EXCERPTS FROM ACTEX STUDY MANUAL FOR SOA EXAM P/CAS EXAM 1

EXCERPTS FROM ACTEX STUDY MANUAL FOR SOA EXAM P/CAS EXAM 1 Table of Contents Introductory Comments Section 2 - Conditional Probability and Independence

TABLE OF CONTENTS INTRODUCTORY COMMENTS SECTION 0 - REVIEW OF ALGEBRA AND CALCULUS Set Theory 1 Graphing an Inequality in To Dimensions 9 Properties of Functions 10 Differentiation 12 Integration 14 Geometric and Arithmetic Progressions 17 Problem Set 0 and Solutions 19 SECTION 1 - BASIC PROBABILITY CONCEPTS Probability Spaces and Events 29 Probability 33 Problem Set 1 and Solutions 43 SECTION 2 - CONDITIONAL PROBABILITY AND INDEPENDENCE Definition of Conditional Probability 49 Bayes' Rule, Bayes' Theorem and the La of Total Probability 51 Independent Events 55 Problem Set 2 and Solutions 59 SECTION 3 - COMBINATORIAL PRINCIPLES Permutations and Combinations 73 Problem Set 3 and Solutions 79 SECTION 4 - RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS Discrete Random Variable 83 Continuous Random Variable 84 Mixed Distribution 85 Cumulative Distribution Function 85 Independent Random Variables 90 Problem Set 4 and Solutions 95 SECTION 5 - EXPECTATION AND OTHER DISTRIBUTION PARAMETERS Expected Value 101 Moments, Variance and Standard Deviation 103 Moment Generating Function 103 Percentiles, Median and Mode 105 Problem Set 5 and Solutions 111

SECTION 6 - FREQUENTLY USED DISCRETE DISTRIBUTIONS Discrete Uniform Distribution 123 Binomial Distribution 123 Poisson Distribution 125 Geometric Distribution 128 Negative Binomial Distribution 129 Hypergeometric Distribution 130 Multinomial Distribution 131 Summary of Discrete Distributions 132 Problem Set 6 and Solutions 133 SECTION 7 - FREQUENTLY USED CONTINUOUS DISTRIBUTIONS Continuous Uniform Distribution 145 Normal Distribution 145 Exponential Distribution 148 Gamma Distribution and Pareto Distribution 151 Lognormal, Beta, Weibull, and Chi-Square Distributions 152 Summary of Continuous Distributions 153 Problem Set 7 and Solutions 155 SECTION 8 - JOINT, MARGINAL, AND CONDITIONAL DISTRIBUTIONS Definition of Joint Distribution 165 Marginal Distributions 168 Independence of Random Variables 1( 1 Conditional Distributions 172 Covariance Beteen Random Variables 176 Moment Generating Function for a Joint Distribution 177 Problem Set 8 and Solutions 183 SECTION 9 - TRANSFORMATIONS OF RANDOM VARIABLES Distribution of a Transformation of \ 205 Distribution of a Transformation of Joint Distribution of \ and ] 206 Distribution of a Sum of Random Variables 207 Distribution of the Maximum of Minimum of Independent Ö\ ß \ ß ÞÞÞß \ 8 211 Order Statistics 212 Mixtures of Distributions 214 Problem Set 9 and Solutions 217 SECTION 10 - RISK MANAGEMENT CONCEPTS Loss Distributions and Insurance 235 The Normal Approximation to Aggregate Claims 240 Insurance Policy Deductible 243 Insurance Policy Limit 245 Proportional Insurance 246 Problem Set 10 and Solutions 249 PRACTICE EXAM 1 271

PRACTICE EXAM 2 291 PRACTICE EXAM 3 313 PRACTICE EXAM 4 339 PRACTICE EXAM 5 359

SECTION 2 - CONDITIONAL PROBABILITY AND INDEPENDENCE Conditional probability of event F given event E: If TÒEÓ!, then TÒFlEÓ TÒF EÓ TÒEÓ is defined to be the conditional probability that event F occurs given that event E has occurred Reriting the equation results in T ÒF EÓ T ÒFlEÓ T ÒEÓ, hich is referred to as the multiplication rule When e condition on event E, e are assuming that event F has occurred so that E becomes the ne probability space, and all conditional events must take place ithin event E (the ne probability space) Dividing by TÒEÓ scales all probabilities so that E is the entire probability space, and T ÒElEÓ To say that event F has occurred given that event E has occurred means that both F and E ( F E) have occurred ithin the probability space E This is the explanation behind the definition of the conditional probability TÒFlEÓ Example 2-1: Suppose that a fair six-sided die is tossed The probability space is W Öß ß ß %ß &ß ' We define the folloing events: E the number tossed is Ÿ Öß ß, F the number tossed is even Öß %ß ', G the number tossed is a or a Öß, H the number tossed doesn't start ith the letters 'f' or 't' Öß ' The conditional probability of E given F is T ÒÖßß Öß%ß' Ó T ÒÖ Ó Î' TÒElFÓ T ÒÖß%ß' Ó T ÒÖß%ß' Ó Î The interpretation of this conditional probability is that if e kno that event F has occurred, then the toss must be 2, 4 or 6 Since the original 6 possible tosses of a die ere equally likely, if e are given the additional information that the toss is 2, 4 or 6, it seems reasonable that each of those is equally, each ith a probability of Then ithin the reduced probability space F, the (conditional) probability that a event E occurs is the probability, in the reduced space, of tossing a 2; this is The conditional probability of Egiven G is TÒElGÓ To say that G has occurred means that the toss is 1 or 2 It is then guaranteed that event E has occurred, since G E The conditional probability of F given G is TÒFlGÓ

& ( Example 2-2: If T ÒEÓ ' and T ÒFÓ, and T ÒElFÓ T ÒFlEÓ!, find TÒE FÓ TÒE FÓ TÒE FÓ Solution: T ÒFlEÓ TÒEÓ 'T ÒE FÓ and T ÒElFÓ TÒFÓ & T ÒE FÓ ( p Ð' & Ñ TÒE FÓ! p TÒE FÓ 2 Þ IMPORTANT NOTE: The folloing manipulation of event probabilities arises from time to time: TÒEÓTÒElFÓ TÐFÑTÒElFÓ TÐFÑ This relationship is a version of the La of Total Probability This relationship is valid since for any events Fand E, e have TÒEÓ TÒE FÓTÒE F Ó We then use the relationships T ÒE FÓ T ÒElFÓ T ÐFÑ and T ÒE F Ó T ÒElF Ó T ÐF Ñ If e kno the conditional probabilities for event E given some other event F and if e also kno the conditional probability of E given the complement F, and if e are given the (unconditional) probability of event F, then e can find the (unconditional) probability of event E An application of this concept occurs hen an experiment has to (or more) steps The folloing example illustrates this idea Example 2-3: Urn I contains 2 hite and 2 black balls and Urn II contains 3 hite and 2 black balls An Urn is chosen at random, and a ball is randomly selected from that Urn Find the probability that the ball chosen is hite Solution: Let F be the event that Urn I is chosen and F is the event that Urn II is chosen The implicit assumption is that both Urns are equally likely to be chosen (this is the meaning of an Urn is chosen at random) Therefore, TÒFÓ and TÒF Ó Let E be the event that the ball chosen in hite If e kno that Urn I as chosen, then there is probability of choosing a hite ball (2 hite out of 4 balls, it is assumed that each ball has the same chance of being chosen); this can be described as TÒElFÓ In a similar ay, if Urn II is chosen, then TÒElF Ó & (3 hite out of 5 balls) We can no apply the relationship described prior to this example T ÒE FÓ T ÒElFÓ T ÒFÓ Ð ÑÐ Ñ %, and T ÒE F Ó T ÒElF Ó T ÒF Ó Ð & ÑÐ Ñ! Finally, TÒEÓ TÒE FÓTÒE F Ó %!! The order of calculations can be summarized in the folloing table E F 1 TÐE FÑTÒElFÓ TÒFÓ F 2 T ÐE F Ñ T ÒElF Ó T ÒF Ó 3 TÐEÑ TÐE FÑTÐE F Ñ

Bayes' rule and Bayes' Theorem (basic form): T ÒF EÓ T ÒElFÓ T ÒFÓ For any events Eand F ith TÒEÓ!, TÒFlEÓ TÒEÓ TÒEÓ The usual ay that this is applied is in the case that e are given the values of TÒFÓ, TÒElFÓ and TÒElF Ó (from TÒFÓe get TÒF Ó TÒFÓ), and e are asked to find TÒFlEÓ (in other ords, e are asked to turn around the conditioning of the events Eand F) We can summarize this process by calculating the quantities in the folloing table in the order indicated numerically (1-2-3-4) (other entries in the table are not necessary in this calculation, but might be needed in related calculations) E E Fß TÒFÓ 1 TÒElFÓ T ÒE FÓ T ÒElFÓ T ÒFÓ TÒElFÓ TÒElFÓ T ÒE FÓ T ÒE lfó T ÒFÓ Fß TÒF Ó TÒFÓ 2 TÒElF Ó T ÒE F Ó T ÒElF Ó T ÒF Ó TÒElF Ó TÒElF Ó T ÒE F Ó T ÒE lf Ó T ÒF Ó TÒF Ó TÒFÓ Step 4: TÒFlEÓ TÒF EÓ TÒEÓ 3 TÒEÓ TÒE FÓTÒE F Ó TÒEÓ TÒEÓ This can also be summarized in the folloing sequence of calculations TÒFÓ, TÒElFÓ, given TÒF Ó TÒFÓ, TÒElF Ó, given Ì Ì TÒE FÓ TÒE F Ó T ÒElFÓ T ÒFÓ T ÒElF Ó T ÒF Ó Ì TÒEÓ TÒE FÓTÒE F Ó Algebraically, e have done the folloing calculation: TÒF EÓ TÒE FÓ TÒElFÓ TÒFÓ TÒFlEÓ T ÒEÓ T ÒE FÓT ÒE F Ó T ÒElFÓ T ÒFÓT ÒElF Ó T ÒF Ó, here all the factors in the final expression ere originally knon Note that the numerator is one of the components of the denominator

Exam questions on this topic (and the extended form of Bayes' rule revieed belo) have occurred quite regularly The key starting point is identifying and labeling unconditional events and conditional events and their probabilities in an efficient ay Example 2-4: Urn I contains 2 hite and 2 black balls and Urn II contains 3 hite and 2 black balls One ball is chosen at random from Urn I and transferred to Urn II, and then a ball is chosen at random from Urn II The ball chosen from Urn II is observed to be hite Find the probability that the ball transferred from Urn I to Urn II as hite Solution: Let F denote the event that the ball transferred from Urn I to Urn II as hite and let E denote the event that the ball chosen from Urn II is hite We are asked to find TÒFlEÓ From the simple nature of the situation (and the usual assumption of uniformity in such a situation, meaning that all balls are equally likely to be chosen from Urn I in the first step), e have TÒFÓ (2 of the 4 balls in Urn I are hite), and TÒF Ó If the ball transferred is hite, then Urn II has 4 hite and 2 black balls, and the probability of choosing a hite ball out of Urn II is ; this is TÒElFÓ If the ball transferred is black, then Urn II has 3 hite and 3 black balls, and the probability of choosing a hite ball out of Urn II is ; this is TÒElF Ó All of the information needed has been identified From the table described above, e do the calculations in the folloing order: 1 T ÒE FÓ T ÒElFÓ T ÒFÓ Ð ÑÐ Ñ 2 T ÒE F Ó T ÒElF Ó T ÒF Ó Ð ÑÐ Ñ % ( 3 TÒEÓ TÒE FÓTÒE F Ó % T ÒF EÓ Î % 4 TÒFlEÓ T ÒEÓ (Î ( Example 2-5: Identical tins come from the same egg and hence are of the same sex Fraternal tins have a 50-50 chance of being the same sex Among tins, the probability of a fraternal set is : and an identical set is ; : If the next set of tins are of the same sex, hat is the probability that they are identical?

Solution: Let E be the event the next set of tins are of the same sex, and let F be the event the next sets of tins are identical We are given T ÒElFÓ ß T ÒElF Ó Þ& TÒE FÓ T ÒFÓ ; ß T ÒF Ó : ; Then T ÒFlEÓ TÒEÓ is the probability e are asked to find But T ÒE FÓ T ÒElFÓ T ÒFÓ ;, and T ÒE F Ó T ÒElF Ó T ÒF Ó Þ&: Thus, TÒEÓTÒE FÓTÒE FÓ;Þ&:;Þ&Ð;ÑÞ&Ð;Ñ ; TÒFlEÓ Þ&Ð;Ñ This can be summarized in the folloing table, and ESame sex Fidentical TÒElFÓ (given), prob ; TÒE FÓ T ÒElFÓ T ÒFÓ ; E not same sex TÒElFÓ T ÒElFÓ! F fraternal T ÒElF Ó Þ& (given), T ÒE lf Ó prob :; TÒE FÓ TÒElFÓÞ& T ÒElF Ó T ÒF Ó Þ&Ð ;Ñ Ì TÒEÓTÒE FÓTÒE FÓ;Þ&Ð;ÑÞ&Ð;Ñ TÒE FÓ ; TÒFlEÓ T ÒEÓ Þ&Ð;Ñ Bayes' rule and Bayes' Theorem (extended form): If F ß F ß ÞÞÞß F8 form a partition of the entire probability space W, then TÒE F4Ó TÒE F4Ó TÒElF4Ó TÒF4Ó T ÒF4lEÓ TÒEÓ 8 8 for each 4 ß ß ÞÞÞß 8 T ÒE F Ó T ÒElF Ó T ÒF Ó 3 3 3 3 3 For example, if the F's form a partition of 8 events, e have TÒF leó T ÒF EÓ T ÒEÓ T ÒElF Ó T ÒF Ó T ÒE F ÓT ÒE F ÓT ÒE F Ó T ÒElF Ó T ÒF Ó T ÒElF Ó T ÒF ÓT ÒElF Ó T ÒF ÓT ÒElF Ó T ÒF Ó The relationship in the denominator, T ÒEÓ T ÒElF Ó T ÒF Ó is the general La of Total 8 3 3 3 Probability The values of TÒF Ó are called prior probabilities, and the value of TÒF leó is 4 4 called a posterior probability The basic form of Bayes' rule is just the case in hich the partition consists of to events, F and F The main application of Bayes' rule occurs in the situation in hich the TÒF Óprobabilities are knon and the TÒElF Óprobabilities are knon, and e are asked to find 4 4 TÒF3lEÓ for one of the 3's The series of calculations can be summarized in a table as in the basic form of Bayes' rule This is illustrated in the folloing example

Example 2-6: Three dice have the folloing probabilities of throing a six: :ß;ß<ß respectively One of the dice is chosen at random and thron (each is equally likely to be chosen) A six appeared What is the probability that the die chosen as the first one? Solution: The event a 6 is thron is denoted by 6 T ÒÐ die Ñ Ð' ÑÓ T Ò ' l die Ó T Ò die Ó : TÒ die l' Ó TÒ ' Ó TÒ ' Ó TÒ ' Ó But TÒ'ÓTÒÐ'Ñ Ð die ÑÓTÒÐ'Ñ Ð die ÑÓTÒÐ'Ñ Ð dieñó TÒ ' l dieó TÒ die ÓTÒ' l dieó TÒ die ÓTÒ' l dieó TÒ dieó :;< : : : : ; < Ê TÒ die l' Ó T Ò ' Ó Ð:;<Ñ :;< These calculations can be summarized in the folloing table Die 1, TÐdie 1) Die 2, TÐdie 2) Die 3, TÐdie 3) (given) (given) (given) Toss 6 TÒ6 ldie 1 Ó : (given) TÒ6 ldie 2 Ó ; (given) TÒ6 ldie 3 Ó < (given) TÒ6 die 1 Ó TÒ6 die 2Ó TÒ6 die 3Ó TÒ6 ldie 1Ó TÒdie 1 Ó TÒ6 ldie 2Ó TÒdie 2Ó TÒ6 ldie 3Ó TÒdie 3Ó : ; < TÒ6 Ó: ; < Ð:;<Ñ In terms of Venn diagrams, the conditional probability is the ratio of the shaded area probability in the first diagram to the shaded area probability in the second diagram

In Example 2-6 there is a certain symmetry to the situation and general reasoning can provide a T ÒÐ die Ñ Ð' ÑÓ shortened solution In the conditional probability TÒ die l' Ó TÒ ' Ó, e can think of the denominator as the combination of the three possible ays a 6 can occur, :;<, and e can think of the numerator as the 6 occurring from die 1, ith probability : Then the : conditional probability is the fraction :;< The symmetry involved here is in the assumption that each die as equally likely to be chosen, so there is a chance of any one die being chosen : This factor of cancels in the numerator and denominator of If e had not had this Ð:;<Ñ symmetry, e ould have to apply different eights to the three dice Another example of this sort of symmetry is a variation on Example 2-3 above Suppose that Urn I has 2 hite and 3 black balls and Urn II has 4 hite and 1 black balls An Urn is chosen at random and a ball is chosen The reader should verify using the usual conditional probability ' rules that the probability of choosing a hite is! This can also be seen by noting that if e consider the 10 balls together, 6 of them are hite, so that the chance of picking a hite out of ' the 20 is! This orked because of to aspects of symmetry, equal chance for picking each Urn, and same number of balls in each Urn Independent events E and F : If events Eand F satisfy the relationship T ÒE FÓ T ÒEÓ T ÒFÓ, then the events are said to be independent or stochastically independent or statistically independent The independence of (non-empty) events E and F is equivalent to T ÒElFÓ T ÒEÓ, and also is equivalent to T ÒFlEÓ T ÒFÓ

Example 2-1 continued: A fair six-sided die is tossed E the number tossed is Ÿ Öß ß, F the number tossed is even Öß %ß ', G the number tossed is a or a Öß, H the number tossed doesn't start ith the letters 'f' or 't' Öß ' We have the folloing probabilities: TÒEÓ ß TÒFÓ ß TÒGÓ ß TÒHÓ Þ Events E and F are not independent since ' T ÒE FÓ Á T ÒEÓ T ÒFÓ % We also see that Eand F are not independent because TÒElFÓ Á TÒEÓÞ Also, E and G are not independent, since T ÒE GÓ Á T ÒEÓ T ÒGÓ (also since TÒElGÓ Á TÒEÓ) Events F and G are independent, since T ÒF GÓ ' T ÒFÓ T ÒGÓ (alternatively, TÒFlGÓ TÒFÓ, so that F and G are independent) The reader should check that both E and F are independent of H Mutually independent events E ß E ß ÞÞÞß E8 : If events E ß E ß ÞÞÞß E8 satisfy the relationship T ÒE E â E Ó T ÒE Ó T ÒE ÓâT ÒE Ó T ÒE Ó, then 8 8 3 3 the events are said to be mutually independent 8 Some rules concerning conditional probability and independence are: (i) T ÒE FÓ T ÒFlEÓ T ÒEÓ T ÒElFÓ T ÒFÓ for any events E and F (ii) If TÒE E â E Ó! 8, then T ÒE E â E Ó T ÒE Ó T ÒE le Ó T ÒE le E ÓâT ÒE le E â E 8 8 8 (iii) T ÒE lfó T ÒElFÓ (iv) T ÒE FlGÓ T ÒElGÓ T ÒFlGÓ T ÒE FlGÓ ; properties (iv) and (v) are the same properties satisfied by unconditional events TÒE FÓ TÒEÓ (v) if E F then T ÒElFÓ TÒFÓ TÒFÓ, and T ÒFlEÓ (vi) if E and F are independent events then E and F are independent events, Eand F are independent events, and E and F are independent events Ó (vii) since TÒgÓ TÒg EÓ! TÒgÓ TÒEÓ for any event E, it follos that g is independent of any event E

Example 2-7: Suppose that events Eand F are independent Find the probability, in terms of TÒEÓand TÒFÓ, that exactly one of the events Eand F occurs Solution: T Òexactly one of E and FÓ T ÒÐE F Ñ ÐE FÑÓ Since E F and F E are mutually exclusive, it follos that TÒexactly one of Eand FÓ TÒE F ÓTÒE FÓ Since Eand F are independent, it follos that E and F are also independent, as are F and E Then TÒÐE FÑ ÐE FÑÓTÒEÓ TÒFÓTÒFÓ TÒEÓ T ÒEÓÐ T ÒFÓÑ T ÒFÓÐ T ÒEÓÑ T ÒEÓ T ÒFÓ T ÒEÓ T ÒFÓ Example 2-8: 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 irons ( M), 55% have electric toasters ( X), 50% have ( MV), 40% have ( VX), 30% have ( MX), and 20% have all three Find the folloing proportions (i) Of those households that have a toaster, find the proportion that also have a radio (ii) Of those households that have a toaster but no iron, find the proportion that have a radio Solution: This is a continuation of Example 1-3 given earlier in the study guide The diagram belo deconstructs the three events (i) This is TÒVlXÓ The language of those households that have a toaster means, given that the household has a toaster, so e are being asked for a conditional probability TÒV XÓ Þ% ) Then, TÒVlXÓ TÒXÓ Þ&& Þ TÒV X M Ó Þ % (ii) This TÒVlX M Ó TÒX MÓ Þ& & Þ

Example 2-9: In a survey of 94 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 folloing proportions (i) Of those ho took Math, the proportion ho took neither English nor Chemistry, (ii) Of those ho took English or Math, the proportion ho also took Chemistry Solution: The folloing diagram illustrates ho the numbers of students can be deconstructed We calculate proportion of the numbers in the various subsets (i) 56 students took Math, and 8 of them took neither English nor Chemistry TÐI G QÑ ) TÐI G lqñ T ÐQÑ &' ( (ii) 82 ( )%')'! in I QÑstudents took English or Math (or both), and 30 of them ( %'! in ÐI QÑ G) also took Chemistry T ÒG ÐI QÑÓ! & TÐGlI QÑ T ÐI QÑ ) % Þ