STA107H1b.pdf

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Department
Statistical Sciences
Course
STA130H1
Professor
V.Radu Craiu
Semester
Fall

Description
STA107H1b.doc Lecture #12 – Wednesday, October 15, 2003 P ARTITIONS 1) The sets{A1, 2 ,..n} Aare mutually exclusive n 2) ▯ Ai= S i=1 n • If l ,…,nA is a partition then for any event B Ì▯S(B ˙ Ai) i=1 Example c • B =(B˙ A )¨ (B˙ A ) L AW OF TOTAL P ROBABILITY • If{B,B c} is a partition, PhAn)= P(A| BP B )+ (A| Bc)P(Bc) • If{B1,B2,B 3} is a partition of SPaBdi> 0 for i =1,2,3,..., then P(A)= P A| B1)P(B1)+ P(A| 2 P B 2)+ PA| B3)P(B3) Generalization n • P(A)= ▯ P(A| BiP B i) i=1 Example A transport company needs to do safety checking for its cars. Suppose there are only two repair stores in town. One of them is owned by a rival firm and the other is owned by an allied firm so the diagnostics will not be objective. The friendly garage will lie with probability 20% if a car is unsafe and will tell the truth if the car is safe. The “rival” garage with probability 35% will declare a car unsafe even if the car is safe and will always declare the truth if the car is unsafe. Assume that 20% of the company’s cars are actually unsafe and assume that the company will obtain a certificate from each garage for each of its cars. What is the percentage of cases in which the two conclusions are contradictory? • G ={Goodcars }, B ={Badcars} • C ={Conclusionsarecontradicto}y • P(C) = PC ˙G + P(C ˙B )becauseG and Bformsa partition of allcars = PC |G P(G )+ PC | )P(B) • PC |G )= 35% because the friendly garage tells the truth and the rival garage tells the lie • PC | B)= 20% because the rival garage tells the truth and the friendly garage tells the lie • P(C) = 35 · 80 + 20 · 20 = 32 100 100 100 100 100 B AYES R ULE • Start with a parti{B1, 2 , 3}, • Take an event A Page 1 of 17 STA107H1b.doc • Instead of being interested in(A| B1), we are now interested in B 1 A ) P B |A = P B 1˙ A) 1 P A) = P A B P1B ( 1 P A ˙ B1)+ P A ˙ B 2)+ P A ˙ B3 ) = P A B P1B ( 1 P A B P1B ( 1 + P A B P2B ( 2 + P A B P3B ( 3 Experiment In 2 Stages • First stage – one of the B’s happens • Second stage – we check whether A happens P(B 1 P(B3) P(B 2 Stage 1 B1 B1 B 1 P(A | B ) P(A | B ) P(A | B ) 1 2 3 c c c Stage 2 A A A A A A • I know A happened. What is the probability tha1 B happened? P B1˙ A ) P A B 1 B( 1 • P B1 |A = P A ) = P A B P B( ) + P A B P B ( ) +P A B P B ( ) 1 1 2 2 3 3 • P(A )= P(A| B1)P(B1 + P (A| B2)P(B2 )+ P(A| B3)P(B 3) • P(B ˙ A ) = P(A˙B )= P (A| B P (B ) 1 1 1 1 Example Say I have a wallet that contains either a $5 bill or a $20 bill (with equal probability), but I don’t know which one. I add a $5 bill. Later, I reach into my wallet (without looking) and remove a bill. It’s a $5 bill. What is the probability that the bill left in the wallet is a $5 bill? Start 1/2 1/2 $5 $20 Add 2 · $5 $5, $20 1 0 1/2 1/2 Extract $5 $20 $5 $20 Page 2 of 17 STA107H1b.doc 1 ×1 2 2 • P start with $5|extract )=$1 1 1 = 3 ×1+ × 2 2 2 Example A lab blood test is 95% effective in detecting a certain disease when it is in fact, present. However the test also yields a “false positive” result for 1% of the healthy people tested. If 0.5% of the population has the disease, what is the probability hat a person who was tested positive actually has the disease? Initial Condition 0.005 0.995 sick healthy 0.95 0.05 0.01 0.95 Test P N P N P sick˙ positive P positive|sickP(sick) • P sick | positi=e P positiv) = P(positiv) • P(positiv)= 0.005·0.95+0.995·0.01 • P(sick˙positive)= 0.005·0.95 • P sick | positi=e 0.005·0.95 = 32.31% 0.005·0.95 +.995 ·.01 Lecture #13 – Wednesday, October 22, 2003 INDEPENDENCE • Two events, A and B are independent P(A˙B )= P(A ×P B ). • In general,P(A˙B )= P(A| B)×P(B . • If A and B are independenP(A| B)= P A ) – The fact that B happened doesn’t change the chance of the occurrence of A. • In general{A , A ,..., } are independent if any twi Aj, A are independent. 1 2 n Example An urn contains five red and seven blue balls. Suppose that two balls are selected at random and with replacement. Let A and B be the events that the first and the second balls are red, respectively. Check whether A and B are independent or not. Redo the calculation for the case of random selection without replacement. • With replacement: • P A˙B =) 5·5 . 12·12 Page 3 of 17 STA107H1b.doc • P A = 5 , P B = 5 . 12 12 • So A, B are independentA ^ B ). • Without replacement: • P A˙B = ) 5·4 = 20 . 12·11 12·11 5 4 5 5 7 55 5 • P A = , P B = P B | A P A + P B | A P Ac = × + × = = . 12 11 12 11 12 11·12 12 • So A and B are not independent. Exercises 1) Show that if E and F are independent then E and F are independent. What about then E and F ? c c • I want to show P(E ˙F )= P E )×P(F . P E ˙ F c)= P E ˙ F c)+ P(E ˙ F)- P(E ˙ F) = P( )- P ( ˙ F ) = P( )- P ( ) ( ) = P( ) (1 P ( )) = P( )×P F ) c c c • E ^ F ▯ E ^ F ▯ E ^ F 2) Show that if E and F are mutually exclusive wi(E )> 0 and P (F )> 0 , then E, F cannot be independent. • P (E ˙ F)= 0 . • If E and F are independent,(E ˙ F)= P E )×P(F) > 0 . • So A and B are not independent if they are mutually exclusive. 3) True or False: If E and F are independent and E and G are independent, then E and F ˙G are independent. • False! 4) True or False: In the case of 3 events E, F, G, the eq(E ˙ F ˙G )= P E )×P(F)×P G ) doesn’t imply that E, F, G are independent. • True! P ROBABILITY AS A C ONTINUOUS FUNCTION • f :R fi R . • x ¾¾¾fi x . n nfi¥ • f is continuous in x lim f xn = f x ) xfi¥ • P :sets from S fi[0,] Page 4 of 17 STA107H1b.doc Increasing Sequence of Events • {E 1E 2...,En} infinite sequence is increasinEkÌ E k+1,"k ‡1 . E 2 E 1 E 3 E 4 ¥ • The “limit” is the E . ▯ k k=1 Decreasing Sequence of Events • {E 1E 2...,En} infinite sequence is decreasinE k E k+1,"k ‡1 . E 3 E4 E 2 E 1 ¥ • The “limit” is the E k. ▯ k 1 • {E n} is decreasing or increasing. I can define nim ▯ =E n ▯ En). • In both cases, lim PE n)= P▯lim E n ▯. nfi¥ ▯ fi¥ ▯ Lecture #14 – Friday, October 24, 2003 10 26 ¥ • If sets are increasing, then lin E▯= En – the smallest set containing all the sets. nfi¥ n=1 ¥ • If sets are increasing, then lin E▯= E n – the smallest set containing all the sets. nfi¥ n=1 Example 2n2 th - 2 If the probability that the entire population will die before having offspring ie the n, what is the probability that it will survive forever? Page 5 of 17 STA107H1b.doc th • Let E n ={population wiped out by tne generatio}. • So E Ì E Ì ...ÌE Ì E . 1 2 n n+1 P (urvivesforeve)= - P(diesout sometime in the fut)re = - P(E1 ¨ E2 ¨ .¨. En) ¥ • = - P ▯ E ▯ ▯▯ n▯ ▯n=1 ▯ = - lim P(En ) nfi¥ 2 ▯ 2n ▯ • P (E n )= exp -▯ 2 6n + 3 ▯ 2n 2 ▯ ▯ -2n 2 ▯ ▯ 2▯ -1 • lim PE n)= lim exp▯- 2 = exp▯lim 2 = exp▯- = e3 nfi¥ nfi¥ 6n +3 nfi¥ 6n +3 6 Lecture #15 – Monday, October 27, 2003 R ANDOM V ARIABLES • In many situations when an experiment is performed the interest is in some numerical function of the outcome rather than the actual outcome itself. • If S is the sample space of an experiment then a map X : S fi R is called a random variable. -1 • Additional Requirement: For any interval I Ì R , (I) is an event in S. Example • Tomorrow: rain, snow, shine. • I bet $5 that it’s rain. • X = my profit ={-$5,$5} • P (X = $5) • Y = profit from betting $100on snow • If S is a discrete sample space, then X : S fi R is a discrete random variable. The set of possible values of X is also discrete – X : {a ,a ,...,a} 1 2 n • Some S’s may not be discrete. Example Suppose that three cards are drawn from an ordinary deck of 52 cards one-by-one at random and with replacement. Let X be the number of spades drawn. Fin(X = ),i = 0,1,2,3. • Possible values for random variable X {0,1,2}. X : S fi {0,1,2}. Page 6 of 17 STA107H1b.doc 39·39·39 13·39·39 13·13·39 • P X = 0) = , P(X =1 )= ·3, P (X = 2)= ·3, 52·52·52 52·52·52 52·52·52 13·13·13 P X = 0 = 52·52·52 R ANDOM S ELECTION OF P OINTS IN NTERVALS • Fix a < b and a, b such that a £ a < b £ b . The probability that a point is randomly selected in the intervala,b ) is b-a . b -a a a b b • Sample space is not discrete but continuous (interval). • Let C be a fixed point in the inta,b). If X is a point randomly selected in the (a,b)vthen the probability that X is selected to be exactly C is)= 0 . • If(a,b) was discrete{a ,a ,...,a}, thenP(X = a )= 1 1 2 n i n • Define E = C▯- 1 ,C + 1▯. n ▯ n n▯ 1 1 C - n+1 C + n+1 C -1 C C + 1 n n ▯ 1 1 ▯ ▯ 1 1 ▯ • EnÉ E n+1 so En+1 =▯C - n+1 ,C + n+1 ▯. Sequence of events n = ▯ - n ,C +n ▯ which is decreasing. 2 ▯¥ ▯ • P(X = C )= P(X ˛ E1and...and X˛ En)= P ▯▯ X ˛ E n▯= lim P(X ˛E n)= lim n = 0 ▯n=1 ▯ nfi¥ nfi¥ b-a Example A train passes through a town at random time between 10:00 am and 10:12 pm. If I drive through town between 9:55 am and 10:05 am, what is the probability that I see the train? • (a,b)= 10:00am,10:12am ) 9:55 10:00 10:05 10:12 5 • P Isee the train = P train goes through between10:00and10:05 = – does not depend on the scale of 12 measurement. Page 7 of 17 STA107H1b.doc Example A patient with flu may have a fever between 39°C and 42°C. Let X be the temperature of a randomly selected flu patient. What is the probability that the temperature is less than 40°C? • S ={allpossible temperatur}s={39°C,40°C } • a,b )= 39,42), a,b )= 39,40 ) • P(X ˛ (39,40))= 40-39 = 1 42-39 3 • P(X = 39.5)= 0 • P X ˛ 39.49,39.51 = 0.02 3 • For continuous random variable, we look atP(interva) instead of P(point). Definition • For a discrete random variableX : S fi {a1,a2,...,n}, we can define a functioP: {a1,a2,...,n}fi 0,1] such that P ai) = P(X = ai). • P is called probability function of the random variable X. Lecture #16 – Wednesday, October 29, 2003 D ISTRIBUTION F UNCTIONS • The distribution function F :R fi 0,1] of a random variable X is defined aF(t)= P(X £ t). • For example, F 1)= P X £1 ). Properties of F t £ t ▯ F t )£ F t ) 1) F is non-decreasing – i.e.1 2 1 2 . So F is increasing or constant. 2) lim F t =1 tfi¥ 3) lim F t = 0 tfi-¥ 4) F is continuous to the right. Proof of (1) 1 £ t2. I want to showF t1)£ F t2 ) • F(t1)= P(X £ t1), F t2) = P(X £ t2). • E = {X £ 1}, F ={X £ t1} F • E Ì F ▯ P E )£ P(F ) E -▯ ▯ P(E )= P(X £ t1)= P(t1)▯ • P(F )= P(X £ t )= P t )▯ ▯ F (1) £ F(t2) 2 2 ▯ • Attention: t1< t2 does not imply F (1) < F(t2); t1< t2 implies F (1 )£ F(t2) Page 8 of 17 STA107H1b.doc Proof of (2) • X : S fi R – real random variable. • Always, there will be t large enough to have P(X £ t) is almost 1. • Let E = {X £ n},n˛ Z . n • lim F t = lim F n = lim P n ( ). tfi¥ nfi¥ nfi¥ • E n E n+1 ▯ the sequence E in increasing. I can use the continuity property of the probability function. ▯ ¥ ▯ ▯ ¥ ▯ • lim P (En )= P ▯im E n▯ = P▯ ▯ En ▯= P ▯▯ X £ n =1 nfi¥ ▯nfi¥ ▯ ▯ n 1 ▯ ▯n 1 ▯ Proof of (3) • The proof of lim F t = 0 is the mirror of the proof oflim F t =1 . tfi-¥ tfi¥ Proof of (4) • Convergence from the left means for all x n x 0 and xnfi x 0 . In fact, for all practical purposes, one can consider only increasing sequences. x n x n+1n˛ Z and xn£ x 0 and nfi¥ x n 0 . • Notation: x ›nx 0 – converges to x0from the left. lim f (xn )= f(x 0 ). n›x0 • If f(x 0 )= f(x 0), then f is continuous to the left. • Continuity to the right. x 0 • xn‡ x n+1▯ "n˛ Z xn‡ x 0 ▯
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