Math 564 Homework 1. Solutions. Problem 1. Prove Proposition 0.2.2. A guide to this problem: start with the open set S = (a; b), for example. First assume that a > −∞, and show that the number a has the properties that it is a lower bound for S, and, for any x > a, x is not a lower bound for S. Does this establish the statement inf S = a? Now, let a = −∞, so S = (−∞; b). Show that inf S = −∞. Solution: Let S = (a; b) with −∞ < a < b ≤ 1. Clearly, a is a lower bound for S. Moreover, if α > a, then choose any a < x < α and since x 2 S this means α is not a lower bound. If a = −∞, then S has no lower bound. The set of lower bounds of S is empty and the infimum of this set is −∞. Problem 2. If S ⊆ T , show that sup T ≥ sup S and inf T ≤ inf S. Solution: Let us define U(A) as the set of all upper bounds for the set A. Let α be an upper bound for T . Since S ⊆ T , then α is also an upper bound for S, i.e. U(T ) ⊆ U(S). From this it follows that the least element of U(T ) is in U(S), so the least element of U(S) cannot be larger. The argument is the same for the infima. Problem 3. Prove Proposition 0.2.10. Solution: Fix a family of sets fAngn. We first consider the statement about the lim sup. Let us assume that x is in infinitely many of 1 the An. We first show that, for any N, x 2 [n=N An. Assume that it is not, which means that 1 !c 1 [ \ c x 2 An = An; n=N n=N so x is in none of the An for n ≥ N. But then this means that x can be in An only for n < N, but 1 then this is a contradiction. Since x 2 [n=N An for all N, this means x 2 lim supn An. Conversely, let x 2 lim supn An. Assume that x is not in infinitely many of the An, i.e. it is in finitely many 0 1 An. Then there must be a largest n such that x 2 An, and call this N . Then x 62 [n=N 0 An and this is a contradiction. We now consider the lim inf. Let us assume that x is in all but finitely many of the An. We will give the set of indices that x is not in the name S(x), i.e. S(x) = fn : x 62 Ang; 0 1 and S(x) is finite by assumption. Then choose N (x) = max S(x), and clearly x 2 \n=N 00 An for 00 0 any N > N , and thus x 2 lim infn An. Similarly, assume that x 2 lim infn An but assume that 0 1 S(x) is infinite. Then, for any N, there is a N > N such that x 62 AN 0 , and therefore x 62 \n=N AN for any N. Problem 4. a. Show that for any set E, the power set 2E, defined as the set of all subsets of E, is a σ-algebra on E. Show also that the set f;;Eg is a σ-algebra on E. b. Show that 2E is the finest σ-algebra on E, i.e. if E is any σ-algebra on E, then E ⊆ 2E. c. Show that f;;Eg is the coarsest σ-algebra on E, i.e. if E is any σ-algebra on E, then f;;Eg ⊆ E. d. Give an example of a set E and two σ-algebras, E; F on E such that E is neither coarser nor finer than F. Solution: a. 2E : since every subset of E is in 2E, then the conclusions of conditions #1|#3 are all trivially satisfied. f;;Eg : It satisfies condition #1 trivally. Notice also that ;c = E and Ec = ;, so it satisfies condition #2. Finally, if we consider any sequence of An such that each An = ; or E, then [An = ; only if all of the An = ;, and is E otherwise. Thus #3 is satisfied. b. This follows from the fact that E is a collection of subsets of E, and as such each of those is in 2E. c. Let E be a σ-algebra. It follows from condition 1 that ; 2 E and from this and condition 2, that E 2 E. Therefore E ⊇ f;;Eg. d. Consider the set E = f1; 2; 3g and define the two partitions P1 = f1g; f2; 3g, and P2 = f1; 2g; f3g. Clearly these define σ-algebras in the manner described above, and just as clearly f1g 2 E1, f1g 62 E2, but on the other hand f3g 2 E2, f3g 62 E1. Therefore neither σ-algebra is finer or coarser than the other. Problem 5. Give examples of the following, all defined on Z: a. A distribution; b. A finite measure that is not a distribution; c. A σ-finite measure that is not finite; d. A measure which is not σ-finite. Solution: a. There are some trivial examples where we let the distribution have finite support. For example, we could choose the distribution λ where λ2 = 1 and λk = 0 for all k 6= 2. A less trivial example could be 3 λ = ; k 6= 0; λ = 0: k π2k2 0 P1 One can show that k=−∞ λk = 1. b. Multiply any distribution by 10, and we are done. For example, we can choose 30 λ = ; k 6= 0; λ = 0: k π2k2 0 P1 One can show that k=−∞ λk = 10, so it is a finite measure but not a distribution. c. The counting measure on Z is such a measure: λk = 1 for all k 2 Z. Clearly it is not finite. But we also see that µ([−M; M] \ Z) = 2M + 1 < 1, and moreover [ ([−M; M] \ Z) = Z: M d. Take the counting measure on R. Problem 6. Let X be a random variable that takes values in N. Prove that 1 X E[X] = P(X ≥ n): n=1 Solution: We write this out: 1 1 1 X X X P(X ≥ n) = P(X = k): n=1 n=1 k=n Rewriting the sum, and noting that the set f(k; n): k ≥ n; n ≥ 1g can also be written f(k; n): n ≤ k; k ≥ 1g; we can reorder the sum to obtain 1 1 1 k 1 k 1 X X X X X X X P(X = k) = P(X = k) = P(X = k) 1 = P(X = k) · k = E[X]: n=1 k=n k=1 n=1 k=0 n=1 k=0 Problem 7. We define the function ': N ! N as the number of distinct σ-algebras defined on the set S = f1; 2; 3; : : : ; ng. a. Let (n) be the number of distinct collections of subsets of S. Show that (n) = 22n and that '(n) ≤ (n). b. Show that for n > 1, '(n + 1) > '(n), so that ' is a strictly increasing function. c. Compute by hand '(k) for k = 0; 1; 2; 3; 4 by enumerating all possible σ-algebras (of course, Theorem 0.3.19 will make your life easier). Do you see a pattern? Perhaps you might find this pattern in the Integer Sequence Database. How does the growth of this sequence compare to the bounds in part (a)? Solution: a. Since every σ-algebra is, by definition, a set of subsets, we have '(n) ≤ (n). Since every collection of subsets is a subsets of 2S, this means it is an element of 22S , and that set has 22n elements. b. We prove this for partitions, using Theorem 0.3.19. We define [n] = f1; 2; : : : ; ng. We define a map from partitions of [n] to partitions of [n + 1] as follows: if Π = fB1;:::;Bkg then let Πe be fB1;:::;Bk; fn + 1gg. Clearly this is a partition of [n + 1]. Moreover, it is not hard to see that this map is one-to-one. Finally, it is not onto since it only hits those partitions where n + 1 is sitting by itself. Therefore '(n + 1) > '(n). c. It is not hard to see that '(1) = 1, since there is only one partition of [1]. If we choose S = [2], then we have two partitions: we can either split the set or not. For [S] = 3, we can split the set as: 123; 1 · 23; 2 · 13 3 · 12; 1 · 2 · 3; so '(3) = 5. Finally, for n = 4, we need to be a bit more systematic. We can split into sets of size 3 and 1, and there are 4 ways to do that: 1 · 234; 2 · 134; 3 · 124; 4 · 123: We can also split into two sets of size two, and there are 3 ways to do that: 12 · 34; 13 · 24; 14 · 23: We can split into one set of size two and two sets of size one, and there are 6 ways to do this: 12 · 3 · 4; 13 · 2 · 4; 14 · 2 · 3; 23 · 1 · 4; 24 · 1 · 3; 34 · 1 · 2: And, of course, there are the two extremal partitions: the maximally fine 1 · 2 · 3 · 4 and the maximally course 1234. This gives '(4) = 4 + 3 + 6 + 1 + 1 = 15: If we look this up in the OEIS database, we get the sequence 1; 1; 2; 5; 15; 52; 203; 877; 4140; 21147; 115975; 678570; 4213597; 27644437; 190899322; 1382958545; 10480142147; 82864869804; 682076806159; 5832742205057; 51724158235372; 474869816156751; 4506715738447323; 44152005855084346; 445958869294805289; 4638590332229999353; 49631246523618756274;::: which has graph so it only grows exponentially, not expo-exponentially! See http: // oeis.