Chase Joyner. 901 Homework 1. September 15, 2017

Transcription

1 Chase Joyner 901 Homework 1 September 15, 2017 Problem 7 Suppose there are different types of coupons available when buying cereal; each box contains one coupon and the collector is seeking to collect one of each in order to win a prize. After buying n boxes, what is the probability p n that the collector has at least one of each type? Solution: Let A k, 1 k, denote the event that the kth coupon was obtained after buying n boxes. Therefore, we are interested in P A k = 1 P where A c k denotes the event that the kth coupon was not obtained. By the inclusion-exclusion principle, we know P = P A c k P A c ia c j+ A c k 1 i<j<k 1 i<j A c k P A c ia c ja c k P A c 1 A c. First note that P A c 1...Ac = 0 since we must obtain at least one coupon. ext, consider the following arguments: 1 n P A c i = 2 n 1 n 2 n P A c ia c j = P A c i A c jp A c j = = 1 3 n 2 n 3 n P A c ia c ja c k = P Ac i A c ja c k P Ac ja c k = = 2 and so on. Lastly, since each pair of events, say A i or A i A j or A i A j A k, etc, are all equally likely of occurring, the summations can be captured through the choose function. Therefore, 1 n 2 n 1 P A c k = k n = 1 k+1. k 1 n

2 and hence the probability p n that the collector has at least one of each type is Problem 16 1 p n = 1 1 k+1 k k n. Suppose B is a σ-field of subsets of Ω and suppose Q: B [0, 1] is a set function satisfying a Q is finitely additive on B. b 0 QA 1 for all A B and QΩ = 1. c If A i B are disjoint and i=1 A i = Ω, then i=1 QA i = 1. Show that Q is a probability measure; that is, show Q is σ-additive. Solution: Recall the following lemma from the notes: Suppose µ is a finitely additive set function on a σ-field B with µω = 1. Then if A 1 A 2... and A n = implies that lim µa n = 0, then µ is countably additive. We will use this lemma. First, note that Q is a finitely additive set function on B and QΩ = 1. ow suppose that A 1 A 2... and A n =. We must show that lim QA n = 0. ote that given the set containments, we have A c 1 Ac 2... and Ac n = Ω. To use property c above, consider B 0 = and for n 1, B n = A c n \ A c n 1 which are disjoint sets. Furthermore, B n = Ac n = Ω and so i=1 QB i = 1, by part c, i.e. lim i=n+1 QB i = 0. otice also that for any n, B k \A c n B k. Indeed, if x B k \A c n, then x A c k \ A c k 1 and x A c n. Therefore, given the set containment of the A c k s, x Ac k \ Ac k 1 for some k n + 1, i.e. x B k. Combining all of this, we have lim QA n = lim Q Ω \ A c n = lim Q B k \A c n lim Q B k lim = 0 QB k which shows that Q is countably additive by the lemma. by sub-additivity 2

3 Problem 19 Let Ω, B, P be a probability space. Call a set null if B and P = 0. Call a set B Ω negligible if there exists a null set such that B. otice that for B to be negligible, it is not required that B be measurable. Denote the set of all negligible subsets by. Call B complete with respect to P if every negligible set is null. Suppose that B is not complete. Define a Show B is a σ-field. B := { A M : A B, M }. Solution: Clearly Ω B since Ω = Ω and Ω B,. Let C B. Then, there exists sets A B and M such that C = A M. Since M, there exists a set B such that M and P = 0. Consider c M c \ c C c = A c M c = A c = A c c A c M c \ c. We know that A c c B. otice that A c M c \ c M c \ c Ω \ c, and since c Ω, P Ω \ c = P Ω P c = 0, i.e. Ω \ c is a null set. Therefore, A c M c \ c, and so C c B. ow suppose C 1, C 2,... B. Then there exists A 1, A 2,... B and M 1, M 2,... such that C n = A n M n. ote C n = A n M n. Clearly A n B since B is a σ-field. Also, since M n, there exists a null set O n B such that M n O n and P O n = 0. Therefore, M n O n and by subadditivity P O n P O n = 0, i.e. O n B and is null. Therefore, M n. This shows that C n B and hence B is a σ-field. b Show that if A i B and M i for i = 1, 2 and then P A 1 = P A 2. A 1 M 1 = A 2 M 2, Solution: otice that A 1 M 1 = A 2 M 2 implies that A 1 M 1 \ M2 A 2, and so we know A 1 \ M 2 A 2. Since M 2, there exists a null set B 2 B such that M 2 B 2 and P B 2 = 0. Therefore, we have A 1 \ B 2 A 1 \ M 2 A 2. ow, consider by inclusion-exclusion P A 1 \ B 2 = P A 1 B c 2 = P A 1 + P B c 2 P A 1 B c 2 = P A = P A 1. Therefore, we have P A 1 = P A 1 \ B 2 P A 2. Reversing the first step to A 1 A2 M 2 \ M 1 gives the result. 3

4 c Define P : B [0, 1] by Show that P is an extension of P to B. P A M = P A, A B, M. Solution: Let B B. Then, P B = P B = P B. This shows P is an extension of P to B. d If B Ω and A i B, i = 1, 2 and A 1 B A 2 and P A 2 \ A 1 = 0, then show B B. Solution: otice that B = A 1 B \ A 1. Also B \ A1 A 2 \ A 1 and A 2 \ A 1 is null. Therefore, B B. e Show B is complete. Thus every σ-field has a completion. Solution: Let A. Then by definition of P, P A = P = 0. Since A = A, we have A is a null set in B. Therefore, B is complete. f Suppose Ω = R and B = BR. Let p k 0 and k p k = 1. Let {a k } be any sequence in R. Define P by P {a k } = p k, P A = p k, A B. What is the completion of B? a k A Solution: Let A R. Define the following two sets: G = { a k : P {a k } > 0 } and B = { a k A: P {a k } > 0 }. First notice that P G = 1 and also A = B A \ B. Clearly B B since it can be written as a countable union of measurable sets {a k }. Also, since A \ B G c and P G c = 0, we have A \ B is a negligible set. Therefore, A is in the completion of B. Since the set A R was arbitrary, this implies that the completion of B must be the power set of R, i.e. PR. g Say that the probability space Ω, B, P has a complete extension Ω, B 1, P 1 if B B 1 and P 1 B = P. The previous problem c showed that every probability space has a complete extension. However, this extension may not be unique. Suppose that Ω, B 2, P 2 is a second complete extension of Ω, B, P. Show that P 1 and P 2 may not agree on B 1 B 2. Solution: Let Ω = {1, 2, 3} and consider the σ-field B = {, {1}, {2, 3}, Ω } where P {1} = 1/2 and P {2, 3} = 1/2. ow, define the extensions B 1 = B 2 = PΩ, which are both complete because the only negligible set is the empty set which is trivially null. However, if we define the probabilities P 1 {1} = 1/2 P 2 {1} = 1/2 P 1 {2} = 1/4 P 2 {2} = 1/3 P 1 {3} = 1/4 P 2 {3} = 1/6 4

5 we see that P 1 and P 2 do not agree on B 1 B 2 since, say, P 1 {2} P 2 {2}. ote that these values were chosen so that P {1} = P 1 {1} = P 2 {1} and also P {2, 3} = P 1 {2, 3} = P 2 {2, 3}, where P i {2, 3} = P i {2} + P i {3} for i = 1, 2. h Is there a minimal extension? Solution: We will show that Ω, B, P is the minimal complete extension. Let Ω, B, P be another complete extension of Ω, B, P. Then any set M is a null set in B, i.e. B. Also, clearly B B since B is an extension. ow, take any B B, i.e. B = A M where A B and M. Since B B and B, we have A, M B and so too is A M. Thus, B B and we conclude that B B. Therefore, Ω, B, P is the minimal extension. 5