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Claims Frequency Distribution Models Claims Frequency Distribution Models Chapter 6 Stat 477 - Loss Models Chapter 6 (Stat 477) Claims Frequency Models Brian Hartman - BYU 1 / 19 Introduction Introduction Here we introduce a large class of counting distributions, which are discrete distributions with support consisting of non-negative integers. Generally used for modeling number of events, but in an insurance context, the number of claims within a certain period, e.g. one year. We call these claims frequency models. Let N denote the number of events (or claims). Its probability mass function (pmf), pk = Pr(N = k), for k = 0; 1; 2;:::, gives the probability that exactly k events (or claims) occur. Chapter 6 (Stat 477) Claims Frequency Models Brian Hartman - BYU 2 / 19 Introduction probability generating function The probability generating function N P1 k Recall the pgf of N: PN (z) = E(z ) = k=0 pkz . Just like the mgf, pgf also generates moments: 0 00 PN (1) = E(N) and PN (1) = E[N(N − 1)]: More importantly, it generates probabilities: dm P (m)(z) = E zN = E[N(N − 1) ··· (N − m + 1)zN−m] N dzm 1 X k−m = k(k − 1) ··· (k − m + 1)z pk k=m Thus, we see that 1 P (m)(0) = m! p or p = P (m)(0): N m m m! Chapter 6 (Stat 477) Claims Frequency Models Brian Hartman - BYU 3 / 19 Some discrete distributions Some familiar discrete distributions Some of the most commonly used distributions for number of claims: Binomial (with Bernoulli as special case) Poisson Geometric Negative Binomial The (a; b; 0) class The (a; b; 1) class Chapter 6 (Stat 477) Claims Frequency Models Brian Hartman - BYU 4 / 19 Bernoulli random variables Bernoulli random variables N is Bernoulli if it takes only one of two possible outcomes: 1; if a claim occurs N = : 0; otherwise q is the standard symbol for the probability of a claim, i.e. Pr(N = 1) = q. We write N ∼ Bernoulli(q). Mean E(N) = q and variance Var(N) = q(1 − q) Probability generating function: PN (z) = qz + (1 − q) Chapter 6 (Stat 477) Claims Frequency Models Brian Hartman - BYU 5 / 19 Binomial random variables Binomial random variables We write N ∼ Binomial(m; q) if N has a Binomial distribution with pmf: m m! p = Pr(N = k) = qk(1 − q)m−k = qk(1 − q)m−k; k k k!(m − k)! for k = 0; : : : ; m. Binomial r.v. is also the sum of independent Bernoulli's with Pm N = k=1 Nk where each Nk ∼ Bernoulli(q). Mean E(N) = mq and variance Var(N) = mq(1 − q) Probability generating function: m PN (z) = [qz + (1 − q)] Chapter 6 (Stat 477) Claims Frequency Models Brian Hartman - BYU 6 / 19 Poisson random variables Poisson random variables N ∼ Poisson(λ) if pmf is λk p = P (X = k) = e−λ ; for k = 0; 1; 2;::: k k! Mean and variance are equal: E(N) = Var(N) = λ Probability generating function of a Poisson: λ(z−1) PN (z) = e : Sums of independent Poissons: If N1;:::;Nn be n independent Poisson variables with parameters λ1; : : : ; λn, then the sum N = N1 + ··· + Nn has a Poisson distribution with parameter λ = λ1 + ··· + λn. Chapter 6 (Stat 477) Claims Frequency Models Brian Hartman - BYU 7 / 19 Poisson random variables decomposition Decomposition property of the Poisson Suppose a certain number, N, of events will occur and N ∼ Poisson(λ). Suppose further that each event is either a Type 1 event with probability p or a Type 2 event with probability 1 − p. Let N1 and N2 be the number of Types 1 and 2 events, respectively, so that N = N1 + N2. Result: N1 and N2 are independent Poisson random variables with respective means E(N1) = λp and E(N2) = λ(1 − p): Proof to be provided in class. This result can be extended to several types, say 1; 2; : : : ; n, with N = N1 + ··· + Nn. Chapter 6 (Stat 477) Claims Frequency Models Brian Hartman - BYU 8 / 19 Poisson random variables example Example Suppose you have a portfolio of m independent and homogeneous risks where the total number of claims from the portfolio has a Poisson distribution with mean parameter λ. Suppose the number of homogeneous risks in the portfolio was changed to m∗. Prove that the total number of claims in the new portfolio still has a Poisson distribution. Identify its mean parameter. Chapter 6 (Stat 477) Claims Frequency Models Brian Hartman - BYU 9 / 19 Negative binomial random variable Negative binomial random variable N has a Negative Binomial distribution, written N ∼ NB(β; r), if its pmf can be expressed as k + r − 1 1 r β k p = Pr(N = k) = ; k k 1 + β 1 + β for k = 0; 1; 2;::: where r > 0; β > 0. Probability generating function of a Negative Binomial: −r PN (z) = [1 − β(z − 1)] : Mean: E(N) = rβ Variance: Var(N) = rβ(1 + β). Clearly, since β > 0, the variance of the NB exceeds the mean. Chapter 6 (Stat 477) Claims Frequency Models Brian Hartman - BYU 10 / 19 Negative binomial random variable Geometric Geometric random variable The Geometric distribution is a special case of the Negative Binomial with r = 1. N is said to be a Geometric r.v. and written as N ∼ Geometric(p) if its pmf is therefore expressed as 1 β k p = Pr(N = k) = ; for k = 0; 1; 2;:::: k 1 + β 1 + β Mean is E(N) = β and variance is Var(N) = β(1 + β). Its pgf is: 1 P (z) = N 1 − β(z − 1) Chapter 6 (Stat 477) Claims Frequency Models Brian Hartman - BYU 11 / 19 Negative binomial random variable mixture of Poissons Negative Binomial as a mixture of Poissons Suppose that conditionally on the parameter risk parameter Λ = λ, the random variable N is Poisson with mean λ. To evaluate the unconditional probability of N, use the law of total probability: Z 1 Z 1 e−λλk pk = Pr(N = k j Λ = λ)u(λ)dλ = u(λ)dλ, 0 0 k! where u(·) is the pdf of Λ. In the case where Λ has a gamma distribution, it can be shown (to be done in lecture) that N has a Negative Binomial distribution. In effect, the mixed Poisson, with a gamma mixing distribution, is equivalent to a Negative Binomial. Chapter 6 (Stat 477) Claims Frequency Models Brian Hartman - BYU 12 / 19 Negative binomial random variable limiting case Limiting case of the Negative Binomial Prove that the Poisson distribution is a limiting case of the Negative Binomial distribution. Proof to be discussed in class. Chapter 6 (Stat 477) Claims Frequency Models Brian Hartman - BYU 13 / 19 Negative binomial random variable SOA question SOA question Actuaries have modeled auto windshield claim frequencies and have concluded that the number of windshield claims filed per year per driver follows the Poisson distribution with parameter λ, where λ follows the gamma distribution with mean 3 and variance 3. Calculate the probability that a driver selected at random will file no more than 1 windshield claim next year. Chapter 6 (Stat 477) Claims Frequency Models Brian Hartman - BYU 14 / 19 Special class of distributions the (a; b; 0) class Special class of distributions The (a; b; 0) class of distributions satisfies the recursion equations of the general form: p b k = a + ; for k = 1; 2;:::: pk−1 k The three distributions (including Geometric as special case of NB) are the only distributions that belong to this class: Binomial, Poisson, and Negative Binomial. It can be shown that the applicable parameters a and b are: Distribution Values of a and b q q Binomial(m; q) a = − 1−q , b = (m + 1) 1−q Poisson(λ) a = 0, b = λ β β NB(β; r) a = 1+β , b = (r − 1) 1+β Chapter 6 (Stat 477) Claims Frequency Models Brian Hartman - BYU 15 / 19 Special class of distributions the (a; b; 0) class Example Suppose N is a counting distribution satisfying the recursive probabilities: p 4 1 k = − ; pk−1 k 3 for k = 1; 2;::: Identify the distribution of N. Chapter 6 (Stat 477) Claims Frequency Models Brian Hartman - BYU 16 / 19 Special class of distributions the (a; b; 0) class SOA question The distribution of accidents for 84 randomly selected policies is as follows: Number of Accidents Number of Policies 0 32 1 26 2 12 3 7 4 4 5 2 6 1 Identify the frequency model that best represents these data. Chapter 6 (Stat 477) Claims Frequency Models Brian Hartman - BYU 17 / 19 Truncation and modification at zero the (a; b; 1) class Truncation and modification at zero The (a; b; 1) class of distributions satisfies the recursion equations of the general form: p b k = a + ; for k = 2; 3;:::: pk−1 k Only difference with the (a; b; 0) class is the recursion here begins at p1 instead of p0. The values from k = 1 to k = 1 are the same up to a constant of proportionality. For the class to be a distribution, the remaining probability must be set for k = 0. zero-truncated distributions: the case when p0 = 0 zero-modified distributions: the case when p0 > 0 The distributions in the second subclass is indeed a mixture of an (a; b; 0) and a degenerate distribution. A zero-modified distribution can be viewed as a zero-truncated by setting p0 = 0. Chapter 6 (Stat 477) Claims Frequency Models Brian Hartman - BYU 18 / 19 Truncation and modification at zero illustrative example Illustrative example Consider the zero-modified Geometric distribution with probabilities 1 p = 0 2 12k−1 p = ; for k = 1; 2; 3;::: k 6 3 Derive the probability generating function, the mean and the variance of this distribution.
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