Important Concepts Read Chapter 2. Experiments. Phenomena. Probability Models. Unpredictable in detail. Examples

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1 Probability Models Important Concepts Read Chapter 2 Probability Models Examples - The Classical Model - Discrete Spaces Elementary Consequences of the Axioms The Inclusion Exclusion Formulas Some Indiscrete Models Experiments Phenomena Unpredictable in detail The set of possible outcomes in known. Examples a) Scientific experiments b) Games of chance c) Human performace d) Financial indices e) The Weather Monotone Sequences and Continuity Events and The Sample Space The Sample Space. Let Ω denote the set of possible outcomes for a given experiment. Events: Susbsets of the sample space, A, B, C Ω. Example: Coin Tossing. Ω = {hh, ht, th, tt } and A = {ht, th}. The Algebra of Events Set theory operations on events for example, A B = {ω : ω A or ω B}, AB = {ω : ω A and ω B}, A c = {ω : ω / A}. B A = BA c Three Elements The Model The sample space: Ω. Events: Subsets of A, B, C, Ω. Probability: Let A be the class of events, and let P : A IR must satisfy P (Ω) = 1, (1) 0 P (A) 1, (2) P (A B) = P (A) + P (B) (3) whenever A and B are events for which AB =. Notes a). Probability is a property of events. b). (1), (2), and (3) are axioms and admit various interpretations.

2 The Birthday Problem The Classical Model Games of Chance The Model. Ω is a finite set; A is the class of all subsets of Ω; and Example: Roulette and P (A) = #A #Ω. Ω = {0, 00, 1, 2, 3, 4, 35, 36} P ({Red Outcome}) = = Q: If n people gather, what is the probability that no two have the same birthday? A: Regard the birthdays of the n people as a sample w.r. from {1, 2, 365} (ignoring leap year). Ω is all lists and #Ω = 365 n. Let ω = (i 1, i n ) A = {ω : i j i k all j k}. A consists of all permutations of n days, #A = (365) n, and P (A) = (365) n 365 n = p n say. Some Values n p n On Infinite Sums Discrete Probability Models Suppose Ω = {ω 1, ω 2 }, finite or infinite; let p : Ω IR, satisfy p(ω) 0 for all ω, p(ω) = 1. ω Ω Let P (E) = p(ω) ω E for E Ω. Notes a) (1), (2), and (3) hold. b) p(ω) = P ({ω}). Example. In the classical model, p(ω) = 1/#Ω. If x 1, x 2, IR, then x k = lim n provided that the limit exists. n x k, Examples a). If 1 < x < 1, then b). For any x, x k 1 = 1 1 x. 1 k! xk = e x. Alternative Notation: If A = {x 1, x 2, }, and f : A [0,, ), write f(x) = f(x k ). x A

3 Let and, intuitively, Waiting for Success Play Roulette Until a You Win Betting on Red r = 9 19, q = 1 r = 10 19, Ω = {1, 2, } p(1) = r, p(2) = qr, p(3) = q 2 r, p(ω) = rq ω 1. Let p(ω) = ω Ω ω=1 = r 1 q = 1. rq ω 1 P (A) = ω A p(ω). Amusing Calculation: Let Odd = {1, 3, }. P (Odd) = = r = r 1 q 2 = rq (2k+1) 1 q 2k The Objective Interpretation What Does Probability Mean? The Subjective Interpretation. Probabilities reflect the opinion of the observer. Strategy: Assess probabilities by imagining bets. Examples a). Peter is willing to give two to one odds that it will rain tomorrow. His subjective probability for rain tomorrow is at least 2/3. b). Paul accepts the bet. His subjective probability for rain tomorrow is at most 1/3. Applications Business. Thought Experiment: Imagine the experiment repeated N times. For an event A, let Example: Coin Tossing N A = # occurrences of A. N A P (A) = lim N N. N N H /N Note: Consistent with P (H) =.5. Example. In many roulette games, about 9/19 will result in red.

4 Consequences of the Axioms Suppose that P satisfies (1), (2), and (3). If A and B are events for which A B, then P (B A) = P (B) P (A). (4) For any event A, P (A c ) = 1 P (A). (5) In particular, P ( ) = 0. (6) For any events A and B, P (A B) = P (A) + P (B) P (AB) (7) If A 1, A m are any m events, then m m P ( A i ) P (A i ), (8) with equality if A i A j = whenever i j. If A B, then Proofs B = A (B A) and A (B A) =. So, by (3) and, therefore, P (B) = P (A) + P (B A), P (B A) = P (B) P (A). (4) For (5), A c = Ω A. So, P (A c ) = P (Ω) P (A) = 1 P (A). (5) For (6). P ( ) = P (Ω c ) = 0. Example. In the birthday problem, the probability that at least two people have the same birthday is A c, and P (A c ) = 1 P (A) = 1 (365) n 365 n. More on Unions For (7), Proofs: Continued A B = A (B AB), and A (B AB) =. So, P (A B) = P (A) + P (B AB) If m = 2, then = P (A) + P (B) P (AB). m P ( A i ) m P (A i ), (8) by (7); and if A 1 A 2 =, then there is equality by (3). The general case follows from mathematical induction. If A 1, A m are events, let m σ 1 = P (A i ), σ 2 = σ 3 = 1 i<j m 1 i<j<k m σ k = 1 i 1 <i k m P (A i A j ), P (A i A j A k ), P (A i1 A ik ), σ m = P (A 1 A 2 A m ). m P ( A i ) = σ 1 σ 2 + ± σ m. Proof. By induction messy.

5 The Matching Problem Let Ω be all permuations of 1, 2, n. Thus, Let by symmetry. σ k = Examples. Gift exhange ω = (i 1, i n ) Ω = n!. A j = {ω : i j = j} n A = A i. ( ) n P (A 1 A k ), k Here 1 (n 1)! P (A 1 ) = = 1 n! n!, (n 2)! P (A 1 A 2 ) = = 1, n! (n) 2 P (A 1 A k ) = (n k)! n! for k = 1, n. So, ( ) n σ k = (n) k = 1 k k!, and = 1 (n) k, P (A) = σ 1 σ 2 + ± σ n n 1 = k! ( 1)k 1, P (A) = 1 n 1 k! ( 1)k 1 1 e. Note: Accurate to three places if n 6. Refinements More on Events. Not all subsets of Ω need be events; but the class of events must be closed under union, intersection, and complementation. More on the Third Axiom. A stronger version of (3) requires P ( A k ) = P (A k ), whenever A 1, A 2, are mutually exclusive events (that is, A i A j = for i j). Remark: (3*) implies (3). Proposition. The discrete probability models satisfy (3*), as well as (3). Proof. Omitted (3 ) Intervals Some Indiscrete Models (a, b) = {x : a < x < b}, (a, b] = {x : a < x b}, [a, b) = {x : a x < b}, [a, b] = {x : a x b}, Densities. Let Ω be an interval and f a function for which f(ω) 0 and f(ω)dω = 1. let Ω P (I) = f(ω)dω for intervals I and extend f to a larger class of events using the axioms. Example The Uniform Spinner. Let Ω = ( π, π] and f(ω) = 1/2π. I P ((a, b)) = = P ([a, b]) = b a 2π.

6 Monotone Sequences Note. For any ω, then Amusing Calculation About the Extension Process P ({ω}) = P ([ω, ω]) = If ω ω C = {ω 1, ω 2, }, P (C) = P ({ω i }) = 0. f(ω )dω = 0. The probability of a rational outcome is zero. Events A 1, A 2, are increasing if and decreasing if A 1 A 2 A 1 A 2. The limit of an increasing (respectively, decreasing) sequence is respectively, A = A = A k, A k. Example. If Ω = IR and A k = (, 1 k ) = {ω : ω < 1 k }, then A k are decreasing and A = {ω : ω < 1 k = (, 0]. for allk} The Monotone Sequences Theorem Suppose that P satisfies (1), (2), and (3 o ). P satisfies (3) iff De Morgan s Laws. For any events A i, i = 1, n, ( n ) c n A i = A c i, ( n ) c n A i = A c i. Also true if n =. Proof-e.g.. ω ( n A i) c iff ω / n A i iff ω / A i for any i iff ω cup n Ac i. Corollary. If A 1, A 2, is increasing or decreasing, then then P (A ) = lim n P (A n), whenever A 1, A 2, is an increasing, or decreasing, sequence of events. Proof. Later, or see the text. Remarks a). Type of continuity. b). Equivalent to (3). c). Useful. (A ) c = (A c ).