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1 Statistics & Data Analysis 2 (Part I) Zhu Huaiqiu University

3 Example49613,983,816l /52=

4 Example n p(n) 12% 41% 70% 97% % % 1 ( ) 1 ( ) 100% Example

5 Example

6 E (Random experiment E ) E (Experiment E ) EE 1, E ω generic i element ω Ω sample space Ω

7 Example 2.1E 1 Example 2.2E 2 Example 2.3E 3 DNA Example 2.4E 4 Example 2.5E 5 5 HBsAg() -HBs()еHBeAg( )е-hbe(е )-HBc() 3 Example 2.6E 6 50% A Random event A A Event A Ω A, B, BCA C, A1, A2, Ω Ω Example 2.1E 1 3

8 Example pe2.2e 2 (H)(T) ) 3H Example 2.3E 3 DNA 3 Example 2.5E 5 5 HBsAg() -HBs()еHBeAg( )е-hbe( )-HBc() () 3 3 ΩA, B, C, A k (k=1, 2, ) Ω 1 Ω 2

9 3 Ω 4 Ω 5 Ω 6 Ω

10 7 Ω 1AB=BAAB=BA 2(AB)C=A(BC)(AB)C=A(BC) 3A(BC)=(AB)(AC)A(BC)=ABAC ( ) ( )( ) ( ) 4(De Morgan) A B AB AB AB 5A 6 A A A 7AA 8A B AB A( AB)

11 Example p 2.7(a)(b)Ai ( ) ( ) ={i} AiA={LR} 1 (a) L 2 3 R 4 (b) L R 2 5 (Complete events) ΩnA i (i=1, 2,, n) 1A i A j ( i j, i, j = 1, 2,, n) j 2A 1 A 2 A n = Ω naa i (i=1, 12,, n)ω, partition A 2 A 1 A 3 An 1 A n

1 3 4 R 2 5 (Complete events) ΩnA i (i=1, 2,, n) 1A i

12 2.1.2 Kolmogorov σ σ σ-field or σ-algebra ΩF Ω Ωσ: F C A F A F ( n N, A F ) A F n n 1 σ σ σω n

13 Kolmogorov Probability Axioms Ω Ω F Ω σf Ω PF Ω Ω AF Ω P(A)[0, 1] P(Ω)=1. A 1, A 2,... P A1 A2 P A i (...) ( ) ΩEF Ω σ A F Ω P(A)R + R + P(A) P(A)[0, 1] P(Ω)=1 1 A 1,A 2,... P ( A ) P ( A ) P(A)A i i 1 i 1 i

14 1 P()=0 0 2 A 1,A 2,... A n n n P ( A ) P ( A ) i i1 i1 i 3 A P ( A) 1 P( A) 4 ABP(A-B)=P(A)-P(B) 5 A, BP(AB)=P(A)+P(B)-P(AB) (John Venn )1923) (Richard Von Mises ) n A naa n A /naf n (A) 10 f n (A) 1 2f n (Ω)1 3AB= f n (A+B)= f n (A)+ f n (B)

15 f n n A f n (A) n n H f n (H) De Morgan Buffon K. Pearson K. Pearson n A f n (A)

50000 3 1 0.33333 4 1 0.25000 5 2 0.40000 10 2 0.20000 20 5 0.25000 100 12 0.12000 200 39 0.

16 f n (A) n EnAf n (A) p[0, 1]ppA P(A)P(A) nf n (A)p P(A) 10 P(A) 1 2P(Ω)1 3AB= P(A+B)=P(A)+P(B) 3k

18 Laplace E Laplace 1ΩΩ{ωΩ{ω 1, ω 2,, ω n } 2 P(ω 1 )=P(ω 2 )= =P(ω n )=1/n EΩnA map(a) PA ( ) m n

19 P(A) 10 P(A) 1 2P(Ω)1 3ABAB=ΦP(A+B)=P(A)+P(B) 3k Example p 2.8Simpson s paradox A B C

20 1 s (s 1)i r i i=1, 2,, s r 1 +r 2 + +r+ s 2 s(s 1)ir s i i=1, 2,, s r 1 r 2 r s 3 n m(m n) m n! A n nn ( 1)( n2)...( nm1) ( n m)! m=n n m (m n)n n m n m(m n) m nn ( 1)( n 2)...( n m 1) n! R n m m ( n m )!

21 4 nm (m n) C m n C m n nn ( 1)( n 2)...( nm 1) n! mm ( 1)...1 m!( nm)! n m C n C C C m m1 m n n n1 C C... C 2 C 0 1 n n n n m m k m k n1 n C 2 nc 1 n2 k 0 n Example

22 Example n 2n Example 2.11 Capture/recapture method n

23 1Ω 2 Ω A D( A) PA ( ) D ( ) D(Ω)ΩΩ D(A)A Ω A ΩAΩA A

24 P(A) P(A) 10 P(A) 1 2P(Ω)11 3ABAB=ΦP(A+B) P(A+B)=P(A)+P(B)P(A)+P(B) 3k ΩA Buffon's needle: D LL<D p=2l/(πd)π 1901 Mario Lazzarini 3408π 355/113 6

25 Example 2.12 [0, T]t (t T) y T t 0 t T x Example 2.13Bertrand paradox 1

27 2.1.3 Example 2.14Digitalis Digitalis DigitalisTBeller DigitalisTDigitalis g g D D Total D T 25 (0.185) 14 (0.104) 39 (0.289) T 18 (0.133) 78 (0.578) 96 (0.711) Total 43 (0.318) 92 (0.682) 135 D T T Beller, G. et al. (1971). Digitalis intoxication: A prospective clinic study with serum level correlations. New England J. Med., 284: Conditional Conditional probability EΩA, BP(A)>0 P ( AB ) P ( B A) P ( A ) AB Ω AB A AB B P(A B)

28 3 1 0 PB ( A ) 1 2P(Ω A)1, P( A)=0 3A1, A2,, P ( Bi A) P( Bi A) A PA ( A B ) PA ( B ) PA ( B ) PAA ( B ) P ( B A ) 1 P ( B A )

29 Multiplication Multiplication law P ( AB ) P ( A ) P ( B A ) ( P ( A ) 0) P ( AB ) P ( B ) P ( A B ) ( P ( B ) 0) P(ABC)=P(A)P(B A)P(C AB) P(A 1 A 2 A n )=P(A 1 )P(A 2 A 2 ) P(A n A 1 A 2 A n-1 ) Example Example

30 Example 2.17Poloya br c nn 1 n 2 =nn 1 Poloya Poloyann y 2? Total probability EΩAA 1, A 2,,A n Ω P(A i )>0 (i=1, 2,, n)b n P ( B ) P ( Ai) P ( B Ai) i 1

31 BΩn A 1,A 2,,AP(B A n i ) A 2 A 1 B A 3 An 1 A n Example 2.18 UML U 1 UU 2 UGlassHall1954EnglandWales Transition probability matrix 10%U40%M50%L UML Transition probability matrix U M L U2 M2 L2 U M L U: upper level M: middle level L: lower level Glass, D., & Hall, J. (1954). A study of intergeneration changes in status. In Social Mobility in Britain.

32 Bayes Bayes Example 2.18 UU Transition probability bilit matrix U2 M2 L2 U M L U: upper level M: middle level L: lower level Bayes Bayes EΩAA 1, A 2,,A n Ω P(A i )>0 (i=1, 2,, n)bp(b)>0 P ( A i ) P ( B A i ) P ( A i B ) n P ( A ) P ( B A ) j 1 j j A 2 B A 1 A 3 An 1 An

33 B A 1, A 2,, A n BBayesBayes B P ( A B ) i n j 1 P ( A ) P ( B A ) i P ( A ) P ( B A ) j i j P(A i ) {A i } (prior probability) P(B) B (prior probability) P(A i B) (posterior probability) Bayes A 2 A 3 B An 1 A 1 An

34 Structure of Bayesian network. Labeled ovals represent nodes; arrows (arcs) represent conditional dependence relationships. Each node is a data structure that contains conditional probability tables to quantify probabilistic relationships between variables. Cacalcifications, FCfibrocystic change, FHxfamily history of breast cancer, HRThormone replacement therapy, LNlymph node, P/A/O present, absent, or obscured. Elizabeth etal., (2006) Bayesian Network to Predict Breast Cancer Risk of Mammographic Micro-calcifications and Reduce Number of Benign Biopsy Results: Initial Experience. Radiology, 240, Example 2.19 X 0123 T 0, T 1, T 2, T 3 PD ( T) i PD ( ) PT ( i D ) PD ( ) PT ( D) PD ( ) PT ( D) i i i P(T i D+) P(T i D-) P(D+) P(D+) 0.92 Diamond, G. & Forrester, J. (1979). Analysis of probability as an aid in the clinical diagnosis of coronary-artery disease. New Eng. J. Med., 300:

35 Example 2.20 Lie detector t Gastwirth1987P(+ L)=0.88 P( T)=0.86LT + P(T)=0.99P(L)=0.01 P(L +) Gastwirth, J. (1987). The statistical precision of medical screening procedures. Statistical Science, 3: Example

36 Example A={}B={ }P(A)P(B)P(AB)P(B A) 1. ABP(AB)=P(A)P(B)AB A,, BP(A)>0A, B P(B A)=P(B) P(B)>0A, BP(A B)=P(A)

37 1AB 2A 3 B AB 4A B 2. A, B, C P(AB)=P(A)P(B); P(A)P(B) P(BC)=P(B)P(C); P(AC)=P(A)P(C); P(A)P(C); P(ABC)=P(A)P(B)P(C) A, B, C

38 n na 1,A 2,,Ak(1<kn) n ki 1, i 2,, i k (1i 1 <i 2 < < i k n) i i... i i i... i P A A A P A P A P A 1 2 k 1 2 na 1, A 2,, A n 1A 1, A 2,, A n k(1<kn) 2k=2 3nn n 4 k n C 2 n 1 k 2 n k Example 2.7(a)(b)Ai={i} p A={LR} 1 (a) L 2 3 R 4 (b) L R 2 5

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untitled Statistics & Data Analysis 2 (Part I) Zhu Huaiqiu @Peking University 2.1 Example49613,983,816 13,983,816/52=268,919 Example232111 23 n 10 20 30 50 100 200 300 350 366 p(n) 12% 41% 70% 97% 99.99996% 99.9999999999999999999999999998%