ACTEX. SOA Exam P Study Manual. With StudyPlus Edition Samuel A. Broverman, Ph.D., ASA. ACTEX Learning Learn Today. Lead Tomorrow.

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1 ACTEX SOA Exam P Study Manual With StudyPlus + StudyPlus + gives you digital access* to: Flashcards & Formula Sheet Actuarial Exam & Career Strategy Guides Technical Skill elearning Tools Samples of Supplemental Textbooks And more! *See inside for keycode access and login instructions 2016 Edition Samuel A. roverman, Ph.D., ASA ACTEX Learning Learn Today. Lead Tomorro.

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3 ACTEX SOA Exam P Study Manual 2016 Edition Samuel A. roverman, Ph.D., ASA ACTEX Learning Ne Hartford, Connecticut

4 ACTEX Learning Learn Today. Lead Tomorro. Actuarial & Financial Risk Resource Materials Since 1972 Copyright 2017 SRooks, Inc. ISN: Printed in the United States of America. No portion of this ACTEX Study Manual may be reproduced or transmitted in any part or by any means ithout the permission of the publisher.

5 Your Opinion is Important to Us ACTEX is eager to provide you ith helpful study material to assist you in gaining the necessary knoledge to become a successful actuary. In turn e ould like your help in evaluating our manuals so e can help you meet that end. We invite you to provide us ith a critique of this manual by sending this form to us at your convenience. We appreciate your time and value your input. Publication: ACTEX P Study Manual, 2016 Edition I found Actex by: (Check one) A Professor School/Internship Program Employer Friend Facebook/Titter In preparing for my exam I found this manual: (Check one) Very Good Good Satisfactory Unsatisfactory I found the folloing helpful: I found the folloing problems: (Please be specific as to area, i.e., section, specific item, and/or page number.) To improve this manual I ould: Name: Address: Phone: (Please provide this information in case clarification is needed.) Send to: Stephen Camilli ACTEX Learning P.O. ox 715 Ne Hartford, CT Or visit our ebsite at.actexmadriver.com to complete the survey on-line. Click on the Send Us Feedback link to access the online version. You can also your comments to Support@ActexMadRiver.com.

6 ACTEX Exam Prep Online Courses! Your Path To Exam Success Full time students are eligible for a 50% discount on ACTEX Exam Preparation courses. Learn from the experts and feel confident on exam day Comprehensive instructor support, video lectures, and suggested exercises/solutions, supplemental resources Responsive, student-focused instructors ith engaging style and helpful support make preparing for your exam efficient and enjoyable Weekly timed practice tests ith video solutions Live discussion forums and real-time chat sessions Emphasizes problem-solving techniques Comparable to a one-semester, undergraduate-level course Students ill have one-on-one support from the instructor(s) from the start date of the course, through the indicated exam sitting. Student Testimonials I took the P exam today and passed! The course has been extremely effective and I anted to thank you for all your help! The instructor as very knoledgeable about all subjects covered. He as also quick to respond hich led to less time aiting for an anser before moving forard. The video lectures ere extremely helpful as ere the office hours, but the problems and solutions for the book ere the most helpful aspects. Rich Oens as a great instructor. He as very knoledgeable, helpful, available, and any other positive adjective you could think of. Visit.ActexMadRiver.com for specific course information and required materials.

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9 iii INTRODUCTORY COMMENTS TALE OF CONTENTS SECTION 0 - REVIEW OF ALGERA AND CALCULUS Set Theory 1 Graphing an Inequality in To Dimensions 9 Properties of Functions 10 Limits and Continuity 14 Differentiation 15 Integration 18 Geometric and Arithmetic Progressions 24 Problem Set 0 and Solutions 25 SECTION 1 - ASIC PROAILITY CONCEPTS Probability Spaces and Events 37 De Morgan's Las 38 Probability 41 Problem Set 1 and Solutions 51 SECTION 2 - CONDITIONAL PROAILITY AND INDEPENDENCE Definition of Conditional Probability 59 ayes' Rule, ayes' Theorem and the La of Total Probability 62 Independent Events 68 Problem Set 2 and Solutions 75 SECTION 3 - COMINATORIAL PRINCIPLES Permutations and Combinations 101 Problem Set 3 and Solutions 107 SECTION 4 - RANDOM VARIALES AND PROAILITY DISTRIUTIONS Discrete Random Variable 117 Continuous Random Variable 119 Mixed Distribution 121 Cumulative Distribution Function 123 Independent Random Variables 125 Problem Set 4 and Solutions 133 SECTION 5 - EXPECTATION AND OTHER DISTRIUTION PARAMETERS Expected Value 143 Moments of a Random Variable 145 Variance and Standard Deviation 146 Moment Generating Function 148 Percentiles, Median and Mode 150 Chebyshev's Inequality 152 Problem Set 5 and Solutions 161

10 iv SECTION 6 - FREQUENTLY USED DISCRETE DISTRIUTIONS Discrete Uniform Distribution 177 inomial Distribution 178 Poisson Distribution 181 Geometric Distribution 183 Negative inomial Distribution 185 Hypergeometric Distribution 187 Multinomial Distribution 188 Summary of Discrete Distributions 189 Problem Set 6 and Solutions 191 SECTION 7 - FREQUENTLY USED CONTINUOUS DISTRIUTIONS Continuous Uniform Distribution 207 Normal Distribution 208 Approximating a Distribution Using a Normal Distribution 210 Exponential Distribution 214 Gamma Distribution 217 Summary of Continuous Distributions 219 Problem Set 7 and Solutions 221 SECTION 8 - JOINT, MARGINAL, AND CONDITIONAL DISTRIUTIONS Definition of Joint Distribution 233 Expectation of a Function of Jointly Distributed Random Variables 237 Marginal Distributions 238 Independence of Random Variables 241 Conditional Distributions 242 Covariance and Correlation eteen Random Variables 246 Moment Generating Function for a Joint Distribution 248 ivariate Normal Distribution 248 Problem Set 8 and Solutions 255 SECTION 9 - TRANSFORMATIONS OF RANDOM VARIALES Distribution of a Transformation of 291 Distribution of a Transformation of Joint Distribution of and 292 Distribution of a Sum of Random Variables, Covolution Method 293 Distribution of the Maximum or Minimum of Independent 297 Order Statistics 298 Mixtures of Distributions 301 Problem Set 9 and Solutions 303 SECTION 10 - RISK MANAGEMENT CONCEPTS Loss Distributions and Insurance 325 Insurance Policy Deductible 326 Insurance Policy Limit 328 Proportional Insurance 329 Problem Set 10 and Solutions 339

11 v TALE FOR THE NORMAL DISTRIUTION PRACTICE EXAM PRACTICE EXAM PRACTICE EXAM PRACTICE EXAM PRACTICE EXAM PRACTICE EXAM PRACTICE EXAM PRACTICE EXAM 8 491

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13 vii INTRODUCTORY COMMENTS This study guide is designed to help in the preparation for the Society of Actuaries Exam P. The study manual is divided into to main parts. The first part consists of a summary of notes and illustrative examples related to the material described in the exam catalog as ell as a series of problem sets and detailed solutions related to each topic. Many of the examples and problems in the problem sets are taken from actual exams (and from the sample question list posted on the SOA ebsite). The second part of the study manual consists of eight practice exams, ith detailed solutions, hich are designed to cover the range of material that ill be found on the exam. The questions on these practice exams are not from old Society exams and may be somehat more challenging, on average, than questions from previous actual exams. eteen the section of notes and the section ith practice exams I have included the normal distribution table provided ith the exam. I have attempted to be thorough in the coverage of the topics upon hich the exam is based. I have been, perhaps, more thorough than necessary on a couple of topics, particularly order statistics in Section 9 of the notes and some risk management topics in Section 10 of the notes. Section 0 of the notes provides a brief revie of a fe important topics in calculus and algebra. This manual ill be most effective, hoever, for those ho have had courses in college calculus at least to the sophomore level and courses in probability to the sophomore or junior level. If you are taking the Exam P for the first time, be aare that a most crucial aspect of the exam is the limited time given to take the exam (3 hours). It is important to be able to ork very quickly and accurately. Continual drill on important concepts and formulas by orking through many problems ill be helpful. It is also very important to be disciplined enough hile taking the exam so that an inordinate amount of time is not spent on any one question. If the formulas and reasoning that ill be needed to solve a particular question are not clear ithin 2 or 3 minutes of starting the question, it should be abandoned (and returned to later if time permits). Using the exams in the second part of this study manual and simulating exam conditions ill also help give you a feeling for the actual exam experience. If you have any comments, criticisms or compliments regarding this study guide, please contact the publisher, ACTEX, or you may contact me directly at the address belo. I apologize in advance for any errors, typographical or otherise, that you might find, and it ould be greatly appreciated if you bring them to my attention. Any errors that are found ill be posted in an errata file at the ACTEX ebsite,.actexmadriver.com. It is my sincere hope that you find this study guide helpful and useful in your preparation for the exam. I ish you the best of luck on the exam. Samuel A. roverman June, 2016 Department of Statistics University of Toronto 2brove@rogers.com or sam@utstat.toronto.edu.sambroverman.com.brovermusic.com

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15 NOTES, EXAMPLES AND PROLEM SETS

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17 SECTION 0 - REVIEW OF ALGERA AND CALCULUS 1 SECTION 0 - REVIEW OF ALGERA AND CALCULUS In this introductory section, a fe important concepts that are preliminary to probability topics ill be revieed. The concepts revieed are set theory, graphing an inequality in to dimensions, properties of functions, differentiation, integration and geometric series. Students ith a strong background in calculus ho are familiar ith these concepts can skip this section. SET THEORY A set is a collection of elements. The phrase " is an element of E" is denoted by E, and " is not an element of E" is denoted by  E. Subset of a set: E F means that each element of the set E is an element of the set F. F may contain elements hich are not in E, but E is totally contained ithin F. For instance, if E is the set of all odd, positive integers, and F is the set of all positive integers, then EœÖ"ß ß &ß ÞÞÞ and F œö"ß ß ß ÞÞÞ. The notation E F also denotes that Eis a subset of Fbut that Emay be equal to F. For these to sets it is easy to see that E F, since any member of E(any odd positive integer) is a member of F (is a positive integer). The Venn diagram belo illustrates E as a subset of F. A Union of sets: E F is the set of all elements in either E or F (or both). E FœÖl Eor F E F If Eis the set of all positive even integers ( EœÖß %ß 'ß )ß"!ß "ß ÞÞÞ ) and Fis the set of all positive integers hich are multiples of 3 ( F œöß 'ß *ß "ß ÞÞÞ ), then E F œ Ößß%ß'ß)ß*ß"!ß"ßÞÞÞ is the set of positive integers hich are either multiples of 2 or are multiples of 3 (or both).

18 2 SECTION 0 - REVIEW OF ALGERA AND CALCULUS Intersection of sets: E F is the set of all elements that are in both E and F. E FœÖl Eand F E F If E is the set of all positive even integers and F is the set of all positive integers hich are a multiple of 3, then E F œö' ß "ß ÞÞÞ is the set of positive integers hich are a multiple of 6. The elements of E F must satisfy the properties of both Eand F. In this example, that means an element of E F must be a multiple of 2 and must also be a multiple of 3, and therefore must be a multiple of 6. The complement of the set F : The complement of F consists of all elements not in F, and is denoted F ß F or µ F. F œ Öl  F. When referring to the complement of a set, it is usually understood that there is some "full set", and the complement of F consists of the elements of the full set hich are not in F. For instance, if Fis the set of all positive even integers, and if the "full set" is the set of all positive integers, then F consists of all positive odd integers. The set difference of "set E minus F" is EFœÖl E+8.ÂF and consists of all elements that are in E but not in F. Note that EFœE F. EF can also be described as the set that results hen the intersection E F is removed from E. A A Example 0-1: Verify the folloing set relationships (DeMorgan's Las): (i) ÐE FÑ œ E F (the complement of the union of E and F is the intersection of the complements of E and F) (ii) ÐE FÑ œ E F (the complement of the intersection of E and F is the union of the complements of E and F)

19 SECTION 0 - REVIEW OF ALGERA AND CALCULUS 3 Solution: (i) Since the union of E and F consists of all points in either E or F, any point not in E F is in neither E nor F, and therefore must be in both the complement of E and the complement of F; this is the intersection of E and F. The reverse implication holds in a similar ay; if a point is in the intersection of E and F then it is not in E and it is not in Fß so it is not in E F, and therefore it is in ÐE FÑ. Therefore, ÐE FÑ and E F consist of the same collection of points, they are the same set. A A A A ( A) A (ii) The solution is very similar to (i). A ( A) A Empty set: The empty set is the set that contains no elements, and is denoted g. It is also referred to as the null set. Sets E and F are called disjoint sets if E F œ g (they have no elements in common). Relationships involving sets: 1. E FœF Eà E FœF Eà E EœEà E EœE 2. If E F, then E FœFand E FœE (this can be seen from the Venn diagram in the paragraph above describing subset) 3. For any set E, 9 E (the empty set is a subset of any other set E) 4. E 9 œ Eà E 9 œ 9à E9 œ E 5. E ÐF GÑ œ ÐE FÑ ÐE GÑ 6. E ÐF GÑ œ ÐE FÑ ÐE GÑ 7. For any sets Eand F, E F E E F and E F F E F 8. ÐE FÑ œe F and ÐE FÑ œe F

20 4 SECTION 0 - REVIEW OF ALGERA AND CALCULUS An important rule (that follos from point 4 above) is the folloing. For any to sets E and F, e have E œ ÐE FÑ ÐE F Ñ. A A Related to this is the property that if a finite set is made up of the union of disjoint sets, then the number of elements in the union is the sum of the numbers in each of the component sets. For a finite set W, e define 8ÐWÑ to be the number of elements in W. To useful relationships for counting elements in a set are 8ÐEÑ œ 8ÐE FÑ 8ÐE F Ñ (true since E Fand E F are disjoint), and 8ÐE FÑ œ 8ÐEÑ 8ÐFÑ 8ÐE FÑ (cancels the double counting of E F). A This rule can be extended to three sets, 8ÐE F GÑ œ 8ÐEÑ 8ÐFÑ 8ÐGÑ 8ÐE FÑ8ÐE GÑ8ÐF GÑ 8ÐE F GÑ. The main application of set algebra is in a probability context in hich e use set algebra to describe events and combinations of events (this appears in the next section of this study guide). An understanding of set algebra and Venn diagram representations can be quite helpful in describing and finding event probabilities. Example 0-2: Suppose that the "total set" W consists of the possible outcomes that can occur hen tossing a six-faced die. Then W œ Ö"ß ß ß %ß &ß '. We define the folloing subsets of W: E œ Ö"ß ß (a number less than % is tossed), F œ Öß %ß ' (an even number is tossed), G œ Ö% (a 4 is tossed). Then E F œ Ö"ß ß ß %ß ' ; E F œ Ö ; E and G are disjoint since E G œ g ; G F ; E œ Ö%ß &ß ' (complement of E) à F œ Ö"ß ß & ; E F œ Ö"ß ß ß %ß ' ; and ÐE FÑ œö& œe F (this illustrates one of DeMorgan's Las). This is illustrated in the folloing Venn diagrams ith sets identified by shaded regions.

21 SECTION 0 - REVIEW OF ALGERA AND CALCULUS 5 Example 0-2 continued: A A A ( A) A Venn diagrams can sometimes be useful hen analyzing the combinations of intersections and unions of sets and the numbers of elements in various. The folloing examples illustrates this. Example 0-3: A heart disease researcher has gathered data on 40,000 people ho have suffered heart attacks. The researcher identifies three variables associated ith heart attack victims: A - smoker, - heavy drinker, C - sedentary lifestyle. The folloing data on the 40,000 victims has been gathered: 29,000 ere smokers ; 25,000 ere heavy drinker ; 30,000 had a sedentary lifestyle ; 22,000 ere both smokers and heavy drinkers ; 24,000 ere both smokers and had a sedentary lifestyle ; 20,000 ere both heavy drinkers and had a sedentary lifestyle ; and 20,000 ere smokers, and heavy drinkers and had a sedentary lifestyle. Determine ho many victims ere: (i) neither smokers, nor heavy drinkers, nor had a sedentary lifestyle; (ii) smokers but not heavy drinkers; (iii) smokers but not heavy drinkers and did not have a sedentary lifestyle? (iv) either smokers or heavy drinkers (or both) but did not have a sedentary lifestyle?

22 6 SECTION 0 - REVIEW OF ALGERA AND CALCULUS Solution: It is convenient to represent the data in Venn diagram form. For a subset W, 8ÐWÑ denotes the number of elements in that set (in thousands). The given information can be summarized in Venn diagram form as follos: Example 0-3 continued: C C C 30 A 29 A 25 A 8ÐEÑ œ *,!!! (smoker) 8ÐFÑ œ &ß!!! (heavy drinker) 8ÐGÑ œ!ß!!! (sedentary lifestyle) C C C A 22 A 24 A 20 8ÐE FÑ œ ß!!! 8ÐE GÑ œ %ß!!! 8ÐF GÑ œ!ß!!! (smoker and heavy drinker) (smoker and sedentary lifestyle) (heavy drinker and sedentary lifestyle) C A 20 ÐE F GÑ œ!ß!!! (smoker and heavy drinker and sedentary lifestyle)

23 SECTION 0 - REVIEW OF ALGERA AND CALCULUS 7 Example 0-3 continued: Working from the inside outard in the Venn diagrams, e can identify the number ithin each minimal subset of all of the intersections: C 6 A A typical calculation to fill in this diagram is as follos. We are given 8ÐE FÑ œ ß!!! ; e use the relationship 8ÐE F GÑ œ!ß!!! and ß!!! œ 8ÐE FÑ œ 8ÐE F GÑ 8ÐE F G Ñ œ!ß!!! 8ÐE F G Ñ to get 8ÐE F G Ñœß!!! (this shos that the 22,000 victims in E Fho are both smokers and heavy drinker can be subdivided into those ho also have a sedentary lifestyle 8ÐE F GÑ œ!ß!!!, and those ho do not have a sedentary lifestyle, 8ÐE F G Ñ, the other œ ß!!!). Other entries are found in a similar ay. From the diagram e can gain additional insight into other combinations of subsets. For instance, 6,000 of the victims have a sedentary lifestyle, but are neither smokers nor heavy drinkers; this is the entry "6", hich in set notation is 8ÐE F GÑ œ 'ß!!!. Also, the number of victims ho ere both heavy drinkers and had a sedentary lifestyle but ere not smokers is 0. We can no find the requested numbers. (i) The number of victims ho had at least one of the three specified conditions is 8ÐE F GÑ, hich, from the diagram can be calculated from the disjoint components: 8ÐE F GÑ œ!ß!!! ß!!! %ß!!!! ß!!! ß!!! 'ß!!! œ )ß!!!. The "total set" in this example is the set of all 40,000 victims. Therefore, there ere 2,000 heart attack victims ho had none of the three specified conditions; this is the complement of 8ÐE F GÑ. Algebraically, e have used the extension of one of DeMorgan's las to the case of three sets, "none of Eor For G" œðe F GÑ œe F G œ"not E" and "not F" and "not G".

24 8 SECTION 0 - REVIEW OF ALGERA AND CALCULUS Example 0-3 continued: (ii) The number of victims ho ere smokers but not heavy drinkers is 8ÐE F Ñ œ ß!!! %ß!!!. This can be seen from the folloing Venn diagram C 4 A 3 (iii) The number of victims ho ere smokers but not heavy drinkers and did not have a sedentary lifestyle is 8ÐE F G Ñ œ ß!!! (part of the group in (ii)). (iv) The number of victims ho ere either smokers or heavy drinkers (or both) but did not have a sedentary lifestyle is 8ÒÐE FÑ G Ó Þ This is illustrated in the folloing Venn diagram. C A ÒÐE FÑ G Ó œ ß!!! ß!!! ß!!! œ )ß!!!.

25 SECTION 0 - REVIEW OF ALGERA AND CALCULUS 9 GRAPHING AN INEQUALITY IN TWO DIMENSIONS The joint distribution of a pair of random variables \ and ] is sometimes defined over a to dimensional region hich is described in terms of linear inequalities involving and C. The region represented by the inequality C +, is the region above the line C œ +, (and C +, is the region belo the line). " * " * satisfies both of the inequalities C and C). Example 0-4: Using the lines Cœ and Cœ), find the region in the C - plane that Solution: We graph each of the straight lines, and then determine hich side of the line is represented by the " * inequality. The first graph belo is the graph of the line Cœ, along ith the shaded region, " * hich is the region C, consisting of all points "above" that line. The second graph belo is the graph of the line C œ ), along ith the shaded region, hich is the region C ), consisting of all points "belo" that line. The third graph is the intersection (first region and second region) of the to regions. Although the boundary lines of the regions in the graphs are solid lines, the ineqaulities are strict inequalities. y.5x 4.5 y.5x4.5 y2x8 y2x 8

26 10 SECTION 0 - REVIEW OF ALGERA AND CALCULUS PROPERTIES OF FUNCTIONS Definition of a function 0 : A function 0ÐÑ is defined on a subset (or the entire set) of real numbers. For each, the function defines a number 0ÐÑ. The domain of the function 0 is the set of -values for hich the function is defined. The range of 0 is the set of all 0ÐÑ values that can occur for 's in the domain. Functions can be defined in a more general ay, but e ill be concerned only ith real valued functions of real numbers. Any relationship beteen to real variables (say and C) can be represented by its graph in the Ðß CÑ-plane. If the function C œ 0ÐÑ is graphed, then for any in the domain of 0, the vertical line at ill intersect the graph of the function at exactly one point; this can also be described by saying that for each value of there is (at most) one related value of C. Example 0-5: (i) Cœ defines a function since for each there is exactly one value. The domain of the function is all real numbers (each real number has a square). The range of the function is all real numbers!, since for any real, the square is!. (ii) C œ does not define a function since if!, there are to values of Cfor hich C œ. These to values are. This is illustrated in the graphs belo y 2 y 2 2 y x y x x 1 1 x 1 2 Functions defined pieceise: A function that is defined in different ays on separate intervals is called a pieceise defined function. The absolute value function is an example of a pieceise defined function: for! ll œ for!.

27 SECTION 0 - REVIEW OF ALGERA AND CALCULUS 11 Multivariate function: A function of more than one variable is called a multivariate function. Example 0-6: Dœ0ÐßCÑœ/ C is a function of to variables, the domain is the entire 2-dimensional plane (the set ÖÐß CÑl, C are both real numbers ), and the range is the set of strictly positive real numbers. The function could be graphed in 3-dimensional CD - - space. The domain ould be the (horizontal) C - plane, and the range ould be the (vertical) D-dimension. The 3-dimensional graph is shon belo. z y x The concept of the inverse of a function is important hen formulating the distribution of a transformed random variable. A preliminary concept related to the inverse of a function is that of a one-to-one function. One-to-one function: The function 0 is called a one-to-one if the equation 0ÐÑ œ C has at most one solution for each C (or equivalently, different -values result in different 0ÐÑ values). If a graph is dran of a one-to-one function, any horizontal line crosses the graph in at most one place. Example 0-7: The function 0ÐÑ œ is one-to-one, since for each value of C, the relation C œ has C exactly one solution for in terms of C; œ. The function 1ÐÑ œ ith the hole set of real numbers as its domain is not one-to-one, since for each C!, there are to solutions for in terms of C for the relation Cœ (those to solutions are œc and œ C; note that if e restrict the domain of 1ÐÑ œ to the positive real numbers, it becomes a one-to-one function). The graphs are belo.

28 12 SECTION 0 - REVIEW OF ALGERA AND CALCULUS y 3x 2 2 y x y x 1 1 x 2 " Inverse of function 0 : The inverse of the function 0 is denoted 0. The inverse exists only if 0 is " one-to-one, in hich case, 0 ÐCÑ œ is the (unique) value of hich satisfies 0ÐÑ œ C (finding the " inverse of Cœ0ÐÑ means that e solve for in terms of C, œ0 ÐCÑ). For instance, for the " "Î function C œ œ 0ÐÑ, if œ " then C œ 0Ð"Ñ œ Ð" Ñ œ ß so that " œ 0 ÐÑ œ ÐÎÑ. For the example just considered, the inverse function applied to Cœ is the value of for hich 0ÐÑœ, or equivalently, œ, from hich e get œ ". Example 0-8: (i) The inverse of the function Cœ&"œ0ÐÑ is the function C" " œ & œ0 ÐCÑ(e solve for in terms of C). (ii) Given the function Cœ œ0ðñ, solving for in terms of Cresults in œ C, so there are to possible values of for each value of C; this function does not have an inverse. Hoever, if the function is defined to be Cœ œ0ðñ for! only, then œ Cœ0 " ÐCÑ ould be the inverse function, since 0 is one-to-one on its domain hich consists of non-negative numbers. Quadratic functions and equations: A quadratic function is of the form :ÐÑ œ +, -. The roots of the quadratic equation +, - œ! are <ß< œ The quadratic equation has: (i) distinct real roots if, %+-!, (ii) distinct complex roots if, %+-!, and (iii) equal real roots if, %+- œ!. ",, %+- +.

29 SECTION 0 - REVIEW OF ALGERA AND CALCULUS 13 Example 0-9: The quadratic equation ' % œ! has to distinct real solutions: œ &. The quadratic equation % % œ! has both roots equal: œ. The quadratic equation % œ! has to distinct complex roots: œ " 3. 4 x 2 2x4 x 2 4x4 x 2 6x4 Exponential and logarithmic functions: Exponential functions are of the form 0ÐÑ œ,, here,!ß,á", and the inverse of this function is denoted 691ÐCÑ. Thus C œ, Í 691 ÐCÑ œ. The log function ith base / is the natural logarithm, 691 ÐCÑ œ 68 C, / Some important properties of these functions are:,!, œ " 691, Ð"Ñ œ! " ".97+38Ð0Ñ œ œ <+81/Ð0 Ñ <+81/Ð0Ñ œ Ð!ß Ñ œ.97+38ð0 Ñ 691 ÐCÑ,, œcfor C! 691Ð,, Ñœfor all 68, 68 C, œ / 691, ÐCÑ œ 68, C C 5 Ð, Ñ œ, 691 ÐC Ñ œ ÐCÑ C C,,,, œ, 691 ÐCDÑ œ 691 ÐCÑ 691 ÐDÑ,,, C C, Î, œ, 691, ÐCÎDÑ œ 691, ÐCÑ 691, ÐDÑ 68C For the function /, e have / œ C for an C!, and for the natural log function, e have 68 / œ for any real number.

30 14 SECTION 0 - REVIEW OF ALGERA AND CALCULUS LIMITS AND CONTINUITY Intuitive definition of limit: The expression lim 0ÐÑ œ P means that as gets close to Ä- (approaches) the number -, the value of 0ÐÑgets close to P. Example 0-10: limðñœ%, lim / œ! and lim œlimðñœ% (for this last limit, note that Ä" Ä Ä" " Ä" ÐÑÐ"Ñ " œ " œ if Á ", but in taking this limit e are only concerned ith hat! happens "near" œ", that fact that " œ! at œ" does not mean that the limit does not exist; it means that the function does not exist at the point œ" ). Continuity: The function 0 is continuous at the point œ - if there is no "break" or "hole" in the graph of Cœ0ÐÑ, or equivalently, if lim0ðñ œ 0Ð-Ñ. In Example 0-10 above, the third function is Ä-! not continuous at œ" because 0Ð"Ñœ! is not defined. Another reason for a discontinuity in 0ÐÑ occurring at œ-is that the limit of 0ÐÑis different from the left than it is from the right. Example 0-11: (i) If 0ÐÑœ68 and -œ! then 0is discontinuous at -œ! since the function 68is not defined at the point œ! (this ould also be the case for the function 0ÐÑœ " and -œ ). "Î if Á!! if œ! (ii) If 0ÐÑ œ, then 0ÐÑis discontinuous at œ! since even though 0Ð!Ñ is defined, lim0ðñ Á 0Ð!Ñ ( lim0ðñ doesn't exist). Ä! Ä! 2 1 y ln( x) y 1 x

31 SECTION 0 - REVIEW OF ALGERA AND CALCULUS 15 DIFFERENTATION Geometric interpretation of derivative: The derivative of the function 0ÐÑ at the point œ! is the slope of the line tangent to the graph of C œ 0ÐÑ at the point Ð! ß 0Ð! ÑÑ. The derivative of 0ÐÑ at œ! is denoted 0ÐÑ! or.0. Þ œ! This is also referred to as the derivative of 0 ith respect to at the point œ!. The algebraic definition of 0ÐÑ! is 0ÐÑœ! 0Ð! 2Ñ0Ð! Ñ 0ÐÑ0Ð! Ñ lim 2 œlim 2Ä! Ä!!. Tangent Line Slope is f( x0) f( x0) x 0!!! The second derivative of 0 at is the derivative of 0 ÐÑ at the point. It is denoted 0 Ð Ñ or ÐÑ 0 Ð! Ñ or.0. Þ The 8 -th order derivative of 0 at! ( 8 repeated applications of differentiation) is œ Ð8Ñ. 0 denoted 0 Ð! Ñ œ.8.! 8 œ! The derivative as a rate of change: Perhaps the most important interpretation of the derivative 0ÐÑ! is as the "instantaneous" rate at hich the function is increasing or decreasing as increases Ðif 0!, the graph of C œ 0ÐÑ is rising, ith the tangent line to the graph having positive slope, and if 0!, the graph of C œ 0ÐÑ is fallingñ, and if 0ÐÑœ! then the tangent line at that point is horizontal (has slope 0). This interpretation is the one most commonly used hen analyzing physical, economic or financial processes.!

32 16 SECTION 0 - REVIEW OF ALGERA AND CALCULUS The folloing is a summary of some important differentiation rules. Rules of differentiation: 0 ÐÑ 0 ÐÑ - (a constant)! 8 8" Poer rule - - ( 8 ) -8 1ÐÑ 2ÐÑ 1 ÐÑ 2 ÐÑ Product rule - 1ÐÑ 2ÐÑ 1 ÐÑ 2ÐÑ 1ÐÑ 2 A?@A Quotient rule - 1ÐÑ 2ÐÑ 2ÐÑ1 ÐÑ1ÐÑ2 ÐÑ Ò2ÐÑÓ Chain rule - 1Ð2ÐÑÑ 1 Ð2ÐÑÑ 2 ÐÑ 1ÐÑ 1ÐÑ / 1 ÐÑ / 1ÐÑ 68Ð1ÐÑÑ 1ÐÑ + Ð+!Ñ / / " 68 " 691, 68, =38-9= -9= =38 Example 0-12: What is the derivative of 0ÐÑ œ %Ð "Ñ? Solution: We apply the product rule and chain rule: here 0ÐÑ œ 1ÐÑ 2ÐÑ, 1ÐÑ œ % ß 2ÐÑ œ Ð "Ñ ß 1 ÐÑ œ % ß 2 ÐÑ œ Ð "Ñ. 0 ÐÑ œ % Ð "Ñ %Ð "Ñ œ %Ð "Ñ Ð( "Ñ. Notice that 2ÐÑ œ Ð "Ñ œ ÒAÐÑÓ œ 2ÐAÐÑÑ, here 2ÐAÑ œ A and AÐÑ œ ". The chain rule tells us that 2 ÐÑ œ 2 ÐAÑ A ÐÑ œ A ÐÑ œ Ð "Ñ ÐÑ.

33 SECTION 0 - REVIEW OF ALGERA AND CALCULUS 17 L'Hospital's rules for calculating limits: A limit of the form lim is said to be in indeterminate Ä- 1ÐÑ form if both the numerator and denominator go to 0, or if both the numerator and denominator go to. L'Hospital's rules are: 0ÐÑ 1. IF (i) lim0ðñ œ lim1ðñ œ!, and (ii) 0Ð-Ñexists, and (iii) 1Ð-Ñexists and is Á! THEN 0ÐÑ lim 1ÐÑ Ä- Ä Ä- 0Ð-Ñ œ 1 Ð-Ñ 2. IF (i) lim0ðñ œ lim1ðñ œ!, and Ä- (ii) 0 and 1 are differentiable near -, 0ÐÑ (iii) lim 1ÐÑ exists and THEN 0ÐÑ lim 1ÐÑ Ä- 0 ÐÑ œ lim 1 ÐÑ In 1 or 2, the conditions lim0ðñ œ! and lim1ðñ œ! can be replaced by the conditions Ä- lim0ðñ œ and lim1ðñ œ, and the point - can be replaced by ith the conclusions Ä- remaining valid. Ä- Ä- Ä- Ä- Ä- Î Example 0-13: Find lim. Ä * Solution:. The limits in both the numerator and denominator are 0, so e apply l'hospital's rule.. œ 68, and Î Î ". Î Î " 68 ". œ 68, so that lim * œ lim 68 œ. This limit can also be found by Ä Ä ' factoring the denominator into Î Î * œ Ð ÑÐ Ñ, and then canceling out the factor Î in the numerator and denominator. Differentiation of functions of several variables - partial differentiation: Given the function 0ÐßCÑ, a function of to variables, the partial derivative of 0 ith respect to at the point Ð! ß C! Ñ is found by differentiating 0 ith respect to and regarding the variable C as constant - then substitute in the values œ! and CœC!. The partial derivative of 0ith respect to is usually `0 denoted `. The partial derivative ith respect to C is defined in a similar ay: "Higher order" partial `0 ` `0 `0 ` `0 derivatives can be defined - ` œ `Ð `Ñ, `C œ `CÐ `CÑ ; and "mixed partial" derivatives can be defined (the order of partial differentiation does not usually matter) - ` 0 ` `0 ` `0 ` 0 ` `C œ `Ð`CÑ œ `CÐ`Ñ œ `C `. Example 0-14: C `0 ` 0 If 0Ðß CÑ œ for ß C! then find ` and `C. " " Solution: `0 C" " "Î " ` œ C œ Ð ÑÐ%Ñ œ %, and " Ð%ß Ñ Ð%ß Ñ Ð%ß Ñ `0 C ` 0 C "Î `C œ Ð68Ñ and `C œ Ð68Ñ œ% Ð68%Ñ œð68%ñ. " Ð%ß Ñ

34 18 SECTION 0 - REVIEW OF ALGERA AND CALCULUS INTEGRATION Geometric interpretation of the "definite integral" - the area under the curve: Given a function 0ÐÑ on the interval Ò+ß,Ó, the definite integral of 0ÐÑover the interval is denoted, + 0ÐÑ., and is equal to the "signed" area beteen the graph of the function and the -axis from œ+ to œ,. Signed area is positive hen 0ÐÑ! and is negative hen 0ÐÑ!. What is meant by signed area here is the area from the interval(s) here 0ÐÑ is positive minus the area from the intervals here 0ÐÑ is negative. Integration is related to the antiderivative of a function. Given a function 0ÐÑ, an antiderivative of 0ÐÑ is any function J ÐÑ hich satisfies the relationship J ÐÑ œ 0ÐÑ. According to the Fundamental Theorem of Calculus, the definite integral for 0ÐÑ can be found by first finding J ÐÑ, an antiderivative of 0ÐÑ. The basic relationships relating integration and differentiation are: (i) If JÐÑœ0ÐÑ for +ŸŸ,, then, + 0ÐÑ.œJÐ,ÑJÐ+Ñ. (ii) If KÐÑ œ + 1Ð>Ñ.> ß then K ÐÑ œ 1ÐÑ Example 0-15: Find the definite integral of the function 0ÐÑ œ on the interval Ò "ß Ó. Solution: The graph of the function is given belo. It is clear that 0ÐÑ! for, and 0ÐÑ! for. An antiderivative for 0ÐÑ is J ÐÑ œ Þ The definite integral ill be " Ð Ñ Ð"Ñ.œJÐÑJÐ"ÑœÐ' ÑÐ Ñœ%ÞNote that the area beteen the graph and the " * -axis from œ " to œ is ÐÑÐÑ œ, and the signed area beteen the graph and the -axis " " * " from œ to œ is Ð"ÑÐ"Ñœ. The total signed area is œ%. 3 2 y 2 x

35 SECTION 0 - REVIEW OF ALGERA AND CALCULUS 19 Antiderivatives of some frequently used functions: 0ÐÑ 0ÐÑ. (antiderivative) 1ÐÑ 2ÐÑ 1ÐÑ. 2ÐÑ. - 8" 8 Ð8 Á "Ñ 8" - " 68 - / / Ð+!Ñ / + + / / =38-9= = =38 - Integration of 0 on Ò+ß,Óhen 0 is not defined at + or,, or hen + or, is : Integration over an infinite interval (an "improper integral") is defined by taking limits:,, 0ÐÑ. œ lim 0ÐÑ., ith a similar definition applying to 0ÐÑ., +,Ä + 0ÐÑ. œ lim + +Ä + 0ÐÑ. and. If 0is not defined at œ+ (also called an improper integral), or if 0is discontinuous at œ+, then, 0ÐÑ. œ lim, 0ÐÑ.. + -Ä+ - A similar definition applies if 0is not defined at œ,, or if 0is discontinuous at œ,. If 0ÐÑ has a discontinuity at the point œ - in the interior of Ò+ß,Ó, then, -, 0ÐÑ. œ 0ÐÑ. 0ÐÑ Example 0-16: " (a) " " "Î "Î!. œ lim -. œ lim œ lim Ò -Ó œ, œ" -Ä! -Ä! œ- -Ä! œ- " - "Î "Î " lim " lim lim -Ä -Ä -Ä œ" œ, " " " " lim,ä lim œ",ä, " " " " "! " " "! " " " " "!. œ lim +Ä! +. œ lim Ò " +Ä! + Ó œ. œ. œ œ Ò - Ó œ Þ (b). œ œ Ò Ð "ÑÓ œ " (c).. Note that has a discontinuity at œ!, so that. œ... The second integral is, thus, the second improper integral does not exist (hen lim is infinite or does not exist, the integral is said to "diverge"). Ä

36 20 SECTION 0 - REVIEW OF ALGERA AND CALCULUS A fe other useful integration rules are: (i) for integer 8! and real number -! 8-8x! /. œ - 8". 2ÐÑ (ii) if KÐÑ œ + 0Ð?Ñ.?, then K ÐÑ œ 0Ò2ÐÑÓ 2 ÐÑ,, (iii) if KÐÑ œ 0Ð?Ñ.?, then K ÐÑ œ 0ÐÑ, (iv) if KÐÑ œ 0Ð?Ñ.?, then K ÐÑ œ 0Ò1ÐÑÓ 1 ÐÑ,, 1ÐÑ 2ÐÑ 1ÐÑ (v) if KÐÑ œ 0Ð?Ñ.?, then K ÐÑ œ 0Ò2ÐÑÓ 2 ÐÑ 0Ò1ÐÑÓ 1 ÐÑ. Double integral: Given a continuous function of to variables, 0ÐßCÑ on the rectangular region bounded by œ+ßœ,ßcœ- and Cœ., it is possible to define the definite integral of 0over the region. It can be expressed in one of to equivalent ays:,.., Ðß CÑ.C. œ 0Ðß CÑ..C The interpretation of the first expression is, [ Ðß CÑ.C ]., in hich the "inside integral" is. - 0ÐßCÑ.C, and it is calculated assuming that the value of is constant (it is an integral ith respect to the variable C). When this definite "inside integral" has been calculated, it ill be a function of alone, hich can then be integrated ith respect to from œ + to œ,. The second equivalent expression has a similar interpretation;, + 0ÐßCÑ. is calculated assuming that C is constant; this results in a function of C alone hich is then integrated ith respect to C from C œ - to C œ.. Double integration arises in the context of finding probabilities for a joint distribution of continuous random variables. Example 0-17: " Find.C..! " Solution: C First e assume that is constant and find " C.C œ Ð68CÑ œ Ð68Ñ. Then e find "! Ò Ð68 ÑÓ. œ Ð68 Ñ Cœ Cœ" œ" " We can also rite the integral as "! C..C, and first find œ" " Cœ " " " "! C. œ C œ C. Then, " C.C œ Ð68 CÑ œ Ð68 Ñ. œ! Cœ" œ! œ 68. For double integration over the rectangular to-dimensional region +ŸŸ,ß -ŸCŸ., as the expression, Ðß CÑ.C. œ., - + 0Ðß CÑ..C indicates, it is possible to calculate the double integral by integrating ith respect to the variables in either order ( C first and second for the integral on the left, and first and Csecond for the integral on the right of the " œ " sign).

37 SECTION 0 - REVIEW OF ALGERA AND CALCULUS 21 Formulations of probabilities and expectations for continuous joint distributions sometimes involve integrals over a non-rectangular to-dimensional region. It ill still be possible to arrange the integral for integration in either order (.C. or..c), but care must be taken in setting up the limits of integration. If the limits of integration are properly specified, then the double integral ill be the same hichever order of integration is used. Note also that in some situations, it may be more efficient to formulate the integration in one order than in the other. Example 0-18: " Which of the folloing integrals is equal to!! 0Ðß CÑ.C. for every function for hich the integral exists? CÎ A) 0Ðß CÑ..C " ) 0Ðß CÑ..C " C) 0Ðß CÑ..C D) E)!!!! C " Î!! 0Ðß CÑ..C "! CÎ 0Ðß CÑ..C Solution: The graph at the right illustrates the region of integration. The region is!ÿÿ"ß C! Ÿ C Ÿ. Writing C œ as œ, e see that the inequalities translate into C!ŸCŸ, and ŸŸ". Anser: E y 3x y x 3 x 1 Example 0-19: The function 0ÐßCÑ is to be integrated over the to-dimensional region defined by the folloing constraints:!ÿÿ" and "ŸCŸ. Formulate the double integration in the.c. order and then in the..c order. Solution: The graph at the right illustrates the region of integration. The region is!ÿÿ"ß "ŸCŸ. The integral can be formulated " in the.c. order as! "0ÐÞCÑ.C. ; for each, the integral in the vertical direction starts on the line Cœ" and continues to the upper boundary Cœ. To use the..c order, e must split the integral into to double " " integrals;! "C0Ðß CÑ..C to cover the " triangular area belo Cœ", and "! 0ÐßCÑ..C to cover the square area above C œ " y y 1x 1 x

38 22 SECTION 0 - REVIEW OF ALGERA AND CALCULUS There are a fe integration techniques that are useful to kno. The integrations that arise on Exam P are usually straightforard, but knoing a fe additional techniques of integration are sometimes useful in simplifying an integral in an efficient ay. The Method of Substitution: Substitution is a basic technique of integration that is used to rerite the integral in a standard form for hich the antiderivative is ell knon. In general, to find 0ÐÑ. e may make the substitution? œ 1ÐÑ for an "appropriate" function 1ÐÑ. We then define the "differential".? to be.? œ 1 ÐÑ., and e try to rerite 0ÐÑ. as an integral ith respect to the variable?. For example, to find %Î Ð "Ñ., e let?œ ", so that.?œ., or equivalently, ".?œ.? %Î " ; then the integral can be ritten as.?, hich has (Î antiderivative %Î " "?.? œ %Î "? " (Î?.? œ (Î œ (? Ð-Ñ. We can then rite the antiderivative in terms of the original variable - %Î " (Î " (Î Ð "Ñ. œ (? œ ( Ð "Ñ. The main point to note in applying the substitution technique is that the choice of?œ1ðñshould result in an antiderivative hich is easier to find than as the original antiderivative. Example 0-20: " Find "..! Solution: " Let?œ" Then.?œ., so that.?œ., and the antiderivative can be ritten as "Î " " Î " Î? Ð Ñ.? œ? œ Ð" Ñ. œ" " The definite integral is then " Î " "! ".œ Ð" Ñ œ!ð Ñœ. Note that once the appropriate substitution has been made, the definite integral may be calculated in terms of the variable?:?ð!ñ œ " and?ð"ñ œ! - œ! "Î " " Î " " ".œ? Ð Ñ.?œ? œ!ð Ñœ. "?Ð"Ñœ!!?Ð!Ñœ"?œ!?œ1

39 SECTION 0 - REVIEW OF ALGERA AND CALCULUS 23 Integration by parts: This technique of integration is based upon the product rule.. Ò0ÐÑ 1ÐÑÓ œ 0ÐÑ 1 ÐÑ 0 ÐÑ 1ÐÑ. This can be reritten as. 0ÐÑ 1 ÐÑ œ. Ò0ÐÑ 1ÐÑÓ 0 ÐÑ 1ÐÑ, hich means that the antiderivative of 0ÐÑ 1 ÐÑ can be ritten as 0ÐÑ 1 ÐÑ. œ 0ÐÑ 1ÐÑ 0 ÐÑ 1ÐÑ.. This technique is useful if 0 ÐÑ 1ÐÑ has an easier antiderivative to find than 0ÐÑ 1 ÐÑ. Given an integral, it may not be immediately apparent ho to define 0ÐÑ and 1ÐÑ so that the integration by parts technique applies and results in a simplification. It may be necessary to apply integration by parts more than once to simplify an integral. Example 0-21: Find + /., here + is a constant. Solution: + / + If e define 0ÐÑ œ and 1ÐÑ œ +, then 1 ÐÑ œ /, and + /. œ 0ÐÑ1 ÐÑ. œ 0ÐÑ1ÐÑ 0 ÐÑ1ÐÑ.. Since 0 ÐÑ œ ", it follos that 0 ÐÑ1ÐÑ. œ / +. œ / + + +, and therefore / / /. œ An alternative to integration by parts is the folloing approach: / +/ /.+ /. œ /. and.+ /. œ.+ + œ so it follos that + +/ / / / /. œ + œ This integral has appeared a number of times on the exam, usually ith +Á!) and it is important to be familiar ith it. +! (it is valid for any An alternative ay to apply integration by parts is via "Tabular Integration". Suppose that e ish to find the We create to columns, a column of the successive derivatives of?ðñ and a column of the successive antiderivatives In order for the method to ork efficiently, e try to choose?ðñ to be a polynomial hich ill eventually have a derivative of 0. The integral in Example 0-21 ill be used to illustrate tabular integration. We make the folloing choices for?ðñ +?ÐÑ œ /. The to columns are Ro Derivatives of?ðñ œ Antiderivatives œ / + 0 / " / Î+ + 2! / Î+

40 24 SECTION 0 - REVIEW OF ALGERA AND CALCULUS We pair up entries in each ro of the "Derivatives of?ðñ" column ith the folloing ro of the "Antiderivatives column and multiply the pairs, alternative " " and " " for each pair, and add + them up. In this example, e pair (Ro 0 of "Derivatives") ith / Î+ (Ro 1 of "Antiderivatives"), + then e pair " (Ro 1 of "Derivatives") ith / Î+ (Ro 2 of "Antiderivatives") and apply " " to this pair. For this example, e can stop at this point, since all higher order derivatives of?ðñ are 0 in + + Ro 2 and higher. The integral / / /. is / Î+" / Î+ œ + +, as in Example NOTE: An extension of Example 0-21 shos that for integer 8! and -! 8 -! /. œ 8x - 8". This is another useful identity for the exam. GEOMETRIC AND ARITHMETIC PROGRESSIONS Geometric progression: +ß +<ß +< ß +< ß ÞÞÞß sum of the first 8terms is 8 8 8" 8" <" "< ++<+< â+< œ+ò"<< â< Óœ+ <" œ+ "<, + if "<" then the infinite series can be summed, ++<+< ✠"< Arithmetic progression: +ß+.ß+.ß+.ßÞÞÞß sum of the first 8terms of the series is 8+. 8Ð8"Ñ, a special case is the sum of the first 8 integers - "â8œ 8Ð8"Ñ Example 0-22: A product sold 10,000 units last eek, but sales are forecast to decrease 2% per eek if no advertising campaign is implemented. If an advertising campaign is implemented immediately, the sales ill decrease by 1% of the previous eek's sales but there ill be 200 ne sales for the eek (starting ith this eek). Under this model, calculate the number of sales for the 10-th eekß 100-th eek and 1000-th eek of the advertising campaign (last eek is eek 0, this eek is eek 1 of the campaign). Solution: Week 1 sales: ÐÞ**ÑÐ"!ß!!!Ñ!! ß Week 2 sales: ÐÞ**ÑÒÐÞ**ÑÐ"!ß!!!Ñ!!Ó!! œ ÐÞ**Ñ Ð"!ß!!!Ñ Ð!!ÑÒ" Þ**Ó Week 3 sales: ÐÞ**ÑÒÐÞ**Ñ Ð"!ß!!!Ñ Ð!!ÑÒ" Þ**ÓÓ!! œ ÐÞ**Ñ Ð"!ß!!!Ñ Ð!!ÑÒ" Þ** Þ** Ó ã "! * Week 10 sales: ÐÞ**Ñ Ð"!ß!!!Ñ Ð!!ÑÒ" Þ** Þ** â Þ** Ó "! "! "Þ** œ ÐÞ**Ñ Ð"!ß!!!Ñ Ð!!ÑÒ "Þ** Ó œ "!ß *&'Þ. "!! "!! "Þ** Week 100 sales: ÐÞ**Ñ Ð"!ß!!!Ñ Ð!!ÑÒ "Þ** Ó œ "'ß *Þ(. "!!! "!!! "Þ** Week 1000 sales: ÐÞ**Ñ Ð"!ß!!!Ñ Ð!!ÑÒ "Þ** Ó œ "*ß ***Þ'.