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Claims Frequency Distribution Models
Chapter 6
Stat 477 - Loss Models
Chapter 6 (Stat 477)
Claims Frequency Models
Brian Hartman - BYU
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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
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Introduction
probability generating function
The probability generating function
Recall the pgf of N : PN (z) = E(z N ) =
P∞
k=0 pk z
k.
Just like the mgf, pgf also generates moments:
PN0 (1) = E(N ) and PN00 (1) = E[N (N − 1)].
More importantly, it generates probabilities:
m
d
(m)
N
z
PN (z) = E
= E[N (N − 1) · · · (N − m + 1)z N −m ]
dz m
∞
X
=
k(k − 1) · · · (k − m + 1)z k−m pk
k=m
Thus, we see that
(m)
PN (0) = m! pm or pm =
Chapter 6 (Stat 477)
Claims Frequency Models
1 (m)
P (0).
m!
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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)
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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)
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Binomial random variables
Binomial random variables
We write N ∼ Binomial(m, q) if N has a Binomial distribution with
pmf:
m k
m!
q k (1 − q)m−k ,
pk = Pr(N = k) =
q (1 − q)m−k =
k!(m − k)!
k
for k = 0, . . . , m.
Binomial
P r.v. is also the sum of independent Bernoulli’s with
N= m
k=1 Nk where each Nk ∼ Bernoulli(q).
Mean E(N ) = mq and variance Var(N ) = mq(1 − q)
Probability generating function:
PN (z) = [qz + (1 − q)]m
Chapter 6 (Stat 477)
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Poisson random variables
Poisson random variables
N ∼ Poisson(λ) if pmf is
pk = P (X = k) = e−λ
λk
, for k = 0, 1, 2, . . .
k!
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 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)
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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)
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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)
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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
r k
k+r−1
1
β
,
pk = Pr(N = k) =
k
1+β
1+β
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.
Chapter 6 (Stat 477)
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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
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) =
Chapter 6 (Stat 477)
1
1 − β(z − 1)
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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 ∞
Z ∞ −λ k
e λ
pk =
Pr(N = k | Λ = λ)u(λ)dλ =
u(λ)dλ,
k!
0
0
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)
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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)
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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)
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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:
b
pk
= 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
Binomial(m, q)
Poisson(λ)
NB(β, r)
Chapter 6 (Stat 477)
Values of a and b
q
q
a = − 1−q
, b = (m + 1) 1−q
a = 0, b = λ
β
β
a = 1+β
, b = (r − 1) 1+β
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Special class of distributions
the (a, b, 0) class
Example
Suppose N is a counting distribution satisfying the recursive probabilities:
4 1
pk
= − ,
pk−1
k 3
for k = 1, 2, . . .
Identify the distribution of N .
Chapter 6 (Stat 477)
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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
0
1
2
3
4
5
6
Number of Policies
32
26
12
7
4
2
1
Identify the frequency model that best represents these data.
Chapter 6 (Stat 477)
Claims Frequency Models
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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:
pk
b
= 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 = ∞ 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)
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Truncation and modification at zero
illustrative example
Illustrative example
Consider the zero-modified Geometric distribution with probabilities
p0 =
pk =
1
2
1 2 k−1
, for k = 1, 2, 3, . . .
6 3
Derive the probability generating function, the mean and the variance of
this distribution.
Chapter 6 (Stat 477)
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