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Chapter 1- Set theory: π΄ β© (π΅ βͺ πΆ) = (π΄ β© π΅) βͺ (π΄ β© πΆ) π΄ βͺ (π΅ β© πΆ) = (π΄ βͺ π΅) β© (π΄ βͺ πΆ) Aο B DeMorganβs Laws: (π΄ βͺ π΅)β² = π΄β² β© π΅ β² (π΄ β© π΅)β² = π΄β² βͺ π΅ β² π(π΅) = π(π΅ β© π΄) + π(π΅ β© π΄β² ) π(π΄ βͺ π΅) = π(π΄) + π(π΅) β π(π΄ β© π΅) π(π΄ βͺ π΅ βͺ πΆ) = π(π΄) + π(π΅) + π(πΆ) β π(π΄ β© π΅) β π(π΅ β© πΆ) β π(πΆ β© π΄) + π(π΄ β© π΅ β© πΆ) Independence: π(π΄ β© π΅) = π(π΄)π(π΅) Chapter 2- Counting Techniques: Permutations: Order matters!! π(π, π) = number of ordered samples of size k from n without replacement. π(5,3) = 5 β 4 β 3 π(10,2) = 10 β 9 Combinations: Order doesnβt matter!! πΆ(π, π) = number of unordered samples of size k from n using sampling without replacement. 5β4β3 10β9 5 10 πΆ(5,3) = 3β2β1 = ( ) πΆ(10,2) = 2β1 = ( ) 2 3 Samples With Replacement: N objects, sample size k Probability= ππ Distinguishable permutations: 11! How many distinct words can you make from the word Mississippi = 4!4!2! Chapter 3- Conditional Probability: π(π΄|π΅) = 1 β π(π΄β² |π΅) π(π΄|π΅) = π(π΄β©π΅) π(π΅) π(π΄ β© π΅) = π(π΄|π΅)π(π΅) Bayesβ Theorem: π(π΄)π(π΅|π΄) π(π΄|π΅) = π(π΅) π(π΅) = π(π΄)π(π΅|π΄) + π(π΄β² )π(π΅|π΄β² ) Independence: π(π΄|π΅) = π(π΄) π(π΄ β© π΅ β© πΆ) = π(π΄)π(π΅|π΄)π(πΆ|π΄ β© π΅) Chapter 4- Random Variables: Continuous random variables: The probability of any single value is zero, since there are an infinite number of values. Probability Density Function (pdf): π(π₯) = π(π = π₯) π π(π β€ π β€ π) = β«π π(π₯)ππ₯ = πΉ(π) β πΉ(π) β β«ββ π(π₯) = 1 Continuous Discrete Function (cdf): πΉ(π) = π(π β€ π₯) π₯ π(π β€ π₯) = β«ββ π(π₯)ππ₯ π(π β€ π₯) = π(π < π₯) πΉ(ββ) = 0 πΉ(β) = 1 Converting between pdf and cdf: π₯ πΉ(π₯) = β«ββ π(π₯)ππ₯ π(π₯) = πΉ β² (π₯) Measures of Central Tendency: πππ(π + π + π) = πππ(π) + πππ(π) + 2πΆππ£(π, π) πππ(ππ β ππ) = π2 πππ(π) + π 2 πππ(π) β 2πππΆππ£(π, π) π(ππ + π) = |π|π(π) πΈ(ππ + π) = ππΈ(π) + π Approximations of discrete random variables: When using an integral to approximate a discrete case (which can only take integer values), you need to change the limits. π+0.5 π(π β€ π β€ π) β π(π β 0.5 < π < π + 0.5) = β« πβ0.5 Chebyshevβs Inequality: π(π¦)ππ¦ X is a random variable with mean π and variance π 2 . 1 π(|π β π| β₯ ππ) β€ 2 π 1 The probability that X is not within 2π of the mean, is π 2 . Percentiles: π(π β€ π₯π ) = πΉ(π₯π ) = π The probability that X is in the 35% percentile: π(π β€ π₯0.35 ) = πΉ(π₯0.35 ) = 0.35 Solve for x. Moment generating function: ππ₯ (π‘) = ππ π‘π₯ ππ₯ (0) = 1 ππ₯ β² (0) = πΈ(π) ππ₯ β²β² (0) = πΈ(π 2 ) π = π + ππ₯ ππ (π‘) = π ππ‘ ππ₯ (ππ‘) = π ππ‘ β π ππ‘π₯ ππ+π (π‘) = ππ₯ (π‘) β ππ (π‘) Mode: The point where π(π₯) reaches its maximum. Set π β² (π₯) = 0 (the maximum value of a function happens with the derivative =0). Solve for x. Median: -If given cdf, set πΉ(π₯) = 0.5, solve for x π₯ -If given pdf, integrate between β«0 π(π₯)ππ₯ and set answer equal to 0.5, solve for x. -For split distributions, draw graphs or substitute in limits to decide where 0.5 falls. Standard deviations: βWhat percent of claims fall within one standard deviation of the mean?β -Find πΈ(π), π -Make range: [πΈ(π) β π, πΈ(π) + π] -add up the percentage of claims in that range. Integrals: 4 |π₯| 4 2 π₯ π₯ β« ππ₯ = β« ππ₯ β β« ππ₯ β2 10 0 10 0 10 β« π₯ β1 ππ₯ = ln π₯ Chapter 5- Discrete Distributions: Name Binomial Negative Binomial Density f ( x) ο½ ο¨ ο©p q f ( x) ο½ ο¨ Geometric Hypergeometric Poisson Uniform MGF n k k nοk ο¨ M x ο¨t ο© ο½ pe t ο« q ο© n n failures before k successes ο¦ 1 ο qe t M x ο¨t ο© ο½ ο§ο§ ο¨ p οΆ ο· ο· οΈ n n failures before the first success ο¦ 1 ο qe t M x ο¨t ο© ο½ ο§ο§ ο¨ p οΆ ο· ο· οΈ c(m1 , x)c(m2 , n ο x) c(m, n) Sample of n taken from total of m n ο« k ο1 n ο©p q k n f ( x) ο½ q p f ( x) ο½ x = 0,1,2β¦n e οο¬ ο¬x f ( x) ο½ x! 1 f ( x) ο½ n Bernoulli f ( x) ο½ p x q n ο x Poisson Approx. to the Binomial e ο np (np ) x f ( x) ο½ x! ο¬ is the rate of a rare event x = 0,1,2β¦n M x (t ) ο½ great : n ο³ 100, np ο£ 10 Var ( X ) ο½ npq kq E( X ) ο½ p Var ( X ) ο½ kq q p Var ( X ) ο½ q E( X ) ο½ E( X ) ο½ t ο1) N (e t ο 1) M x ο¨t ο© ο½ pe t ο« q good : n ο³ 20, p ο£ 0.05 E ( X ) ο½ np ο1 e t (e Nt ο 1) ο¨ Var(X) οk m ο½ m1 ο« m2 M x (t ) ο½ e ο¬ (e E(X) ο© nm1 m E( X ) ο½ ο¬ E( X ) ο½ N ο«1 2 E( X ) ο½ p p2 p2 ο¦ m οΆο¦ m οΆο¦ m ο n οΆ Var ( X ) ο½ nο§ 1 ο·ο§ 2 ο·ο§ ο· ο¨ m οΈο¨ m οΈο¨ m ο 1 οΈ Var ( X ) ο½ ο¬ N 2 ο1 Var ( X ) ο½ 12 Var ( X ) ο½ pq Chapter 6- Continuous Distributions: Name Uniform Density π(π₯) = 1 πβπ π(π₯) = Beta Weibull Pareto Exponential (π + π β 1)! π₯ πβ1 (1 β π₯)πβ1 (π β 1)! (π β 1)! π(π₯) = ππ₯ π(π₯) = Mx ο¨t ο© CDF π₯ πβ1 β( βπ) π₯βπ πβπ πβ€π₯β€π πΉ(π₯) = π, π 0β€π₯β€1 No set formula ππ₯ (π‘) = π π, π πΉ(π₯) = 1 β π β( πΌπ πΌ (π₯ + π)πΌ+1 πΌ, π π πΌ πΉ(π₯) = 1 β ( ) π₯+π 1 βπ₯β π π π π πΉ(π₯) = 1 β π π₯β ) π βπ₯β π π¬(πΏπ ) = Gamma Normal Lognormal π(π₯) = π (β (π₯βπ)2 ) 2π2 πΌ, π (π, π ) Must be calculated with table of values (π, π 2 ) not equal to mean, variance Must be calculated with table of values 2 πβ2π Not worth it Not worth it π+π 2 πΈ(π) = N/A π πΈ(π) = π+π N/A 1 πΈ(π) = π β ( ) ! π π½π π! (πΆ β π). . (πΆ β π) ππ₯ (π‘) = πΈ(π) = π πΌβ1 Var ο¨X ο© πππ(π) = (π β π)2 12 πππ(π) = ππ (π + π)2 (π + π + 1) Not worth it use ππ₯ (π‘) to calculate 1 1 β ππ‘ πΈ(π) = π πππ(π) = π 2 1 (1 β ππ‘)πΌ πΈ(π) = πΌπ πππ(π) = πΌπ 2 πΈ(π) = π πππ(π) = π 2 ππ₯ (π‘) = βπ₯ π₯ πΌβ1 π βπ π(π₯) = πΌ π (π β 1)! π ππ‘ β π ππ‘ π‘(π β π) π π ππ π(π₯) = E ο¨X ο© ππ₯ (π‘) = π (ππ‘+ π2 π‘ 2 ) 2 2 N/A πΈ(π) = π (π+12π2 ) πππ(π) = π (2π+2π ) 2 β π (2π+π ) Chapter 7- Normal Distribution: The normal distribution: Has a complicated pdf and an unsolvable cdf. Use the table of values to look up the cdf for certain values. The Standard Normal Distribution has π = 0 and π = 1. π~π(π, π 2 ) Evaluating the standard normal: π(π β€ π₯) = Ξ¦(x) π(π < π₯) = 1 β π(π > π₯) π(π > π₯) = π(π β€ βπ₯) = Ξ¦(βπ₯) = 1 β Ξ¦(π₯) Ξ¦(π₯) = 1 β Ξ¦(βπ₯) π(π β€ βπ₯) = Ξ¦(βπ₯) = 1 β Ξ¦(π₯) π(π > βπ₯) = π(π < π₯) = Ξ¦(π₯) *Values in the table are for STANDARD normal distribution with π = 0 and π = 1.* What if π = 20 and π = 15? π~π(20,225) π₯βπ π₯βπ π (π < ) = Ξ¦( ) π π 12 β 20 π(π < 12) = π ( ) = π(π < β0.53) = 1 β Ξ¦(0.53) 15 π(π > 5) = 1 β π(π < 5) = 1 β π (π < 5 β 20 ) = 1 β π(π < β1) = 1 β (1 β Ξ¦(1)) = Ξ¦(1) 15 Adding independent distributions: π1 ~π(30,100), π2 ~π(40,150) π = π1 + π2 π~π(70,250) π~π(π, π 2 ) and π = ππ + π π~π(ππ + π, π2 π 2 ) Subtraction of distributions: π1 ~π(10,20) π2 ~π(30,10) π = π1 β π2 π~π(π1 β π2 , π12 + π22 )~π(β20,30) The Central Limit Theorem: π1 , π2 , β¦ , ππ are independent and identically distributed with π and π 2 . When n is large (π β₯ 30), then the sum: π1 + π2 + β― + ππ is ~π(ππ, ππ 2 ) So even if π1 , π2 , β¦ , ππ arenβt normally distributed, we can use the normal distribution on the sum. Sample mean: 1 π = (π1 + π2 + β― + ππ ) π π2 π~π (π, π ) 1 (π + π ) 2 1 2 22 π +π π~π (π, 1 22 2 ) π= Using normal distribution to estimate a discrete distribution: Continuous can take any value, discrete can take only integers. π(π < π₯ < π) β π(π β 0.5 < π₯ < π + 0.5) If Y follows a normal distribution π βΌ π(π, π 2 ), then: π + 0.5 β π π β 0.5 β π π(π < π₯ < π) = Ξ¦ ( )βΞ¦( ) π π Binomial: π~π(ππ, πππ) Poisson: π~π(π, π) Lognormal Distribution: -parameters π and π are not the mean and standard deviation. -if X follows a lognormal with parameters π and π 2 , and π = ln π₯, then Y follows a normal distribution with mean π and standard deviation π. -if X follows a normal distribution with mean π and variance π 2 , and π = π π₯ , then Y follows a lognormal distribution with parameters π and π 2 . Chapter 8- Multivariate Distribution: Joint pdf: ππ₯,π¦ (π₯, π¦) Find π(π₯ = π₯) by adding row π₯ = π₯ in table. π(π = π₯) = ππ₯ (π₯) π(π = π¦) = ππ¦ (π¦) ππ₯ (π₯|π = π¦) = ππ₯,π¦ (π₯, π¦) ππ¦ (π¦) π(π₯, π¦) = π(π¦|π₯)π(π₯) Independence: ππ₯,π¦ (π₯, π¦) = ππ₯ (π₯)ππ¦ (π¦) = marginal probability functions πΈ(ππ) = πΈ(π)πΈ(π) Joint continuous pdf: ππ₯,π¦ (π₯, π¦) β₯ 0 Double integral: Total area must sum to 1 β β β« β« ππ₯,π¦ (π₯, π¦)ππ₯ππ¦ = 1 ββ ββ Marginal Continuous probability functions: β ππ₯ (π₯) = β« ππ₯,π¦ (π₯, π¦)ππ¦ ββ β ππ¦ (π¦) = β« ππ₯,π¦ (π₯, π¦)ππ₯ ββ β πΈ(π) = β π₯ππ₯ (π₯) = β« π₯ππ₯ (π₯)ππ₯ ββ πΈ(π + π) = πΈ(π) + πΈ(π) Moment generating function: ππ,π (π , π‘) = πΈ(π π π₯+π‘π¦ ) πΈ(π|πΉ) = β π₯ππ₯ (π₯|π) πππ(π|π) = πΈ(π 2 |π) β πΈ(π|π)2 Covariance: πΆππ£(π, π) = πΈ(π, π) β πΈ(π)πΈ(π) πππ(π) = πΆππ£(π, π) πΆππ£(π, π) = πΆππ£(π, π) πΆππ£(ππ + ππ, π + π) = ππΆππ£(π, π) + ππΆππ£(π, π) + ππΆππ£(π, π) + ππΆππ£(π, π) Independence: πΆππ£(π, π) = 0 Correlation Coefficient: πΆππ£(π, π) π= ππ₯ ππ¦ Chapter 9- Transformations of Random Variables: Where x is a random variable, π = π(π₯) is a function of x. Method of Transformations: π(π¦) = ππ₯ (πβ1 (π¦)) β |[πβ1 (π¦)]β² | 1. ππ₯ (π₯) is usually given, as well as π = π(π₯) 1 βπ₯β ππ₯ (π₯) = π 100 100 π = 1.1π 2. Find πβ1 (π¦) (aka solve for x in second equation) π = 1.1π π = πβ1.1 3. Use equation π(π¦) = ππ₯ (πβ1 (π¦)) β |[πβ1 (π¦)]β² | = ( 1 β(πβ1.1)β 1 1 βπβ 100 ) β ( π )= π 110 100 1.1 110 Method of Distribution Functions: Use πΉπ¦ (π¦) = π(π β€ π¦) and πΉπ₯β² (π₯) = ππ₯ (π₯) 1. ππ₯ (π₯) and π = π(π₯) given πΉπ¦ (π¦) = π(π β€ π¦) Substitute Y ο π(π β€ π¦) = π(1.1π β€ π¦) Solve for x ο π(π β€ πβ1.1) Means: πΉπ₯ (πβ1.1) 2. Use cdf of ππ₯ (π₯) = 1 β π β( Notes on SOA 127 Packet: Average of two variances: π1 +π2 πππ ( 2 2 ) = (12 )πππ(π1 + π2 ) Average of n variances: ππ ππππ(π) πππ(π) πππ ( ) = = π π2 π πβ )/100 1.1 = 1 β πβ πβ 110 ππ‘ π(π¦) = π(π‘(π¦)) β | | = π(π‘) β π‘β² ππ¦ Integration by parts: π π β« π’ ππ£ = π’π£|ππ β β« π£ ππ’ π π Example 1: π π(π₯) = π₯π π₯ , β« π₯π π₯ ππ₯ = β« π’ ππ£ π π’=π₯ ππ£ = π π₯ ππ₯ ππ’ = ππ₯ π£ = ππ₯ implies π = π’π£|ππ π β β« π£ ππ’ = π₯π π₯ |ππ β β« π π₯ ππ₯ π π Example 2: π(π₯) = π₯π βπ₯βπ π , β« π₯π β π₯β π ππ₯ = β« π’ ππ£ π π’=π₯ π₯ ππ£ = π β βπ ππ₯ implies π = π’π£|ππ β β« π£ ππ’ = βππ₯π β π Example 3: π₯ π₯ π(π₯) = π β βπ , π π’=π₯ π₯ 1 ππ£ = π π β βπ ππ₯ ππ’ = ππ₯ π₯ π£ = βππ β βπ π β« π₯π β π₯β π π| π π₯β π π + β« ππ β π₯β π ππ₯ π ππ₯ = β« π’ ππ£ π implies π = π’π£|ππ β β« π£ ππ’ = βπ₯π β π π₯β π π| π ππ’ = ππ₯ π₯ π£ = βπ β βπ π + β« πβ π₯β π ππ₯ π Deductibles: An insurance policy has a deductible of d. This means that if x is less than d, the payout is 0, and if x is more than d, the payout is x-d. Payout = { 0 π₯<π π₯βπ π₯ β₯π β πΈ(π) = β«π (π₯ β π)π(π₯)ππ₯ β πΈ(π 2 ) = β«π (π₯ β π)2 π(π₯)ππ₯