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Lecture 4
! Conditionally Independent Events
! Bayes’ Theorem
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Review 1: Conditional Probability
! The probability of an event A changes after it has been learned that some other event B has occurred. This new probability of A is called the conditional probability of the event A given that the event B has occurred. It is denoted by
)|Pr( BA
)Pr()Pr()|Pr(
BABBA = , if Pr(B)>0.
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Review 2: Independence of Several Events
! The k events are independent if, for every subset of j of these events (j=2,3,...,k),
! E.g. Two conditions must be satisfied in order for three events A, B and C to be independent.
kAA ,...,1
jii AA ,...,1
)2( )Pr()Pr()Pr()Pr(
)Pr()Pr()Pr()1()Pr()Pr()Pr(
)Pr()Pr()Pr(
CBAABC
CBBCCAACBAAB
=
===
)1()2()2()1( ÞÞ
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Conditionally Independent Events
! A1,…,Ak are conditionally independent given Bif, for every subset Ai1,…,Aij (j=2,3,…,k),
Question: Can conditional independence imply independence.
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Are A,B conditionally independent given C?
Are A,B independent?
Question: can independence imply conditional independence?
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Theorem: Suppose that A1, A2 and B are events suchthat Pr(A1B)>0. Then A1 and A2 are conditionallyindependent given B if and only if
Pr(A2|A1B)=Pr(A2|B)
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2.3 Bayes’ Theorem虚假阳性病例
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由于治疗这种疾病有严重的副作用,所以医生请教概率专家乔·贝叶斯,乔引用了一个法则给以答复,此定理最早由他的祖先牧师托马斯·贝叶斯(1744年~1809年)证明过。
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( ) ( | )
( )From the statement of the problem, the following information is available: ( ) .001 ( ) .999
Solution:
C
P A BP B AP A
P B P BP
Ç=
= =
( | ) .99 ( | ) .02
By the Multiplicative Probability Rule, ( ) ( ) ( | ) (.001)(.99) .00099, ( ) ( ) ( | ) (.999)(.02) .01998.We ca
C
C C C
A B P A B
P A B P B P A BP A B P B P A B
= =
Ç = = =
Ç = = =n also get
( ) ( ) ( ) .00099 .01998 .02097.Therefore,
( ) .00099 ( | ) .0472 ( ) .02097
CP A P A B P A B
P A BP B AP A
= Ç + Ç = + =
Ç= = =
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Prior and Posterior Probabilities
! Pr(B) is called the prior probability.• Before the experiment.
! Pr(B|A) is called the posterior probability.• After the experiment’s result is known.
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Bayes’ Theorem! Probability revision
Prior Probabilities
NewInformation
Bayes’ Theorem
Posterior Probabilities
initial probabilityestimates for specificevents of interest
additional informationabout the events
revised probabilities
Pr(Bj) event A happened Pr(Bj|A)=?
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Probability and Partitions! A partition of the sample space S:B1,…,Bk are disjoint and
! A is any other event in S, we will have a partition of A:
( ) ( ) ( )1 2 kA B A B A B A= È È ÈL
SBiki =È =1
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1
1
Pr( ) Pr( )
Pr( ) 0, 1, Pr( ) Pr( )Pr( | )
Pr( ) Pr( )Pr( | )
k
jj
j j j j
k
j jj
A B A
B j k B A B A B
A B A B
=
=
=
> = Þ =
Þ =
å
å
L
Theorem 2.3.1 Law of total probability. Suppose that the eventsB1,…,Bk form a partition of the sample space S and Pr(Bj)>0.Then, for any event A in S,
å=
=k
jjj BABA
1)|Pr()Pr()Pr(
Proof:
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Bayes’ Theorem! Let the events B1, ...,Bk form a partition of the space S
such that Pr(Bi)>0 for j=1,...,k, and let A be any event such that Pr(A)>0. Then, for j=1,...,k,
Proof.
å=
= k
jjj
iii
BAB
BABAB
1)|Pr()Pr(
)|Pr()Pr()|Pr(
å=
=
=
=
k
jjj
iii
ii
BABA
BABABAABAB
1)|Pr()Pr()Pr(
)|Pr()Pr()Pr()Pr()Pr()|Pr(
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Example: Identifying the Source of a Defective Item
! Three different machines: M1, M2, M3. 20% of the items were produced by M1, 30% by machine M2, and 50% by machine M3.
! 1% of the items produced by M1 are defective; 2% of the items produced by M2 are defective; 3% of the items produced by M3 are defective. The outcomes from each machine are independent.
! One item is selected at random from the entire batch and it is found to be defective. What is the probability that this item was produced by M2?
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! Bi: the selected item was produced by machine Mi, i=1,2,3A: the selected item is defective
We need to calculate Pr(B2|A)Pr(B1)=0.2, Pr(B2)=0.3, Pr(B3)=0.5Pr(A|B1)=0.01, Pr(A|B2)=0.02, Pr(A|B3)=0.03
2 22 3
1
Pr( )Pr( | )Pr( | )Pr( )Pr( | )
(0.3)(0.02) 0.261(0.2)(0.01) (0.3)(0.02) (0.5)(0.03)
j jj
B A BB AB A B
=
=
= =+ +
å
1 3Similarly, Pr( | ) 0.087, Pr( | ) 0.652B A B A= =
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Computation of Posterior Probabilities in More Than One StageA box contains one fair coin and one coin with a head on each side. One coin is selected at random and when it is tossed, a head is obtained. What is the probability that the selected coin is the fair coin?
B1: the coin is fairB2: the coin has two headsH1: a head is obtained when the coin is tossed
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)1)(5.0()5.0)(5.0()5.0)(5.0(
)|Pr()Pr()|Pr()Pr()|Pr()Pr()|Pr(
212111
11111
=+
=
+=
BHBBHBBHBHB
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The same coin is tossed again and another head is obtained. What is the new posterior probability that the coin is fair?
H2: a head is obtained on the second toss
51
)1)(5.0()25.0)(5.0()25.0)(5.0(
)|Pr()Pr()|Pr()Pr()|Pr()Pr()|Pr(
22121211
1211211
=+
=
+=
BHHBBHHBBHHBHHB
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遗传风险
在人类遗传学中,某种坏的基因会引起夭折。设隐性基因 a是这样一个基因,基因型 aa不能长大成人,基因型 Aa的人为携带者。假定在一般总体中携带者的概率为 p.
已知某成人有一个哥哥或一个姐姐夭折,求该人为携带者的概率。
解:首先,双亲都必须是携带者 Aa,那么,他们的孩子是 AA,Aa,aa的概率为1/4, 1/2, 1/4.
P(Aa|成人)=P(Aa)/P(AA或 Aa)=2/3