Solutions to Assignment 1 ACTSC 433/833, Winter 2012
1. (a) We know that M SEX () = V ar(X ) = n and M SES 2 () = V ar(S 2 ) = E (S 4 ) 2 .
Note that (n 1)S 2 = n=1 (Xi X )2 = n=1 Yi2 nY 2 , where Yi = Xi
i
i
and Y = X . By using the mgf of the Poisson d
ACTSC 433 - WINTER 2009 (SECTION 1)
MID-TERM EXAM #1FEBRUARY 32008
'Su.~e~
~nlu~lI'~
NAME: _
Aids:
Time: Examiner:
Calculator 1 hour Johnny S.H. Li
Question 1 2 3 4 Total
Maximum 12 8 8 13 41
Mark
1. In a study of survival times (X), you are given the fol
SOLUTIONS TO TEST # 1 ACTSC 433/833, WINTER 2012
1. (a) The distribution function of X(n) is
0,
FX(n) (x) =
1,
x2n
,
2n
x < 0,
0 x < ,
x .
Thus, E (n ) = 0 (1 FX(n) (x)dx = 1
asymptotically unbiased for .
(b) The probability density function of Tn =
1
2
ACTSC 433/833 Winter 09 Assignment 2 Due: 10 March 2009 (hand in to the instructor at the beginning of class)
1. You are given the following censored and truncated data: I 1 2 3 4 5 6 7 8 di 0 0 0 2.5 2 0 3 0 xi 1 2 4 5 8 ui 3 4.5 10 -
(a) Create a table
Assignment 1 ACTSC 433/833, Winter 2012
(Due at the beginning of the class on January 19)
1. Let X1 , ., Xn be a sample of the random variable X that has a Poisson distribution with
mean > 0.
(a) Calculate M SEX () and M SES 2 (), where X and S 2 are the
Assignment 2 ACTSC 433/833, Winter 2012
(Due at the beginning of the class on February 9)
1. Let X1 , ., Xn be a random sample from the distribution function F (x) = 1 S (x) and
Fn (x) be the empirical estimator for F (x).
(a) Show that for any > 0, Pr |F
Solutions to Assignment 2 ACTSC 433/833, Winter 2012
1. (a) Pr |Fn (x) F (x)| >
2n
4nV ar(Fn (x)
2
=
4F (x)(1F (x)
2
1
.
2
(b) Note that = Prcfw_a < X b = F (b) F (a). Thus n = Fn (b) Fn (a) is consistent
(1)
for since E n = and V ar(n ) = n 0 as n .
(c)
Outline
ACTSC 433/833 Winter 09 Chapter 11 Contemporary Issues Part II Stochastic Mortality Models
Johnny Li
Department of Statistics and Actuarial Science, University of Waterloo
March, 2009
Outline
Outline
1
Older Methods for Estimating Future Mortality
Assignment 3 ACTSC 433/833, Winter 2012
(Due at the beginning of the class on March 1)
1. You are given the following ages at time of death for 10 individuals: 38, 45, 52, 52, 56,
59, 68, 70, 74, and 82. Determine the kernel density estimate for the proba
Assignment 1 ACTSC 433/833, Winter 2013
(Due at the beginning of the class on January 22)
1. Let X1 , X2 , X3 be a sample of size 3 from the uniform distribution U ( 2, + 2), where
2. Let X(2) be the sample median.
(a) Derive the distribution function FX
Assignment 2 ACTSC 433/833, Winter 2013
(Due at the beginning of the class on January 31)
1. Let X1 , ., Xn be a random sample from the distribution function F (x) = 1 S (x) and
Fn (x) be the empirical estimator for F (x).
cfw_
(a) Show that for any > 0,
SOLUTIONS TO MIDTERM TEST # 2 ACTSC 433/833,
WINTER 2012
1. (a) The kernel density estimate for the probability that an individual aged 90 will survive
at least 3 years using the uniform kernel with bandwidth b = 0.5 is S (3) = 1 (0 +
8
0 + 0.9 + 1 + 2 1
Solutions to Assignment 5 ACTSC 433/833, Winter 2012
1. (a) For Line 1: S1 (x) = S0 (x) and f1 (x) = 100 , x > 100. For Line 2: S2 (x) = (S0 (x)b
x+1
b100b
and f2 (x) = xb+1 , x > 100, where b = e . Thus,
L(, b) = f1 (135)f1 (160)f1 (240)f1 (320)f1 (380)f
Outline
ACTSC 433/833 Winter 2009 Chapter 13 Contemporary Issues Part III Mortality Derivatives
Johnny Li
Department of Statistics and Actuarial Science, University of Waterloo
April, 2009
Outline
Outline
1
Introduction Four Examples Swiss Re Mortality Bo
Outline
ACTSC 433/833 Winter 09 Chapter 11 Contemporary Issues Part I What Is Longevity Risk?
Johnny Li
Department of Statistics and Actuarial Science, University of Waterloo
March, 2009
Outline
Outline
1
The Problem Mortality on the Move The Need for an
ACTSC 433/833 Analysis of Survival Data
Johnny S.H. Li
Department of Statistics and Actuarial Science, University of Waterloo
January 6, 2009
Why ACTSC 433/833? About the Course About the Instructor
A Required Course?
I need ACTSC 433 to graduate!
Why ACT
ACTSC 433 - WINTER 2009 (SECTION 2)
~.n~t" ~ \el
'S.\
MID-TERM EXAM #1 - FEBRUARY 32008
NAME:
_
Aids:
Time: Examiner:
Calculator 1 hour Johnny S.H. Li
Question 1 2 3 4 Total
Maximum 12 8 8 13 41
Mark
1. In a study of survival times (X), you are given the
Assignment 4 ACTSC 433/833, Winter 2012
(Due at the beginning of the class on March 15)
1. Four losses are observed with values 96, 135, 278, and 396. Six other losses are known to
be less than 90. Losses have an inverse exponential distribution with dist
Solutions to Assignment 4 ACTSC 433/833, Winter 2012
1. (a) We have f (x) = F (x) =
/x
e
,x
x2
> 0. Thus,
L() = f (96)f (135)f (278)f (396)(F (90)6 4 e0.0906131 .
It follows from
d ln L()
d
= 0 that = 44.1437.
(b) It follows from F (x) = 0.5 that the max
Assignment 5 ACTSC 433/833, Winter 2012
(Due at the beginning of the class on March 29)
For this assignment, if needed, you can use any software such as Maple, Mathematica, and MATLAB to nd roots of an equation.
1. A one-covariate Cox proportional hazards
Assignment 3 ACTSC 433/833, Winter 2013
(Due at the beginning of the class on February 26)
1. Let S (t) be the Product-Limit estimator for the survival function S (t).
Calculate Cov (S (yj ), S (yj +1 ) for j = 1, ., k 1.
[8 marks]
2. A mortality study is