ARCONES MANUAL FOR THE SOA EXAM P/CAS EXAM 1, PROBABILITY, SPRING 2010 EDITION.

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1 A self published manuscript ARCONES MANUAL FOR THE SOA EXAM P/CAS EXAM 1, PROBABILITY, SPRING 21 EDITION. M I G U E L A R C O N E S Miguel A. Arcones, Ph. D. c 28. All rights reserved.

2 Author Miguel A. Arcones Department of Mathematical Sciences Binghamton University Binghamton, NY 1392 E mail: arcones@math.binghamton.edu Web page: Title Arcones Manual for the SOA Eam P/CAS Course 1, Probability. Spring 21 Edition. Printed on September 19, 29. Published by Miguel A. Arcones. c Miguel A. Arcones. 29. All rights reserved. The whole and parts of this work are subject to copyright in any and all forms of media.

3 iii Contents Preface i Table of Contents iii 1 Introduction to insurance and risk management Insurance Actuaries Deductible, benefit limits, inflation Some mathematical models Probability theory Definition and main properties of probabilities Sample space, events Definition of probability Problems Problems from actuarial eams Solutions to problems Solution to problems from actuarial eams: Counting techniques Counting techniques Permutations Combinations Problems Problems from actuarial eams Solutions to problems Solution to problems from actuarial eams: Conditional probability, Bayes Theorem Conditional probability Independent events Bayes theorem Problems Problems from actuarial eams Solutions to problems

4 iv 4.7 Solutions to problems from actuarial eams Random variables Random variables Discrete random variables Continuous random variables Mied distributions Mode, median, quartiles, percentiles Problems Problems from actuarial eams Solutions to problems Solutions to problems from actuarial eams Epected values Mean of a random variable Epectation of a function of a random variable Variance Problems Problems from actuarial eams Solutions to problems Solutions to problems from actuarial eams Common discrete distributions Binomial Geometric Negative binomial Poisson Hypergeometric Problems Problems from actuarial eams Solutions to problems Solutions to problems from actuarial eams

5 v 8 Common continuous distributions Uniform Eponential Gamma Beta Normal distribution Further distributions Problems Problems from actuarial eams Solutions to problems Solutions to problems from actuarial eams Multivariate distributions, covariance Multivariate distributions Multivariate discrete distributions Jointly continuous distributions Independent random variables Epectation of a function of several random variables Covariance Multinomial distribution Problems Problems from actuarial eams Solutions to problems Solutions to problems from actuarial eams Conditional epectations Conditional probabilities for a unique r.v Conditional probability mass functions and conditional densities Conditional epectations Conditional variances Some properties of the conditional epectations Problems Problems from actuarial eams Solutions to problems Solutions to problems from actuarial eams

6 vi 11 Moment generating functions Moment generating functions Problems Problems from actuarial eams Solutions to problems Solutions to problems from actuarial eams Sums of independent random variables Sums of independent random variables Sums of independent normal random variables Multivariate normal distribution Problems Problems from actuarial eams Solutions to problems Solutions to problems from actuarial eams Transformations of random variables One dimensional transformations Multivariate transformations Maimums and minimums Order statistics Transformations of discrete r.v. s Problems Problems from actuarial eams Solutions to problems Solutions to problems from actuarial eams Central limit theorem Central limit theorem Chebychev s inequality Problems Problems from actuarial eams Solutions to problems Solutions to problems from actuarial eams

7 vii 15 Practice Eams Practice Eam Practice Eam Practice Eam Practice Eam Practice Eam Practice Eam Practice Eam Practice Eam Practice Eam Practice Eam Answer key Solutions to Practice Eam Number Solutions to Practice Eam Number Solutions to Practice Eam Number Solutions to Practice Eam Number Solutions to Practice Eam Number Solutions to Practice Eam Number Solutions to Practice Eam Number Solutions to Practice Eam Number Solutions to Practice Eam Number Solutions to Practice Eam Number A Appendi 451 A.1 Cheat sheet A.2 Mean, variance and m.g.f. of some common distributions A.3 Greek letters A.4 Tables given in the first actuarial eamination Subject Inde About the Author 459

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9 CHAPTER 8 Common continuous distributions 8.1 Uniform A uniform distribution in the interval (a, b) has a density which is a constant in the interval (a, b) and it is zero outside this interval. The constant value of this density in the interval (a, b) is such that the total probability is one. y Figure 8.1: Uniform density Definition 8.1. A r.v. X has a uniform distribution on the interval (a, b) if its density is { 1 if a b b a f() = else If X has a uniform distribution on the interval (a, b), we say that X d Unif(a, b). Net, we find the mean and the variance of an uniform distribution: Theorem 8.1. Let X be a r.v. with uniform distribution on the interval (a, b), then E[X] = a+b and Var(X) = (b a) and Proof. We have that E[X] = b a 1 b a d = E[X 2 ] = b a 2 1 b a d = 2 2(b a) 3 3(b a) 145 b a b a = b2 a 2 = a+b 2(b a) 2 = b3 a 3 = a2 +ab+b 2 3(b a) 3

10 146 CHAPTER 8. COMMON CONTINUOUS DISTRIBUTIONS So, Var(X) = E[X 2 ] (E[X]) 2 = a2 +ab+b 2 ( ) a+b = 4a2 +4ab+4b 2 3(a 2 +2ab+b 2 ) = a2 2ab+b 2 = (b a) As one should have epected the average of a uniform distribution is the midpoint of its support. Notice that the variance depends on the length b a of the support of the distribution. Eample 8.1. A car owner insures a $25 worth car. He estimates that the repair costs can be modeled by a uniform random variable on the interval (, 25). The insurance policy has a deductible of 25. Find the mean and the variance of the repair costs not paid by the insurance policy. Solution: Let X be the repair costs. X has p.d.f. { 1 if 25, 25 f() = otherwise. We have to find the epectation and the variance of { X if X 25, g(x) = X (X 25) + = 25 if 25 < X. We have that and E[g(X)] = = 2 5 E[(g(X)) 2 ] = = d (25) d = = , 25 d + 25 (25) 2 1 d = (25 25) 2 = , Var(g(X)) = (248.75) 2 = Eponential Theorem 8.2. For each integer n, n n! e d = n j= j e j! + c. Proof. We proceed by induction on n. If n =, we need to prove that e d = e + c,

11 SECTION 8.2. EXPONENTIAL 147 which is obvious. Suppose that the claim holds for n 1, i.e. suppose that Then, n 1 n 1 (n 1)! e j d = e j! + c. n n! e d = n n! d( e ) = n n! ( e ) ( e )d( n ) n! = n n! e + n 1 n e d = (n 1)! n! e n 1 j j= e + c = n j j! j= e + c. j! j= Eample 8.2. Find: (i) 2 e d. (ii) 3 e d. (iii) 4 24 e d. Solution: (i) By the previous theorem, ( ) 2 2 e d = e c. So, 2 e d = e ( ) + c. So, (ii) By the previous theorem, 3 ( 6 e d = e ) + c. 3 e d = e ( ) + c. (iii) By the previous theorem, ( ) 4 24 e d = e c. Corollary 8.1. For each n, Proof. By the previous theorem, n n! e d = n e d = n!. n j= j e j! = 1.

12 148 CHAPTER 8. COMMON CONTINUOUS DISTRIBUTIONS y Figure 8.2: eponential density Eample 8.3. Find: (i) 4 e d. (ii) 12 e d. (iii) 23 e 2 d. (iv) 24 e /3 d. Solution: (i) 4 e d = 4! = 24. (ii) 12 e d = 12! = (iii) By the change of variables y = 2, 23 e 2 d = (iv) By the change of variables y = 3, 24 e /3 d = ( y 2 ) 23 e y 1 2 dy = 23! y 24 e y 3 dy = 24!(3) 25. Definition 8.2. A r.v. X has an eponential distribution with parameter λ > if its density is { e λ if λ f() = else The above function f defines a bona fide density because it is nonnegative and e λ f() d = d = e λ λ = 1. If X has an eponential distribution with parameter λ >, we say that X d Ep(λ).

13 SECTION 8.2. EXPONENTIAL 149 Theorem 8.3. Let X be a r.v. with an eponential distribution with parameter λ >, then E[X] = λ and Var(X) = λ 2. and Proof. By the change of variables y = λ, E[X] = E[X 2 ] = So, Var(X) = E[X 2 ] (E[X]) 2 = λ 2. e λ λ 2 e λ λ d = λ d = λ2 ye y dy = λ y 2 e y dy = 2λ 2. Let X be an eponential r.v. with mean λ. Then, for each, e t λ t P[X ] = dt = e λ λ = e λ. Thus, for each s, t >, P[X s + t X s] = P[X s + t, X s] P[X s] = P[X s + t] P[X s] So, an eponential r.v. X satisfies that for each s, t >, = s+t e λ e s λ = e t λ = P[X t]. P[X s + t X s] = P[X t]. This property is called the memoryless property of the eponential distribution. Suppose that an electronic component follows an eponential distribution. By memoryless property the probability that a s year old component will last t or more years is the same as the probability that a new component will last t or more years. Eample 8.4. A certain HMO pays 8% of the medical bills sustained by a member. An actuary estimates that the annual medical epenses of a member has an eponential distribution with a mean of 3. What is the 95 percentile of medical epenses not reimbursed to a member of this HMO in a year? Solution: Let X be the annual medical epenses of a member of this HMO. The p.d.f. of X is { 1 3 f() = e 3 if, else. The medical epenses not reimbursed to a member are.2x. Let q be the 95 percentile of.2x. Then,.95 = P[.2X q] = P[X 5q] = 5q 1 3 e 3 d 5q = e 3 = 1 e q 6 and q = (6) ln(.5) =

14 15 CHAPTER 8. COMMON CONTINUOUS DISTRIBUTIONS Eample 8.5. Let X be a random variable with density function { 1.4e 2 +.9e 3 for > f() = elsewhere Find E[X]. Solution: By the change of variables u = 2 in the first integral and the change of variables 3 = v in the second integral, we get that E[X] = (1.4e 2 +.9e 3 ) d = (1.4)e 2 d + (.9)e 3 d = (1.4) u 2 e u du + (.9) v 2 3 e v dv = (.35) ue u du + (.1) ve v dv 3 = = Gamma Definition 8.3. The gamma function is defined by Γ(α) = t α 1 e t dt, α >. Theorem 8.4. The gamma function satisfies the following properties: (i) For each α > 1, Γ(α) = (α 1)Γ(α 1). (ii) For each integer n 1, Γ(n) = (n 1)!. Proof. (i) For each α > 1, by an integration by parts Γ(α) = t α 1 e t dt = t α 1 d( e t ) dt = t α 1 ( e t ) = t α 1 ( e t ) + (α 1)tα 2 e t dt = (α 1)Γ(α 1). (ii) follows from Corollary 8.1. We also have that Γ(1/2) = π (see problem 22 in Section 8.7). Eample 8.6. Let X have probability density function given by Compute c and the epected value of X. Solution: First, we find c, 1 = So, c = 1. The epected value of X is 9! E[X] = f() = c 9 e, >. c 9 e d = cγ(1) = c 9!. 1 9! 9 e d = 1 9! 1 e d = 1! 9! ( e t )d(t α 1 ) = 1.

15 SECTION 8.3. GAMMA 151 Definition 8.4. A r.v. X is said to have a gamma distribution with parameters α > and β > if its density is α 1 e β if >, Γ(α)β f() = α otherwise. α is called the shape parameter of the gamma distribution. β is called the scale parameter of the gamma distribution. The function above is really a pdf because by the change of variable /β = y, As a consequence, we have that (8.1) α 1 e β Γ(α)β α d = y α 1 e y Γ(α) dy = 1. α 1 e β d = Γ(α)β α. If X has a gamma distribution with parameters α > and β >, we say that X d Gam(α, β). Theorem 8.5. If X has a gamma distribution with parameters α > and β >, then Proof. By (8.1), E[X] = αβ and Var(X) = αβ 2. and So, E[X] = α 1 e β d = 1 α e Γ(α)β α Γ(α)β α β d = Γ(α+1)β α+1 Γ(α)β α E[X 2 ] = 2 α 1 e β Γ(α)β α d = 1 Γ(α)β α = Γ(α+2)βα+2 Γ(α)β α = β2 Γ(α+2) Γ(α) = β2 (α+1)γ(α+1) Γ(α) α+1 e β d = βαγ(α) Γ(α) = αβ, = β2 (α+1)(α)γ(α) Γ(α) = (α + 1)αβ 2. Var(X) = E[X 2 ] (E[X]) 2 = (α + 1)αβ 2 (αβ) 2 = αβ 2. Eample 8.7. Let X have p.d.f. f() = 83 3 e 2, >. Find the mean and the standard deviation of X. Solution: The r.v. X has a gamma distribution with α = 4 and β = 1 2. Hence, E[X] = αβ = 2, Var(X) = αβ 2 = 1 and the standard deviation of X is 1. Eample 8.8. Let X be a r.v. with p.d.f. f() = Find the mean and the variance of 1 X e 5, >.

16 152 CHAPTER 8. COMMON CONTINUOUS DISTRIBUTIONS Solution: By the change of variables y = 5, and So, = E[ 1 X ] = e 5 d = y 2 3 e y dy = 2! 3 = 1 15 E[ 1 ] = X 2 = y 15 e y dy = e 5 d = e 5 d = (5y) e y 5 dy 375 e 5 d = 5y 375 e y 5 dy ( ) [ ( ) ] 2 ( [ ]) Var = E E = 1 ( ) 2 1 X X X 15 = Beta We will need to use the following theorem regarding a change of variables for bivariate integrals: Theorem 8.6. Let A, B R 2, let f : A R be a continuous function and let g : A B be a one to one onto continuous function having continuous partial derivatives. We denote g(u, v) = (g 1 (u, v), g 2 (u, v)). Let ) Then, B J(u, v) = det f(, y) d dy = ( g1 (u,v) u g 2 (u,v) u A g 1 (u,v) v g 1 (u,v) v f(g(u, v)) J(u, v) du dv. J(u, v) is called the Jacobian of the transformation g. If (, y) = g(u, v), then J(u, v) is also denoted by (,y) (u,v). Theorem α 1 (1 ) β 1 d = Γ(α)Γ(β) Γ(α + β). Proof. We apply Theorem 8.6 with A = (, 1) (, ), B = (, ) (, ) and (, y) = g(u, v) = (uv, (1 u)v). It is easy to see that g is a one to one onto function. The Jacobian of the transformation is ( ) (, y) (u, v) = det v u = v v 1 u So, = 1 α 1 y β 1 e y d dy = 1 (uv)α 1 ((1 u)v) β 1 e uv (1 u)v v dv du u α 1 (1 u) β 1 v α+β 1 e v dv du = Γ(α + β) 1 uα 1 (1 u) β 1 du. Γ(α)Γ(β) =.

17 SECTION 8.4. BETA 153 Definition 8.5. X is said to have a beta density with parameters α > and β > if its density function is { Γ(α+β) Γ(α)Γ(α) f() = α 1 (1 ) β 1 if 1 otherwise By Theorem 8.7, the density of a beta distribution integrates one. If X has a beta density with parameters α > and β >, we say that X d Beta(α, β). The beta distribution is used to model proportions. Theorem 8.8. Suppose that X has a beta distribution with parameters α > and β >. αβ Then, E[X] = and Var(X) = and α α+β (α+β) 2 (α+β+1). Proof. Using that Γ(α + 1) = αγ(α), we get that = E[X] = 1 Γ(α+β) = Γ(α+β) Γ(α)Γ(β) E[X 2 ] = 1 Γ(1+α)Γ(β) Γ(1+α+β) 2 Γ(α+β) = Γ(α+β) Γ(2+α)Γ(β) Γ(α)Γ(β) Γ(2+α+β) (α+1)α = (α+β+1)(α+β) Var(X) = αβ (α+β) 2 (α+β+1). 1 Γ(α+β)αΓ(α) Γ(α)Γ(β) = Γ(α)(α+β)Γ(α+β) 1 Γ(α+β)(1+α)Γ(1+α) Γ(α)Γ(β) = Γ(α)(1+α+β)Γ(1+α+β) Γ(α)Γ(β) α 1 (1 ) β 1 d = Γ(α+β) = Γ(α+β)Γ(1+α) = Γ(α)Γ(1+α+β) Γ(α)Γ(β) α 1 (1 ) β 1 d = Γ(α+β) = Γ(α+β)Γ(2+α) = Γ(α)Γ(2+α+β) ( (α+1)α α (α+β+1)(α+β) α+β 1+α 1 (1 ) β 1 d α, α+β 2+α 1 (1 ) β 1 d Γ(α+β)(1+α)(α)Γ(α) Γ(α)(1+α+β)(α+β)Γ(α+β) ) 2 = α ((α + 1)(α + β) α(α + β + 1)) (α+β) 2 (α+β+1) Eample 8.9. Find 1 2 (1 ) 7 d. Solution: Using Theorem 8.7 with α = 3 and β = 8, 1 2 (1 ) 7 d = Γ(3)Γ(8) Γ(11) = (2!)(7!) 1! = Eample 8.1. The proportion of time that a machine is down for repairs can be looked upon as a random variable having a beta distribution with α = 2 and β = 7. What is the probability that the machine is down more than 2% of the time? Solution: Let X be proportion of time that a machine is down. X has p.d.f. { Γ(α+β) Γ(α)Γ(α) f() = α 1 (1 ) β 1 = Γ(2+7) Γ(2)Γ(7) 2 1 (1 ) 7 1 = 56(1 ) 6 if 1, otherwise. By the change of variables y = 1, P[X >.2] = (1 )6 d =.8 56(1 y)y 6 d = (8y 7 7y 8 ) = 8(.8) 7 7(.8) 8 =

18 154 CHAPTER 8. COMMON CONTINUOUS DISTRIBUTIONS y Figure 8.3: Normal density 8.5 Normal distribution Theorem 8.9. e 2 2 d = 2π. Proof. We apply Theorem 8.6 with A = R 2 {(, ) R 2 : }, B = (, ) (, 2π) and (, y) = g(r, θ) = (r cos θ, r sin θ). The Jacobian of the transformation is ( ) (, y) (r, θ) = det cos θ r sin θ = r sin θ r cos θ Thus, ( = 2π e 2 e r 2 ) 2 2 d = 2 r dr = 2π( e r2 2 ) 2 e 2 y2 2 d dy = = 2π. 2π e r 2 2 r dθ dr Definition 8.6. A r.v. X has a normal distribution with parameters µ 1 and σ 2, where µ R and σ >, if its density is f() = 1 2πσ e ( µ)2 2σ 2. The previous definition makes sense because f is a nonnegative function and by the change of variables µ = t σ 1 2πσ e ( µ)2 2σ 2 d = 1 2π e t2 2 dt. If X has a normal distribution with parameters µ and σ 2, where µ R and σ >, we say that X d N(µ, σ 2 ). Theorem 8.1. If X has a normal distribution with parameters µ and σ 2, then E[X] = µ and Var(X) = σ 2. 1 µ is a the (small) lower case Greek letter mu. σ is the lower case Greek letter sigma.