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LECTURE 1: Probability models and axioms • Sections Sample 1.1, space Readings:• 1.2 Sample space • Probability Probability laws laws • – Axioms Axioms – Lecture outline – Properties Some properties – that follow from the axioms Some properties • Examples Examples • – Discrete Discrete – – Continuous Continuous – • Discussion Discussion • – Countable Countable additivity additivity – – Mathematical Mathematical subtleties subtleties – Interpretations of of probabilities probabilities ••• Interpretations Interpretations Sample space • List (set) of possible outcomes, Ω • List must be: – Mutually exclusive – Collectively exhaustive Sample space – At the “right” granularity ssible outcomes • Two steps: – Describe possible outcomes sive – Describe beliefs about likelihood of outcomes haustive granularity Sample space Sample space Sample space ssible outcomes • List (set) of possible outcomes, Ω • List (set) of possible outcomes, Ω • List must be: sive • – List must be: Mutually exclusive haustive Mutually exclusive – Collectively exhaustive granularity – At Collectively exhaustive the “right” granularity – At the “right” granularity Sample space: discrete/finite example Sample space: discrete/finite example Sample space: discrete/finite example space: die discrete/finite example • Two rolls of aSample tetrahedral • Two rolls of a tetrahedral die rolls of a tetrahedral die • Two rolls of a vs. tetrahedral –dieSample space vs. sequential description – space Sample space sequential description mple vs. sequential description Die roll example – Sample space vs. sequential description 1,1 1,2 1,3 1,4 1 4 4 2 44 Y = Second 3 3 Y = roll Second 2 roll 1 1 Y = Second 2 1,1 1,2 4 1,3 1,4 3 1 Y = Second 33 2 3 12 2 X = First roll 4 3 3 1 3 2 1 4 1 2 1 3 44,4 3 4 4 X = First roll 4,4 4 X = First roll 4,4 3 •1 A 2 3 4 continuous 2 ntinuous sample space: 1 4 sample space: 4 (x, y) such that 0 ≤ x, y ≤ 1 • A continuous sample space: ) such that 0 ≤ x, y ≤ 1 X = First roll X = First roll (x, y) such y y that 0 ≤ x, y ≤ 1 y Y)=2 • A continuous space: • Let B be1thesample event: min(X, (x, •y)Let such 0 Y≤) x, y1≤ 1 M =that max(X, 4,4 1 y • P(M = 1 | B) = 1 x1 1,1 1,2 1,3 1,4 2 2 3 Y = Second roll roll 1 2 roll 2 1 1 1,1 1,2 1,3 1,4 1 x mple space: continuous example mple space: Sample space: continuous example Sample space: sample continuous • A continuous space:example ntinuous sample space: Sample space: continuous example • (x, y) such that 0 ≤ x, y ≤ 1 inuous sample space: such that 0space: ≤ x, y ≤continuous 1 Sample example 0 ≤ x, y ≤ 1 y us sample space: uch that 0 ≤ x, y ≤ 1 1y that 0 ≤ x, y ≤ 1 y 1 1 Sample space: continuous example example Sample space: continuous x sample space: • A continuous sample • A continuous 1 space: 1 x • (x, y) 0such • (x, y) such that ≤ x, that y ≤y10 ≤ x, y ≤ 1 y y11 1 1 xy 1 1 x Probability axioms Probability axioms et of the•sample space Event: a subset of the sample space assigned – to Probability events Probability axioms is assignedProbability to events axioms Probability Probability axioms axioms • •Event: a subset of of thethe sample space Event: a subset subset sample space • Event: a of the sample Axioms: • Event: a subset of the sample space space – –Probability is is assigned to to events Probability assigned events Probability is 1. P– ≥0 –(A) Probability is assigned assigned to to events events • •Axioms: Axioms: • Axioms: 1 Axioms: = 1 2. •P(universe) 1. P (A) ≥0 – Nonnegativity: P(A) (A) ≥ ≥ 00 – Nonnegativity: P – Nonnegativity: P (A) ≥ 0 3. If A ∩ B = Ø, P (universe) =1 – Normalization: P(Ω) = 11 = P(A)2. + Pthen (B) Normalization: P = P (A ∪ B) = P(A) +P – Normalization: P(Ω) (Ω) =(B) 1 3. If– A(Finite) ∩ B = Ø, additivity: additivity: – (Finite) additivity: (to be strengthened later) thenIf P (A ∪ B) = P (A) + P (B) Ø, then then P P(A (A ∪ ∪ B) B) = =P P(A) (A)+ +P P(B) (B) If A A∩ ∩B B= = Ø, Ø, then P (A ∪ B) = P (A) + P (B) k }) = P({s + 1· ,· s· 2+ }) = P({s1}) + · · · + P({s }) • 1}) P({s , .P . .({s , skk}) k = P(s1) + · · · + P(sk ) = P(s1) + · · · + P(sk ) • P({s == P({s ++ · · ·· + P({s 1 , s2 1 })}) k })}) ({s , ,s.22., .. ,. s. k, }) skk}) P ({s ····+ P ({s •• P }) = P ({s }) + · + P ({s }) 1 1 1 1 P({s1, s2, . . . , sk }) = P({s1}) + · · · + P({skkk}) == P(s )+ · · ·· + P(s ) ) 1 k P (s ) + · · + P (s 1 = P (s ) + · · · + P (s = P(s11) + · · · + P(skkk)) ds strengthening – Axiom 3 needs strengthening • Axioms • Axioms • Consequences • Axioms • Consequences • Consequences • Axioms For disjoint sets: Probability axioms axioms • Axioms For disjoint sets: Probability • Consequences• Axioms For disjoint sets: • Consequences Probability axioms • Consequences a subset subset of of the sample sample space P(A ∪ B) = P(A) + P(B) a the Some simple space consequences of the axioms P(A ∪ B) = P(A) + P(B) • Axioms For disjoint sets: ility assigned to events ∪ B) = P (A) + P(B) Pility ({s1is ,issof , .the . . , ssample =P ({s }) disjoint 2assigned 1}) + · · · + P({sP For sets: k }) to k(A bset space events For disjoint sets: • Consequences • Axioms Some simple consequences of the ax Some simple consequences of the a is assigned to events P (A ∪ ·B) =PP(s (A) + P (B) = P (s ) + · · + ) 1 k • Consequences Some consequences of the axioms P(A ∪•B)P= P(A) (B) (A) ≤ 1 + Psimple gativity: P (A) ≥ 0 • P (A) ≤ 1 ativity: •P(A) ≥ 0 P(A ∪For B) disjoint = P(A) sets: + P(B) Axioms • P(A) 1 consequences Some simple of the axioms • ≤ P (Ø) =0 ization: Pdisjoint (Ω) =1 vity: PFor (A) ≥ 0= sets: • P (Ø) = 0 ization: P (Ω) 1 • Consequences Some simple consequences of the axioms consequences of+the axioms • P(Ø) = P (A ∪0 B)CSome =disjoint: P(A)simple +P (B) P(A) ≤ 1 )ion: additivity: (to be• strengthened later) • A, B, P (A ∪ B ∪ C) = P (A) P (B) + P(C P (Ω) = 1 additivity: (to be strengthened later) • ≤ A,1B, C disjoint: P(A ∪ B ∪ C) = P(A) + P(B) + P( • P(A) c ) = 1 for k disjoint events B = Ø,Pthen P (A= ∪P B) =+ P(A) + P(B)• P(A) (A ∪P B) (A) P(B) and similarly (A) + ≤ 1 (A For disjoint events: P = Ø, then (A ∪ B) = P (A) + P (B) • P (Ø) = 0 and similarly for k disjoint events ditivity: (to be strengthened later) Some simple consequences of the ax • P(Ø) = 0 Ø, then P(A ∪ B) =• PA, (A) +CPdisjoint: (B) • ∪CP ({s , .P .P .(A) ,(A sk }) = ({s + · ·+ · +PP ({s+ • A, P(A (Ø) = 0 C) B, disjoint: ∪ B ∪P C) = P (A) (B) P(C) 1,,ss2 1}) k }) B, P B ∪ = + P (B) + P (C) • P ({s , . . . , s }) = P ({s }) + · · · + P ({s }) 1 2 1 k k Some simple consequences of the axioms P(A ∪ B) = P(A) + P(B) • and P(A) ≤1 A, • B,similarly C disjoint: Pdisjoint (A ∪ B ∪events C) = P(A) + P(B) + P(C) for kP and similarly for disjoint events • kA, B, C disjoint: (A ∪ B ∪ C) = P·(A) +PP (B) + P(C) = P (s ) + · · + (s ) and similarly for k disjoint events 1 k • P(A) ≤ 1 = P(s1) + · · · + P(sk ) • similarly P (Ø) = 0})k=disjoint and for P ({s , s , . . . , s P ({s + · · · + P({sk }) 1 2 1 })events • PSome ({s1, s2simple , . . . , sk•}) = P ({s }) + · · · + P ({s k 1 consequences of the axioms k }) • P({s1, s2, . . . , sk }) = P({s1}) + · · · + P({sk }) • P(Ø) = 0 • A, B, C disjoint: P(A}) ∪ B ·∪· ·C) P(A) + P(B) + P(C • P({s = P({s += P({s 1 , s2 , . . . , sk }) = 1+ + k }) P (s ) · · · + P (s ) • P(A) ≤ 1 = P(s · · · + P(s k 1) + k )k 1disjoint events and for • A, B, C disjoint: P(A ∪ B ∪ C) = P(A) +similarly P(B) += P (C) P(s1) + · · · + P(sk ) = P(s1) + · · · + P(sk ) •and P(Ø) = 0 for k disjoint events similarly • P({s1, s2, . . . , sk }) = P({s1}) + · · · + P({sk }) – Axiom 3 needs strengthening P(A) + P(Ac) = 1 Axiom • •P({s P({s 1 , s2 , . . . , sk }) = P({s1 }) + · · · +– k }) 3 needs strengthening = P(s – Axiom 3 needs strengthening 1 ) + · · · + P(sk ) – Do weird sets have probabilities? Do+weird • A, B, C disjoint: P(A ∪ B ∪ C) = – P(A) P(B)sets + Phave (C) probabilities? = P(s1) + ·–· · + P(s ) Do weird sets have probabilities? k – Axiom 3 needsevents strengthening and similarly for k disjoint – Axiom 3 needs strengthening sets}) have – Axiom needs strengthening • P({s1, s2, .–. . ,Do sk })weird = P({s · · probabilities? ·+ P3({s weird sets 1 – +Do k }) have probabilities? Axiom 3 needs strengthening – Do weird sets have probabilities? Probability axioms axioms Probability Probability axioms a subset subset of of the sample space a the sample space Some simple consequences of the axioms ility assigned to events Pility ({s1is ,issof , .the . . , ssample =P ({s 2assigned 1}) + · · · + P({sk }) k }) to bset space events • Axioms is assigned to events = P(s1) + · · · + P(sk ) • Consequences gativity: P (A) ≥ 0 ativity: •P(A) ≥ 0 Axioms ization: Pdisjoint (Ω) =1 vity: PFor (A) ≥ 0= ization: 1 sets: •P(Ω) Consequences )ion: additivity: (to be strengthened later) P(Ω) = (to 1 be additivity: strengthened later) B = Ø,Pthen P (A= ∪P B) = P(A) + P(B) (A ∪P B) (A) P(B) For disjoint events: = Ø, then (A B) =+ P(A) +later) P(B) ditivity: (to be∪strengthened Ø, then P(A ∪ B) = P(A) + P(B) Some P(A ∪ B) = P (A) +simple P(B) consequences of the axioms • P(A) ≤ 1 Some simple consequences of the axioms • P(Ø) = 0 • P(A) ≤ 1 • A, B, C disjoint: P(A ∪ B ∪ C) = P(A) + P(B) + P(C) •and P(Ø) = 0 for k disjoint events similarly P(A) + P(Ac) = 1 • •P({s 1 , s2 , . . . , sk }) = P({s1 }) + · · · + P({sk }) • A, B, C disjoint: P(A ∪ B ∪ C) = P(A) + P(B) + P(C) = P(s1) + · · · + P(s ) and similarly for k disjoint events k • P({s1, s2, . . . , sk }) = P({s1}) + · · · + P({sk }) Axiom 3 needs strengthening P(A ∪ B) = P(A) + P(B) • Axioms Some simple consequences of the axioms • Consequences • P(A) ≤ 1 •ForP(Ø) = 0sets: disjoint Some simple consequences of the axioms c =1 P({s1, s2•, . .P . ,(A) sk })+=PP(A ({s)1}) + · · · + P({sk }) P(A ∪ B) = P(A) + P(B) • A, B, C disjoint: P(A ∪ B ∪ C) = P(A) + P(B) + P(C) = P(s1) + · · · + P(sk ) and similarly for k disjoint events Some simple consequences of the axioms • P({s1, s2, . . . , sk }) = P({s1}) + · · · + P({sk }) • P(A) ≤ 1 • P(Ø) = 0 = P(s1) + · · · + P(sk ) • P(A) + P(Ac) = 1 • A, B, C disjoint: P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – Axiom 3 needs strengthening and similarly for k disjoint events – Do weird sets have probabilities? • P({s1, s2, . . . , sk }) = P({s1}) + · · · + P({sk }) = P(s1) + · · · + P(sk ) – Axiom 3 needs strengthening – Do weird sets have probabilities? Axiom 3 needs strengthening More consequences of the axioms More consequences of the axioms , then P(A) ≤ P(B) • If A ⊂ B, then P(A) ≤ P(B) ) = P(A) + P(B) − P(A ∩ B) • P(A ∪ B) = P(A) + P(B) − P(A ∩ B) ) ≤ P(A) + P(B) • P(A ∪ B) ≤ P(A) + P(B) ∪ C) = P(A) + P(Ac ∩ B) + P(Ac ∩ B c ∩ C) More consequences of the axioms • P(A ∪ B ∪ C) = P(A) + P(Ac ∩ B) + P(Ac ∩ B c ∩ C) • If A ⊂ B, then P(A) ≤ P(B) • P(A ∪ B) = P(A) + P(B) − P(A ∩ B) • P(A ∪ B) ≤ P(A) + P(B) More consequences of the axioms • P(A ∪ B ∪ C) = P(A) + P(Ac ∩ B) + P(Ac ∩ B c ∩ C) • If A ⊂ B, then P(A) ≤ P(B) • P(A ∪ B) = P(A) + P(B) − P(A ∩ B) • P(A ∪ B) ≤ P(A) + P(B) • P(A ∪ B ∪ C) = P(A) + P(Ac ∩ B) + P(Ac ∩ B c ∩ C) More consequences of the axioms • If A ⊂ B, then P(A) ≤ P(B) • P(A ∪ B) = P(A) + P(B) − P(A ∩ B) More consequences of the axioms • P(A ∪ B) ≤ P(A) + P(B) , then P(A) ≤ P(B) • P(A ∪ B ∪ C) = P(A) + P(Ac ∩ B) + P(Ac ∩ B c ∩ C) ) = P(A) + P(B) − P(A ∩ B) ) ≤ P(A) + P(B) ∪ C) = P(A) + P(Ac ∩ B) + P(Ac ∩ B c ∩ C) Example 4 4 3 Y = Second roll 3 Y = Second 2 roll 1 2 11 Probability calculation: discrete/finite example Sample space: discrete/finite example 2 3 2 X =1First roll 4 3 4 X = First roll Exampledie • Two rolls of a tetrahedral • Let possible Letevery Z = min(X, Y )outcome have probability 1/16 every – Sample space vs. 4sequential description • P•(XLet = 1) = possible outcome have probability 1/16 Die roll example • P(Z = 1) = • P(X = 1) 1,1 = 3 Y = Second 1 2 1,4 4 1 Y = Second 3 1,2 • P(Z = 2) = 1,3 roll Let Z =2 min(X, Y ) • P(Z = 3) = Z ==min(X, •LetP(Z 1) = Y ) 3 4) = • P(Z = •• P P(Z (Z = = 1) 2) = = 4 Y = Second 31 2 3 4 roll roll 2 2 X = First roll 1 t every possible outcome have 3 2 1 1 X = First roll bability 1/16 3 2 1 4 X = 1)•=A continuous sample space: (x, y) such that X 0 =≤First x, yroll ≤1 y 4 • Let B be the event: min(X, Y ) = 2 • Let M = max(X, Y ) • P(M = 1 | B) = 1 4 •• P P(Z (Z = = 2) 3) = = 4,4 •• P P(Z (Z = = 3) 4) = = • P(Z = 4) = Discrete uniform law Discrete uniform law Discrete uniform law − Assume Ω consists of n equally Discrete uniform law likely element Discrete uniform law − Assume A consists of m elements number of elements of A Then−: Assume P(A) = e points be equally likely Ωnumber consistsofofelements n equallyoflikel Ω − Assume Ω consists of Discrete uniform law − Assume Ω consists of n equally likely elements − Assume A consists of m elements − Assume A consists of Assume Aofconsists of of klikely elements • Let all − sample points be equally number elements A • Just count. . . number of eleme P(A) = num number of elements of A k total number of sample points Then = : P(A) = Then : P (A) = Then : P(A) = • Then, number of eleme number of elements of Ω n num number of elements of A P(A) = • Just count. . . total number of sample points 1 • Just count. . . = count. . . •prob Just n • Just count. . . Continuous uniform law prob = m” numbers in [0, 1]. y Continuous uniform law 1 • Two “random” numbers in [0, 1]. : Probability = Area y 1 1 x 1 n Probabilitynumbers calculation: example • Two “random” in [0,continuous 1]. Probability calculation: continuous exam y • Two “random” numbers in [0, 1]. 1 1]. y in [0, • Two “random” numbers mple space: continuous example mple space: Sample space: continuous 1 1 example Probability calculation: continuous example • A continuous sample space: Two “random” numbers in [0, 1]. Sample space: continuous example y 0 ≤ x, y ≤ • space: (x, y) such that uous sample Sample space: continuous example ≤ x, y ≤ 1 y 1y hat 0 ≤ x, y ≤ 1 1 x 1 x • Uniform probability law: Probability =x Area 1 1 • Uniform probabilty law: Probability = Area 1 s sample space: h that 0 ≤ x, y ≤ 1 y � � P {(x, y) |probability x + y ≤ 1/2} � law:=Probability = Area • �Uniform P {(x, y) | x + y ≤ 1/2} = y 1 � � P {0.5, P� {(x, y) | 0.3)} x �+ y x 1 � P {0.5, 0.3)} = 1 � � Uniform law: Probability = Area Sample space: continuous example Sample space: continuous example P {(0.5, 0.3)} = x 1 • A continuous sample space: • A continuous sample space: 1 x • (x, y)0such that • (x, y) such that ≤ x, y ≤ 10 ≤ x, y ≤ 1 y1 1 xy y1 1 1 x x 1 1 x � ≤=1/2} = Probability calculation steps Probability calculation steps he sample space • Specify the sample space probability law • Specify a probability law n event of interest • Identify an event of interest ... • Calculate... ability calculation: discrete but infinite sample space Probability calculation: discrete but infinite sample space pace: {1, 2, . . .} • Sample space: {1, 2, . . .} 1 given P(n) = n , n = 1, 2, . . . 1 2 – We are given P(n) = n , n = 1, 2, . . . 2 me is even) = • P(outcome is even) = p p 1/2 1/2 1/4 1/8 1/16 1/4 Probability calculation: discrete Probability discrete but but infinite infinite sample sample space space Sample space: space: {1, 2, . . .} •• Sample We are are given P(n) = 2−n, n –– We n= = 1, 1,2, 2,...... Find P P(outcome (outcome Find is even) Probability calculation: ••discrete but infinite sample space Probability calculation: discrete but infinite sample space discrete but pinfinite sample space • Sample Probability space: {1, 2, calculation: . . .} • Sample space: {1, 2,−n . . .} • are Sample – We givenspace: P(n) ={1, 2 2, ,. .n.}= 1, 2, . . . 1/2 1/2 – We are given P(n) = 2−n , n = 1, 2, . . . 1 • Find –P(outcome is even) We are given P(n) = n , n = 1, 2, . . . • Find P(outcome is even)2 discrete but infinite sample Probability calculation: space 1/4 1/4 p • Find P(outcome is even) • Sample space: {1,p 2, . . .} 1 p1/2 – We are given P(n) = 1/2 , n = 1, 2, . . . 2n • P(outcome is even) = 1 22 33 1/16 1/16 ….. ….. 44 1/4 1/2 1/4 Solution: 1/8 •• Solution: 1/16 1/4 1/8 p 1 1/8 1/8 2 1 1/2 3 2 41/8 3 ….. P({2, 4, 6, . . .}) ….. ….. 1/16 4 1/16 = = P P(2) (2) + +P P(4) (4)+ +······ 11 11 11 11 = = 22 + + 44 + + 66 + +······= = 22 22 22 33 1 2 3 4 • Solution: • Solution: 1/4 needed: •• Axiom Axiom needed: P({2, 4, 6, . . .}) = P(2) ··· If ,, .P . .(4) are+disjoint events, then: then: If A A A22+ events, 1/8 • Solution: 1,, A P({2, 4, 6, . . .}) 11= P1(2) +1/16 P (4) + ··1 · 1 + 4 + 16 + ·1· · = 2=1 P({2, 4, 6,= . . .}) (2)2+ P(4) P(A · )) = =P P(A (A11))+ +P P(A (A22))+ +· ·· ·· · A2·3·∪· · · ·1 1 ∪+ 2= 2P+ 1 2 321 4 1+ 61+ · · · = 2 2 2 31 = + + + · · · = • Axiom needed: 22 24 26 3 • Axiom needed: If •A1Solution: , A2, . . . are disjoint events, then: • If Axiom A1, A2,needed: . . . are disjoint events, then: ({2, 4,∪6,· ·. ·. ).}) =(AP P(4) +· · · · ∪ Adisjoint =P )(2) ++ P(A If A1, A2P , .(A . .P1are events, 2 1then: 2) + ….. Countable additivity axiom Countable additivity axiom Countable additivity axiom ens the finite additivity axiom • Strengthens the finite additivity axiom • Strengthens the finite additivity axiom A3,. . . is an infinite sequence of events, A3),.+ . . Pis(Aan) infinite sequence of events, 2,(A ∪ A2 ∪ A3 ∪If· ·A · )1,=AP 1 2 + P(A3 ) + · · · Countable Axiom: then P(A1 ∪ AAdditivity 2 ∪ A3 ∪ · · · ) = P(A1) + P(A2) + P(A3) + · · · If A1, A2, A3,. . . is an infinite sequence of disjoint events, then P(A1 ∪ A2 ∪ A3 ∪ · · · ) = P(A1) + P(A2) + P(A3) + · · · Countable additivity axiom •Mathematical Strengthens subtleties the finite additivity axiom Countable Additivity Axiom: If A1, A2, A3,. . . is an infinite sequence of disjoint events, then P(A1 ∪ A2 ∪ A3 ∪ · · · ) = P(A1) + P(A2) + P(A3) + · · · Mathematical •• • •• • • • • Mathematical subtleties Mathematical subtleties Mathematical subtleties • Additivity holds only for “countable” Additivity sequences of events Additivity holds holdsonly onlyfor for“countable” “countable” sequences of events Additivity holds only for “countable” sequences of events • The unit square (simlarly, the real line (itscountable elements cannot be arranged in a The thethe real line, etc.) is not countable The unit unit square square(simlarly, (similarly, real line, etc.) is not The unit square (simlarly, the real line, etc.) are not countable (its in in a sequence) (its elements elementscannot cannotbebearranged arranged a sequence) • “Area” is a legitimate probability law “Area” as long as we do not try to assign pro “Area” is is a a legitimate legitimate probability probability law law on on the the unit unit square, square, “Area” is a legitimate probability law on the unit square, as toto assign probabilities/areas to to “very strange” “very strange” sets as long long as aswe wedo donot nottry try assign probabilities/areas as long as we do not try to assign probabilities/areas sets to “very strange” sets to “very strange” sets Interpretations of probability theor Interpretations ofInterpretations probability theory of theory • probability A narrow view: a branch of math Interpretations theory Interpretationsofofprobability probability theory – Axioms ⇒ theorems ow view: •a branch of math A narrow view: a branch of math • –A narrow view: aabranch ms ⇒ theorems “Thm:” “Frequency” of event A “is” P(A) AAxioms narrow⇒ view: branchofofmath math theorems – ⇒ – Axioms Axioms ⇒Atheorems theorems “Frequency” of event “is” P(A) “Thm:” “Frequency” of event A “is” P(A) • Are probabilities frequencies? “Thm:” “Frequency” of event A “is” P(A) “Thm:” “Frequency” of event A “is” P (A) toss yields heads) = 1/2 – P (coin obabilities•frequencies? Are probabilities frequencies? • Are probabilities frequencies? n toss yields heads) = 1/2yields •– probabilities frequencies? – Are P(coin (coin toss heads) = = 1/2 1/2 – P(sole shooter of JFK) = 0.92 P toss yields heads) e shooter of =shooter 0.92 – P(a piece of equipment aboard the space shuttle fa – JFK) P(the (coin toss yields = 1/2 (sole of = be 0.92 – P president of JFK) .heads) . . will reelected) = 0.7 iece of equipment aboard the of space shuttle fails) = 10−8 −8 (the president . . . will be reelected) = 0.7 – P(a piece of equipment aboard the space shuttle fails) =a10 • Probability models are framework for • Probabilities are often intepreted as: describing uncertainty ility models are a framework for Betting preferences Probabilities are often •– Probability models are aintepreted frameworkas: for ing uncertainty – Use for consistent reasoning – Description of uncertainty –describing Description ofbeliefs beliefs or consistent reasoning – Use for predictions, decisions for consistent reasoning •– framework for dealing with experiments that have uncertain outcomes –AUse Betting preferences or predictions, decisions Use for – Rules forpredictions, consistent decisions reasoning •– AUsed framework for analyzing phenomena with uncertain outcomes for predictions and decisions Rules consistent reasoning – P (sole for shooter of JFK) = 0.92 – P Used for predictions and decisions – (a piece of equipment aboard the space shuttle fails) = 10−8 – P(sole shooter of JFK) = 0.92 • Probability models are a framework for –describing P(a pieceuncertainty of equipment aboard the space shuttle fails) = 10−8 ory The role of probability th • Are probabilities frequencies? – P(coin toss yields heads) = 1/2 The role of probability theory world – P(the president of . . . will be reelected) Real = 0.7 The role of proba The role of probability theory • Probabilities are often intepreted Real worldas: Data Real world – Betting preferences The role of probabili Real world The role of probability theory – Description of beliefs Data Inference/Statistics Data Real world • A frameworkData for analyzing phenomena with uncertain outcomes Inference/Statistics Analysis – Rules for consistent reasoning Inference/Statistics Data – Used for predictions and decisions Inference/Statistics (Analysis) Predictions – P(sole shooter of JFK) = 0.92 Analysis The role of probability theory −8 Inference/Statistics – P(a piece of equipment aboard the space shuttle fails) = 10 (Analysis) Predictions Probability theory Predictions The role theory • Probability models are a framework for of probability Real world The role of probability theory (Analysis) Predictions Decisions describing uncertainty Probability theoryModel building – Use for consistent Probability theory Real world reasoning Data Real world Predictions Probability theory – Use for predictions, decisions Models building Models building Data Inference Data Probability theory Models building Inference/Statistics Analysis Inference/Statistics Analysis Predictions Analysis Predictions Probability theoryPredictions Probability Probability theory Model buildingtheory Models building