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
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
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
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) Determin
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
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
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 distribu
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
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
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
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. More
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 th
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
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 t
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
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
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 se
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 surviv
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
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
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
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 divisib
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
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 (
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 th
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. Actua