Actex TAM-ACT-18SSMP Revised Sample

SECTION 1 - PRELIMINARY REVIEW - PROBABILITYSTAM-1SECTION 1 - PRELIMINARY REVIEW - PROBABILITYBasic Probability, Conditional Probability and IndependenceA significant part of the STAM Exam involves probability and statistical methods applied to variousaspects of loss modeling and model estimation. A good background in probability and statistics isnecessary to fully understand models and the modeling that is done. In this section of the study guide, wewill review fundamental probability rules.1.1 Basic Probability ConceptsSample point and probability spaceA sample point is the simple outcome of a random experiment. The probability space (also called samplespace) is the collection of all possible sample points related to a specified experiment. When theexperiment is performed, one of the sample points will be the outcome. An experiment could beobserving the loss that occurs on an automobile insurance policy during the course of one year, orobserving the number of claims arriving at an insurance office in one week. The probability space is the"full set" of possible outcomes of the experiment. In the case of the automobile insurance policy, it wouldbe the range of possible loss amounts that could occur during the year, and in the case of the insuranceoffice weekly number of claims, the probability space would be the set of integers Ö!ß "ß #ß ÞÞÞ .EventAny collection of sample points, or any subset of the probability space is referred to as an event. We say"event E has occurred" if the experimental outcome was one of the sample points in E.Union of events E and FE F denotes the union of events E and F , and consists of all sample points that are in either E or F .A BUnion of events E" ß E# ß ÞÞÞß E88E" E# â E8 œ E3 denotes the union of the events E" ß E# ß ÞÞÞß E8 , and consists of all sample3œ"points that are in at least one of the E3 's. This definition can be extended to the union of infinitely manyevents.Actex LearningSOA Exam STAM - Short-Term Actuarial Mathematics

STAM-2SECTION 1 - PRELIMINARY REVIEW - PROBABLITYIntersection of events E" ß E# ß ÞÞÞß E88E" E# â E8 œ E3 denotes the intersection of the events E" ß E# ß ÞÞÞß E8 , and consists of all3œ"sample points that are simultaneously in all of the E3 's.A BMutually exclusive events E" ß E# ß ÞÞÞß E8Two events are mutually exclusive if they have no sample points in common, or equivalently, if they haveempty intersection. Events E" ß E# ß ÞÞÞß E8 are mutually exclusive if E3 E4 œ g for all 3 Á 4, where gdenotes the empty set with no sample points. Mutually exclusive events cannot occur simultaneously.Exhaustive events F" ß F# ß ÞÞÞß F8If F" F# â F8 œ W , the entire probability space, then the events F" ß F# ß ÞÞÞß F8 are referred to asexhaustive events.Complement of event EThe 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 Ew or E- and is equal to ÖB À B  E . When the underlying randomexperiment is performed, to say that the complement of E has occurred is the same as saying that E hasnot occurred.Subevent (or subset) E of event FIf event F contains all the sample points in event E, then E is a subevent of F , denoted E § F . Theoccurrence of event E implies that event F has occurred.Partition of event E8Events G" ß G# ß ÞÞÞß G8 form a partition of event E if E œ G3 and the G3 's are mutually exclusive.3œ"DeMorgan's Laws(i) ÐE FÑw œ Ew F w , to say that E F has not occurred is to say that E has not occurredand F has not occurred ; this rule generalizes to any number of events;w88 E3 œ ÐE" E# â E8 Ñw œ E"w E#w â E8w œ E3w3œ"3œ"(ii) ÐE FÑw œ Ew F w , to say that E F has not occurred is to say that either E has notoccurred or F has not occurred (or both have not occurred) ; this rule generalizes to anyw88number of events, E3 œ ÐE" E# â E8 Ñw œ E"w E#w â E8w œ E3w3œ"3œ"Actex LearningSOA Exam STAM - Short-Term Actuarial Mathematics

SECTION 1 - PRELIMINARY REVIEW - PROBABILITYSTAM-3Indicator function for event EThe function ME ÐBÑ œ " if B E! if BÂEis the indicator function for event E, where B denotes a sample point.ME ÐBÑ is 1 if event E has occurred.Some important rules concerning probability are given below.(i) T ÒWÓ œ " if W is the entire probability space (when the underlying experiment isperformed, some outcome must occur with probability 1).(ii) T ÒgÓ œ ! (the probability of no face turning up when we toss a die is 0).(iii) If events E" ß E# ß ÞÞÞß E8 are mutually exclusive (also called disjoint) then88T Ò E3 Ó œ T ÒE" E# â E8 Ó œ T ÒE" Ó T ÒE# Ó â T ÒE8 Ó œ T ÒE3 Ó3œ"3œ"(1.1)This extends to infinitely many mutually exclusive events.(iv) For any event E, ! Ÿ T ÒEÓ Ÿ "(v) If E § F then T ÒEÓ Ÿ T ÒFÓ(vi) For any events E, F and G , T ÒE FÓ œ T ÒEÓ T ÒFÓ T ÒE FÓ(1.2)(vii) For any event E, T ÒEw Ó œ " T ÒEÓ(1.3)(viii) For any events E and F , T ÒEÓ œ T ÒE FÓ T ÒE F w Ó(1.4)8(ix) For exhaustive events F" ß F# ß ÞÞÞß F8 , T Ò F3 Ó œ "3œ"(1.5)If F" ß F# ß ÞÞÞß F8 are exhaustive and mutually exclusive, they form a partition of the entireprobability space, and for any event E,T ÒEÓ œ T ÒE F" Ó T ÒE F# Ó â T ÒE F8 Ó œ T ÒE F3 Ó8(1.6)3œ"(x) The words "percentage" and "proportion" are used as alternatives to "probability". As an example, ifwe 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. Asanother example, suppose that we are tossing a fair die. In the long-run frequency interpretation ofprobability, to say that the probability of tossing a 1 is "' is the same as saying that if we repeatedlytoss the die, the proportion of tosses that are 1's will approach "' .Actex LearningSOA Exam STAM - Short-Term Actuarial Mathematics

STAM-4SECTION 1 - PRELIMINARY REVIEW - PROBABLITY1.2 Conditional Probability and Independence of EventsConditional probability arises throughout the STAM Exam material. It is important to be familiar andcomfortable with the definitions and rules of conditional probability.Conditional probability of event E given event FIf T ÐFÑ !, then T ÐElFÑ œT ÐE FÑT ÐFÑis the conditional probability that event E occurs given thatevent F has occurred. By rewriting the equation we get T ÐE FÑ œ T ÐElFÑ † T ÐFÑ.Partition of a Probability SpaceEvents F" ß F# ß ÞÞÞß F8 are said to form a partition of a probability space W if(i) F" F# â F8 œ W and (ii) F3 F4 œ g for any pair with 3 Á 4.A partition is a disjoint collection of events which combines to be the full probability space.A simple example of a partition is any event F and its complement F w .If E is any event in probability space W and ÖF" ß F# ß ÞÞÞß F8 is a partition of probability space W , thenT ÐEÑ œ T ÐE F" Ñ T ÐE F# Ñ â T ÐE F8 Ñ.A special case of this rule is T ÐEÑ œ T ÐE FÑ T ÐE F w Ñ for any two events E and F .B2B1A B A BAA B1A B2A B3B3BA B4B4Bayes rule and Bayes TheoremFor any events E and F with T ÐEÑ !, T ÐFlEÑ œT ÐElFÑ‚T ÐFÑT ÐEÑ(1.7)If F" ß F# ß ÞÞÞß F8 form a partition of the entire sample space W , thenT ÐF4 lEÑ œT ÐElF4 Ñ‚T ÐF4 Ñ T ÐElF3 Ñ‚T ÐF3 Ñ8for each 4 œ "ß #ß ÞÞÞß 8(1.8)3œ"The values of T ÐF4 Ñ are called prior probabilities, and the value of T ÐF4 lEÑ is called a posteriorprobability. Variations on this rule are very important in Bayesian credibility.Actex LearningSOA Exam STAM - Short-Term Actuarial Mathematics

SECTION 1 - PRELIMINARY REVIEW - PROBABILITYSTAM-5Independent events E and FIf events E and F satisfy the relationship T ÐE FÑ œ T ÐEÑ ‚ T ÐFÑ, then the events are said to beindependent or stochastically independent or statistically independent. The independence of (non-empty)events E and F is equivalent to T ÐElFÑ œ T ÐEÑ or T ÐFlEÑ œ T ÐFÑ.Mutually independent events E" ß E# ß ÞÞÞß E8The events are mutually independent if(i) for any E3 and E4 , T ÐE3 E4 Ñ œ T ÐE3 Ñ ‚ T ÐE4 Ñ, and(ii) for any E3 , E4 and E5 , T ÐE3 E4 E5 Ñ œ T ÐE3 Ñ ‚ T ÐE4 Ñ ‚ T ÐE5 Ñ,and so on for any subcollection of the events, including all events:T ÐE" E# â E8 Ñ œ T ÐE" Ñ ‚ T ÐE# Ñ ‚ â ‚ T ÐE8 Ñ œ T ÐE3 Ñ8(1.9)3œ"Here are some rules concerning conditional probability and independence. These can beverified in a fairly straightforward way from the definitions given above.(i)T ÐE FÑ œ T ÐEÑ T ÐFÑ T ÐE FÑ for any events E and F(ii) T ÐE FÑ œ T ÐFlEÑ ‚ T ÐEÑ œ T ÐElFÑ ‚ T ÐFÑ for any events E and F(1.10)(1.11)(iii) If F" ß F# ß ÞÞÞß F8 form a partition of the sample space W , then for any event ET ÐEÑ œ T ÐE F3 Ñ œ T ÐElF3 Ñ ‚ T ÐF3 Ñ883œ"3œ"(1.12)As a special case, for any events E and F , we haveT ÐEÑ œ T ÐE FÑ T ÐE F w Ñ œ T ÐElFÑ ‚ T ÐFÑ T ÐElFw Ñ ‚ T ÐFw Ñ(1.13)(iv) If T ÐE" E# â E8 " Ñ !, thenT ÐE" E# â E8 Ñ œ T ÐE" Ñ ‚ T ÐE# lE" Ñ ‚ T ÐE lE" E# Ñ ‚ â ‚ T ÐE8 lE" E# â E8 " Ñ(v) T ÐEw Ñ œ " T ÐEÑ and T ÐEw lFÑ œ " T ÐElFÑ(vi) if E § F then T ÐElFÑ œT ÐE FÑT ÐFÑ(1.14)T ÐEÑœ T ÐFÑ , and T ÐFlEÑ œ "(vii) if E and F are independent events then Ew and F are independent events, E and F w are independentevents, and Ew and F w are independent events(viii) since T ÐgÑ œ T Ðg EÑ œ ! œ T ÐgÑ † T ÐEÑ for any event E, it follows that g is independent ofany event EActex LearningSOA Exam STAM - Short-Term Actuarial Mathematics

STAM-6SECTION 1 - PRELIMINARY REVIEW - PROBABLITYExample 1-1:Suppose a fair six-sided die is tossed. We define the following events:E œ "the number tossed is Ÿ " œ Ö"ß #ß , F œ "the number tossed is even" œ Ö#ß %ß ' G œ "the number tossed is a " or a #" œ Ö"ß # H œ "the number tossed doesn't start with the letters 'f' or 't'" œ Ö"ß ' The conditional probability of E given F isT ÐElFÑ œT ÐÖ"ß#ß Ö#ß%ß' ÑT ÐÖ#ß%ß' ÑT ÐÖ# Ñ"Î'œ T ÐÖ#ß%ß' Ñ œ "Î# œ " .Events E and F are not independent, since "' œ T ÐE FÑ Á T ÐEÑ ‚ T ÐFÑ œ "# † "# œ "% ,or alternatively, events E and F are not independent since T ÐElFÑ Á T ÐEÑ.T ÐElGÑ œ " Á "# œ T ÐEÑ, so that E and G are not independent.T ÐFlGÑ œ "# œ T ÐFÑ , so that F and G are independent(alternatively, T ÐF GÑ œ T ÐÖ# Ñ œ "' œ "# † " œ T ÐFÑ † T ÐGÑ ).It is not difficult to check that both E and F are independent of H. IMPORTANT NOTE: The following manipulation of event probabilities arises from time to time:T (E) œ T (ElF ) † T ÐFÑ T (ElF w ) ‚ T ÐF w Ñ. If we know the conditional probabilities for event Egiven some other event F and its complement F w , and if we know the (unconditional) probability of eventF , then we can find the probability of event E. One of the important aspects of applying this relationshipis the determination of the appropriate events E and F .Example 1-2:Urn I contains 2 white and 2 black balls and Urn II contains 3 white and 2 black balls. An Urn is chosen atrandom, and a ball is randomly selected from that Urn. Find the probability that the ball chosen is white.Solution:Let F be the event that Urn I is chosen and F w is the event that Urn II is chosen. The implicit assumptionis that both Urns are equally likely to be chosen (this is the meaning of "an Urn is chosen at random").Therefore, T ÐFÑ œ "# and T ÐF w Ñ œ "# . Let E be the event that the ball chosen in white. If we know thatUrn I was chosen, then there is "# probability of choosing a white ball (2 white out of 4 balls, it isassumed that each ball has the same chance of being chosen); this can be described as T ÐElFÑ œ "# .In a similar way, if Urn II is chosen, then T ÐElF w Ñ œ & (3 white out of 5 balls). We can now apply therelationship described prior to this example. T ÐE FÑ œ T ÐElFÑ ‚ T ÐFÑ œ Ð "# ÑÐ "# Ñ œ "% , and T ÐE F w Ñ œ T ÐElF w Ñ ‚ T ÐF w Ñ œ Ð & ÑÐ "# Ñ œ "!. Finally, ""T ÐEÑ œ T ÐE FÑ T ÐE F w Ñ œ "% "!œ #!.The order of calculations can be summarized in the following tableFE1. T ÐE FÑ œ T ÐElFÑ ‚ T ÐFÑFw2. T ÐE F w Ñ œ T ÐElF w Ñ ‚ T ÐF w Ñ3. T ÐEÑ œ T ÐE FÑ T ÐE F w ÑActex Learning SOA Exam STAM - Short-Term Actuarial Mathematics

SECTION 1 - PRELIMINARY REVIEW - PROBABILITYSTAM-7Example 1-3:Urn I contains 2 white and 2 black balls and Urn II contains 3 white and 2 black balls. One ball is chosenat random from Urn I and transferred to Urn II, and then a ball is chosen at random from Urn II. The ballchosen from Urn II is observed to be white. Find the probability that the ball transferred from Urn I toUrn II was white.Solution:Let F denote the event that the ball transferred from Urn I to Urn II was white and let E denote the eventthat the ball chosen from Urn II is white. We are asked to find T ÐFlEÑ .From the simple nature of the situation (and the usual assumption of uniformity in such a situation,meaning all balls are equally likely to be chosen from Urn I in the first step), we haveT ÐFÑ œ "# (2 of the 4 balls in Urn I are white), and by implication, it follows that T ÒF w Ó œ "# .If the ball transferred is white, then Urn II has 4 white and 2 black balls, and the probability ofchoosing a white ball out of Urn II is # ; this is T ÐElFÑ œ # .If the ball transferred is black, then Urn II has 3 white and 3 black balls, and the probability ofchoosing a white ball out of Urn II is "# ; this is T ÐElF w Ñ œ "# .All of the information needed has been identified. We do calculations in the following order:1. T ÒE FÓ œ T ÒElFÓ ‚ T ÒFÓ œ Ð # ÑÐ "# Ñ œ " 2. T ÒE F w Ó œ T ÒElF w Ó ‚ T ÒF w Ó œ Ð "# ÑÐ "# Ñ œ "%(3. T ÒEÓ œ T ÒE FÓ T ÒE F w Ó œ " "% œ "#T ÒF EÓ"Î 4. T ÒFlEÓ œ T ÒEÓ œ (Î"# œ %( Example 1-4:Three dice have the following probabilities of throwing a "six": : ß ; ß ßrespectively. One of the dice is chosen at random and thrown (each is equally likely to be chosen). A"six" appeared. What is the probability that the die chosen was the first one?Solution:The event " a 6 is thrown" is denoted by "6"T Òdie "l"'"Ó œButT ÒÐdie "Ñ Ð"'"ÑÓT Ò"'" ÓœT Ò"'"ldie "Ó‚T Òdie "ÓT Ò"'"Ó:† "œ T Ò"' "ÓT Ò"'"Ó œ T ÒÐ"'"Ñ Ðdie "ÑÓ T ÒÐ"'"Ñ Ðdie #ÑÓ T ÒÐ"'"Ñ Ðdie ÑÓœ T Ò"'"ldie "Ó ‚ T Òdie "Ó T Ò"'"ldie #Ó ‚ T Òdie #Ó T Ò"'"ldie Ó ‚ T Òdie Óœ :‚Actex Learning" ;‚" ‚" œ: ; :‚ ":‚ ": p T Òdie "l"'"Ó œ T Ò"' "Ó œ Ð: ; ц" œ : ; SOA Exam STAM - Short-Term Actuarial Mathematics

STAM-8Actex LearningSECTION 1 - PRELIMINARY REVIEW - PROBABLITYSOA Exam STAM - Short-Term Actuarial Mathematics

SECTION 1 PROBLEM SETSTAM-9SECTION 1 PROBLEM SETPreliminary Review - Probability1.A survey of 1000 people determines that 80% like walking and 60% like biking, and all like at leastone of the two activities. How many people in the survey like biking but not walking?A) 02.B) .1C) .2D) .3E) .4A life insurer classifies insurance applicants according to the following attributes:Q - the applicant is maleL - the applicant is a homeownerOut of a large number of applicants the insurer has identified the following information:40% of applicants are male, 40% of applicants are homeowners and20% of applicants are female homeowners.Find the percentage of applicants who are male and do not own a home.A) .13.B) .2C) .3D) .4E) .5Let Eß Fß G and H be events such that F œ Ew ß G H œ g, andT ÒEÓ œ "% , T ÒFÓ œ % , T ÒGlEÓ œ "# , T ÒGlFÓ œ % , T ÒHlEÓ œ "% , T ÒHlFÓ œ ")Calculate T ÒG HÓ .&A) #4.D) % E) "B) .05C) .10D) .15E) .20A test for a disease correctly diagnoses a diseased person as having the disease with probability .85.The test incorrectly diagnoses someone without the disease as having the disease with a probabilityof .10. If 1% of the people in a population have the disease, what is the chance that a person fromthis population who tests positive for the disease actually has the disease?A) Þ!!)&6.#(C) #You are given that T ÒEÓ œ Þ& and T ÒE FÓ œ Þ(.Actuary 1 assumes that E and F are independent and calculates T ÒFÓ based on that assumption.Actuary 2 assumes that E and F mutually exclusive and calculates T ÒFÓ based on that assumption.Find the absolute difference between the two calculations.A) 05.B) %"B) Þ!(*"C) Þ"!(&D) Þ"&!!E) Þ*!!!Two bowls each contain 5 black and 5 white balls. A ball is chosen at random from bowl 1 and putinto bowl 2. A ball is then chosen at random from bowl 2 and put into bowl 1. Find the probabilitythat bowl 1 still has 5 black and 5 white balls.A) # Actex LearningB) &'C) ""D) "#'E) " SOA Exam STAM - Short-Term Actuarial Mathematics

STAM-107.People passing by a city intersection are asked for the month in which they were born. It is assumedthat the population is uniformly divided by birth month, so that any randomly passing person has anequally likely chance of being born in any particular month. Find the minimum number of peopleneeded so that the probability that no two people have the same birth month is less than .5.A) 28.C) 4D) 5E) 6B) .3C) .5D) .7E) .9In the game show "Let's Make a Deal", a contestant is presented with 3 doors. There is a prizebehind one of the doors, and the host of the show knows which one. When the contestant makes achoice of door, at least one of the other doors will not have a prize, and the host will open a door(one not chosen by the contestant) with no prize. The contestant is given the option to change hischoice after the host shows the door without a prize. If the contestant switches doors, what is theprobability that he gets the door with the prize?A) !10.B) 3In a T-maze, a laboratory rat is given the choice of going to the left and getting food or going to theright and receiving a mild electric shock. Assume that before any conditioning (in trial number 1)rats are equally likely to go the left or to the right. After having received food on a particular trial,the probability of going to the left and right become .6 and .4, respectively on the following trial.However, after receiving a shock on a particular trial, the probabilities of going to the left and righton the next trial are .8 and .2, respectively. What is the probability that the animal will turn left ontrial number 2?A) .19.SECTION 1 PROBLEM SETB)"'C)" D)"#E)# A supplier of a testing device for a type of component claim

Actex Learning SOA STAM Exam - Short Term Actuarial Mathematics INTRODUCTORY COMMENTS This study guide is designed to help in the preparation for the Society of Actuaries STAM Exam. The first part of this manual consists of a summary of notes, illustrative examples and problem sets with detailed solutions. The second part consists of 5 practice .