chapter5
Statistical and probabilistic concepts, Implementation
to Insurance
Subjects of the Unit
1.
2.
3.
4.
1
Counting
Probability concepts
Random Variables
Used distributions In Insurance
1. Counting
Since many (not all) probability problems will be solved by
counting outcomes, we will develop a number of counting
principles which will prove useful in solving probability
problems.
In counting, we will use the convenient notation
n A :the number of elements in the set (or event) A.
Example 1: A neighborhood association has 100 families on its
membership list. 78 of the families have a credit card and 50 of the
families are currently paying off a car loan. 41 of the families have both
a credit card and a car loan. A financial planner intends to call on one
of the 100 families today.
The planner's sample space consists of the families in the association.
The events of interest to the planner are the following:
Counting (2)
C: the family has a credit card
L: the family has a car loan
We are given the following information:
nC   78 , nL  50 , nL  C   41
The planner is also interested in the answers to some other
questions
For example, she would first like to know how many families do not
have credit cards. Since there are 100 families and 78 have credit
cards, the number of families that do not have credit cards is:
100 – 78 = 22
This can be written using the counting notation as
nC   nS   nC 
Rules
For any finite sample space S and events A and B
nA   nS   n A
n A  B   n A  n B   n A  B 
If A and B are mutually exclusive, then
n A  B   n A  n B 
Definition :
The empty set is the set which has no elements.
It is denoted by the symbol  .
A combination
A combination of n objects taken r at a lime is an r-element
subset of the original n elements (or, equivalently, an
unordered selection of r of the original n elements).
Notation: The number of combinations of n elements taken r
at a time is denoted by  n .
r 
Counting Principle for Combination
n
n!

1. 

r 
r!n  r !
 
n n
2. 
n

0
 1
   
3. n!  n  n  1  n  2 .....1
4. 0!  1
Example 2. A company has ten management trainees. The company will
test a new training method on four of the ten trainees. In how many
ways can four trainees be selected for testing?
Solution. 10  10! 10.9.8.7
  

4
4
!
6
!
 
4!
 210
2. Probability concepts
Approach 1. Equally likely outcomes
Approach 2. Relative frequency
Approach 1. Equally likely outcomes
Probability by Counting for equally likely outcomes:
Propabilit y of an event 
Number of oucomes in the event
Total number of possible outcomes
Example 3: Suppose you are rolling a single six-sides die whose
sides bear the numbers 1,2,3,4,5 and 6. You wish to bet on the
event that you will roll a number less than 4.
The probability of this since the outcomes 1,2,3 are less than 4.
event is 3/6,
Definition.
Let E be an event from a sample space S in which all
outcomes are equally likely. The probability of E
PE  , is defined by denotednE 
P E  
nS 
Example 4. A company has 200 employees. 50 of these employees are
smokers. One employee is selected at random. What is the probability
that the selected employee is a smoker (Sm)?
Solution
PSm  
nSm 
50

 0.25
nS 
200
Approach 2. Relative frequency
Relative Frequency estimate of the probability of an event
Number of times the event occurs in n trials
Probabilit y of an event 
n
Example.
Assume that you are tossing a coin and suspect that it is not
fair. Then the probability of tossing a head can not be
determined by counting, we have to estimate the probability by
tossing the coin a large number of times and count the number
of heads. If you toss the coin 1000 times and observe 750
heads, your best estimate of the probability of a head of one
toss is 750/1000=0.75
The language of Probability
Definition 1. A set is a collection of objects such as the
numbers 1,2,3,4,5 and 6. These objects are called the
elements or members of the set. If the set is finite and
small enough that we can easily list all of its elements,
we can describe the set by listing all of its elements in
braces.
Example 1. The set of all positive real numbers may be written as:
S  x / x is a real number and x  0 
Definition 2. The sample space ,S for a probability
experiment is the set of all possible outcomes of experiment.
Example 2. A coin is tossed and the side facing up is recorded.
The sample space is :
S  H , T 
A set
Example 3. (Death of an insured) An insurance company is
interested in the probability that an insured will die in the
next year.
The sample space is: S  Death, Survival
Example 4. An insurance company has sold 100 individual
life insurance policies. When an insured individual dies, the
beneficiary named in the policy will file a claim for the
amount of the policy. You wish to observe the number of
claims filed in the next year. The sample space consists
of all integers from 0 to 100, so ,
S  0,1,2.....,100
An event
Example 5. An insurance company sells life insurance to a 30year-old female. The company is interested in the age of the
insured when she eventually dies. If the company assumes that
the insured will not live to 110, the sample space is
S  30,31,32,....109
Definition 3. An event is a subset of the sample space S.
Example 6. A coin is tossed. You wish to find the probability of the
event "toss a head." The sample space is S  H , T 
The event is the subset E  H 
Example 7. An insurance company has sold 100 individual life
policies. The company is interested in the probability that at most
5 of the policies have death benefit claims in the next year. The
sample space is S  0,1,2,....100
The event is the subset E  0,1,2,3,4,5
Negation: The event not E is written as E
The Compound Events A or B:
Example: Rolling a single die. Suppose that we have the opportunity
to bet on two different events:
A: an even number is rolled A={2,4,6}
B: a number less than 5 is rolled B={1,2,3,4}
If we bet that A or B occurs, we will win if any element of the two
sets above is rolled. A or B={1,2,3,4,6 }
In forming the set for A or B we have combined the sets A and B
by listing all outcomes which appear in either A or B. The
resulting set is called the union of A and B, and is written as:
A U B.
It should be clear that for any two events A and B;
A Or B  A  B
The Compound Events A and B:
When one decide to bet on the event A and B. In that case, both the
event A and the event B must occur on the single roll. This can
happen only if an outcome occurs which is common to both
events. A and B={2,4}
In forming the set for A and B we have listed all outcomes which
are in both sets simultaneously. This set is referred to as the
intersection of A and B, and is written as A  B.
For any two events A and B: A and B  A  B
Probability rules
1. P( S )  1
2. P( )  0
3. P(C )  1  C
4. Disjunctio n Rule : P A  B   P A  PB   P A  B 
Definition: two events are mutually exclusive if
Addition rule for mutually exclusive event:
5. P A  B  P A  PB
A B 
Conditional probability
1. Conditional probability by counting
P A / B  
n A  B 
n B 
Example. A health insurance pool includes 200 individuals.
The insurer is interested in the number of smokers in the pool among both
males and females. The following table (called a contingency table) shows the
desired numbers.
Males (M)
Females (F) Total
Smokers
28
22
50
Non Smokers
72
78
150
Total
100
100
200
Suppose one individual is to be chosen at random. Counting can be used
to find the probability that the individual is a male, a female, a smoker, or
both.
100
100
50
PM  
200
 0.5; PF  
200
 0.5 PS  
200
 0.25
28
22
PM  S  
 0.14; PF  S  
 0.11
200
200
Suppose you were told that the selected individual was a male, and asked
for the probability that the individual was a smoker, given that the
individual was a male.
nS  M  28
PS / M  

 0.28
nM 
100
Definition For any two events A and B, the conditional probability of A given B
is defined as follows:
P A / B  
P A  B 
P B 
Example . An automobile insurance company does a study to find the probability
for the number of claims that a policyholder will file in a year. Their study gives
the following probabilities for the individual outcomes 0,1,2,3.
Number of claims
0
1
2
3
Probability
0.72
0.22
0.05
0.01
Find the probability that a policyholder files exactly 2 claims, given that
the policyholder has filed at least one claim.
Solution. Let C be the event that at least one claim is filed.
Then C  1,2,3 and PC   0.22  0.05  0.01  0.28
We also need the value P2  C   P2  0.05.
Then
P2  C  0.05
P2 / C  
PC 

0.28
 0.179
This tells us that approximately 17.9% of the policyholders who file claims
will file exactly 2 claims.
3. Random Variables
Definition. A random variable is a numerical quantity
Variable whose value depends on chance.
Example. You are tossing a coin twice and will bet on the
number of heads. The outcome is a number (0, 1 or 2) which
depends on chance. The number of heads is a random
variable.
The Probability Function of a Discrete
Random variable
Definition. Let X be a discrete random variable. A probability function for
p (x
)
isX a function
which
assigns a probability to each
value of the random variable. Such that
1. px   0 for all x, and
2.  px   1
The probability function is also referred to as the probability mass
function or the discrete density function for X .
Example. an automobile insurer studied the number of claims filed by a
policyholder in a given year. The probability function was as follows.
Number of claims
p (x )
0
1
0.72
2
0.22
3
0.05
0.01
The cumulative distribution
Definition. Let X be a random variable. The cumulative
distribution function F (x) for X is defined by
F x  P X  x
Example. The cumulative distribution function for the previous example is given
by the following table:
Number of claims
F(X)
0
1
2
3
0.72
0.94
0.99
1
This tells us that 94% of policyholders file one claim or less in a year
leaving 6% who file more than one claim
4. Used distributions In Insurance
• Discrete distributions
- Binomial distribution
- Poisson distribution
• Continuous distributions
- Exponential distribution
a. Binomial distribution
• If X is a binomial random variable with n trials and PS   p ,
n
nk
P X  k     p k 1  p  , for k  0,1,..., n.
k 
• The mean and variance of binomial distribution
Number of success (x)
P(x)
The mean
E X   np
0
1
q= 1-p
p
The Variance
V  X   np(1  p)  npq
Example
• The probability of car accident within one year for novice
policyholders equals to 0.2.
Assume that you are an insurance agent with 10 novice
clients. You are willing to suppose that car accidents are
independent events.
1. What is the probability of no accident in the next year?
2. What is the probability that that one accident will occur?
3. Calculate the mean and the variance of non accident.
Solution
• If accidents are independent, the number of non accident X
will be a binomial random variable with parameter n=10 and
p  1  0.2  0.8
10 
10
1. p X  10   0.8  0.1073
10 
10 
9
1
2. p X  9   0.8 0.2  0.268
9 
3. The mean and the variance of non accident
E  X   n. p  10 * 0.2  2
V  X   n. p.q  10 * 0.2 * 0.8  1.6
b. Poisson Distribution
• The random variable X follows the Poisson distribution with
parameter  (or average rate) ) if
e   k
P X  k  
, k  1,2,3,....
k!
• The mean and variance of binomial distribution
The mean

The Variance

c. Exponential distribution
• The random variable X follows an exponential distribution
with parameter  if
 .e  . x si x  0
f ( x)  
0
si x  0
• The cumulative and survival function
Cumulative
Survival
F ( x)  1  e  . x
S ( x)  e  . x
• The mean and the variance are:
The mean
E( X ) 
1

The Variance
V (X ) 
1
2
Example
• An insurance company is studying the reliability of a
system. The time T (in hours) from installation to failure
of the system is a random variable. The study shows that T
follows an exponential distribution with   0.001 .
1. What is the probability that the system fails within 100
hours?
2. Find the probability that the part lasts for more than
500 hours.
3. Calculate the mean and the variance of the time failure of
the system.
Solution
1. The probability that a part fails within 100 hours is
P0  T  100  F (100)  F (0)

 1  e 0.001*100  1  e 0.001*0

  e 0.1  1  0.095
2. The probability that the part lasts for more than 500
hours S 500  e500*0.01  e20  0.607
3. The mean and the variance of the time failure of the system.
1
1
1
EX   
 1000; V  X   2  1,000,000

0.001

Note. The system has an expected life of 1000 hours, you might not want
to use it for 1000 hours if your life depended on it.