternative solution is based on the folloing Venn diagrams PA [ B] 70 PA [ B] 90 P[( AB ) ] P[ AB] 10 In the third diagram, the shaded area is the complement of that in the second diagram (using De Morgan's La, e have ÐE FÑœE F œe F) Then it can be seen from diagrams 1 and 3 that E œ ÐE FÑ ÐE FÑ, so that TÒEÓœTÒE FÓTÒE FÓœÞ(Þ"œÞ' Anser: D 7 We identify the folloing events: K - atched gymnastics, F - atched baseball, W - atched soccer We ish to find TÒK F W Ó By DeMorgan's rules e have TÒK F WÓœ"TÒK F WÓ We use the relationship TÒK F WÓ œ TÒKÓTÒFÓTÒWÓ T ÒK FÓ T ÒK WÓ T ÒF WÓ T ÒK F WÓ We are given T ÒKÓ œ Þ) ß T ÒFÓ œ Þ* ß T ÒWÓ œ Þ"* ß T ÒK FÓ œ Þ"% ß T ÒK WÓ œ Þ"! ß T ÒF WÓ œ Þ" ß T ÒK F WÓ œ Þ!) Then TÒK F WÓœÞ%) and TÒK F WÓœ"Þ%)œÞ& Anser: D

23 58 PROBLEM SET 1 8 The probability in question is :% : & We kno that :! : " : : $ : % : & œ" We are given that : : œ : : œ : : œ : : œ : :! " " $ $ % % &, and :! : " œ Þ% If e let > be equal to the common difference : 8 : 8", then :" œ :! >ß : œ :! >ß : $ œ :! $>ß :% œ :! %> and :& œ :! &> Then :! : " : : $ : % : & œ ':! "&> œ " When e combine this equation & " ith :! : " œ :! > œ Þ%, e can solve the to equations to get :! œ %, and > œ '! "( "& $ Then :% œ :! %> œ "!, and :& œ :! &> œ "!, so that :% : & œ "! œ Þ( Anser: C 9 We define the folloing events: Pœfile a liability claim, T œfile a property claim We are given TÐPÑ œ!þ!%ß TÐTÑ œ!þ"!ß TÐP T Ñ œ!þ!" We ish to find TÐP T Ñ TÐTÑœ"TÐTÑ!œ*! It follos that and TÐP T Ñ œ!þ)* Anser: E!Þ*!œTÐTÑœTÐP TÑTÐP TÑœ!Þ!"TÐP TÑÞ 10 We define X to be the event that the next mattress sold is tin-size, and similarly e define O and U as the events that the next mattress sold is king-size and queen-size, respectively We define " TÐXÑ œ - Then TÐOÑ œ $- and TÐUÑ œ % ÒTÐOÑTÐXÑÓ œ - Since TÐXÑTÐUÑTÐOÑ œ ", e have &- œ ", so that - œ!þ Then T ÐO UÑ œ %- œ!þ)þ Anser: C

24 SECTION 8 - JOINT, MARGINAL, AND CONDITIONAL DISTRIBUTIONS 233 SECTION 8 - JOINT, MARGINAL, AND CONDITIONAL DISTRIBUTIONS Joint distribution of random variables \ and ] A random variable \ is a numerical outcome that results from some random experiment, such as the number that turns up hen tossing a die It is possible that an experiment may result in to or more numerical outcomes A simple example ould be the numbers that turn up hen tossing to dice \ could be the number that turns up on the first die and ] could be the number on the second die Another example could be the folloing experiment A coin is tossed and if the outcome is head then toss one die, and if the outcome is tails then toss to dice We could set \œ" for a head and \œ for a tail and ]œtotal on the dice thron In both of the examples just described, e have a pair of random variables \ and ], that result from the experiment \ and ] might be unrelated or independent of one another (as in the example of the toss of to independent dice), or they might be related to each other (as in the coindice example) We describe the probability distribution of to or more random variables together as a joint distribution As in the case of a single discrete random variable, e still describe probabilities for each possible pair of outcomes for a pair of discrete random variables In the case of a pair of random variables \ and ], there ould be probabilities of the form T ÒÐ\ œ BÑ Ð] œ CÑÓ for each pair ÐBß CÑ of possible outcomes For a pair of continuous random variables \ and ], there ould be a density function to describe density over a to dimensional region A joint distribution of to random variables has a probability function or probability density function 0ÐBß CÑ that is a function of to variables (sometimes denoted 0 ÐBß CÑ) It is defined over a to- dimensional region For joint distributions of continuous random variables \ and ], the region of probability (the probability space) is usually a rectangle or triangle in the BC - plane \ß] If \ and ] are discrete random variables, then 0ÐBß CÑ œ T ÒÐ\ œ BÑ Ð] œ CÑÓ is the joint probability function, and it must satisfy (i)!ÿ0ðbßcñÿ" and (ii) 0ÐBßCÑ œ " B C If \ and ] are continuous random variables, then 0ÐBßCÑmust satisfy (i) 0ÐBßCÑ! and (ii) 0ÐBß CÑ C B œ "

25 234 SECTION 8 - JOINT, MARGINAL, AND CONDITIONAL DISTRIBUTIONS In the to dice example described above, if the to dice are tossed independently of one another then " " " 0ÐBß CÑ œ T ÒÐ\ œ BÑ Ð] œ CÑÓ œ T Ò\ œ BÓ T Ò] œ CÓ œ ' ' œ $' for each pair ith B œ "ß ß $ß %ß &ß ' and C œ "ß ß $ß %ß &ß ' The coin-die toss example above is more complicated because the number of dice tossed depends on hether the toss is head or tails If the coin toss is a head then \ œ" and ] œ"ßß$ß%ß&ß' so " " " 0Ð"ß CÑ œ T ÒÐ\ œ "Ñ Ð] œ CÑÓ œ ' œ " for C œ "ß ß $ß %ß &ß ' If the coin toss is tail then \ œ and ] œ ß $ß ÞÞÞß " ith " " " 0Ðß Ñ œ T ÒÐ\ œ Ñ Ð] œ ÑÓ œ $' œ (, " " 0Ðß $Ñ œ T ÒÐ\ œ Ñ Ð] œ $ÑÓ œ $' œ $', etc It is possible to have a joint distribution in hich one variable is discrete and one is continuous, or either has a mixed distribution The joint distribution of to random variables can be extended to a joint distribution of any number of random variables If E is a subset of to-dimensional space, then TÒÐ\ß]Ñ EÓis the summation (discrete case) or double integral (continuous case) of 0ÐBßCÑ over the region E Example 8-1: \ and ] are discrete random variables hich are jointly distributed ith the probability function 0ÐBßCÑ defined in the folloing table: \ "! " " " " " ") * ' From this table e see, for example, that " " " ]! *! ' TÒ\œ!ß] œ "Óœ0Ð!ß"Ñœ * " " " " ' * * Find (i) TÒ\ ] œ "Ó, (ii) TÒ\ œ!ó and (iii) TÒ\ ]Ó Solution: (i) We identify the ÐBß CÑ -points for hich \ ] œ ", and the probability is the sum of 0ÐBß CÑ over those points The only Bß C combinations that sum to 1 are the points Ð!ß "Ñ and Ð"ß!Ñ " " & Therefore, T Ò\ ] œ "Ó œ 0Ð!ß "Ñ 0Ð"ß!Ñ œ * ' œ ") Þ (ii) We identify the ÐBß CÑ -points for hich \ œ! These are Ð!ß "Ñ and Ð!ß "Ñ (e omit Ð!ß!Ñ " " since there is no probability at that point) T Ò\ œ!ó œ 0Ð!ß "Ñ 0Ð!ß "Ñ œ * * œ * (iii) The ÐBß CÑ -points satisfying \ ] are Ð "ß!Ñ ß Ð "ß "Ñ and Ð!ß "Ñ " " " & Then T Ò\ ] Ó œ 0Ð "ß!Ñ 0Ð "ß "Ñ 0Ð!ß "Ñ œ * ") * œ ") Þ

26 SECTION 8 - JOINT, MARGINAL, AND CONDITIONAL DISTRIBUTIONS 235 Example 8-2: Suppose that 0ÐBßCÑ œ OÐB C Ñ is the density function for the joint distribution of the continuous random variables \ and ] defined over the unit square bounded by the points Ð!ß!Ñ ß Ð"ß!Ñ ß Ð"ß "Ñ and Ð!ß "Ñ, find O Find T Ò\ ] "Ó Solution: In order for 0ÐBßCÑ to be a properly defined joint density, the (double) integral of the density function over the region of density must be 1, so that " " " œ $!! OÐB C Ñ C B œ O $ Ê O œ $ Ê 0ÐBß CÑ œ ÐB C Ñ for! Ÿ B Ÿ " and! Ÿ C Ÿ " In order to find the probability \] TÒ\ ] "Ó, e identify the to dimensional region representing " This is generally found by draing the boundary line for the inequality, hich is BCœ" (or Cœ"B) in this case, and then determining hich side of the line is represented in the inequality We can see that BC " is equivalent to C "B This is the shaded region in the graph belo y 1 y 1x 1 x The probability TÒ\ ] "Óis found by integrating the joint density over the to-dimensional region It is possible to represent to-variable integrals in either order of integration In some cases one order of integration is more convenient than the other In this case there is not much advantage of one direction of integration over the other " " TÒ\ ] "Ó œ $ " ÐB C Ñ C B œ " Cœ" $! "B! Ð$B C C Ñ B Cœ"B " œ " $ $! Ð$B "$B Ð"BÑÐ"BÑ ÑBœ % Þ Reversing the order of integration, e have B "C, so that " " TÒ\ ] "Ó œ $ $! "C ÐB C Ñ B C œ % Þ

27 236 SECTION 8 - JOINT, MARGINAL, AND CONDITIONAL DISTRIBUTIONS Example 8-3: Continuous random variables \ and ] have a joint distribution ith density function BC 0ÐBßCÑœB $ for!b" and!c " " Find the conditional probability TÒ\ l] Ó Solution: We use the usual definition TÒElFÓ œ TÒE FÓ TÒFÓ " " " " T ÒÐ\ Ñ Ð] ÑÓ TÒ\ l] Ó œ " TÒ] Ó These regions are described in the folloing diagram 2 y 2 y x X 1 1 Y 2 2 Y x " " T ÒÐ\ Ñ Ð] ÑÓ œ " BC %$ "Î "Î ÒB $ Ó C B œ '% " " " TÒ] Óœ Ò BC "$ 0ÐBßCÑBÓC œ "Î! "Î! ÒB $ ÓBC œ "' " " %$Î'% %$ ptò\ l] Óœ "$Î"' œ & Cumulative distribution function of a joint distribution: If random variables \ and ] have a joint distribution, then the cumulative distribution function is J ÐBß CÑ œ T ÒÐ\ Ÿ BÑ Ð] Ÿ CÑÓ B C In the continuous case, JÐBßCÑ œ 0Ð=ß>Ñ>=, B C and in the discrete case, J ÐBß CÑ œ ` 0Ð=ß >Ñ In the continuous case, `B`C J ÐBß CÑ œ 0ÐBß CÑ =œ >œ Example 8-4: The cumulative distribution function for the joint distribution of the continuous random variables \ and $ " " ] is J ÐBß CÑ œ ÐÞÑÐ$B C B C Ñ, for! Ÿ B Ÿ " and! Ÿ C Ÿ " Find 0Ð ß Ñ ` " " "( Solution: 0ÐBß CÑ œ `B`C J ÐBß CÑ œ ÐÞÑÐ*B )BCÑ p 0Ð ß Ñ œ! Þ

28 PRACTICE EXAM PRACTICE EXAM 7 1 A study of the relationship beteen blood pressure and cholesterol level shoed the folloing results for people ho took part in the study: (a) of those ho had high blood pressure, 50% had a high cholesterol level, and (b) of those ho had high cholesterol level, 80% had high blood pressure Of those in the study ho had at least one of the conditions of high blood pressure or high cholesterol level, hat is the proportion ho had both conditions? " % & ( A) $ B) * C) * D) $ E) * 2 A study of international athletes shos that of the to performance-enhancing steroids Dianabol and Winstrol, 5% of athletes use Dianabol and not Winstrol, 2% use Winstrol and not Dianabol, and 1% use both A breath test has been developed to test for the presence of the these drugs in an athlete Past use of the test has resulted in the folloing information regarding the accuracy of the test Of the athletes that are using both drugs, the test indicates that 75% are using both drugs, 15% are using Dianabol only and 10% are using Winstrol only In 80% of the athletes that are using Dianabol but not Winstrol, the test indicates they are using Dianabol but not Winstrol, and for the other 20% the test indicates they are using both drugs In 60% of the athletes that are using Winstrol but not Dianabol, the test indicates that they are using Winstrol only, and for the other 40% the test indicates they are using both drugs For all athletes that are using neither Dianabol nor Winstrol, the test alays indicates that they are using neither drug Of those athletes ho test positive for Dianabol but not Winstrol, find the percentage that are using both drugs A) 12% B) 24% C) 36% D) 48% E) 60% 3 The random variable R has the folloing characteristics: (i) With probability :, R has a binomial distribution ith ; œ!þ& and 7 œ (ii) With probability ":, R has a binomial distribution ith ;œ!þ& and 7œ% Which of the folloing is a correct expression for Prob ÐR œ Ñ? A)!Þ"&: B)!Þ$(&!Þ"&: C)!Þ$(&!Þ"&: D)!Þ$(&!Þ"&: E)!Þ$(&!Þ"&:

29 470 PRACTICE EXAM 7 4 An insurance company does a study of claims that arrive at a regional office The study focuses on the days during hich there ere at most 2 claims The study finds that for the days on hich there ere at most 2 claims, the average number of claims per day is 12 The company models the number of claims per day arriving at that office as a Poisson random variable Based on this model, find the probability that at most 2 claims arrive at that office on a particular day A) 62 B) 64 C) 66 D) 68 E) 70 5 An actuarial trainee orking on loss distributions encounters a special distribution The student reads a discussion of the distribution and sees that the density of \ is 0ÐBÑ œ α)α Bα " on the region α) \), here α and ) must both be!, and the mean is α" if α " The student is analyzing loss data that is assumed to follo such a distribution, but the values of α and ) are not specified, although it is knon that )!! The data shos that the average loss for all losses is 180, and the average loss for all losses that are above 200 is 300 Find the median of the loss distribution A) Less than 100 B) At least 100, but less than 120 C) At least 120, but less than 140 D) At least 140, but less than 160 E) At least An insurance claims administrator verifies claims for various loss amounts For a loss claim of amount B, the amount of time spent by the administrator to verify the claim is uniformly distributed beteen 0 and "Bhours The amount of each claim received by the administrator is uniformly distributed beteen 1 and 2 Find the average amount of time that an administrator spends on a randomly arriving claim A) 1125 B) 1250 C) 1375 D) 1500 E) A husband and ife have a health insurance policy The insurer models annual losses for the husband separately from the ife \ is the annual loss for the husband and ] is the annual loss for the ife \ has a uniform distribution on the interval Ð!ß &Ñ and ] has a uniform distribution on the interval Ð!ß (Ñ, and \ and ] are independent The insurer applies a deductible of 2 to the combined annual losses, and the insurer pays a maximum of 8 per year Find the expected annual payment made by the insurer for this policy A) 2 B) 3 C) 4 D) 5 E) 6

30 PRACTICE EXAM PRACTICE EXAM 7 - SOLUTIONS 1 We ill use F to denote the event that a randomly chosen person in the study has high blood pressure, and G ill denote the event high cholesterol level The information given tells us that TÐGlFÑ œ Þ&! and TÐFlGÑ œ Þ)! We ish to find TÐF GlF GÑ This is T ÒÐF GÑ ÐF GÑÓ TÐF GÑ T ÒF GÑÓ œ TÐFÑTÐGÑTÐF GÑ " " œ T ÐFÑT ÐGÑT ÐF GÑ œ " " Ò "Ó TÐF GÑ TÐGlFÑ TÐGlFÑ " " % " Þ& * œ œ œ " " Þ& Þ) Anser: B 2 We define the folloing events: H - the athlete uses Dianabol [ - the athlete uses Winstrol XH- the test indicates that the athlete uses Dianabol X[ - the test indicates that the athlete uses Winstrol We are given the folloing probabilities TÐH [ Ñ œ Þ!& ß TÐH [Ñ œ Þ! ß TÐH [Ñ œ Þ!", TÐXH X[lH [ÑœÞ(&ß TÐXH X[ lh [ÑœÞ"&ß TÐXH X[lH [ÑœÞ"ß TÐXH X[lH [ÑœÞß TÐXH X[lH [ÑœÞ)ß TÐXH X[lH [ÑœÞ%ß TÐXH X[lH [ÑœÞ' TÐH [ XH X[ Ñ We ish to find TÐH [lxh X[ Ñ œ TÐXH X[ Ñ The numerator is T ÐH [ X H X [ Ñ œ T ÐX H X [ lh [ Ñ T ÐH [ Ñ œ ÐÞ"&ÑÐÞ!"Ñ œ Þ!!"& The denominator is TÐXH X[ Ñœ TÐXH X[ H [Ñ TÐXH X[ H [Ñ TÐXH X[ H [ Ñ TÐXH X[ H [ Ñ We have used the rule TÐEÑœTÐE FÑTÐE FÑâ ", here FßFßÞÞÞ " forms a partition The partition in this case is F" œ H [ ß F œ H [ ß F$ œ H [ ß F% œ H [, since an athlete must be using both, one or neither of the drugs

31 478 PRACTICE EXAM 7 2 continued We have just seen that T ÐX H X [ H [ Ñ œ Þ!!"& In a similar ay, e have TÐXH X[ H [ÑœTÐXH X[ lh [Ñ TÐH [ÑœÐ!ÑÐÞ!Ñœ!, and TÐXH X[ H [ÑœTÐXH X[ ' lh [ Ñ TÐH [ ÑœÐÞ)ÑÐÞ!&ÑœÞ!%, and TÐXH X[ H [ ÑœTÐXH X[ lh [ Ñ TÐH [ ÑœÐ!ÑÐÞ*Ñœ! (note that TÐH [ Ñ œ "TÐH [Ñ œ "TÐH [ ÑTÐH [Ñ TÐH [Ñ œ Þ* Þ!!"& Then, T ÐH [ lx H X [ Ñ œ Þ!!"&!Þ!%! œ Þ!$', 36% Anser: C % 3 T ÐR œ Ñ œ :T ÐR" œ Ñ Ð" :ÑT ÐR œ Ñ œ :ÐÞ&Ñ Ð" :Ñ'ÐÞ&Ñ œ Þ$(& Þ"&: We have used the binomial probabilities ;Ð";Ñ Anser: E 5 4 Suppose that the mean number of claims per day arriving at the office is - Let \ denote the number of claims arriving in one day Then the probability of at most 2 claims in one day is TÐ\ Ÿ Ñ œ / -/ - / The conditional probability of 0 claims arriving on a day given that there are at most 2 for the day is TÐ\œ!Ñ - / " TÐ\œ!l\ŸÑœ TÐ\ŸÑ œ œ / - / "- The conditional probability of 1 claim arriving on a day given that there are at most 2 for the day is TÐ\œ"Ñ - -/ - TÐ\œ"l\ŸÑœ TÐ\ŸÑ œ œ / -/ "- The conditional probability of 2 claims arriving on a day given that there are at most 2 for the day is TÐ\œÑ TÐ\œl\ŸÑœ TÐ\ŸÑ œ œ / / - "- / / -/ / - - The expected number of claims per day, given that there ere at most 2 claims per day is " - -- Ð!ÑÐ Ñ Ð"ÑÐ Ñ ÐÑÐ Ñ œ "- "- "- " We are told that this is 12 - Therefore - - œ Ð"ÞÑÐ" - Ñ, hich becomes the quadratic equation Þ% - Þ - "Þ œ! Solving the equation results in - œ or "Þ&, but e ignore the negative root The probability of at most 2 claims arriving at the office on a particular day is / T Ð\ Ÿ Ñ œ / / œ Þ'('( Anser: D

32 PRACTICE EXAM The distribution function ill be JÐCÑ œ C 0ÐBÑB œ C α) ) ) B œ " ) Bα" Cα The median 7 occurs here JÐ7Ñ œ " If α and ) ere knon, e could find the median α) The average loss for all losses is α" œ ")!, but both ) and α are not knon The conditional distribution of loss amount B given that \!! is 0ÐBÑ α α α α) ) α!! 0ÐBl\!!Ñ œ T Ð\!!Ñ œ B α"!! α œ Bα "!! α This random variable has a mean of α" We are given that this mean is 300,!! α so α" œ $!!, and therefore α œ $ α) $ ) Then, from α" œ ")!, e get œ ")!, so that ) œ "! " ) The median 7 satisfies the relation œ J Ð7Ñ œ " α "! 7α œ " Ð 7 Ñ $, so that 7 œ "&"Þ Anser: D α α 6 \ œ amount of loss claim, uniformly distributed on Ð"ß Ñ, so 0\ ÐBÑ œ " for " B ]œamount of time spent verifying claim We are given that the conditional distribution of ] given \ œ B is uniform on Ð!ß"BÑ, " so 0ÐClBÑœ "B for!c"b We ish to find IÒ] ÓÞ The joint density of \ and ] is " 0ÐBßCÑœ0ÐClBÑ 0\ ÐBÑœ "B for!c"b and "B There are a couple of ays to find IÒ] Ó : (i) IÒ] Ó œ C0ÐBßCÑ C B or IÒ] Ó œ C0ÐBßCÑ B C, ith careful setting of the integral limits, or ] ] (ii) IÒ]Ó œ C0 ÐCÑC, here 0 ÐCÑ is the pdf of the marginal distribution of ] "B "! IÒ] Ó œ C (iii) The double expectation rule, IÒ] Ó œ IÒ IÒ] l\ó Ó If e apply the first approach for method (i), e get "B IÒ]Óœ " Ð"BÑ C CBœ Cœ "B & "! "B " Ð"BÑ " Bœ % If e apply the second approach for method (i), e must split the double integral into IÒ] Ó œ " C B C $ "! " "B C" C "B B C The first integral becomes $ $! C68 Ð ÑC œ68ð Ñ The second integral becomes $ & C Ò68$ 68 CÓ C œ $ 68$ C 68 C C The integral $ C68CC is found by integration by parts

33 480 PRACTICE EXAM 7 6 continued Let C68CC œ EÞ Let? œ C œ 68 C C, œ C 68C C (antiderivative of 68 C), and then EœC68CC œcðc68c CÑ C ÐC68CCÑC œc 68C C E, so that Eœ " C C68CCœ C 68C % $ $ Then " C C68CC œ C 68C * * % * & % œ 68$ Ð 68 "Ñ œ 68$ 68 % % $ & $ Finally, IÒ] Ó œ 68Ð Ñ 68$ C 68 C C & * & & œ 68 $ 68 68$ Ð 68$ 68 % Ñ œ % The first order of integration for method (i) as clearly the more efficient one (ii) This method is equivalent to the second approach in method (i) because e find 0] ÐCÑ from the relationship 0] ÐCÑ œ 0ÐBßCÑB The to-dimensional region of probability for the joint distribution is "B and!c"b This is illustrated in the graph belo y 3 y 1z x For!C, 0] ÐCÑœ " 0ÐBßCÑBœ " $ " "B Bœ68Ð Ñ, and for ŸB$, 0] ÐCÑœ C" 0ÐBßCÑBœ " C" "B Bœ68$68C Then IÒ] Ó œ $ C 68 C! Ð $ Ñ C Ò68$ 68CÓC, hich is the same as the second part of method (i) (iii) According to the double expectation rule, for any to random variables Y and [, e have IÒY Ó œ IÒ IÒY l[ Ó Ó Therefore, IÒ] Ó œ IÒ IÒ] l\ó Ó We are told that the conditional distribution of ] given \ œ B is uniform on the interval "\ Ð!ß " BÑ, so IÒ] l\ó œ "\ " " " " $ & Then IÒIÒ] l\óó œ IÒ Ó œ IÒ\Ó œ Ð Ñ œ %, since \ is uniform on $ Ð"ß Ñ and \ has mean Anser: B

34 PRACTICE EXAM " " " 7 The joint distribution of \ and ] has pdf 0ÐBßCÑ œ & ( œ $& on the rectangle!b& and!c( The insurer pays \] if the combined loss \] is The maximum payment of 8 is reached if \] ), or equivalently, if \] "! Therefore, the insurer pays \] if \] Ÿ"! (the lighter shaded region in the diagram belo), and the insurer pays ) if \ ] "! (the darker shaded region in the diagram belo) The expected amount paid by the insurer is a combination of to integrals: " ÐBCÑ $& CB, here the integral is taken over the region BCŸ"! (the lightly shaded region), plus " ) $& C B, here the integral is taken over the region \ ] "! (the darker region) ) "' The second integral is $& ÐÑœ $&, since the area of the darkly shaded triangle is 2 (it is a 2 right triangle) y X Y X Y x The first integral can be broken into three integrals: ( " $ ( " & "!B "! B ÐBCÑ $& CB! ÐBCÑ $& CB $! ÐBCÑ $& CB " œ Ò ÐB&Ñ $ (ÐB$Ñ & '!%BB $& B B! $ BÓ " "!* "(* "% œ $& Ò $ ) $ Ó œ $& The total expected insurance payment is "' "% "%! $& $& œ $& œ % Anser: C