STAT 571 Assignment 1 solutions 1. If Ω is a set and a collection of subsets of Ω, let be the intersection of all σ-algebras that contain . ProveC that is the σ-algebra generatedA by . C A C Solution: Let α α A be the collection of all σ-algebras that contain , and fA j 2 g C set = α. We first show that is a σ-algebra. There are three things to A \ A A α A prove. 2 (a) For every α A, α is a σ-algebra, so Ω α, and hence Ω α A α = . 2 A 2 A 2 \ 2 A A (b) If B , then B α for every α A. Since α is a σ-algebra, we have c 2 A 2 A 2 A c B α. But this is true for every α A, so we have B . 2 A 2 2 A (c) If B1;B2;::: are sets in , then B1;B2;::: belong to α for each α A. A A 2 Since α is a σ-algebra, we have 1 Bn α. But this is true for every A [n=1 2 A α A, so we have 1 Bn . 2 [n=1 2 A Thus is a σ-algebra that contains , and it must be the smallest one since α for everyA α A. C A ⊆ A 2 2. Prove that the set of rational numbers Q is a Borel set in R. Solution: For every x R, the set x is the complement of an open set, and hence Borel. Since there are only2 countablyf g many rational numbers1, we may express Q as the countable union of Borel sets: Q = x Q x : Therefore Q is a Borel set. [ 2 f g 3. Prove that the countable union of countable sets is countable. Solution: First we note that a subset of a countable set must be countable. If F is countable there is a function c : F N that is one-to-one. If E F , then the ! ⊆ restriction c E : E N is also one-to-one, so E is countable. j ! n 1 Now if E1;E2;::: are countable sets, define F1 = E1 and Fn = En ( − Fi) for n [i=1 n 2. Then the Fn's are countable, disjoint, and 1 En = 1 Fn. For every ≥ [n=1 [n=1 f 1 Fn, let n(f) denote the unique integer so that f F . Also for n N 2 [n=1 2 n(f) 2 with Fn = , let cn be a one-to-one function from Fn into N. 6 ; Now define a map c from 1 Fn into N N by c(f) = (n(f); c (f)). Let's [n=1 × n(f) convince ourselves that c is one-to-one. Suppose that f1; f2 1 Fn and that 2 [n=1 c(f1) = c(f2). Taking the first coordinate of c(f1) = c(f2), we find that n(f1) = n(f2); let's call the common value n. This means that f1; f2 Fn. The second 2 component of c(f1) = c(f2) tells us that cn(f1) = cn(f2) and since cn is one-to-one, we conclude that f1 = f2. That is, c is one-to-one. 1 See Proposition 1.3.13 on page 15 of our text. We know2 that N N is countable, so there is a one-to-one map φ from N N to × × N. The composition φ c is a one-to-one map from 1 Fn into N. ◦ [n=1 4. Let be the σ-algebra in R generated by the singletons. That is = σ( ), where = x A: x R . Show that is a proper subset of the Borel sets on RA. C C ff g 2 g A Solution: The solution depends on the fact that we have a concrete way to identify sets in . Define = E R E is countable, or Ec is countable ; we claim A F f ⊆ j g that = . If E is a countable set, then E = x E x is the countable union of singletons,A F and so belongs to σ( ) = . If Ec is[ countable,2 f g then Ec, and hence E, belongs to σ( ) = . This showsC thatA . To prove the other inclusion, we note that C , soA it suffices to prove thatF ⊆ Ais a σ-algebra. C ⊆ F F (a) The empty set is countable, so R = c . (b) If E is countable,; then Ec has countable; complement,2 F while if E has countable complement, then Ec is countable. Either way, E implies Ec . 2 F 2 F (c) Suppose that E1;E2;::: belong to . If all of the En's are countable, then so is the union3, and hence belongsF to . On the other hand, if one of the F c c c E's, say EN , has countable complement, then ( nEn) = nE E is [ \ n ⊆ N countable, so that nEn . Either way, nEn . [ 2 F [ 2 F Since singletons are Borel sets, so is every member of σ( ) = . However, the Borel set (0; 1) is not countable4 and neither is its complementC A( ; 0] [1; ). Thus (0; 1) is an example of a Borel set that does not belong to −∞. [ 1 A 5. Prove the following, where (Ω; ;P ) is a probability space and all sets are assumed to be in . F F (i) If A B, then P (A) P (B). ⊆ ≤ (ii) P ( 1 An) 1 P (An). [n=1 ≤ Pn=1 (iii) If An+1 An for all n, then P (An) P ( 1 An). ⊆ ! \n=1 Solution: (i) B is the disjoint union B = A (B A), so P (B) = P (A)+P (B A) P (A): [ n n 1 n ≥ (ii) Define A0 = A1 and for n 2, A0 = An ( − A0 ). Then the A0 's are 1 ≥ n n [i=1 i n disjoint, A0 An for each n, and nAn = nA0 . Therefore n ⊆ [ [ n P ( A ) = P ( A ) = P (A ) P (A ): n n n n0 X n0 X n [ [ n ≤ n (iii) For every n we can write An as the disjoint union An = (An An+1) (An+1 An+2) ::: ( nAn); n [ n [ [ \ 2 Remember that we proved this in the first lecture? 3 I just knew that Exercise 3 would come in handy! 4 See Proposition 1.3.14 on page 15 of the text. to obtain P (An) = P (An An+1) + P (An+1 An+2) + + P ( nAn): n n ··· \ This shows that P (An) P ( nAn) is the tail of a convergent series, and thus converges to zero as n − .\ ! 1 STAT 571 Assignment 2 solutions 6. Prove that a simple function s (as in Definition 2.1.1) is a random variable (as in Definition 2.1.6). n Solution: Write s = k=1 ak1Ak where the Ak's are disjoint members of F. Then for any λ R we haveP 2 ! s(!) < λ = Ak: f j g [ k ak<λ f j g Since this set belongs to F, s is a random variable. 1 7. Suppose that Ω = 0; 1; 2;::: , F = all subsets of Ω, and P( n ) = e− =n! for n Ω. Calculate E(X) where fX(n) = n3g for all n Ω. f g 2 2 3 1 Solution: We need to calculate the infinite sum n1=0 n e− =n!. Let's begin with 1 P a simpler problem: n1=0 ne− =n!. Here the factor of n cancels nicely with part of the factorial on theP bottom to give 1 1 1 1 1 1 ne− =n! = e− =(n 1)! = e− =k! = 1: X X − X n=0 n=1 k=0 Attempting the same trick with n2 shows that we will not get the desired cancel- lation unless we write n2 = n(n 1) + n: − 1 1 n2e 1=n! = [n(n 1) + n]e 1=n! X − X − n=0 n=0 − 1 1 = n(n 1)e 1=n! + ne 1=n! X − X − n=0 − n=0 1 1 = e 1=(n 2)! + ne 1=n! X − X − n=2 − n=0 1 1 = e 1=k! + ne 1=n! X − X − k=0 n=0 = 1 + 1 = 2: To solve the original question, write n3 = n(n 1)(n 2) + 3n(n 1) + n and 3 1 − − − repeat the method above to get 1 n e− =n! = 1 + 3 + 1 = 5. Pn=0 8. Show, by example, that X need not be a random variable even if ! X(!) = λ F for every λ R. f j g 2 2 Solution: Let Ω = R and F be the σ-algebra generated by the singletons. From the previous assignment, we know that A F if and only if A or Ac is countable. Therefore the map X from (R; F) to (R; B2(R)) given by X(!) = ! (the identity mapping) is not a random variable. For example, the interval A = (0; 1) is a Borel 1 set, but X− (A) = A F. 62 1 On the other hand, for every singleton we have X− ( λ ) = λ F. This gives the counterexample. f g f g 2 9. Prove that E(X)2 E(X2) for any non-negative random variable X. Hint: First look at simple functions. ≤ n 2 Solution: If s = k=1 ak1Ak is a simple function, then so is its square s = n 2 P n 2 n 2 k=1 ak1Ak , and E(s) = k=1 akP (Ak) and E(s ) = k=1 akP (Ak).