A&J Study Manual for SOA Exam P/CAS Exam 1

St dy Manual fo Exa

P/1, 1st Edit ion

Electronic Product No Returns

Part 3: Conditiona l Probability and Bayes Formula

SOA Exam P/ CAS Exam 1

Part 3. Conditional Probability and Bayes Formula This chapter discusses about: 1. Conditional Probability 2. Bayes Formula 3.1 Conditional Probability Conditional probability refers to probability given that some other event has occurred. For example, the probability that a 10-year old boy passes away within 5 years and the probability that a 20-year old young man passes away within 5 years are not the same. This is because the attained age is not the same. For the boy, 20 years of living has not occurred, but for the young man, he has already attained 20 years old. The “events that has occurred” are not the same for both persons. Consider another example: in a class of 50 students, 20 are male students, 30 are female students. Out of the 20 male students, 3 of them are smokers. Out of the 30 female students, 2 of them are smokers. The probability that a male student who smokes is chosen out from the 3 whole class is , whereas the the probability that a male student who smokes is chosen out 50 3 from the 20 male students is . 20 As we can see, the total number of possible event (the denominator) determines the probability. The greater the total number of possible event, the smaller the probability is.

P( A | B) =

n ( AB ) n ( A)

=

P ( AB ) P ( A)

, where

P( A | B ) can be interpreted as the probability that A occurs, given that B has already occured Many students are confused between P( AB) and P( A | B ) . The difference between these 2 lies at the time of the occurrence of B : AB implies that A and B happen at the same time, while A | B implies that B has already occurred before A ; of course, | B also implies that both A and B happen, but the timing of occurrence are not the same. Example 3.1 You are given that P ( A) = 0.3 and P ( B ) = 0.6 . If A and B are independent of each other, determine the probability that B occurs, given that A does not occur. Solution 3.1

P( B| A ) = C

P( AC B ) P ( AC )

1 − P ( A)  P ( B ) = = P( B) = 0.6. 0.6 . 1 − P ( A )

Copyright © 2010 A&J Study Manual. This study material was purchased at www.anjstudymanual.com (Order number [000000] on [June 1, 2000]) for the exclusive use of Exam P/1 Candidate. It is a violation of international copyright law to resell, reproduce or otherwise distribute this product without the express written permission of the authors |

Part 3: Conditiona l Probability and Bayes Formula


Example 3.2 You are given that P ( A ) = 0.45 , P ( C ) = 0.3 , P( BC ) = 0.22 , and P( AC ) = 0.32 . If A and B are mutually exclusive, find P ( B | A∪ C ) . Solution 3.2

P ( B | A ∪ C ) =

P ( B ∩ ( A ∪ C ) ) P ( A ∪ C)

=

P ( AB ) + P ( BC ) P ( A) + P ( C ) − P ( AC)

=

0 + 0 .2 2 0.4 0.45 + 0.3 − 0.32

0.5116 . = 0.5116.

Example 3.3 In urn A , there are 3 red balls and 5 blue balls. In urn B , there are 5 red balls, and n blue balls. Determine n if the probability of drawing a blue ball from urn A is twice greater than the probability of drawing a red ball from urn B . Solution 3.3 Let A represents the event that a ball is to be drawn from urn A . Let R represents the event that red ball is drawn. As such,

(

)

(

)

P RC | A = 2 P R | AC ⇒

 5  11 . = 2  ⇒ 25 + 5 n = 80 ⇒ n = 11. + 3+5 5 n   5

Example 3.4 2 urns are selected at random. The probability that urn A is selected is 0.3. In urn A , there are 3 white balls and 7 black balls. In the other urn, there are 2 while balls and 5 black balls. A person is to select an urn, and draw a ball. The ball is then returned to the urn. This is repeated for the 2 nd time. Determine the probability that white balls are drawn for both draws. Solution 3.4 Let A represents the event that a ball is to be drawn from urn A . Let W represents the event that white ball is drawn. The first and second draws are independent of each other, since the result of the first draw does not affect the 2 nd draw. Hence, we need to find the probability that a white ball is drawn, and square it to obtain the answer. Since white ball can be drawn from urn A and other urn, we need to consider both urns: 2

2 P (W ) =  P ( AW ) + P ACW  =  P (W | A ) P ( A ) + P W | AC P AC 



(

)

(



) ( )

2

2

 3    2  0 . 3 1 0 . 3 =  + − ( ) ( )   2+5  = 0.0841    3 + 7   Copyright © 2010 A&J Study Manual. This study material was purchased at www.anjstudymanual.com (Order number [000000] on [June 1, 2000]) for the exclusive use of Exam P/1 Candidate. It is a violation of international copyright law to resell, reproduce or otherwise distribute this product without the express written permission of the authors |

Part 5: Discrete Probability Distribution


Part 5. Discrete Probability Distribution This chapter discusses about: 1. Uniform distribution 2. Binomial distribution 3. Poisson distribution 4. Negative binomial distribution 5. Hypergeometric distribution Consider this illustration: You are a human resource manager, considering 2 fresh graduates who have applied for an actuarial executive position in your company. Their GPAs are not yet known because they have just taken their final exams yesterday. You are to hire one of them based on their statistic exam mark that they took yesterday, given that one of them is from Harvard University whereas another one is from XXX University (a university which is not in the top 5000 world university list). No doubt, common sense tells you to recruit the Harvard graduate, since the probability that Harvard graduate has higher exam mark is much higher. Now, if you are given that the statistic exam mark of both students are in range between a and a + 9 , discretely and equally distributed, is the value of a the same for both students? The answer is no. The Harvard graduate probably has higher value of a , say a = 85 , whereas for XXX graduate, maybe the value of a is lower, say a = 65 . This variable a is known as the parameter of the distribution of the statistic mark of the student. Let us look at some commonly used discrete probability mass distribution with their properties.

5.1 Uniform Distribution If a random variable N follows discrete uniform distribution, mathematically, such relationship is represented by N ∼ U ( a, b) , which means:

P( N = n) =

1

b−a

, {( a, b) | a∈ ℤ, b∈ ℤ} ,

where a and b are the parameters of this distribution. As you can see, every value of n between a and b are equally distributed at the probability of

1

b−a

, therefore it is known as uniform distribution. As such, the expected value of random is si simply E ( N ) =

variable

Var ( N ) =

(b − a ) 12

a+b 2

; the simple average of the boundaries. The variance of N is

2

.

For a random variable

following uniform distribution distribution with parameter a and b ,

E ( N ) =

a+b 2

, Var ( N ) =

(b − a ) 12

2

.




Example 5.1 You are given that the annual claim frequency of an insurance company is uniformly distributed between 101 and 150. Find: 1.

P( N ≤ n) ;

2. E ( N ) ; 3. Var ( N ) ; 4. P ( N < 123) ; 5.

P( N< 136 | N > 120 ) ;

6.

E( N| 135 < N ≤ 140 ) ;

140 ) . 7. Var ( N | 135 < N ≤ Solution 5.1 1.

(f )n=

1 150 − 100

P( N≤ n) =

n

=

1 50

1

∑ 50 =

= 101,102,…150 . Hence, , n

n − 1 00 50

i =101

2. E ( N ) =

150 + 100

3. Var ( N ) =

2

=

250

(150 − 100)

2 2

12

=

, n= 10 1,10 2, …15 0 .

= 125 . 502 12

4.

P( N < 123) = P( N ≤ 122) =

5.

P( N< 136 | N> 120 ) =

= 208.33 .

122 − 101

= 0.42 . 50 135 100 120 100 P (121 ≤ N ≤ 135) ( 5−0 ) − ( 5−0 ) 1 − P ( N ≤ 120 )

=

1−

(120−100) 50

6. If N constraine d between 136 and 140, this means that N is constrained

=

15 30

2

is uniformly distributed distribute d from

136 to 140. Hence, the expected value is E( N| 135 < N≤ 140 ) =

N is constrained 7. If N constraine d between 136 and 140, this means that

1

= . 140 140 + 136 2

= 138 .

is uniformly distributed distribute d from

140) = 136 to 140. Hence, the expected value is Var ( N | 135 < N ≤

140 − 136 136) (140 12

2

4

= . 3




Example 5.2 If you are given that the discrete random variable variable N is uniformly distributed from 1 to t o 100, derive the moment-generating moment-generating function of . Solution 5.2

( t ) = E ( etN ) 100

= ∑ etn f ( n ) n =1 100

= f ( n) ∑ etn n =1

=

1

100

∑e 100

tn

n =1

You should be able to do the reverse: recognize what distribution a random variable follows given the moment-generating moment-generating function.


Author’s Biography


Author’s Biography Alvin Soh was born in Penang, Malaysia. He has been keen in sharing and helping his peers in actuarial exams. His strong passion in mathematics and actuarial career is reflected in his fast progress in actuarial exams. He has passed all the preliminary exams in the mere 1.5 years and he is more than willing to share his study methods and experience to every actuarial student striving to progress in exams. He has also passed an advanced level exam and is pursuing his CERA and FSA in Finance and Enterprise Risk Management. He is now working as an actuarial analyst in Prudential Financial Inc. The following is the progress of his exams: Exam Passed P/1- Probability Probability FM/2- Financial Mathematics MLC- Models for Life Contingencies MFE/3F- Models for Financial Economics C/4- Construction Constructi on and Evaluation of Actuarial Models AFE- Advanced Finance and Enterprise Enterpris e Risk Management

Sitting Spring 2007 Fall 2007 Spring 2008 Fall 2008 Fall 2008 Spring 2009


A&J Study Manual for SOA Exam P/CAS Exam 1

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