FINA2220A

Quantitative Methods for

Actuarial Analysis I

Chapter 3
Conditional Probability and
Independence
 Introduction
 An important concept – Conditional Probability
 The importance of this concept is twofold:

 Calculating probabilities when some partial information

concerning the result of the experiment is available
• Desired probabilities are conditional
 Computing the desired probabilities more easily, even when
no partial information is available

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 Conditional Probabilities
 Suppose that we toss 2 dice
 Suppose that each of the 36 possible outcomes is equally
likely to occur and hence has probability 1/36
 Suppose further that we observe that the first die is a 3
 Given this information, what is the probability that the sum of
the 2 dice equals 8?
 Now, given that the initial die is a 3, it follows that there can be
at most 6 possible outcomes of our experiment, namely, (3,1),
(3,2), (3,3), (3,4), (3,5), (3,6)
 Since each of these outcomes originally had the same
probability of occurring, the outcomes should still have equal
probabilities

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 Conditional Probabilities
 Given that the first die is a 3, the (conditional) probability of
each of the outcomes (3,1), (3,2), (3,3), (3,4), (3,5), (3,6) is 1/6
 Whereas the (conditional) probability of the other 30 points in
the sample space is 0
 Hence, the desired probability will be 1/6

 Let E denote the event that the sum of the dice is 8

 Let F denote the event that the first die is a 3

 The probability just obtained is called the conditional

probability that E occurs given that F has occurred
 It is denoted by

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 Conditional Probabilities
 A general formula for P(E |F) that is valid for all events E and
F is derived in the same manner
 If the event F occurs, then in order for E to occur, it is
necessary that the actual occurrence be a point both in E
and in F
• That is, it must be in E  F
 Now, as we know that F has occurred, it follows that F
becomes our new or reduced sample space
 Hence, the probability that the event E  F occurs will equal
the probability of E  F relative to the probability of F
 Definition 3.1:

 If P(F) > 0, then

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 Exercise 3.1
 A coin is flipped twice
 Assume that all four points in the sample space S are equally
likely
 What is the conditional probability that both flips result in
heads,
 given that the first flip does?
 given that at least one flip lands on heads?

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 Exercise 3.2
 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 and 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

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 Multiplication Rule for Two Events
 By multiplying both sides of the conditional probability
equation by P(F), we obtain

 It states that the probability that both E and F occur is equal to

 The probability that F occurs multiplied by
 The conditional probability of E given that F occurred

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 Exercise 3.3
 A doctor is studying the relationship between blood pressure
(high, low, or normal) and heartbeat abnormalities (regular or
irregular) in her patients
 She tests a random sample of her patients and finds that

 14% have high blood pressure

 22% have low blood pressure
 15% have an irregular heartbeat
 Of those with irregular heartbeat, one-third have high blood
pressure
 Of those with normal blood pressure, one-eighth have an
irregular heartbeat
 What portion of the patients selected have a regular heartbeat
and low blood pressure?
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 Exercise 3.4
 Suppose an urn contains 8 red balls and 4 white balls
 We draw 2 balls from the urn without replacement

 Assume that at each draw, each ball in the urn is equally likely
to be chosen
 What is the probability that both balls drawn are red?

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 Multiplication Rule for General Case
 An expression for the probability of the intersection of an
arbitrary number of events, is sometimes referred to as the
multiplication rule

 Proof:
 To prove the multiplication rule, just apply the definition of
conditional probability to its right-hand side
 This gives

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 Conditional Probability as Relative Frequency
 The definition of P(E | F) is consistent with the interpretation of
probability as being a long-run relative frequency
 Suppose that n repetitions of the experiment are to be
performed, where n is large
 If we consider only those experiments in which F occurs, then
P(E | F) will equal the long-run proportion of them in which E
also occurs
 Since P(F) is the long-run proportion of experiments in which F
occurs, it follows that in the n repetitions of the experiment F
will occur approximately nP(F) times
 Similarly, in approximately nP(E  F) of these experiments
both E and F will occur

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 Conditional Probability as Relative Frequency
 Hence, out of the approximately n P(F) experiments in which F
occurs, the proportion of them in which E also occurs is
approximately equal to

 As this approximation becomes exact as n becomes larger and

larger
 We see that we have the appropriate definition of P(E | F)

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 Total Probability
 Let E and F be events
 We may express E as

 In order for an outcome to be in E, it must either

• Be in both E and F, or
• Be in E but not in F

E F

E  Fc E  F

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 Total Probability
 As E  F and E  Fc are mutually exclusive, we have by Axiom
3 that

 The probability of the event E is a weighted average of

 The conditional probability of E given that F has occurred,
and
 The conditional probability of E given that F has not
occurred

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 Total Probability
 The formula enables us to determine the probability of an event
by first “conditioning” upon whether or not some second event
has occurred
 There are many instances where it is difficult to compute the
probability of an event directly, but it is straightforward to
compute it once we know whether or not some second event has
occurred

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 Exercise 3.5
 In answering a question on a multiple-choice (MC) test, a
student either knows the answer or guesses
 Let p be the probability that the student knows the answer

 Let 1 – p be the probability that the student guesses

 Assume that a student who guesses at the answer will be correct

with probability 1/m, where m is the number of MC alternatives
 What is the conditional probability that a student knew the
answer to a question, given that he or she answered it correctly?

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 Exercise 3.6
 A blood test indicates the presence of a particular disease 95%
of the time when the disease is actually present
 The same test indicates the presence of the disease 0.5% of the
time when the disease is not present
 One percent of the population actually has the disease

 Calculate the probability that a person has the disease given

that the test indicates the presence of the disease

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 Odds
 The change in the probability of a hypothesis when new
evidence is introduced can be expressed compactly in terms of
the change in the odds of this hypothesis
 Definition 3.2:

 The odds of an event A is defined by

 The odds of an event A tells how much more likely it is that the
event A occurs than it is that it does not occur
 For example, if P(A) = 2/3, then P(A) = 2P(Ac), so the odds is 2

 If the odds is equal to , then it is common to say that the odds

are “ to 1” in favor of the hypothesis

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 Odds
 Consider now a hypothesis H that is true with probability P(H)
 Suppose that new evidence E is introduced

 The conditional probabilities, given the evidence E, that H is

true and that H is not true are given by

 Therefore, the new odds after the evidence E has been

introduced is

 That is, the new value of the odds of H is its old value multiplied
by the ratio of the conditional probability of the new evidence
given that H is true to that given that H is not true
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 Odds
 Note that the odds, and thus the probability of H, increases
whenever the new evidence is more likely when H is true than
when it is false
 Similarly, the odds decreases whenever the new evidence is
more likely when H is false than when it is true

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 Exercise 3.7
 When coin A is flipped it comes up heads with probability 1/4
 When coin B is flipped it comes up heads with probability 3/4

 Suppose that one of these coins is randomly chosen and is

flipped twice
 If both flips land heads, what is the probability that coin B was
the one flipped?

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 Bayes’ Formula
 The total probability formula on Page 16 can be generalized
 Suppose that F1, F2, …, Fn are mutually exclusive events such
that

In other words, exactly one of the events F1, F2, …, Fn must


occur
 Now, writing the event E as

 Here, the event E  Fi, i = 1, …, n are mutually exclusive

 Hence we have

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 Bayes’ Formula
 Thus, given events F1, F2, …, Fn of which one and only one
must occur, we can compute P(E) by first conditioning on which
one of the Fi occurs
 That is, P(E) is equal to a weighted average of P(E | Fi), each
term being weighted by the probability of the event on which it
is conditioned
 Suppose now that E has occurred and we are interested in
determining which one of the Fj also occurred
 We have the following proposition

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 Bayes’ Formula
 Proposition 3.1:

 This equation is known as Bayes’ formula, after the English

philosopher Thomas Bayes (1701 – 1761)
 If we think of the event Fj as being possible “hypotheses” about
some subject matter
 Bayes’ formula may be interpreted as showing us how
opinions about these hypotheses held after the experiment,
that is the P(Fj), should be modified by the evidence of the
experiment

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 Exercise 3.8
 An insurance company issues life insurance policies in three
separate categories: standard, preferred, and ultra-preferred
 Of the company’s policyholders, 50% are standard, 40% are
preferred, and 10% are ultra-preferred
 Each standard policyholder has a probability 0.010 of dying in
the next year
 Each preferred policyholder has a probability 0.005 of dying in
the next year
 Each ultra-preferred policyholder has probability 0.001 of
dying in the next year
 A policyholder dies in the next year

 What is the probability that the deceased policyholder was

ultra-preferred?
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 Exercise 3.9
 It is assumed that there are 80% good drivers and 20% bad
drivers in a population
 For a good driver, there is a 10% chance of accident(s) in a year

 For a bad driver, there is a 50% chance of accident(s) in a year

 Suppose that a new automobile insurance customer has an

accident in the first year, what is the probability that he is a
good driver?

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 Independent Events
 In general, knowing that F has occurred changes the chance of
E’s occurrence
 As P(E | F) is not generally equal to P(E)
 In the special cases where P(E | F) does in fact equal P(E)

 We say that E is independent of F

 That is, E is independent of F if knowledge that F has occurred
does not change the probability that E occurs
 Since P(E | F) = P(E  F)/P(F), we see that E is independent of
F if

 The above equation is symmetric in E and F

 Whenever E is independent of F, F is also independent of E

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 Independent Events
 Definition 3.3:
 Two events E and F are said to be independent if the
following holds

 Two events E and F that are not independent are said to be

dependent

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 Example 3.1
 A card is selected at random from an ordinary deck of 52
playing cards
 Let E denote the event that the selected card is an ace

 Let F denote the event that it is a spade

 Then we have

 As P(E  F) = P(E)  P(F), this shows that E and F are

independent

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 Example 3.2
 Suppose that we toss 2 fair dice
 Let E1 denote the event that the sum of the dice is 6

 Let E2 denote the event that the sum of the dice is 7

 Let F denote the event that the first die equal 4

 Then we have

Hence, E1 and F are not independent


 However, we have

 Hence, E2 and F are independent

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 Independent Events
 Proposition 3.2:
 If E and F are independent, then so are E and Fc
 Proof:

 Assume that E and F are independent

 Since E = (E  F)  (E  Fc), and E  F and E  Fc are
obviously mutually exclusive
 Hence we have

 Thus, if E is independent of F, then the probability of E’s

occurrence is unchanged by information as to whether or not F
has occurred

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 Independent Events
 Suppose now that E is independent of F and is also independent
of G
 Is E then necessarily independent of F  G?

 The answer is NO!

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 Example 3.3
 Two fair dice are thrown
 Let E denote the event that the sum of the dice is 7

 Let F denote the event that the first die equals 4

 Let G denote the event that the second die equals 3

 From Example 3.2, we know that E is independent of F

 It is easy to show that E is also independent of G

 But clearly E is not independent of F  G, as

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 Independent Events
 Definition 3.4:
 The three events E, F and G are said to be independent if

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 Independent Events
 If E, F and G are independent, then E will be independent of
any event formed from F and G
 For example, E is independent of F  G

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 Exercise 3.10
 A dental insurance policy covers three procedures:
orthodontics, fillings and extractions
 During the life of the policy, the probability that the
policyholder needs:
 orthodontic work is 1/2
 orthodontic work or a filling is 2/3
 orthodontic work or an extraction is 3/4
 A filling and an extraction is 1/8
 The need for orthodontic work is independent of the need for a
filling and is independent of the need for an extraction
 Calculate the probability that the policyholder will need a
filling or an extraction during the life of the policy

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 Exercise 3.11
 An urn contains 10 balls: 4 red and 6 blue
 A second urn contains 16 red balls and an unknown number of
blue balls
 A single ball is drawn from each urn

 The probability that both balls are the same color is 0.44

 Calculate the number of blue balls in the second urn

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 Independent Events
 The definition of independence can be extended to more than
three events
 The events E1, E2, …, En are said to be independent if, for every
subset E1, E2, …, Er of these events

 We also define an infinite set of events to be independent if

every finite subset of these events is independent

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 Independent Trials
 Sometimes, the probability experiment under consideration
consists of performing a sequence of subexperiments
 In many cases, it is reasonable to assume that the outcomes of
any group of the subexperiments have no effect on the
probabilities of the outcomes of the other subexperiments
 If such is the case, we say that the subexperiments are
independent
 More formally, we say that the subexperiments are independent
if E1, E2, …, En, … is necessarily an independent sequence of
events whenever Ei is an event whose occurrence is completely
determined by the outcome of the ith subexperiment
 If each subexperiment is identical, then the subexperiments are
called trials

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 Exercise 3.12
 An infinite sequence of independent trials is to be performed
 Each trial results in a success with probability p and a failure
with probability 1 – p
 What is the probability that

 At least 1 success occurs in the first n trials

 Exactly k successes occur in the first n trials
 All trials result in successes

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 P(  |F) is a Probability
 Conditional probabilities satisfy all of the properties of
ordinary probabilities
 Proposition 3.3:

 P(E | F) satisfies the three axioms of a probability

• 0  P(E | F)  1
• P(S | F) = 1
• If Ei, i = 1, 2, … are mutually exclusive events, then

 Other useful equations can be proved

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 Conditional Independence
 An important concept in probability theory is that of the
conditional independence of events
 We say that the events E1 and E2 are conditionally independent
given F if
 Given that F occurs, the conditional probability that E1
occurs is unchanged by information as to whether or not E2
occurs
 More formally, E1 and E2 are said to be conditionally
independent given F if

 It is easy to extend the notion of conditional independence to

more than two events

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 Exercise 3.13 (Exercise 3.9)
 It is assumed that there are 80% good drivers and 20% bad
drivers in a population
 For a good driver, there is a 10% chance of accident(s) in a year

 For a bad driver, there is a 50% chance of accident(s) in a year

 What is the conditional probability that a new policyholder will

have an accident in his second year of policy ownership, given
that the policyholder has had an accident in the first year?
 Suppose that the same customer has an accident in the second
year, what is the probability that he is a good driver?

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