04_CondProbBayesIndep.ppt [相容模式]

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1 Conditional Probability & Bayes Rule p. 4-1 Q: Should the following probabilities be different? Event = a pitcher wins at least 16 games in a season P=?? in the beginning of the season P=?? in the middle of the season P=?? close to the end of the season Event = rain tomorrow P=?? if no information about where you are staying P=?? if you are staying in a desert P=?? if a typhoon will hit the place you stay tomorrow Q: What causes the differences? For an event, new information (i.e., some other event has occurred) could change its probability We call the altered probability a conditional probability Mathematical Definition: If A and B are two events in a sample space and P(A)>0, then p. 4-2 is called the conditional probability of B given A. In the classical probability, P(A) = #A/# and P(A B)=#(A B)/# Example: A family is known to have 2 children, at least one of whom is a girl. Q: Probability that the other is a boy=?? = {bb, bg, gb, gg} A={bg, gb, gg} and B={bb, bg, gb} P(B A) = #(A B)/#A = 2/3 Note: # is reduced to #A.

2 In effect by conditioning, we are restricting the sample space from to A, i.e., Ω A, and, for an arbitrary event B (in ) to occur when A has occurred, we need that both A and B occur together, i.e., B B A. The division by P(A) in the definition above rescales all probabilities from the entire sample space to being relative to the new sample space A P(B A) is a probability measure defined on B, but not A. P( A) satisfies the 3 axioms of probability (exercise) p. 4-3 P() P( A) Any propositions developed in Chapter 2 for probability measures can be applied on P( A). (For example, P(B c A)=1P(B A).) 3 Useful Formulas for Calculating Probabilities (for 2-events case) 1.If P(A)>0, then p If 0<P(A)<1, then 3.If 0<P(A)<1 and P(B)>0, then

3 Example (Urn Problem) The Story. n balls sequentially and randomly chosen, without replacement, from an urn containing R red and NR white balls (nn). Q: Given that k of the n balls are red (kr), probability that the 1 st ball chosen is red =?? Let A{k of the n balls are red} Method 1: Method 2: B{1 st ball chosen is red} p. 4-5 p. 4-6 Example (Sampling Experiments): An urn contains R red balls and NR white balls. Sample 2 balls from the urn. A = {red on the first draw} B = {red on the second draw} Sampling Without Replacement: P (A) = R(N 1) N(N 1) = R R(R 1), P(A B) = N N(N 1), P (B A) = P (A B) P (A) Similarly, P (B A c )= = (R(R 1))/(N(N 1)) R/N R N 1. (exercise) = R 1 N 1.

4 P (B) = P (A)P (B A)+P(A c )P (B A c ) = R N R 1 N 1 + N R N R N 1 = R2 R + NR R 2 R(N 1) = N(N 1) N(N 1) = R N. P (A B) = Notes: P (A B) P (B) = (R(R 1))/(N(N 1)) R/N p. 4-7 = R 1 N 1. The probabilities are proportional to # of red balls left P(A B) = P(B A) Symmetry. Sampling With Replacement: P (A) = R N N N = R N, P (A B) = R N R N = R2 N 2, P (B) =N N R N = R N, P (B A) = R2 /N 2 R/N = R N = P (B). p. 4-8 Example (Diagnostic Tests) The Story. A diagnostic test for a rare disease (e.g., an Xray for lung cancer) is part of a routine physical exam. Let {the whole population of Taiwan} D {disease is present} E {test indicates disease present} Suppose that P(D)=0.001, P(E D)=0.98, P(E D c )=0.01 Q: Do you think the test is effective? Let us examine it from an alternative viewpoint. Suppose that you are tested as positive. What is the probability that you actually have the disease? P (D c E) =1 P (D E) =0.9107

5 Q: Now, do you still think the test is effective? The probability of D increased by a factor of roughly 90 ( ) when E occurs, but is still small The P(E D c )(0.01) and P(E c D) (1P(E D)0.02) are called the false positive and false negative rates, respectively. Q: Why the 2 interpretations so different? What causes it? p. 4-9 Example ( 引自快思慢想, Kahneman) 在 1970 年代, 湯姆是美國某州裡重要大學的研究生, 請將下面九個研究所領域排序, 標出湯姆就讀這些領域的可能性,1 代表可能性最高,9 代表最低 A. 企業管理, B. 電腦, C. 工程, D. 人文與教育, E. 法律, F. 醫學, G. 圖書館學, H. 物理和生命科學, I. 社會學和社會工作下面是湯姆念高三時, 心理學家根據心理測驗結果對湯姆的人格素描 : 湯姆是個很聰明的學生, 但缺少真正的創造力他喜歡整潔和秩序, 他的每一樣東西, 不管多少, 都有條有理的擺放在恰當位置上他的作文有點無趣呆板和機械式, 偶爾會出現一些陳舊過時的雙關語和類似科幻想像的句子他的好勝心很強, 對人冷淡, 没什麼同情心, 也不喜歡跟別人來往雖然以自我為中心, 卻有很強的道德感把剛剛那幾個領域再排序一下, 你認為湯姆最可能是哪個領域的研究生,1 代表可能性最高,9 代表最低在 1970 年代, 受試者的排序大致如下 : 1. 電腦, 2. 工程, 3. 企業管理, 4. 物理和生命科學, 5. 圖書館學, 6. 法律, 7. 醫學, 8. 人文與教育, 9. 社會學和社會工作 p. 4-10

6 Extensions of the 3 Useful Formulas (for m-events case) 1.(Multiplication Law) If A 1,, A m are events for which P (A 1 A m 1 ) > 0, then P (A 1 A m ) p = P (A 1 )P (A 2 A 1 )P (A 3 A 1 A 2 ) P (A m A 1 A m 1 ). Example (Birthday Problem, LNp.3-4). Let A k be the event that the k th birthday differs from the first k1. Then, P(A 1 )1, p Example (Matching Problem, LNp.3-14). Q: Probability that exactly k of n members (kn) have matches =?? Let be all permutations = (i 1,, i n ) of 1, 2,, n. Let A j = {: i j = j} and By symmetry, Let and Then, where and,

7 when n is large p (Law of Total Probability) Let A 1,, A m be a partition of and P(A i )>0, i=1,, m, then for any event B, P (B) = m i=1 P (A i)p (B A i ). 3. (Bayes Rule) Let A 1,, A m be a partition of and P(A i )>0, i=1,, m. If B is an event such that P(B)>0, then for 1 j m, P (A j B) = P (A j )P (B A j ) m i=1 P (A i)p (B A i ). From Bayesians viewpoint, p P(A j )probability of A j before B occurs prior prob. P(A j B)probability of A j after B occurs posterior prob. The Bayes rule tells how to update the probabilities of A j in light of the new information (i.e., B occurs) An Application of Bayes Rule. Suppose that a random experiment consists of two random stages 1 st 2 nd The probabilities of the 2 nd -stage results depend on what happened in the 1 st stage We never see the result of the 1 st stage, only the final result We may be interested in finding the probability for outcomes in the 1 st stage given the final result

8 Example (Gold Coins): The Story. Box 1 contains 2 silver coins. Box 2 contains 1 gold and 1 silver coin. Box 3 contains 2 gold coins. 1 st 2 nd Experiment: (i) Select a box at random and, (ii) Examine the 2 coins in order (assuming all choices are equally likely at each stage) Q: Given that 1 st coin is gold, what is the probability that Box k is selected, k=1, 2, 3? Let A k {Box k is selected}, B {1 st coin is gold}, P (B A k )= 0, if k =1, 1/2, if k =2, 1, if k =3. P (B C A k )= P (B C) = = 1 3. p P (B) = = 1 2. p P (A 1 B) = (1/3) 0 =0. 1/2 Similarly, P (A 2 B) =1/3, P (A 3 B) =2/3. Q: Given that 1 st coin is gold, what is the probability that 2 nd coin is gold? Let C = {2 nd coin is gold}. P (C B) = P (B C) P (B) = 1/3 1/2 = 2 3. Example (TV Game Show: Let s Make A Deal) The story. 0, if k =1, 0, if k =2, 1, if k =3. 1. The contestant is given an opportunity to select one of three doors. 2. Behind one of the doors is a great prize (say, a car) and there is nothing behind the other two doors. 3. The host knows which door contains the car, but the contestant does not.

04_CondProbBayesIndep_wHWN.ppt [相容模式]

04_CondProbBayesIndep_wHWN.ppt [相容模式] Conditional Probability & Bayes Rule p. 4-1 Q: Should the following probabilities be different? Event a pitcher wins at least 16 games in a season P?? in the beginning of the season P?? in the middle of