STAT 571 Assignment 1 solutions

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1 STAT 571 Assignment 1 solutions 1. If Ω is a set and C a collection of subsets of Ω, let A be the intersection of all σ-algebras that contain C. rove that A is the σ-algebra generated by C. Solution: Let {A α α A} be the collection of all σ-algebras that contain C, and set A A α. We first show that A is a σ-algebra. There are three things to prove. α A (a) For every α A, A α is a σ-algebra, so Ω A α, and hence Ω α A A α A. (b) If B A, then B A α for every α A. Since A α is a σ-algebra, we have B c A α. But this is true for every α A, so we have B c A. (c) If B 1, B 2,... are sets in A, then B 1, B 2,... belong to A α for each α A. Since A α is a σ-algebra, we have n1b n A α. But this is true for every α A, so we have n1b n A. Thus A is a σ-algebra that contains C, and it must be the smallest one since A A α for every α A. 2. rove that the set of rational numbers Q is a Borel set in. Solution: For every x, the set {x} is the complement of an open set, and hence Borel. Since there are only countably many rational numbers 1, we may express Q as the countable union of Borel sets: Q x Q {x}. Therefore Q is a Borel set. 3. rove 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. Now if E 1, E 2,... are countable sets, define F 1 E 1 and F n E n \ ( n 1 i1 F i) for n 2. Then the F n s are countable, disjoint, and n1e n n1f n. For every f n1f n, let n(f) denote the unique integer so that f F n(f). Also for n N with F n, let c n be a one-to-one function from F n into N. Now define a map c from n1f n into N N by c(f) (n(f), c n(f) (f)). Let s convince ourselves that c is one-to-one. Suppose that f 1, f 2 n1f n and that c(f 1 ) c(f 2 ). Taking the first coordinate of c(f 1 ) c(f 2 ), we find that n(f 1 ) n(f 2 ); let s call the common value n. This means that f 1, f 2 F n. The second component of c(f 1 ) c(f 2 ) tells us that c n (f 1 ) c n (f 2 ) and since c n is one-to-one, we conclude that f 1 f 2. That is, c is one-to-one. 1 See roposition on page 15 of our text.

2 We know 2 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 n1f n into N. 4. Let A be the σ-algebra in generated by the singletons. That is A σ(c), where C {{x} : x }. Show that A is a proper subset of the Borel sets on. Solution: The solution depends on the fact that we have a concrete way to identify sets in A. Define F {E E is countable, or E c is countable}; we claim that A F. If E is a countable set, then E x E {x} is the countable union of singletons, and so belongs to σ(c) A. If E c is countable, then E c, and hence E, belongs to σ(c) A. This shows that F A. To prove the other inclusion, we note that C F, so it suffices to prove that F is a σ-algebra. (a) The empty set is countable, so c F. (b) If E is countable, then E c has countable complement, while if E has countable complement, then E c is countable. Either way, E F implies E c F. (c) Suppose that E 1, E 2,... belong to F. If all of the E n s are countable, then so is the union 3, and hence belongs to F. On the other hand, if one of the E s, say E N, has countable complement, then ( n E n ) c n E c n E c N is countable, so that n E n F. Either way, n E n F. Since singletons are Borel sets, so is every member of σ(c) A. However, the Borel set (0, 1) is not countable 4 and neither is its complement (, 0] [1, ). Thus (0, 1) is an example of a Borel set that does not belong to A. 5. rove the following, where (Ω, F, ) is a probability space and all sets are assumed to be in F. (i) If A B, then (A) (B). (ii) ( n1a n ) n1 (A n). (iii) If A n+1 A n for all n, then (A n ) ( n1a n ). Solution: (i) B is the disjoint union B A (B \A), so (B) (A)+ (B \A) (A). (ii) Define A 1 A 1 and for n 2, A n A n \ ( n 1 i1 A i ). Then the A n s are disjoint, A n A n for each n, and n A n n A n. Therefore ( n A n ) ( n A n) n (A n) n (A n ). (iii) For every n we can write A n as the disjoint union A n (A n \ A n+1 ) (A n+1 \ A n+2 )... ( n A n ), 2 emember that we proved this in the first lecture? 3 I just knew that Exercise 3 would come in handy! 4 See roposition on page 15 of the text.

3 to obtain (A n ) (A n \ A n+1 ) + (A n+1 \ A n+2 ) + + ( n A n ). This shows that (A n ) ( n A n ) is the tail of a convergent series, and thus converges to zero as n. STAT 571 Assignment 2 solutions 6. rove that a simple function s (as in Definition 2.1.1) is a random variable (as in Definition 2.1.6). Solution: Write s n k1 a k1 Ak where the A k s are disjoint members of F. Then for any λ we have {ω s(ω) < λ} A k. {k a k <λ} Since this set belongs to F, s is a random variable. 7. Suppose that Ω {0, 1, 2,...}, F all subsets of Ω, and ({n}) e 1 /n! for n Ω. Calculate E(X) where X(n) n 3 for all n Ω. Solution: We need to calculate the infinite sum n0 n3 e 1 /n!. Let s begin with a simpler problem: n0 ne 1 /n!. Here the factor of n cancels nicely with part of the factorial on the bottom to give ne 1 /n! e 1 /(n 1)! e 1 /k! 1. n0 n1 Attempting the same trick with n 2 shows that we will not get the desired cancellation unless we write n 2 n(n 1) + n: n 2 e 1 /n! [n(n 1) + n]e 1 /n! n0 n0 k0 n(n 1)e 1 /n! + n0 e 1 /(n 2)! + n2 e 1 /k! + k ne 1 /n! n0 ne 1 /n! n0 ne 1 /n! n0

4 To solve the original question, write n 3 n(n 1)(n 2) + 3n(n 1) + n and repeat the method above to get n0 n3 e 1 /n! Show, by example, that X need not be a random variable even if {ω X(ω) λ} F for every λ. Solution: Let Ω 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 A c is countable. Therefore the map X from (, F) to (, B()) given by X(ω) ω (the identity mapping) is not a random variable. For example, the interval A (0, 1) is a Borel set, but X 1 (A) A F. On the other hand, for every singleton we have X 1 ({λ}) {λ} F. This gives the counterexample. 9. rove that E(X) 2 E(X 2 ) for any non-negative random variable X. Hint: First look at simple functions. Solution: If s n k1 a k1 Ak is a simple function, then so is its square s 2 n k1 a2 k 1 A k, and E(s) n k1 a k (A k ) and E(s 2 ) n k1 a2 k (A k). Applying the Cauchy-Schwarz inequality to the vectors x (a 1 (A 1 ) 1/2,..., a n (A n ) 1/2 ) and y ( (A 1 ) 1/2,..., (A n ) 1/2 ) gives E(s) 2 x, y 2 x 2 y 2 E(s 2 )1. Now, for a general non-negative random variable X, let s k F + s so that s k X. Then s 2 k X2 so E(X) 2 lim k E(s k ) 2 lim k E(s 2 k ) E(X2 ). Here s an even better proof that uses the variance of a random variable. 0 E((X E(X)) 2 ) E(X 2 2XE(X) + E(X) 2 ) E(X 2 ) 2E(X)E(X) + E(X) 2 E(X 2 ) E(X) An important concept in statistics is the variance of a random variable, defined as { Var (Y ) E[(Y E(Y )) 2 ] if E(Y 2 ) <, otherwise. Show that if X n (ω) X(ω) for every ω Ω, then Var (X) lim inf n Var (X n ). Solution: We may as well assume that lim inf n Var (X n ) <, otherwise the conclusion is trivial. At the same time, let s extract a subsequence X n so that

5 Var (X n ) lim inf n Var (X n ) as n. In other words, without loss of generality we may assume that sup n Var (X n ) <, let s call this value K. Since Var (X n ) <, we have E(X 2 n) < and by the previous exercise, this implies E( X n ) E(X 2 n) 1/2 <. In other words, X n is integrable. Our next job is to show that E(X n ) is a bounded sequence of numbers. The triangle inequality X(ω) E(X n ) X(ω) X n (ω) + X n (ω) E(X n ) implies the following set inclusion for any value M > 0, and hence {ω : X(ω) E(X n ) > 2M} {ω : X(ω) X n (ω) > M} {ω : X n (ω) E(X n ) > M}, ( X E(X n ) > 2M) ( X X n > M) + ( X n E(X n ) > M). Take expectations over the inequality 1 { Xn E(X n ) >M} (X n E(X n )) 2 /M 2 to give ( X n E(X n ) > M) Var (X n )/M 2 K/M 2. Combined with the previous inequality we obtain ( X E(X n ) > 2M) ( X X n > M) + K/M 2. The pointwise convergence of X n to X implies that the sets ( X X n > M) decrease to as n, so that ( X X n > M) 0. Fix M > 0 so large that K/M 2 < 1/8. The pointwise convergence of X n to X implies that the sets ( X X n > M) decrease to as n, so that ( X X n > M) 0. Therefore we can choose N so large that n N implies ( X X n > M) 1/8 and thus ( X E(X n ) > 2M) 1/4. The sets ( X > N) decrease to as N, so that for some large N we have ( X > N) 1/4. Now lets define a set of good points: G {ω : X(ω) N} {ω : X(ω) E(X n ) 2M}. If G is not empty, then for ω g G we have E(X n ) 2M + X(ω g ) 2M + N. Our bounds show us that (G c ) (( X > N) X E(X n ) > 2M) 1/4+1/4 1/2 so that G is non-empty, for all n N. In other words, E(X n ) 2M + N for n N, which implies that E(X n ) is bounded. Now we have that E(X 2 n) Var (X n ) + E(X n ) 2 is a bounded sequence. Applying Fatou s lemma to the non-negative random variables X 2 n, we conclude that X 2 is integrable and E(X 2 ) lim inf n E(X 2 n). From problem 4, this also shows that X is integrable since E( X ) E(X 2 ) 1/2 <. For any random variable Y and constant c > 0, let s define the truncated random variable Y c Y 1 { c Y c}. For any c > 0, we have E(X n ) E(X) E(X n ) E(X c n) + E(X c n) E(X c ) + E(X c ) E(X) E(X n 1 { Xn >c}) + E(X c n) E(X c ) + E(X1 { X >c} ) E(X 2 n/c) + E(X c n) E(X c ) + E(X 2 /c) sup n E(X 2 n) c + E(X c n) E(X c ) + E(X2 ) c

6 Now for every c, the sequence X c n is dominated by the integrable random variable c1 Ω, and converges pointwise to X c. Therefore the dominated convergence theorem tells us E(X c n) E(X c ). Letting n and then c in the above inequality shows that, in fact, E(X n ) E(X). Finally, we may apply Fatou s lemma to the sequence (X n E(X n )) 2 to obtain Whew! Var (X) E[(X E(X)) 2 ] E[lim inf(x n E(X n )) 2 ] n lim inf n E[(X n E(X n )) 2 ] lim inf n Var (X n ). STAT 571 Assignment 3 solutions 11. (Exercise 3.3.2, page 98) Let be a finitely additive probability on a Boolean algebra U. Show that the following are equivalent. (1) is σ-additive on U. (2) (A n ) n 1 U, A n A n+1, and n1a n A imply that (A n ) (A) if A U. (3) (A n ) n 1 U, A n A n+1, and n1a n imply that (A n ) 0. (4) If (A n ) n 1 U, A n A n+1, and for all n 1, (A n ) δ for some δ > 0, then n1a n (5) (A n ) n 1 U, A n A n+1, and n1a n A imply that (A n ) (A) if A U. Solution: (1) (2) For every n we can write A n as the disjoint union of U sets A n (A n \ A n+1 ) (A n+1 \ A n+2 )... A, and use σ-additivity to obtain (A n ) (A n \ A n+1 ) + (A n+1 \ A n+2 ) + + (A). This shows that (A n ) (A) is the tail of a convergent series, and thus converges to zero as n. (2) (3) (3) is a special case of (2). (3) (1) Suppose that (3) holds and that E m U are disjoint and E m1e m U. For each n N define the U set A n E \ ( n m1e m ) mne m. We have A n A n+1 and n1a n so we know (A n ) 0. On the other hand by finite additivity we have (E) (E 1 ) + + (E n 1 ) + (A n ), so letting n we obtain (E) m1 (E m), which says that is σ-additive. (2) (5) This follows since U is closed under complementation and is a finitely additive probability so that (A c ) 1 (A). (3) (4) These statements are contrapositives of each other. 12. (Exercise parts (1) (4), page 135)

7 (Integration by parts) Let µ and ν be two probability measures on B() with distribution functions F and G. Let µ(dx) and ν(dx) also be denoted by df (x) and dg(x). Let B (a, b] (a, b], and set B + {(x, y) B x < y} and B {(x, y) B x y}. (1) Use Fubini s theorem to express (µ ν)(b ) as two distinct integrals. (2) Let F (x ) lim y x F (y). Show that µ((a, c)) F (c ) F (a). (3) Use (1) and (2) to show that (4) Deduce that (µ ν)(b) {F (b) F (a)}{g(b) G(a)} {F (u ) F (a)} dg(u) + {F (b)g(b) F (a)g(a)} F (u ) dg(u) + {G(u) G(a)} df (u). G(u) df (u). Solution: (1) From Fubini s theorem we can write (µ ν)(b ) 1 B (x, y) (µ ν)(dx, dy) 1 (x)1 (y)1 [y, ) (x) µ(dx)ν(dy) µ(dx)ν(dy) [y,b] µ([y, b]) ν(dy). Noting that 1 [y, ) (x) 1 (,x] (y) we can use a similar argument to obtain (µ ν)(b ) ν((a, x]) µ(dx). (2) µ((a, c)) lim n µ((a, c 1/n]) lim n F (c 1/n) F (a) F (c ) F (a). (3) We have two different ways to calculate (µ ν)(b). The first is by definition (µ ν)(b) µ((a, b])ν((a, b]) (F (b) F (a))(g(b) G(a)). The second is by adding (µ ν)(b ) and (µ ν)(b + ). We already know that (µ ν)(b ) and a similar argument shows that (µ ν)(b + ) ν((a, x]) µ(dx) µ((a, x)) ν(dx) (G(x) G(a)) F (dx), (F (x ) F (a)) G(dx).

8 Therefore we have (F (b) F (a))(g(b) G(a)) (F (x ) F (a)) G(dx) + (G(x) G(a)) F (dx). (4) Starting with the equation above and multiplying out where possible gives F (b)g(b) F (a)g(b) F (b)g(a) + F (a)g(a) F (x ) G(dx) F (a)(g(b) G(a)) + F (x ) G(dx) + G(x) F (dx) G(a)(F (b) F (a)) G(x) F (dx) F (a)g(b) + F (a)g(a) G(a)F (b) + G(a)F (a). Cancelling like terms on both sides, this reduces to F (b)g(b) F (a)g(a) F (x ) G(dx) + G(x) F (dx). STAT 571 Assignment 4 solutions L 13. rove that if X a.s. n X, then X n X. Solution: For every k N choose n k so that n n k implies X n X 1/k. Then for n n k we have ( X n X > 1/k) 1, and taking the intersection over such n gives (sup n nk X n X > 1/k) 1. Now we can also take the intersection over k to obtain ( k [sup n nk X n X > 1/k]) 1. Since X n (ω) X(ω) for a.s. every ω k [sup n nk X n X > 1/k], we have X n X. w 14. rove that if X n X, then Xn X. Solution: Since X n X, we have φ(xn ) φ(x) for any continuous bounded φ. Using the dominated convergence below, we have E(φ(X n )) E(φ(X)), which w by definition means X n X. 15. (Dominated convergence theorem) rove that if X n X, and Xn Y L 1, then E(X n ) E(X). Solution: If E(X n ) E(X), then there is ɛ > 0 and a subsequence n k so that E(X nk ) E(X) ɛ for every k. But X nk X so there is a further

9 a.s. subsequence so that X nkj X. By the usual dominated convergence theorem, we have E(X nkj ) E(X), which contradicts E(X nk ) E(X) ɛ. Therefore E(X n ) E(X) is impossible, so we have E(X n ) E(X). 16. rove or disprove the following implication for convergence in L p, almost surely, and in probability: X n X implies 1 N N n1 X n X. Solution: (a.s.) It suffices to show that X n (ω) X(ω) implies 1 N N n1 X n(ω) X(ω). Suppose X n (ω) X(ω), and pick ɛ > 0. Let n ɛ so that sup n nɛ X n (ω) X(ω) ɛ. Choose N ɛ so large that n ɛ n1 X n(ω) X(ω) N ɛ ɛ. Then for N N ɛ we get 1 N N X n (ω) X(ω) 1 N N X n(ω) X(ω) n1 ( 1 nɛ ) ( N ɛ X n(ω) X(ω) + 1 N N n1 ɛ + Nɛ N 2ɛ, n1 which proves 1 N N n1 X n(ω) X(ω). nn ɛ +1 ) X n(ω) X(ω) Solution: (L p ) ick ɛ > 0 and let n ɛ so that sup n nɛ X n X p ɛ. Choose N ɛ so large that n ɛ n1 X n X p N ɛ ɛ. Then for N N ɛ we get 1 N N X n X 1 N Xn X p N p n1 ( 1 nɛ X n X ) ( N p + 1 N ɛ N n1 ɛ + Nɛ N 2ɛ, which proves 1 N N n1 X n n1 L p X. nn ɛ +1 X n X ) p Solution: (in probability) Let X n be independent random variables with (X n n) 1/n and (X n 0) 1 1/n. For any ɛ > 0 we have ( X n > ɛ) 1/n 0 so X n 0. On the other hand, for any N we have ( sup X n N ) ( ) (X n 0) 1 n N 2 N/2<n N N/2<n N ( 1 1 ) N/2 n N 1 2.

10 ( ) 1 N This gives N n1 X n > 1 2 that (1/N) N n1 X n 0 in probability. ( 1 N sup 1 n N X n > 1 2 ) 1 2, so we conclude 17. rove that if sup n E(X 2 n) <, then (X n ) n N is uniformly integrable. Solution: sup n { X n >c} Xn 2 X n d sup n { X n >c} c d sup n E(Xn) 2 c 0 as c. STAT 571 Assignment 5 solutions 18. rove that if X 0, then E[X G] 0. Solution: Let G {ω : E[X G](ω) < 0}. Then E[X G]1 G 0 and X1 G 0, but G G so 0 G X d E[X G] d 0. A negative random variable G with a zero integral must be zero: thus E[X G]1 G 0, and we conclude that 1 G 0, that is (G) (Dominated convergence theorem) rove that if X n X, and Xn Y L 1, then E[X n G] E[X G]. Solution: Since X n X 0 and X n X 2Y, dominated convergence tells us that E( X n X ) 0 as n. Since E[X n X G] X n X, this shows us that E[X n G] E[X G] in L 1 and hence also in probability. 20. If X, Y L 2, show that E[XE[Y G]] E[Y E[X G]]. Solution: Integrate over the equation E[XE[Y G] G] E[Y G]E[X G] E[Y E[X G] G]. 21. True or false: If X and Y are independent, then E[X G] and E[Y G] are independent for any G. Solution: This is false. Let X, Y be independent and identically distributed with non-zero variance, and set G σ(x + Y ). Then E[X + Y G] X + Y, so by symmetry E[X G] E[Y G] (X + Y )/2. That is, E[X G] and E[Y G] are equal!

11 22. If (X n ) n N is a submartingale, then so is (Y n ) n N where Y n (X n a) + and a is any constant. Solution: Since Y n+1 X n+1 a, we have E[Y n+1 G n ] E[X n+1 a G n ] which is greater than or equal to X n a as (X n ) n N is a submartingale. Also Y n+1 0 implies E[Y n+1 G n ] 0, so combining the two inequalities we get E[Y n+1 G n ] sup{0, X n a} (X n a) + Y n, which shows that (Y n ) n N is a submartingale. STAT 571 Term exam I solutions. 1. Determine the σ-algebra F of -measurable subsets of for the measure whose distribution function is { 1 if x 0, F (x) 0 if x < 0. Solution: Before we try to determine F, let s find out as much as we can about and. Define A n ( 1/n, 0] so that A n+1 A n for every n and n A n {0}. This implies that (A n ) ({0}). Now (A n ) F (0) F ( 1/n) 1 for every n and so we conclude that ({0}) 1, and also ( \ {0}) 0. For any subset E with 0 E we have (E) ({0}) ({0}) 1. On the other hand, if 0 E, then E ( \ {0}) and so (E) ( \ {0}) ( \ {0}) 0. Therefore we have (E) { 1 if 0 E, 0 if 0 E. Let E and Q be arbitrary subsets of. If 0 E, then 0 is contained in exactly one of the sets E Q and E Q c. Therefore 1 (E) (E Q) + (E Q c ) 1. If 0 E, then 0 doesn t belong to either E Q or E Q c, so that 0 (E) (E Q) + (E Q c ) 0. Thus any subset Q is a good splitter and so F contains all subsets of. 2. If is a probability measure on (, B()) and F its distribution function, show that F is continuous at x if and only if ({x}) 0.

12 Solution: Since F is non-decreasing, it has a left limit at x given by F ( x) lim n F (x 1/n) lim n ((, x 1/n]) ((, x)). Subtracting gives F (x) F ( x) ((, x]) ((, x)) ({x}), which shows that ({x}) 0 if and only if F (x) F ( x). This is the same as continuity at x, since F is right-continuous. 3. Show that X is a random variable on (Ω, F, ) if {ω X(ω) > λ} F for all λ. Solution: If {X > λ} F for all λ, then {X λ} n {X > λ 1/n} F. Therefore {X < λ} {X λ} c F, which shows that X is a random variable. 4. Let be a probability measure on (, B()). If Q and A, B are Borel sets so that A Q B and (B \ A) 0, then Q is -measurable. Solution: Since every Borel set is -measurable, and since Q A (Q \ A), it suffices to show that Q \ A is -measurable. Let s see how well Q \ A splits E. For any E we have (E (Q \ A)) (Q \ A) (B \ A) (B \ A) 0. Consequently we obtain (E) (E (Q \ A) c ) (E (Q \ A) c ) + (E (Q \ A)). Subadditivity of gives the reverse inequality, and shows that (Q \ A) is - measurable. 5. Give an example of a probability space and integrable random variables X n so that X n (ω) 0 for every ω Ω, but E[X n ] 0 as n. Solution: Let Ω N, F all subsets of Ω, and ({n}) 2 n for n 1. Define random variables by X n 2 n 1 {n}. For fixed ω, we have X n (ω) 0 for all n > ω so X n (ω) 0. On the other hand, E(X n ) 2 n ({n}) 2 n 2 n 1, for all n. STAT 571 Term exam II solutions 1. Give an example where X n (ω) X(ω) for every ω Ω, but E(X) < lim inf n E(X n ).

13 Solution: Let Ω N, F all subsets of Ω, and ({n}) 2 n for n 1. Define random variables by X n 2 n 1 {n}. For fixed ω, we have X n (ω) 0 for all n > ω so X n (ω) 0. On the other hand, E(X n ) 2 n ({n}) 2 n 2 n 1, for all n. 2. Show that if E[(X a) 2 ] < for some a, then E[(X a) 2 ] < for all a. Solution: If E[(X a) 2 ] <, then the result follows for any b by integrating the inequality (X b) 2 2(X a) 2 + 2(b a) rove that X d sup{ s d 0 s X, s simple} for non-negative X. Solution: If s X, then s d X d and so sup{ s d 0 s X, s simple} X d. On the other hand, X d is defined as lim k sk d where s k is a sequence of simple functions that increases to X. Therefore X d lim k s k d sup{ s d 0 s X, s simple}. Combining the two inequalities gives the result. 4. Let, Q be probabilities on (, B()) where has the density function f. rove that h(z) f(z x) Q(dx) is the density of the convolution Q. Solution: For any Borel set B, we have by the definition of convolution and Fubini s theorem ( Q)(B) 1 B (x + y)( Q)(dx, dy) 2 1 B (x + y)(dx)q(dy). Since has density f and using the change of variables z x + y, this is which gives the result. 1 B (x + y)f(x) dx Q(dy) B 1 B (z)f(z y) dz Q(dy) [ ] f(z y) Q(dy) dz,

14 5. rove that Var (X) (x y) 2 ( )(dx, dy), where is the distribution of X. Solution: First we use Fubini s theorem to rewrite the integral on 2 as an iterated integral: (x y) 2 ( )(dx, dy) 2 (x y) 2 (dx)(dy) E[(X y) 2 ] (dy). This is true whether or not the value is finite. Now if E[(X y) 2 ] for all y, then both Var (X) E[(X E(X)) 2 ] and E[(X y)2 ] (dy) are infinite. Let s suppose that E[(X y) 2 ] < for some y, and hence by problem 2, for all y. Then expanding the square is justified and we obtain E[(X y) 2 ] (dy) E[X 2 2yX + y 2 ] (dy) (E[X 2 ] 2yE[X] + y 2 ) (dy) E[X 2 ] (dy) 2E[X] y (dy) + E[X 2 ] 2E[X]E[X] + E[X 2 ] 2Var (X). y 2 (dy)

15 STAT 571 Final Exam April Instructions: This an open book exam. You can use your text, your notes, or any other book you care to bring. You have three hours. 1. If you roll a fair die, how long on average before the pattern appears? 2. Let Ω (0, 1], F B((0, 1]), and be Lebesgue measure on (0, 1]. Define the sub σ-algebra G σ{(0, 1/4], (1/4, 1/2], (1/2, 1]} and the random variable X(ω) ω 2. Write out an explicit formula for E(X G)(ω). 3. rove that E(X X 2 ) X 2, where X has density function f(x) { (1 + x)/2, if 1 x 1 0, otherwise. 4. For X L 2, prove that the random variable E(X G) has smaller variance than X. 5. Without using any mathematical notation, explain in grammatical English the meaning of a stopping time. Why are they defined in this way, and why do we only consider random times with this special property? 6. Let T be a stopping time and define F T {A F : A {T n} F n }. rove that F T is a σ-algebra. 7. Let S be a stopping time. rove that T (ω) inf{n > S(ω) : X n (ω) 3} is a (F n )-stopping time, where (X n ) n N is adapted to the filtration (F n ) n N. 8. For X L 1, prove that the collection {E(X G) : G is a sub σ algebra of F} is uniformly integrable. This is hard and undeserved measure, my lord. arolles Act 2, Scene 3: All s Well That Ends Well