Claims Frequency Distribution Models, Schemes and Mind Maps of Law

Here we introduce a large class of counting distributions, which are discrete distributions with support consisting of non-negative integers. Generally used for ...

Typology: Schemes and Mind Maps

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Claims Frequency Distribution Models
Chapter 6
Stat 477 - Loss Models
Chapter 6 (Stat 477) Claims Frequency Models Brian Hartman - BYU 1 / 19
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Claims Frequency Distribution Models

Chapter 6

Stat 477 - Loss Models

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.

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

Bernoulli random variables

Bernoulli random variables

N is Bernoulli if it takes only one of two possible outcomes:

N =

1 , if a claim occurs 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)

Poisson random variables

Poisson random variables

N ∼ Poisson(λ) if pmf is

pk = P (X = k) = e−λ^ λk k! , for k = 0, 1 , 2 ,...

Mean and variance are equal: E(N ) = Var(N ) = λ Probability generating function of a Poisson:

PN (z) = eλ(z−1).

Sums of independent Poissons: If N 1 ,... , Nn be n independent Poisson variables with parameters λ 1 ,... , λn, then the sum

N = N 1 + · · · + Nn

has a Poisson distribution with parameter λ = λ 1 + · · · + λn.

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 N 1 and N 2 be the number of Types 1 and 2 events, respectively, so that N = N 1 + N 2. Result: N 1 and N 2 are independent Poisson random variables with respective means

E(N 1 ) = λp and E(N 2 ) = λ(1 − p).

Proof to be provided in class. This result can be extended to several types, say 1 , 2 ,... , n, with N = N 1 + · · · + Nn.

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

pk = Pr(N = k) =

k + r − 1 k

1 + β

)r( β 1 + β

)k ,

for k = 0, 1 , 2 ,... where r > 0 , β > 0. Probability generating function of a Negative Binomial:

PN (z) = [1 − β(z − 1)]−r.

Mean: E(N ) = rβ Variance: Var(N ) = rβ(1 + β). Clearly, since β > 0 , the variance of the NB exceeds the mean.

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

pk = Pr(N = k) =

1 + β

β 1 + β

)k , for k = 0, 1 , 2 ,....

Mean is E(N ) = β and variance is Var(N ) = β(1 + β). Its pgf is: PN (z) =

1 − β(z − 1)

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.

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.

Special class of distributions the (a, b, 0) class

Example

Suppose N is a counting distribution satisfying the recursive probabilities: pk pk− 1

k

for k = 1, 2 ,... Identify the distribution of N.

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.

Truncation and modification at zero illustrative example

Illustrative example

Consider the zero-modified Geometric distribution with probabilities

p 0 =

pk =

)k− 1 , for k = 1, 2 , 3 ,...

Derive the probability generating function, the mean and the variance of this distribution.