Math 564 Homework 1. Solutions.

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1 Math 564 Homework 1. Solutions. Problem 1. Prove Proposition 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. 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 Solution: Fix a family of sets {A n } n. We first consider the statement about the lim sup. Let us assume that x is in infinitely many of the A n. We first show that, for any N, x n=n A n. Assume that it is not, which means that ( ) c x A n = A c n, n=n so x is in none of the A n for n N. But then this means that x can be in A n only for n < N, but then this is a contradiction. Since x n=n A n for all N, this means x lim sup n A n. Conversely, let x lim sup n A n. Assume that x is not in infinitely many of the A n, i.e. it is in finitely many A n. Then there must be a largest n such that x A n, and call this N. Then x n=n A n and this is a contradiction. We now consider the lim inf. Let us assume that x is in all but finitely many of the A n. We will give the set of indices that x is not in the name S(x), i.e. n=n S(x) = {n : x A n },

2 and S(x) is finite by assumption. Then choose N (x) = max S(x), and clearly x n=n A n for any N > N, and thus x lim inf n A n. Similarly, assume that x lim inf n A n but assume that S(x) is infinite. Then, for any N, there is a N > N such that x A N, and therefore x n=n A N for any N. Problem 4. a. Show that for any set E, the power set 2 E, defined as the set of all subsets of E, is a σ-algebra on E. Show also that the set {, E} is a σ-algebra on E. b. Show that 2 E is the finest σ-algebra on E, i.e. if E is any σ-algebra on E, then E 2 E. 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. 2 E : since every subset of E is in 2 E, then the conclusions of conditions #1 #3 are all trivially satisfied. {, E} : It satisfies condition #1 trivally. Notice also that c = E and E c =, so it satisfies condition #2. Finally, if we consider any sequence of A n such that each A n = or E, then A n = only if all of the A n =, 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 2 E. 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 P 1 = {1}, {2, 3}, and P 2 = {1, 2}, {3}. Clearly these define σ-algebras in the manner described above, and just as clearly {1} E 1, {1} E 2, but on the other hand {3} E 2, {3} E 1. 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:

3 a. There are some trivial examples where we let the distribution have finite support. For example, we could choose the distribution λ where λ 2 = 1 and λ k = 0 for all k 2. A less trivial example could be One can show that k= λ k = 1. λ k = 3 π 2 k 2, k 0, λ 0 = 0. b. Multiply any distribution by 10, and we are done. For example, we can choose λ k = 30 π 2 k 2, k 0, λ 0 = 0. One can show that k= λ k = 10, so it is a finite measure but not a distribution. c. The counting measure on Z is such a measure: λ k = 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. d. Take the counting measure on R. M Problem 6. Let X be a random variable that takes values in N. Prove that E[X] = P(X n). n=1 Solution: We write this out: 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 we can reorder the sum to obtain {(k, n): n k, k 1}, P(X = k) = n=1 k=n k P(X = k) = k=1 n=1 k P(X = k) 1 = k=0 n=1 P(X = k) k = E[X]. k=0

4 Problem 7. We define the function 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) = 2 2n 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 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 ϕ(n) ψ(n). Since every collection of subsets is a subsets of 2 S, this means it is an element of 2 2S, and that set has 2 2n elements. b. We prove this for partitions, using Theorem We define [n] = {1, 2,..., n}. We define a map from partitions of [n] to partitions of [n + 1] as follows: if Π = {B 1,..., B k } then let Π be {B 1,..., B k, {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, , 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, We can also split into two sets of size two, and there are 3 ways to do that: 12 34, 13 24, We can split into one set of size two and two sets of size one, and there are 6 ways to do this: , , , , , And, of course, there are the two extremal partitions: the maximally fine and the maximally course This gives ϕ(4) = = 15.

5 If we look this up in the OEIS database, we get the sequence 1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, , , , , , , , , , , , , , , , , ,... which has graph so it only grows exponentially, not expo-exponentially! See http: // oeis. org/ A 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 {E i } any partition. Define the relation on E by x y if they are in the same E i, and x y if they are in different E i. Prove that this defines an equivalence relation. Solution: This is clearly reflexive, since if x R i, then x R i. Similarly, it will be symmetric, since if x y, then x, y are both in the same R i, 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 B i be a partition of Ω. We can write ( ) A = A Ω = A B i = (A B i ). i i

6 Moreover, the sets A B i must be disjoint, since the B i are. By the definition of measure, this means that ( ) P(A) = P (A B i ) = P(A B i ). i i Now recall that P(A B i ) = P(A B i )P(B i ) (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 B c, 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, Therefore P(B) = P(B \ A) + P (A B), 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 = 2 Y 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} = i B {x X : f(x) = i} = i B f 1 (i), 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.

7 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 F 1 as the information gained after one flip. a. Show that E[X F 1 ]: Ω R only takes two values, and determine the sets on which they are constant. b. Compute these values using Theorem , 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 F 1 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 F 1 ]; as argued above, this is constant on H and T. But we then see that E[X H ] = E[X 1 + X i H ] = E[X 1 H ] + E[ X i H ] = 1 + E[X i ] = = 9 2. i=2 Similarly, E[X T ] = 7/2. i=2 i=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 binomial distribution 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 probability distribution B(n, k). Solution: Let X be a Bernoulli trial, then by definition we have φ X (t) = E[t X ] = t 0 P(X = 0) + t 1 P(X = 1) = pt + (1 p). Let us choose n independent Bernoulli trials, and write n Y = X i. By Proposition 0.8.3, we have φ Y = (pt + (1 p)) n = n k=0 i=1 ( ) n (pt) k (1 p) n k = k so we see that Y can only take the values 0, 1, 2,..., k, and ( ) n P(Y = k) = p k (1 p) n k. k n k=0 ( ) n p k (1 p) n k t k, k

8 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 P(X (a, b]) = P(X (, b] (, a] c ) = 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.