MATH 529 Probability Axioms Here we shall use the geeral axioms of a probability measure to derive several importat results ivolvig probabilities of uios ad itersectios. Some more advaced results will come later. First, the basic axioms are I. Let Ω be a o-empty set ad let F be a algebra of subsets (i.e., evets) such that (i) Ω F; (ii) F is closed uder complemets (i.e., if A F, the A c F). (iii) F is closed uder uios (i.e., if A, B F, the A B F); Note: It follows that = Ω c F. Also, if A, B F, the A B = (A c B c ) c F, ad B A = B A c F. Moreover, by iductio, a algebra is closed uder fiite uios ad fiite itersectios. I geeral, a algebra may ot be closed uder all ifiite uios or all ifiite itersectios. Here are some examples of algebras: (i) The power set of Ω (which cosists of all subsets of Ω ) is clearly a algebra. (ii) Let G be a collectio of subsets of Ω. The algebra geerated by G is give by the itersectio of all algebras F that cotai G : A(G) = F, where G F. We ote that Ω F for all ; thus Ω F = A(G). Also, if A, B A(G) the A, B F for all ; thus A c F = A(G) ad A B F = A(G). I other words, A(G) itself is a algebra of sets. (iii) Let Ω be ucoutable. Let F = {A Ω : A is coutable or A c is coutable}. As a exercise, you may verify that F is a algebra. Questio: Is this F closed uder ifiite uios? (iv) Let Ω = [0, 1). Let F be the collectio of all fiite uios of disjoit itervals of the form [a, b) [0, 1). The F is a algebra. Moreover, F is ot closed uder ifiite uios. For example, we ca costruct a ifiite sequece of disjoit itervals [a 1, b 1 ), [a 2, b 2 ),..., [a, b ),... withi [0, 1). We could make these itervals uio to [0, 1), which is i F, by makig a 1 = 0 ; b i = a i+1 ; ad lim b =1. But if there is separatio betwee each b i ad a i+1, the there is o way to write this ifiite collectio of disjoit itervals as a fiite uio of disjoit itervals. Thus the ifiite uio is ot i F.
Propositio 0. Give a sequece of evets A i, we ca create a sequece of disjoit evets { B i } { }, such that Bi = Ai for 1. Proof. We disjoitify the evets as follows: Let B 1 = A 1, B 2 = A 2 A 1, B 3 = A 3 (A 1 A 2 ), i 1..., B i = A i Ak for i >1. By costructio, B i A i for each i ; thus, Bi Ai for all. Ad if x Ai, the we ca choose the first idex i such that x A i. If, i 1 the x B 1 Bi. If i >1, the x A i but ot i Ak. So x B i Bi. Thus, we i 1 have Ai Bi. Note that if x B i, the x is ot i Ak. So x caot be i ay B k for k < i. Thus, the { B i } are disjoit. Lastly, we ote that the { B i } are ideed evets due to the closure properties of a algebra. Next we discuss the axioms of a probability measure. II. A probability measure P is a real-valued fuctio defied o F that satisfies the followig properties: (i) P(A) 0 for all evets A F (ii) P(Ω) = 1 (iii) If { A i } the P( Ai ) = i= 1 is a deumerable sequece of disjoit evets i F such that Ai F, P(A i ) ; i.e., P is coutably additive. From these axioms, we ca prove quickly prove some basic results: Propositio 1. P( ) = 0. Proof. (a) Let A 1 = Ω ad let A i = for i 2. The A i { } evets from F ad Ai = Ω F. So by Axiom (iii), we have P(Ω) = P( Ai ) = P(A i ) = P(Ω) + P( ). i = 2 is a collectio of disjoit By subtractig P(Ω), we have 0 = P( ). The because P( ) 0, we must have i= 2 P( ) = 0.
Propositio 2. A probability measure is fiitely additive. Proof. Let { A i } be a collectio of disjoit evets i F. The the fiite uio Ai is still a evet i F. Lettig A i = for i >, we have Ai = Ai F. Moreover { A i } is still a collectio of disjoit evets. So by coutable additivity ad because P( ) = 0, we have P( Ai ) = P( Ai ) = P( A i ) = P(A i ) + P( ) = P( A i ). i= 1 i= +1 Propositio 3. If A F, the P(A) = 1 P(A c ). Proof. Because evets A ad A c are disjoit ad uio to Ω, Propositio 2 gives 1 = P(Ω) = P(A A c ) = P(A) + P( A c ). Theorem 1. Give a ested icreasig sequece of evets A 1 A 2 A 3... A... such that Ai is also a evet, we have lim P(A ) = P( Ai ). Proof. We first disjoitify the evets as follows: Let B 1 = A 1, B 2 = A 2 A 1,..., B i = A i A i 1 for i >1. The B i are disjoit evets with Bi = Ai = A, for all, ad Bi = Ai. Hece, lim P(A ) = lim P( Bi ) = lim P(B i ) = P(B i ) = P( Bi ) = P( Ai ).
Corollary 1. Give a ested decreasig sequece of evets A 1 A 2 A 3... A... such that Ai is also a evet, we have lim P(A ) = P( Ai ). Proof. The complemets of the evets are ested icreasig: (A 1 ) c (A 2 ) c (A 3 ) c... ad (Ai ) c = ( Ai ) c is a evet. Thus, by Propositio 2 ad De Morga s Law, we have lim P(A ) = lim 1 P((A )c ) =1 lim P((A )c ) =1 P (Ai ) c c =1 P Ai = P Ai. Theorem 2. Give a sequece of evets { A i }, the P( Ai ) P(A i ) for 1. Proof. We ow disjoitify the evets as follows: Let B 1 = A 1, B 2 = A 2 A 1, i 1 B 3 = A 3 (A 1 A 2 ),..., B i = A i Ak for i >1. The evets B i are disjoit evets with Bi = Exercise 1). Hece, for 1, we have Ai, for 1. Moreover, B i A i for all i, which implies P(B i ) P(A i ) (see P( Ai ) = P( Bi ) = P(B i ) P(A i ). Example. Let Ω = [ a, b ] o the real lie, where a < b. Let F be the algebra geerated by closed itervals [ c, d ] with a c < d b. Defie P ([ c, d ]) = (d c) / (b a). The for a < c < b, [ c ] F ad P ([ c ]) = 0. Proof. By the algebra axioms, we have (c, b] = [a,c] c F. Thus [c] = [c, b] (c, b] F. The for N large eough, [c 1/ N, c +1/ N] [a, b] ad [c] = [c 1/ i, c +1/ i]. Thus by i= N 2 / Corollary 1, P([c]) = lim = P([c 1/, c +1/ ]) = b a = 0. Usig Exercise 1(b), we ote that P((c, d]) = P([c, d] [c]) = P([c, d]) P([c]) = P([c, d]).
Exercises 1. Let A, B be evets with B A. (a) Prove that P(B) P(A). (b) Prove that P(A B) = P(A) P(B). 2. Let A be a evet. Prove that P(A) 1. 3. For [c, d] [0, 20], let P([c, d]) = (d c) / 20. I each case below, determie if {A i } is ested icreasig or ested decreasig. If icreasig, determie Ai ad verify that Theorem 1 holds with these evets. If decreasig, determie Ai ad verify that Corollary 1 holds with these evets. (a) A i = (10 1/ i, 15+1/ i) for i 1. (b) A i = [10 +1/ i, 15 1/ i] for i 1.