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1、Lecture 4 : Conditional Probability and Bayes Theorem 0/ 26 1/ 26 The conditional sample space Motivating examples 1. Roll a fair die once S= 123 456 Let A=6 appears B=an even number appears So P(A) = 1 6 P(B) = 1 2 Lecture 4 : Conditional Probability and Bayes Theorem 2/ 26 Now what about P 6 appea
2、rs given an even number appears ! Philosophical Remark (Ignore this remark unless you intend to be a scientist) At present the above probability does not have a formal mathematical defi nition but we can still compute it. Soon we will give the formal defi nition and our computation “will be justifi
3、ed”. This is the mysterious way mathematics works. Somehow there is a deeper reality underlying the formal theory. Back to Stat 400 The above probability will be written written P(A|B)to he read P(A given B). Lecture 4 : Conditional Probability and Bayes Theorem 3/ 26 Now we know an even number occu
4、rred so the sample space changes 1 2 3 4 5 6 the conditional sample space So there are only 3 possible outcomes given an even number occurred so P(6 given an even number occurred) = 1 3 The new sample space is called the conditional sample space. Lecture 4 : Conditional Probability and Bayes Theorem
5、 4/ 26 2. Very Important example Suppose you deal two cards (in the usual way without replacement). What is P()i.e., P(two hearts in a row). Well, P(fi rst heart)= 13 52. Now what about the second heart? Many of you will come up with12/51and P() = (13/52)(12/51) Lecture 4 : Conditional Probability a
6、nd Bayes Theorem 5/ 26 There are TWO theoretical points hidden in the formula. Lets fi rst look at P(on 2nd | z this isnt really correct ) =12/51 What we really computed was the conditional probability P(on 2nddeal| on fi rst deal) =12/51 Why? Given we got a heart on the fi rst deal the conditional
7、sample space is the “new deck” with 51 cards and 12 hearts so we get P(on 2nd| on 1st) =12/51 Lecture 4 : Conditional Probability and Bayes Theorem 6/ 26 The second theoretical point we used was that in the following formula we multiplied the two probablities P(on 1st)and P(on 2nd| on 1st) together.
8、 This is a special case of a the formula for the probability of the intersection of two events that we will state below. P() =P(on 1st)P(on 2nd| on 1st) = 13 52 ! 12 51 ! The general formula, the multiplicative formula, we will give as a defi nition shortly is P(AB) =P(A)PB|A) Compare this to the ad
9、ditive formula which we already proved P(AB) =P(A) +P(B) P(AB) Lecture 4 : Conditional Probability and Bayes Theorem 7/ 26 Three Basic Questions These three examples will occur repeatedly in todays lecture. The fi rst is the example we just discussed, the second is the reverse of what we just discus
10、sed and the third is a tricky variant of fi nding the probability of a heart on the fi rst with no other information. 1What is P(on 2nd| on 1st | z reverse of pg. 5 ) 2What is P(on 1st| on 2nd | z reverse of pg. 5 ) 3What is P(on 2ndwith no information on what happened on the 1st). Lecture 4 : Condi
11、tional Probability and Bayes Theorem 8/ 26 The Formal Mathematical Theory of Conditional Probability (S) =n, (A) =a, (B) =b, (AB) =C Problem Let S be a fi nite set with the equally - likely probability measure and A and B be events with coordinalities shown in the picture. Lecture 4 : Conditional Pr
12、obability and Bayes Theorem 9/ 26 Problem Compute P(A|B). We are given B occurs so the conditional sample space is B Only part of A is allowed since we know B occurred namely the part AB. So counting elements we get P(A|B) = (AB) (B) = c b Lecture 4 : Conditional Probability and Bayes Theorem 10/ 26
13、 We can rewrite this as P(A|B) = c b = c/n b/n = P(AB) P(B) so P(A|B) = P(AB) P(B) (*) This formula for the equally likely probability measure leads to the following. Lecture 4 : Conditional Probability and Bayes Theorem 11/ 26 Formal Mathematical Defi nition Let A and B be any two events in a sampl
14、e space S with P(B) ,0. The conditional probability of A given B is written P(A|B) and is defi ned by P(A|B) = P(AB) P(B) (*) so, reversing the roles of A and B (so we get the formula that is in the text) if P(A) ,0 then P(B|A) = P(BA) P(A) = P(AB) P(A) (*) Since ABBA. Lecture 4 : Conditional Probab
15、ility and Bayes Theorem 12/ 26 We wont prove the next theorem but you could do it and it is useful. Theorem Fix B with P(B) ,0. P(|B), ()so P(A|B) as a function of A), satisfi es the axioms (and theorems) of a probability measure - see Lecture 1. For example 1P(A1A2|B) =P(A1|B) +P(A2|B) P(A1A2|B) 2P
16、(A0|B) =1P(A|B) P(A|)(so P(A|B)as a function of B) does not satisfy the axioms and theorems. Lecture 4 : Conditional Probability and Bayes Theorem 13/ 26 The Multiplicative Rule for P(AB) Rewrite (*) as P(AB) =P(A)P(B|A)() ()is very important, more important then (*). It complement the formula P(AB) =P(A) +P(B) P(AB) Now we know how P interacts with the bas
