GUBNER SOLUTIONS PDF

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Generally, an eBook can be downloaded in five minutes or less Show More. Han Nguyen. Ahmed Ganz , No Downloads. Views Total views. Actions Shares. Embeds 0 No embeds. No notes for slide. Chapter 1 Problem Solutions 3 8. We must show that C A. But then w 2 A B [C , which implies w 2 A. Therefore, w 2 B, and then w 2 AB. We must show that w 2 I too. In other words, we must show that w 2 A w 2 B. But we already have w 2 B. Y invertible. Let f :X! Y be a function such that f takes only n distinct values, say y1,.

But f x must be one of the values y1,. We must show that B is countable. Otherwise, there is at least one element S of B in A, say ak. Let A B where A is uncountable. We must show that B is uncountable.

We prove this by contradiction. Suppose that B is countable. Then by the previous problem, A is countable, contradicting the assumption that A is uncountable. Suppose A is countable and B is uncountable. Wemust show that A[B is uncountable. Suppose that A [ B is countable. Then since B A [ B, we would have B countable as well, contradicting the assumption that B is uncountable.

We must check the four axioms of a probability measure. First, P? Third, for disjoint An, suppose w0 2 S n An. Chapter 1 Problem Solutions 7 The hard part is to show the reverse inclusion. Then w 2 Fn for some n in the range 1,. However, w may belong to Fn for several values of n since the Fn may not be disjoint. For arbitrary events Fn, let An be as in the preceding problem.

Now suppose the union bound holds for some N 2. To establish the union bound for a countable sequence of events, we proceed as fol-lows. In this problem, the probability of an interval is its length. We first show that this collection is a s-field. First, it contains? Second, since A1,. Hence, the collection is closed under complementation. This shows that the collection is a s-field. Finally, since every element in our collec-tion must be contained in every s-field that contains A1,.

We claim that A is not a s-field. Our proof is by contradiction: We assume A is a s-field and derive a contradiction. Now, any set in A must belong to some An. By the Suppose that a s-field A contains an infinite sequence Fn of sets. If the sequence is not disjoint, we can construct a new sequence An that is disjoint with each An 2 A.

Furthermore, since the Ai are disjoint, each sequence a gives a different union, and we know from the text that the number of infinite sequences a is uncountably infinite. Hence, Ac 2 T a Aa. Third, if An 2 A for all n, then for each n and each a, An 2 Aa.

Finally, if D is any s-field that contains C , then D is one of the A s in the intersection. Chapter 1 Problem Solutions 11 The union of two s-fields is not always a s-field. Here is an example. Let W denote the positive integers, and let A denote the collection of subsets A such that either A or Ac is finite. Then E does not belong to A since neither E nor Ec the odd integers is a finite set.

First suppose that A1,. Then In the second case, suppose that some Acj is finite. Let W be an uncountable set. Let A denote the collection of all subsets A such that either A is countable or Ac is countable.

We show that A is a s-field. First, the empty set is countable. Second, if A 2 A , we must show that Ac 2 A. There are two cases. If Ac is countable, then Ac 2 A. Third, let A1,S A2,. There are two cases to consider. If all An are countable, then n An is also countable by an earlier problem. Let I denote the collection of open intervals, and let O denote the collection of open sets.

AP9335TH PDF

Solutions Gubner

Vanilson Santos flag Denunciar. If two disjoint events both have positive probability, then they cannot be independent. LetW denote the event that the decoder outputs the wrong message. Of course,W c is the event that the decoder outputs the correct message. Now, W c occurs if only the first bit is flipped, or only the second bit is flipped, or only the third bit is flipped, or if no bits are flipped. Denote these disjoint events by F, F, F, and F, respectively.

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