Download Chapter 3.1 Random Experiment, Outcomes and Events O O O ]O,O

Chapter 3.1 Random Experiment, Outcomes and Events In the analysis of economic data any conclusions involve some uncertainty. To incorporate ideas of uncertainty into the study of economic data an understanding of probability theory is needed. A random experiment has a set of basic outcomes. O1 increase O2 decrease O3 no change This is the set of basic outcomes in the sample space S that belong to both A and B. If events A and B have no common basic outcomes then A  B is the empty set. In this case A and B are mutually exclusive. Example: View the random experiment as investing in the stock market. For a selected company the possible outcomes for the daily change in the closing price of one share are: Consider two events A and B. The intersection is denoted by A  B . The union is denoted by A  B . This is the set of basic outcomes that are in either A or B or both. If the union of several events gives the whole sample space S then the events are collectively exhaustive. Clearly, at the start of the business day, there is uncertainty as to which outcome will occur. The complement of A is denoted by A . This is the set of basic outcomes in the sample space S that do not belong to the event A . Therefore: The sample space S is the set of all basic outcomes. A  A  S A and A are collectively exhaustive, and A  A  empty, A and A are mutually exclusive An event is a subset of basic outcomes from the sample space S. Example: For the daily change in stock market price define the events: 1
A  [ O1 , O 3 ] no loss B  [ O2 , O3 ] no gain Econ 325 – Chapter 3
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Econ 325 – Chapter 3
A Venn diagram gives an illustration. Sample Space S
A Venn diagram for the complement of event A. Sample Space S
AB
A
B
A
A
AB
In the next Venn diagram the events A and B are mutually exclusive. Sample Space S
A
As an alternative to the above, the next diagram demonstrates that A and A are collectively exhaustive and mutually exclusive. Sample Space S A B
A 3
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 A General Result  Some Results Consider two events A and B. The events A  B and A  B are mutually exclusive, and Let E1 , E2 , . . . , EK be K mutually exclusive and collectively exhaustive events. That is, is the empty set for all i  j , and ( A  B )  ( A  B )  B . That is, the union is B . Ei  E j The events A and A  B are mutually exclusive, and E1  E 2    EK  S A  ( A  B )  A  B . That is, the union is A  B . These results are shown with a Venn diagram. Suppose A is some other event in the sample space S. Then the K events: E1  A , E2  A , . . . , EK  A are mutually exclusive and their union is A . Sample Space S
This result is shown in a diagram with K = 4. A
E1 B
E2 E3 E1  A E 2  A E3  A AB
AB
E4  A Sample Space S  Event A 5
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Chapter 3.2 Probability For a random experiment probabilities can be assigned to outcomes and events. Denote: P(Oi ) the probability that basic outcome Oi occurs for i = 1, 2, . . . P(A) For an event A let n A be the number of times the event is observed in n trials of a random experiment. The relative frequency approach to probability asserts that as the number of trials n becomes “large” then: the probability that event A occurs. How can probability be viewed ? Consider an event A in the sample space S. (1) 0  P( A )  1 Example: The random experiment is a coin toss. The outcomes are: O1 head O2 tail P(A )  0 means the event is impossible, and P(A )  1 means the event is certain to occur. The higher the probability the more likely the event is to occur. (2) P(A )   P(Oi ) A computer program was used to simulate the experiment. In 1000 repetitions of the coin toss the results gave: n1  507 the number of heads, and n2  493 the number of tails. A
The summation notation means that the summation is over all basic outcomes in event A . (3) P(S)   P(Oi )  1 S
For the outcome of heads, the relative frequencythe proportion of occurrences of outcome O1 in n trials is: nA
 P( A ) n
 Probability Facts (or Postulates) A probability assessment can be made by observing the frequency of individual outcomes in n trials of a random experiment. The summation is over all basic outcomes in the sample space S. n1 507

 0.507 n 1000
This can be viewed as the probability of getting a head in a fair toss of a coin. 7
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 A Special Case – Equally Likely Outcomes  Some More Probability Results Suppose the sample space S is made up of m equally likely basic outcomes O1 , O2 , . . . , Om . If A and B are mutually exclusive events then: Examples of random experiments with equally likely basic outcomes are the coin toss and the dice toss. Results are: 1
(1) P(Oi ) 
m
for i = 1, 2, . . . , m A
B
 P( A )  P(B)
If E1 , E2 , . . . , EK are collectively exhaustive events then: m
This gives P(A  B)   P(Oi )   P(Oi )
P(E1  E 2    EK )  P(S)  1  P(Oi )  1 i1
Example: the random experiment is the dice toss. The basic outcomes are: Oi = face i for i = 1, 2, 3, 4, 5, 6 The probabilities are: P(Oi ) 
1
6
for i = 1, 2, 3, 4, 5, 6 (2) If event A is made up of m A basic outcomes then P(A )   P(Oi )
A
m
 A
m
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Example Counting Rules The application is offshore drilling for oil in the Atlantic Ocean. The probability of oil reserves that can be recovered is estimated as: more than 1 billion barrels more than 2 billion barrels Evaluating probabilities of events may make use of counting rules. probability 0.5 0.1  Orderings What is the estimated probability of reserves between 1 and 2 billion barrels ? Consider x objects. How many different orderings are possible, repetitions not being allowed ? The number of ways of arranging the x objects are: To get an answer define the events: x A more than 2 billion barrels x1 possible ways to fill position 2 B reserves between 1 and 2 billion barrels. x2 possible ways to fill position 3 
The information available is: P(A )  0.1 and 1 P( A  B )  0.5 
 way to fill the last position (only one object is left to place in the last position). The calculation formula for the number of possible orderings is: The problem is to find the probability P(B) . The events A and B are mutually exclusive, therefore: possible ways to fill position 1 (x)( x1)( x2) . . . (2)(1) = x ! (x factorial) P( A  B )  P(A )  P(B) Rearrange the expression and calculate: P(B)  P( A  B )  P( A )  0.5  0.1
 0.4
Recall that valid probabilities must be between 0 and 1. 11
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Example: A consumer is asked to rank in order of preference the taste of 3 brands of bottled water with the labels A, B, C. Assume that any particular ordering is equally likely. What is the probability that the selected ordering will be C A B ? There are 3! = (3)(2)(1) = 6 different possible orderings. A list is: From n objects choose x objects (n > x), repetitions not being allowed. The number of possible orderings of x chosen from n is called the number of permutations. The possibilities are: n possible ways to fill position 1 n1 possible ways to fill position 2 

nx+1 ABC
ACB
BAC
BCA
CAB
CBA
 possible ways to fill position x n
The number of permutations is denoted by Px . The calculation formula is: Since each is equally likely the probability of selecting C A B is: 13
 Permutations 1
6
Pxn  (n) (n  1) (n  2) (n  x  1) This can be expressed as: Pxn
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 ( n ) ( n  1 ) ( n  2 ) ( n  x  1 )
n!

(n  x) !
(n  x) !
(n  x) !
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Example: From 5 brands of bottled water with the labels A, B, C, D, E a consumer is asked to rank in order of preference the top 3 brands. Assume that any particular ordering is equally likely. What is the probability of selecting C A B for the first three places ?  Repetitions allowed From n objects choose x objects (n > x), and allow repetitions. The possibilities are now: n possible ways to fill position 1 n possible ways to fill position 2 The number of permutations is: P35 
5!
120

 60 ( 5  3) !
2
Since each is equally likely the probability is: 
n 
 possible ways to fill position x The calculation formula for the number of different arrangements is: nx Example: How many different numbers can be generated for the first 5 digits of a UBC student number ? 1
60
The first digit must be a number from 1 to 9. The other 4 digits can also include the number 0. The number of different 5‐digit numbers is: (9)(10)(10)(10)(10) = 9  10 = 90,000 4
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 Combinations Example: From 5 brands of bottled water what is the probability that a consumer will select labels A, B, C as the top three brands in any order ? Another counting problem: from n objects choose x objects (n > x), repetitions not being allowed and different orderings of the same x objects not being counted separately. The number of possible selections is called the number of combinations. The number of combinations is: How is this calculated ? x objects can be ordered in x ! different ways. n
C nx 
n!
1 n
Px 
x!
x ! (n  x)!
ABC
ABD
ABE
ACD
ACE
ADE
BCD
BCE
BDE
CDE
By using the assumption that all selections are equally likely the probability that the selection is labels A, B, C is: A standard notation for this formula, stated in words as “n choose k”, is: 1 5 60
P3 
 10 3!
6
A list is: Therefore, the number of combinations, denoted by C x , is calculated as: C53 
 n
  k
1
10
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Chapter 3.3 Probability Rules For two events A and B in the sample space S define: Intersection A  B Union A  B Complement When A and B are mutually exclusive then P(A  B)  0 and P( A  B )  P(A )  P(B) In general, when events A and B are not mutually exclusive, the addition rule of probabilities is: A P( A  B )  P(A )  P(B)  P(A  B) This result follows from earlier work. Recall: A and A are mutually exclusive and collectively exhaustive. This gives: P( A  A )  P(A)  P( A ) (1) P( A  A )  P(S)  1 (2) P(A)  P( A )  1 A  ( A  B )  A  B . Therefore, the complement rule is: 19
( A  B )  ( A  B )  B . (2) The events A and A  B are mutually exclusive, and From (1) and (2): (1) The events A  B and A  B are mutually exclusive, and From (2): P( A  B )  P(A)  P( A  B ) (3) From (1): P( A  B )  P(B)  P( A  B ) (4) Substitute (4) into (3) to get: P( A )  1  P(A) Econ 325 – Chapter 3
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P( A  B )  P(A )  P(B)  P(A  B)
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 Conditional Probability The addition rule of probabilities can be seen with a Venn diagram. Sample Space S
The current state of the economy is conditioned by what is known from the past. Suppose information is available about event B. What is the probability of event A given that event B has occurred? AB
A
The probability of A given B is called conditional probability and is denoted by: B
AB
P( A B) How can conditional probability be viewed ? Repeat a random experiment n times and observe: In the above diagram, the events A and B are not mutually exclusive. It should be clear that, P(A )  P(B)  P( A  B ) That is, the sum P( A )  P(B) will count P(A  B) twice. nB number of occurrences of event B, n B  0 nA B number of times both A and B occur The proportion of times event A has occurred when B has also occurred is: So subtract one to get: P(A )  P(B)  P(A  B)  P( A  B ) nAB
nB
As n becomes “large”: 21
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nA B
 P(A B) nB
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Example: For household heads in Canada consider the events: A reads newspaper on a weekly basis B actively trades stocks n A  750 nB  320 n A B  280 P(A B) 
P( A  B )
P(B)
for P(B)  0 P(B A ) 
P( A  B )
P(A)
for P(A )  0 From the above, the multiplication rule of probabilities is stated as: The probability that a household head reads the newspaper on a weekly basis is calculated as the relative frequency: P(A ) 
Also, A survey of 1000 household heads found: For conditional probability define: P( A  B )  P(A B) P(B)
 P(B A ) P(A )
n A 750

 0.75 n
1000
The probability that a household head who actively trades stocks reads the newspaper on a weekly basis is: 23
P(A B) 
n A  B 280

 0.875 nB
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A special case arises when events A and B are statistically independent. This occurs if and only if: Example A shop finds that 90% of computers sold will not require a repair in the first year. Three computers are purchased. Find the probability that none of the three computers will be returned for a repair in the first year. P( A  B )  P(A ) P(B) Statistical independence gives: P(A B) 
P( A  B ) P(A) P(B)

P(B)
P(B) Define the events and probabilities: for P(B)  0  P(A)
Also, P(B A )  P(B) E1 computer 1 does not need a repair, P(E1 )  0.9 E2 computer 2 does not need a repair, P(E2 )  0.9 E 3 computer 3 does not need a repair, P(E 3 )  0.9 Assume computer failures are independent and calculate the answer as: for P(A )  0 P(E1  E 2  E 3 )  P(E1 ) P(E 2 ) P(E 3 )  (0.9 )3  0.729 That is, information about event B is no use in determining the probability of event A and vice versa. Now find the probability that all three computers will be returned for a repair. From the complement rule, the probability that computer 1 requires a repair is: In general, events E1 , E 2 , . . . , E K are statistically independent if and only if: P(E1  E2    EK )  P(E1 ) P(E2 ) P(EK ) Assuming independence, the answer is: Problem: If two events are mutually exclusive, are they independent ? Econ 325 – Chapter 3
P( E1  E2  E3 )  P( E1 ) P( E2 ) P( E3 )  (0.1)3  0.001 The probability that at least one computer requires a repair is: 25
P( E1 )  1  P(E1 )  0.1 26
1  P(none require a repair )  1  0.729  0.271 Econ 325 – Chapter 3
A store owner finds: In the above set‐up, are events A and B statistically independent ? 30% of customers ask for assistance, Statistical independence follows if: P( A  B )  P( A ) P(B) 20% of customers make a purchase, and Example But, P( A ) P(B)  (0.3)(0.2)  0.06  0.15  P( A  B ) 15% of customers both ask for assistance and make a purchase. Therefore, the two events are not statistically independent. Find the probability that a customer does at least one of these two things. Events and probabilities can be set as: Problems: Are the two events mutually exclusive ? Are the two events collectively exhaustive ? A customer asks for assistance, P(A )  0.30 B customer makes a purchase, P(B)  0.20 P( A  B )  0.15 By the addition rule of probabilities the answer is found as: P( A  B )  P(A )  P(B)  P(A  B)  0.30  0.20  0.15
 0.35
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Example Chapter 3.4 A city survey finds: 18% of adults watch business programs on TV, 12% of adults read business magazines, and 10% of adults do both. Consider a set of H mutually exclusive and collectively exhaustive events denoted by: Events and probabilities can be set as: A watch TV business programs, P(A )  0.18 B read business magazines, P(B)  0.12 A 1 , A 2 , . . . , A H Also, consider a second set of K mutually exclusive and collectively exhaustive events denoted by: P( A  B )  0.10 B1 , B2 , . . . , BK Example: For a labour force survey define the events: Find the probability that an adult who watches TV business programs reads business magazines. The answer is the conditional probability: A 1 female 
 A 2 male

Gender 

B2 medium ‐ income  
B3 high ‐ income

Income B1 low ‐ income
P( A  B ) 0.10
P(B A) 

 0.556 P(A)
0.18
Now find the probability that an adult who reads business magazines watches TV business programs. The answer is: The basic outcomes can be stated as: P( A  B ) 0.10
P(A B) 

 0.833 P(B)
0.12
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A1  B1 female, low‐income A1  B2 female, medium‐income A 1  B3 female, high‐income A 2  B1 male, low‐income A 2  B2 male, medium‐income A 2  B3 male, high‐income Econ 325 – Chapter 3
The probabilities P( A i  B j ) for i = 1, 2, . . . , H and j = 1, 2, . . ., K Example are called bivariate probabilities or joint probabilities. A store survey collects information on the events: The probabilities of individual events are called marginal probabilities and are calculated as: A 1 frequent 
 Frequency of Visit A 2 infrequent 
K
P(A i )   P( A i  B j ) j 1
for i = 1, 2, . . . , H , and H
P(B j )   P( A i  B j ) i 1
for j = 1, 2, . . . , K The bivariate probabilities (obtained as the proportion of people surveyed in each of the six joint classifications) and marginal probabilities are presented in the table: A result is: H
 P(A i )  1 i1


B2 sometimes  Purchase of Generic Products 
B3 never

B1 often
K
and  P(B j )  1 j 1
B1 B2 B3 Totals A1 0.12 0.48 0.19 0.79  P( A 1 ) A2 0.07 0.06 0.08 0.21  P( A 2 ) P(A 1 )  P( A 1  B1 )  P( A 1  B2 )  P( A 1  B3 ) and Totals 0.19 0.54 0.27 P(B1 )  P( A 1  B1 )  P( A 2  B1 ) That is, the marginal probabilities must sum to one because the events are mutually exclusive and collectively exhaustive. Example: For the labour force survey presented above,  P(B1 )  P(B2 )  P(B3 ) Note: it can be verified that the probabilities sum to one. 31
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(c) Are the events B3 and A 1 independent ? Questions and answers. With independence, P( A 1 B3 )  P( A 1 ) (a) What is the probability that a customer both is a frequent shopper and often purchases generic products ? But P( A 1 B3 )  0.7037  0.79  P( A 1 ) P( A 1  B1 )  0.12 Therefore, the events are not independent. (b) What is the probability that a customer who never buys generic products visits the store frequently ? An equivalent approach is to recognize that independence guarantees: Calculate the conditional probability: P(A 1 B3 ) 
P(A 1  B3 )  P(A 1 ) P(B3 ) P(A 1  B3 ) 0.19

 0.7037 0.27
P(B3 )
But P( A 1 ) P(B3 )  (0.79) (0.27)  0.2133  0.19 As an extra comment, the probability a customer who never buys generic products visits the store infrequently is: P(A 2 B3 ) 
This supports the previous finding that the two events are not independent. P(A 2  B3 ) 0.08

 0.2963 0.27
P(B3 )
(d) What is the probability that a customer either visits the store frequently or never buys generic products or both ? Note: P( A 1 B 3 )  P( A 2 B3 )  1 Use the addition rule of probabilities: Conditional probabilities with respect to a particular event always add up to one. P(A 1  B3 )  P(A 1 )  P(B3 )  P(A 1  B3 )
 0.79  0.27  0.19
 0.87
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Chapter 3.5 Bayes’ Theorem Now suppose that E1 , E 2 , . . . , E K are K mutually exclusive and collectively exhaustive events with known probabilities. A result is: For two events A and B the multiplication rule gives: K
P(B A ) P( A )  P( A B) P(B) Divide both sides by P(A ) to get Bayes’ Theorem: P(B A ) 
P(A B) P(B)
P(A )
 P(E j )  1 j 1
Let A be some other event and consider that conditional probabilities are known for: This has an interpretation for modifying probability assessments when new information becomes available. This can be described as follows. Suppose the interest is event B. At the start, the probability assessment is P(B) . This is called the prior probability. If new information becomes available then make use of this to modify the likelihood that event B will occur. If event A is known to have happened then the updated probability P(B A ) is called the P(A E j ) for j = 1, 2, . . . , K This information can be used to calculate P(E i A ) , the conditional probability of Ei given A . This is done as follows. From Bayes’ theorem: P(Ei A ) 
P(A Ei ) P(Ei )
P( A )
(1) (2) Now find P( A ) as: posterior probability. K
P(A )   P( E j  A )
j 1
K
  P(A E j ) P(E j )
j 1
Substitute (2) into (1) to get a different form of Bayes’ theorem: P(Ei A) 
P(A Ei ) P(Ei )
K
for i = 1, 2, . . . , K  P(A E j ) P(E j )
j 1
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Example Questions and answers. A survey of grades of students completing a first‐year university course is available. Consider the events: A (a) Find P( A ) , the probability that a randomly chosen student graduated in the top 10% in high school. student graduated in top 10% in high school P(A )  P(A E1 ) P(E1 )  P(A E2 ) P(E2 )  P(A E3 ) P(E 3 )


E 2 middle half  student rank in university 
E 3 bottom 25% 
E1 top 25%
 (0.7) (0.25)  (0.5) (0.5)  (0.2) (0.25)
 0.475
(b) Find P(E 1 A ) , the probability that a randomly chosen It is clear that the events E1 , E 2 , and E 3 are mutually student who graduated in the top 10% in high school will rank in the top 25% of the university class. exclusive and collectively exhaustive. Also, they are defined so that: P(E1 )  0.25 , P(E2 )  0.5 and P(E3 )  0.25 Note: the probabilities sum to one. Use Bayes’ theorem: Information is available for the conditional probabilities: 37
P(E1 A ) 
P(A E1 ) P(E1 )
P(A )

(0.7) (0.25)
 0.368 0.475
P( A E1 )  0.7 P( A E 2 )  0.5 Problem: The results show P(E1 A )  P(E1 ) . P(A E3 )  0.2 Does this result seem reasonable for this problem ? Does this result always apply ? Econ 325 – Chapter 3
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(c) Find P( E1 A ) , the probability that a randomly chosen student who did not graduate in the top 10% in high school will not rank in the top 25% of the university class. First, note: P( E1  A )  1  P(E1  A) This result can be demonstrated with a Venn diagram. Calculate: P(E1  A )  P(E1 )  P(A )  P(E1  A )
 P(E1 )  P(A )  P(A E1 ) P(E1 )
 0.25  0.475  (0.7) (0.25)
 0.55
Now find: P( E1 A ) 
P( E1  A ) 1  0.55

 0.857 1  0.475
P( A )
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