Project ACTSC 832, Fall 2012
Deadline: 5:30pm, Thursday, December 20. Please email a PDF le of your project
to [email protected] by the deadline.
Project requirement: Projects must be typed using LaTex or Word with a cover page
that shows your name and st
ACTSC 432/832 - Loss Models II
Lecture 8
June 4, 2014
Lecture 8,
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Frequentist estimation for (a, b, 0) class
k frequency
nk number of observations
pk () = Pr(N = k|)
Moment matching
k=0 knk
k=0 nk
2
k=0 k nk
k=0 nk
E(
ACTSC 432/832 - Loss Models II
Lecture 5
May 21, 2014
Lecture 5,
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1/9
Method of percentile matching
Df of the model: F (x|)
= (1 , . . . , p ): the parameter to be estimated
Available data x1 , . . . , xn
p arbitrarily ch
ACTSC 432/832 - Loss Models II
Lecture 9
June 6, 2014
Lecture 9,
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Empirical distribution
Parametric distribution: a set of distribution functions, each
member of which is determined by specifying one or more
values calle
ACTSC 432/832 - Loss Models II
Lecture 10
June 9, 2014
Lecture 10,
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Graphic comparison
F (x): the model distribution function
Fn (x): the empirical distribution
Model versus data cdf plot
Model versus data density plot
ACTSC 432/832 - Loss Models II
Lecture 11
June 11, 2014
Lecture 11,
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Chi-square goodness-of-t test
t = c0 < c1 < < ck =
pj
= F (cj ) F (cj1 )
pnj
= Fn (cj ) Fn (cj1 )
Ej
= npj (number of expected observations)
Oj
= npnj
ACTSC 432/832 - Loss Models II
Lecture 12
June 16, 2014
Lecture 12,
ACTSC 432/832 - Loss Models II
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Likelihood ratio test
H0 : The data came from a population with distribution A
H1 : The data came from a population with distribution B
Distribution A m
ACTSC 432/832 - Loss Models II
Lecture 13
June 18, 2014
Lecture 13,
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Credibility
Credibility Theory a set of quantitative tools that allows an
insurer to perform prospective experience rating (adjust future
premiums based
ACTSC 432/832 - Loss Models II
Lecture 14
June 23, 2014
Lecture 14,
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Example
Ex14.1 An insurance company has decided to establish its
full-credibility requirements for an individual state rate ling.
the full-credibility s
ACTSC 432/832 - Loss Models II
Lecture 7
June 2, 2014
Lecture 7,
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Frequentist estimation for discrete distributions
Poisson
Negative binomial
Binomial
The (a, b, 1) class
k frequency
nk number of observations
pk pr
ACTSC 432/832 - Loss Models II
Lecture 4
May 14, 2014
Lecture 4,
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1/10
Conjugate prior distributions
A prior distribution is said to be a conjugate prior distribution
for a given model if the resulting posterior distribution
SOLUTIONS TO MIDTERM TEST 1 ACTSC 432/832
2
2
1. (a) The mgf of Xj is MXj (t) = E (etXj ) = E (E (etXj |) = E (et+t ) = et M (t) = e
which means that Xj has the normal distribution N (0, 6).
(b) The posterior distribution of , given X1 = x1 , ., Xn = xn ,
ASSIGNMENT 4, ACTSC 432/832, FALL 2012
Due at the beginning of the class on Friday, November 9.
1. Past claims data on a portfolio of policyholders are given below:
Policyholder Year 1
1
750
2
625
3
900
Year 2
800
600
950
Year 3
650
675
850
(a) Calculate
ASSIGNMENT 3, ACTSC 432/832, FALL 2012
Due at the beginning of the class on Friday, October 26.
1. Let Xj be the loss in year j for j = 1, 2, ., n, n + 1. Assume that
2
E (Xj ) = j , V ar(Xj ) = j , and Cov (Xi , Xj ) = ( 1) i j ,
i = j,
where i, j = 1, .
ASSIGNMENT 2, ACTSC 432/832, FALL 2012
Due at the beginning of the class on Friday, October 5.
1. The amounts of claims in the past n exposure units were X1 , X2 , , Xn , which are
independent and have a common distribution as X = N Yj , where N, Y1 , Y2
ASSIGNMENT 1 ACTSC 432/832, FALL 2012
Due at the beginning of the class on Wednesday, September 26.
1. Let X be the number of claims in a portfolio. Given = > 0, the conditional
distribution of X is the negative binomial distribution N B (2, ). The distri
ACTSC432/832
Spring 2014
Homework Set III (Due June 16, 2014)
13. Solve the unselected questions of Midterm 1.
4. An automobile insurance policy provides benets for accidents caused by both underinsured
and uninsured motorists. Data on 1,00 policies revea
ACTSC432/832
Spring 2014
Homework Set II (Due June 2, 2014)
1. A random sample of ve claims from a lognormal distribution is given as follows:
500
1, 000
1, 500
2, 500
4, 500
Estimate and by the method of moments. Estimate the probability that a loss will
ACTSC432/832
Spring 2014
Homework Set IV (Due July 2, 2014)
1. You are given:
(a) A sample of claim payments:
29
64
90
135
182
(b) Claim sizes are assumed to follow an exponential distribution.
(c) The mean of the exponential distribution is estimated usi
ACTSC432/832
Spring 2014
Homework Set I (Due May 21, 2014)
1. Let X have the uniform distribution over the range ( 2, + 2). That is,
2 < x < + 2.
fX (x) = 0.25,
Show that the median from a sample of size 3 is an unbiased estimator of .
2. You are given t
ACTSC 432/832 - Loss Models II
Lecture 3
May 12, 2014
Lecture 3,
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Bayesian estimation: Denition
Prior distribution () - probability distribution over the
space of possible parameter values. Represents our opinion
concern
ACTSC 432/832 - Loss Models II
Lecture 2
May 7, 2014
Lecture 2,
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Examples
Ex2.1 The maximum observation from a uniform distribution on
the interval (0, ),
n = maxcfw_X1 , . . . , Xn ,
is an asymptotically unbiased and co
ACTSC 432/832 - Loss Models II
Lecture 16
July 2, 2014
Lecture 16,
ACTSC 432/832 - Loss Models II
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The Bayesian methodology
Past claim r.v: X = (X1 , . . . , Xn )T
Past observations: x = (x1 , . . . , xn )T
Goal: set a rate to cover Xn+1 .
Xj has c
ACTSC 432/832 - Loss Models II
Lecture 17
July 16, 2014
Lecture 17,
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The Buhlmann credibility
The credibility premium is:
n
j Xj = Z X + (1 Z)
0 +
j=1
with
Z =
k
Lecture 17,
=
n
Buhlmann credibility factor
n+k
v
E[Var(Xj
Practice Questions 2 ACTSC 432/832, Fall 2014
1. The amounts of claims in the past n exposure units were X1 , X2 , , Xn , which are
independent and have a common distribution as X = N Yj , where N, Y1 , Y2 , ., are
j=1
independent, N has a negative binomi
Solutions to Practice Questions 1 ACTSC 432/832, Fall 2014
1. (a) The expected amount of a claim is E(X) = E(E(X|) = E(/2) = 25.
(b) V ar(X) = E(V ar(X|) + V ar(E(X|) = E(32 /4) + V ar(/2) = 2078.33.
(c) For x > 0, F (x) =
0.025x 0.03x log
3
100
1
1 +x
d