A Probability Path.pdf

# The monotonicity property the measure p is non

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The monotonicity property: The measure P is non-decreasing: For events A,B If A C B then P(A) ,:::: P(B), since P(B) = P(A) + P(B \A) ::: P(A). 6. Subadditivity: The measure Pis a-subadditive: For events An, n ::: 1, To verify this we write 00 UAn =A1 ... , n=l and since P is a-additive, 00 P(U An) =P(AI) + + + · · · n=l ,::::P(AI) + P(A 2 ) + P(A3) + · · · by the non-decreasing property of P. 7. Continuity: The measure P is continuous for monotone sequences in the sense that (i) If An t A, where An E B, then P(An) t P(A). (ii) If An .J, A, where An E B, then P(An) .J, P(A). To prove (i), assume A 1 c Az c A3 c · · · c An c · · · and define Then {B;} is a disjoint sequence of events and n 00 UB; =An, UB; =UA; =A. i=l i=l i

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32 2. Probability Spaces By a -additivity oo oo n P(A) =P(UB;) = LP(B;) = t LP(B;) i=I i=I i=I n = lim t P(U B;) = lim t P(An). n-+oo n-+oo i=I To prove (ii}, note if An .J.. A, then t Ac and by part (i) = 1- P(An) t P(Ac) = 1- P(A) so that PAn .J.. PA. 0 8. More continuity and Fatou's lemma: Suppose An E B, for n 2: 1. (i) Fatou Lemma: We have the following inequalities P(liminfAn) liminfP(An) n-+oo n-+oo lim sup P(An) P(limsupAn). n-+00 n-+oo (ii) If An -+ A, then P(An) -+ P(A). Proof of 8. (ii) follows from (i) since, if An -+ A, then limsupAn = liminfAn =A. n-+oo n-+oo Suppose (i) is true. Then we get P(A) = P(lim inf An) lim inf P(An) n-+oo n-+oo limsupP(An) P(limsupAn) = P(A), n-+oo n-+oo so equality pertains throughout. Now consider the proof of (i): We have P(liminfAn) =P( lim t <n Ak)) n-+00 n-+oo = lim t P<n Ak) n-+oo (from the monotone continuity property 7) inf P(An) n-+oo
2.1 Basic Definitions and Properties 33 P(limsupAn) = P( lim t <U Ak)) n-H>O n-+00 k?!_n = lim t P(U Ak) n-+00 k?!_n (from continuity property 7) :::: lim sup P(An). n-+00 completing the proof. D Example 2.1.1 Let Q = IR, and suppose P is a probability measure on JR. Define F(x) by F(x) = P((-oo,x]), x e JR. Then (i) F is right continuous, (ii) F is monotone non-decreasing, (iii) F has limits at ±oo F(oo) := lim F(x) = 1 xtoo F(-oo) := lim F(x) = 0. x.j.-oo (2.3) Definition 2.1 .1 A function F : lR [0, 1] satisfying (i), (ii), (iii) is called a (probability) distribution function. We abbreviate distribution function by df. Thus, starting from P, we get F from (2.3). In practice we need to go in the other direction: we start with a known df and wish to construct a probability space (Q, B, P) such that (2.3) holds. See Section 2.5. Proof of (i), (ii), (iii). For (ii), note that if x < y, then (-oo,x] c (-oo,y] so by monotonicity of P F(x) = P((-oo,x]):::; P((-oo,y]):::; F(y).

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34 2. Probability Spaces Now consider (iii). We have F(oo) = lim F(xn) Xntoo (for any sequence Xn t oo) = lim t P((-OO,Xn]) Xntoo = P( lim t ( -oo, Xn]) Xntoo (from property 7) =P(U(-oo,xn]) = P((-oo , oo)) n = P(IR) = P(Q) = 1. Likewise, F(-oo) = lim F(Xn) = lim -1. P((-OO , Xn]) Xni-oo Xni-oo =P( lim (-OO,Xn]) Xni-oo (from property 7) =P<n<-oo.xnD = P(eJ) = o. n For the proof of (i), we may show F is right continuous as follows: Let Xn -1. x . We need to prove F(xn) -1. F(x) . This is immediate from the continuity property 7 of P and (-OO,Xn] -1. (-oo , x] . 0 Example 2.1.2 (Coincidences) The inclusion·exclusion formula (2.2) can be used to compute the probability of a coincidence. Suppose the integers 1, 2, . . . , n are randomly permuted. What is the probability that there is an integer left un· changed by the permutation?
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