The monotonicity property the measure p is non

Info icon This preview shows pages 45–49. Sign up to view the full content.

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
Image of page 45

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

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
Image of page 46
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).
Image of page 47

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

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?
Image of page 48
Image of page 49
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern