ACTSC 431 - Loss Models 1 FALL 2007 TEST #2
Name : ID Number : 1. (8 marks) Suppose that the per payment r.v. Yp is PAR( = 5; = 200) distributed. In addition, the number of payments Np has a zero-modi.ed BIN(n = 4; q = 0:4) distribution with a probability
ACTSC 431/831: Loss Models 1 Spring 2009 - TEST #2
1. (12 marks) Let S1 ; S2 and S3 be three independent compound Poisson random variables: S1 has a Poisson parameter S2 has a Poisson parameter S3 has a Poisson parameter
1 2 3
= 2 and secondary distributi
ACTSC 431/831: Loss Models 1 Fall 2009 - TEST #2
1. (12 marks) The ground-up loss X has density fX (x) = p for 0 < p < 1. (a) (3 marks) Find E [X] and V ar (X). Solution: First, we remark that X is a mixture of an exponential distribution and an Erlang-2
ACTSC 431 - Loss Models 1 TEST #1
1. (20 marks) Suppose that the ground-up loss random variable X has density function fX (x) = e for a 2 (0; 1) and > 0.
x
a + 2 (1
a) e
x
,
x
0,
(a) (1 mark) Determine the survival distribution of the random variable X. S
ACTSC 431 - Loss Models 1 TEST #1
1. (13 marks) Let X be a random variable with cumulative distribution function FX (x) = for > 1. De.ne Y =
1 X 1
x
,
0<x< ,
.
(a) (3 marks) Find the density function of Y . Solution: By de.nition, FY (y) = Pr = Pr 1 X 1 X
LOSS MODELS 1 - ACTSC 431/831, FALL 2008
Instructor: Jun Cai, MC 6014, 519-888-4567 x 36990, [email protected] Instructor's Office Hours: 1:30 - 2:30, Tuesday and Thursday Teaching Assistants and Their Office Hours: To be announced on the course webp
Continuous distributions
1. X GAM( ; ): Gamma distribution with parameters
x
> 0 and
>0
(x) e density fX (x) = x ( ) , mean E [X] = variance V ar (X) = 2 m.g.f. MX (s) = 1 1 s , For the gamma distribution de.ned above, the parameter the parameter is a sca
Formula Sheet for ACTSC 431/831 Fall 2008
1. Discrete Distributions (a) Poisson with parameter > 0: A random variable X is said to have a Poisson distribution denoted by X P () if X has the following probability function (pf): Prcfw_X = k = with E(X) = V
During the final class on 3rd December, it was mentioned that the Cramer's asymptotic expression corresponds to the exact ruin probability when claim sizes are exponential with mean . Here we give the example that the expression 1 -Ru (u) = e 1+ solves th
ACTSC 431 - Loss Models 1 FALL 2007 TEST #2
Name : ID Number : 1. (8 marks) Suppose that the per payment r.v. Yp is PAR( = 5; = 200) distributed. In addition, the number of payments Np has a zero-modi.ed BIN(n = 4; q = 0:4) distribution with a probability
ACTSC 431/831 QUESTION SET 3
1. A ground-up model of the losses has cdf FX (x) = 1 - 1458000(45 + 2x) (90 + x)4 for x 0.
The insurance policy calls for an ordinary deductible of 30 to be imposed. Moreover, the number of payments NP has a logarithmic distr
ACTSC 431/831 QUESTION SET 2
1. Suppose that GAM(r, ). Given = , let N POI( + ) where > 0 is a constant. (a) Show that N has pgf
P (z) =
n=0
pn z n = e(z-1) [1 - (z - 1)]-r .
(b) Find p0 . (c) Show that P (z) = [ + r(1 + - z)-1 ]P (z), and hence (1 + )P (
Review Notes for Loss Models 1 - ACTSC 431/831, FALL 2008 Part 6 The Classical Continuous-Time Ruin Model
In this part, times are measured in years, unless stated otherwise. 1. Poisson Process: Let Nt be the number of claims up to time t or the number of
Review Notes for Loss Models 1 - ACTSC 431/831, FALL 2008 Part 5 Aggregate Loss Models
Roughly speaking, an aggregate loss model is used to describe the total loss of an insurance portfolio in a xed time period. 1. Individual Risk Model: There are n polic
Review Notes for Loss Models 1 - ACTSC 431/831, FALL 2008 Part 4 Frequency and Severity with Coverage Modications
Let X be the ground-up loss for an insurance policy or an insurer and assume that X is a continuous r.v. with cdf F (x), sf S (x), and pdf f
Review Notes for Loss Models 1 - ACTSC 431/831, FALL 2008 Part 3 Frequency Models
1. Frequency models are used to model the number of events or claims. 2. A counting random variable N is a nonnegative integer-valued random variable with pf pk = Pcfw_N = k
Review Notes for Loss Models 1 - ACTSC 431/831, FALL 2008 Part 2 Severity Models
1. Severity models are distributions that are used to model the amount of a claim. 2. A parametric distribution is a set of distribution functions. Any member in the set is d
Review Notes for Loss Models 1 - ACTSC 431/831, Fall 2008 Part 1 Random Variables and Distributional Quantities
1. The distribution function (df ) or cumulative distribution function (cdf ) and survival function (sf ) of a random variable (rv) X are dened
SOLUTIONS TO MIDTERM TEST ACTSC 431/831, FALL 2008 1. (a) Let Y L be the per loss r.v. for the policy. Then
0, X 20,
X 20; 20 < X 100;
YL =
80 + 0.5(X 100), 100 < X 120; 90, X > 120.
Thus, the mean of the loss for the policy is E (Y L ) =
120 1 1 dx + (
ASSIGNMENT 5 ACTSC 431/831, FALL 2008 Due at the beginning of the class on Thursday, November 27. 1. Assume that the aggregate losses in a portfolio are a collective risk model S = N Xi , i=1 k 0 1 2 3 4 5 where N has the following pf , and X1 , X2 , . Pr
SOLUTIONS TO ASSIGNMENT 4 ACTSC 431/831, FALL 2008 1. The conditional distribution of X , given = , is a Poisson distribution with mean . The pgf of X is given by PX (z ) = M (z 1), where M (t) is the mgf of . Since is innitely divisible, for any n = 1, 2
ASSIGNMENT 4 ACTSC 431/831, FALL 2008 Due at the beginning of the tutorial on Wednesday, November 12 1. Prove that if X has a mixed Poisson distribution and the mixing distribution is innitely divisible, then X is innitely divisible. [6 marks] 2. Let N L
SOLUTIONS TO ASSIGNMENT 3 ACTSC 431/831, FALL 2008
a 1. (a) We have 1 = fX (x)dx = 0.06 5 + 0.1 5 + 3000 , which gives a = 600. Then E (X ) = xfX (x)dx = 7.5. Furthermore, we have E (X 2 ) = x2 fX (x)dx = 91.6667. Hence, V ar(X ) = 35.42.
(b) We have
0,
ASSIGNMENT 3 ACTSC 431/831, FALL 2008 Due at the beginning of the class on Tuesday, October 21. 1. Let loss X have a three-component spliced distribution. The pdf of X has the following form:
f X ( x) =
0, 0.06, 0.1, a , x4
x < 0, 0 x < 5, 5 x < 10, x 1
SOLUTIONS TO ASSIGNMENT 2 ACTSC 431/831, FALL 2008 1. (a)
1 i. E (YA ) = 1 (E (X1 ) + E (X2 ) = 30 and E (YB ) = 2 (E (X1 ) + E (X2 ) = 30. 2 1 2 2 2 ii. E (YA ) = 2 (E (X1 ) + E (X2 ) = 1306.5 and thus V ar(YA ) = 1306.5 302 =
406.5. V ar(YB ) = 1 (V ar(
ASSIGNMENT 2 ACTSC 431/831, FALL 2008 Due at the beginning of the tutorial on Wednesday, October 8. 1. Let two independent random variables X1 and X2 have distributions F1 and F2 , respectively. Actuary A assumes that the loss of a portfolio is YA and the