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Math 564 Homework 1. Solutions.

Problem 1. Prove Proposition 0.2.2. A guide to this problem: start with the 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 ≤ ∞. Clearly, a is a lower bound for S. Moreover, if α > a, then choose any a < x < α and since x ∈ 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 {An}n. We first consider the statement about the lim sup. Let us assume that x is in infinitely many of ∞ the An. We first show that, for any N, x ∈ ∪n=N An. Assume that it is not, which means that

∞ !c ∞ [ \ c x ∈ 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 ∞ then this is a contradiction. Since x ∈ ∪n=N An for all N, this means x ∈ lim supn An. Conversely, let x ∈ lim supn An. Assume that x is not in infinitely many of the An, i.e. it is in finitely many 0 ∞ An. Then there must be a largest n such that x ∈ An, and call this N . Then x 6∈ ∪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) = {n : x 6∈ An}, 0 ∞ and S(x) is finite by assumption. Then choose N (x) = max S(x), and clearly x ∈ ∩n=N 00 An for 00 0 any N > N , and thus x ∈ lim infn An. Similarly, assume that x ∈ lim infn An but assume that 0 ∞ S(x) is infinite. Then, for any N, there is a N > N such that x 6∈ AN 0 , and therefore x 6∈ ∩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 of E, is a σ-algebra on E. Show also that the set {∅,E} 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 {∅,E} is the coarsest σ-algebra on E, i.e. if E is any σ-algebra on E, then {∅,E} ⊆ 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 of E is in 2E, then the conclusions of conditions #1—#3 are all trivially satisfied. {∅,E} : It satisfies condition #1 trivally. Notice also that ∅c = E and Ec = ∅, so it satisfies

condition #2. Finally, if we consider any 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 ∅ ∈ E and from this and condition 2, that E ∈ E. Therefore E ⊇ {∅,E}.

d. Consider the set E = {1, 2, 3} and define the two partitions P1 = {1}, {2, 3}, and P2 = {1, 2}, {3}. Clearly these define σ-algebras in the manner described above, and just as clearly {1} ∈ E1, {1} 6∈ E2, but on the other hand {3} ∈ E2, {3} 6∈ 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 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 . 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 P∞ 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 P∞ 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 ∈ Z. Clearly it is not finite. But we also see that µ([−M,M] ∩ Z) = 2M + 1 < ∞, and moreover [ ([−M,M] ∩ Z) = Z. M

d. Take the counting measure on R.

Problem 6. Let X be a that takes values in N. Prove that ∞ X E[X] = P(X ≥ n). n=1

Solution: We write this out:

∞ ∞ ∞ X X X P(X ≥ n) = P(X = k). n=1 n=1 k=n Rewriting the sum, and noting that the set

{(k, n): k ≥ n, n ≥ 1} can also be written {(k, n): n ≤ k, k ≥ 1}, we can reorder the sum to obtain

∞ ∞ ∞ k ∞ k ∞ 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 ϕ: N → N as the number of distinct σ-algebras defined on the set S = {1, 2, 3, . . . , n}.

a. Let ψ(n) be the number of distinct collections of subsets of S. Show that ψ(n) = 22n and that ϕ(n) ≤ ψ(n).

b. Show that for n > 1, ϕ(n + 1) > ϕ(n), so that ϕ is a strictly increasing function.

c. Compute by hand ϕ(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 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 ϕ(n) ≤ ψ(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] = {1, 2, . . . , n}. We define a map from partitions of [n] to partitions of [n + 1] as follows: if Π = {B1,...,Bk} then let Πe be {B1,...,Bk, {n + 1}}. 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 ϕ(n + 1) > ϕ(n).

c. It is not hard to see that ϕ(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 ϕ(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

ϕ(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. org/ A000110 .

Problem 8. Show that Z with the counting measure is not finite, but it is σ-finite.

Solution: See the solution to 5c).

Problem 9. Let E be a set and {Ei} any partition. Define the relation ∼ on E by x ∼ y if they are in the same Ei, and x 6∼ y if they are in different Ei. Prove that this defines an equivalence relation.

Solution: This is clearly reflexive, since if x ∈ Ri, then x ∈ Ri. Similarly, it will be symmetric, since if x ∼ y, then x, y are both in the same Ri, and thus y ∼ x. Finally, if x ∼ y and y ∼ z, then x is in the same set as y, but so is z, and thus x ∼ z.

Problem 10. Prove the Law of Total Probability.

Solution: Let Bi be a partition of Ω. We can write ! [ [ A = A ∩ Ω = A ∩ Bi = (A ∩ Bi). i i Moreover, the sets A ∩ Bi must be disjoint, since the Bi are. By the definition of measure, this means that ! [ X P(A) = P (A ∩ Bi) = P(A ∩ Bi). i i

Now recall that P(A ∩ Bi) = P(A|Bi)P(Bi) (where we are abusing notation slightly by declaring that “undefined” · 0 = 0), and we are done.

Problem 11. Prove the Inclusion–Exclusion Principle for two sets, i.e.

P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

Solution: If we write A \ B = A ∩ Bc, then clearly

A ∪ B = (A \ B) ∪ (B \ A) ∪ (A ∩ B), and this union is disjoint, so

P(A ∪ B) = P(A \ B) + P(B \ A) + P(A ∩ B). On the other hand, clearly A = (A \ B) ∪ (A ∩ B) is also a disjoint union, so

P(A) = P(A \ B) + P (A ∩ B), and similarly, P(B) = P(B \ A) + P (A ∩ B), Therefore

P(A) + P(B) = P(A \ B) + P (A ∩ B) + P(B \ A) + P (A ∩ B) = P(A ∪ B) + P(A ∩ B), and we are done.

Problem 12. Consider the function f : X → Y , where Y is a countable space. Choose the σ- algebra Y = 2Y and choose any σ-algebra X on X. Show that f is (Y/X )-measurable if and only if f −1(i) ∈ X . How does this imply the assertion of Remark 0.5.9?

Solution: Let us assume that f −1(i) ∈ X for all i. Choose B ⊂ Y , then [ [ f −1(B) = {x ∈ X : f(x) ∈ B} = {x ∈ X : f(x) = i} = f −1(i), i∈B i∈B and since each of those sets are in X , so is their union. Conversely, if f −1(B) ∈ X for all B ⊂ Y , then just restrict to the singletons. Problem 13. In this problem, you will prove (7). Recall the context: we will flip eight coins, and use Ω = {H,T }8 as our probability space. We define X as the total number of heads on the eight flips, and define F1 as the information gained after one flip.

a. Show that E[X|F1]:Ω → R only takes two values, and determine the sets on which they are constant. b. Compute these values using Theorem 0.5.15, by writing everything out explicitly.

Solution: If we can show that the filtration 1 is generated by a partition with only two elements, then we are done. But by definition 0.6.4, the filtration F1 is generated by the two sets H∗ and T ∗ (where we are using the notation that H∗ is any string of length eight starting with H, etc. Moreover, these are the sets on which this function is constant.

We now want to compute E[X|F1]; as argued above, this is constant on H∗ and T ∗. But we then see that 8 8 8 X X X 7 9 [X|H∗] = [X + X |H∗] = [X |H∗] + [ X |H∗] = 1 + [X ] = 1 + = . E E 1 i E 1 E i E i 2 2 i=2 i=2 i=2 Similarly, E[X|T ∗] = 7/2.

Problem 14. A Bernoulli trial is a random variable X that takes values zero and one, and P(X = 1) = p (implying, of course,that P(X = 0) = 1 − p). This can be thought of as an experiment that has probability p of success and 1 − p of failure. The B(n, k) is defined as the probability of having k successes if we consider n independent Bernoulli trials. Compute the probability generating function of a Bernoulli trial, and use this to compute the B(n, k).

Solution: Let X be a Bernoulli trial, then by definition we have

X 0 1 φX (t) = E[t ] = t P(X = 0) + t P(X = 1) = pt + (1 − p). Let us choose n independent Bernoulli trials, and write n X Y = Xi. i=1 By Proposition 0.8.3, we have n n X n X n φ = (pt + (1 − p))n = (pt)k(1 − p)n−k = pk(1 − p)n−ktk, Y k k k=0 k=0 so we see that Y can only take the values 0, 1, 2, . . . , k, and n (Y = k) = pk(1 − p)n−k. P k Problem 15. If X ∼ U(0, 1), show that, for any 0 ≤ a < b ≤ 1,

P(X ∈ [a, b]) = P(X ∈ [a, b)) = P(X ∈ (a, b]) = P(X ∈ (a, b)) = b − a.

Solution: We first compute

c P(X ∈ (a, b]) = P(X ∈ (−∞, b] ∩ (−∞, a] ) = P(X ∈ (−∞, b]) − P(X ∈ (−∞, a]) = b − a, thus establishing the identity for the third set. But now notice that {b} ⊆ (a, b] for any a < b, so that P({b}) ≤ P((a, b]) = b − a for all a < b; taking the limit a → b− shows that P({b}) ≤ 0, and of course it is nonnegative by definition, so P({b}) = 0. From this, the identities for the other three sets follow.