Exam (elaborations) TEST BANK FOR Probability and Random Processes for Electrical and Computer Engineers By John A. Gubner (Solution Guide)

Exam (elaborations) TEST BANK FOR Probability and Random Processes for Electrical and Computer Engineers By John A. Gubner (Solution Guide) Solutions Manual for Probability and Random Processes for Electrical and Computer Engineers John A. Gubner University of Wisconsin–Madison File Generated July 13, 2007 CHAPTER 1 Problem Solutions 1. W = {1,2,3,4,5,6}. 2. W = {0,1,2, . . . ,24,25}. 3. W = [0,¥). RTT 10 ms is given by the event (10,¥). 4. (a) W = {(x,y) 2 IR2 : x2+y2  100}. (b) {(x,y) 2 IR2 : 4  x2+y2  25}. 5. (a) [2,3]c = (−¥,2)[(3,¥). (b) (1,3)[(2,4) = (1,4). (c) (1,3)[2,4) = [2,3). (d) (3,6] (5,7) = (3,5]. 6. Sketches: x y 1 1 1 B 0 B x y −1 B −1 −1 x y x y x y 3 x y 3 C1 H3 J3 1 2 Chapter 1 Problem Solutions x y x y 3 3 3 3 H3 J3 U = M3 H U 3 J3 N3 = x y 2 M2 N3 U = M2 2 x y 4 3 M4UN3 4 3 7. (a) [1,4]  [0,2][[3,5]  =  [1,4][0,2]  [  [1,4][3,5]  = [1,2][[3,4]. (b)  [0,1][[2,3] c = [0,1]c [2,3]c = h (−¥,0)[(1,¥) i h (−¥,2)[(3,¥) i =  (−¥,0) h (−¥,2)[(3,¥) i [  (1,¥) h (−¥,2)[(3,¥) i = (−¥,0)[(1,2)[(3,¥). (c) ¥ n=1 (−1 n , 1 n ) = {0}. (d) ¥ n=1 [0,3+ 1 2n ) = [0,3]. (e) ¥[ n=1 [5,7− 1 3n ] = [5,7). (f) ¥[ n=1 [0,n] = [0,¥). Chapter 1 Problem Solutions 3 8. We first let C  A and show that for all B, (AB)[C = A(B[C). Write A(B[C) = (AB)[(AC), by the distributive law, = (AB)[C, since C  A)AC =C. For the second part of the problem, suppose (AB)[C = A(B[C). We must show that C  A. Let w 2 C. Then w 2 (A B) [C. But then w 2 A (B [C), which implies w 2 A. 9. Let I := {w 2 W : w 2 A)w 2 B}. We must show that AI = AB. : Let w 2 AI. Then w 2 A and w 2 I. Therefore, w 2 B, and then w 2 AB. : Let w 2 AB. Then w 2 A and w 2 B. 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. 10. The function f : (−¥,¥)![0,¥) with f (x) = x3 is not well defined because not all values of f (x) lie in the claimed co-domain [0,¥). 11. (a) The function will be invertible if Y = [−1,1]. (b) {x : f (x)  1/2} = [−p/2,p/6]. (c) {x : f (x) 0} = [−p/2,0). 12. (a) Since f is not one-to-one, no choice of co-domain Y can make f : [0,p] !Y invertible. (b) {x : f (x)  1/2} = [0,p/6][[5p/6,p]. (c) {x : f (x) 0} =?. 13. For B  IR, f −1(B) = 8 : X, 0 2 B and 1 2 B, A, 1 2 B but 0 /2 B, Ac, 0 2 B but 1 /2 B, ?, 0 /2 B and 1 /2 B. 14. Let f :X !Y be a function such that f takes only n distinct values, say y1, . . . ,yn. Let B Y be such that f −1(B) is nonempty. By definition, each x 2 f −1(B) has the property that f (x) 2 B. But f (x) must be one of the values y1, . . . ,yn, say yi. Now f (x) = yi if and only if x 2 Ai := f −1({yi}). Hence, f −1(B) = [ i:yi2B Ai. 15. (a) f (x) 2 Bc , f (x) /2 B,x /2 f −1(B),x 2 f −1(B)c. (b) f (x) 2 ¥[ n=1 Bn if and only if f (x) 2 Bn for some n; i.e., if and only if x 2 f −1(Bn) for some n. But this says that x 2 ¥[ n=1 f −1(Bn). 4 Chapter 1 Problem Solutions (c) f (x) 2 ¥ n=1 Bn if and only if f (x) 2 Bn for all n; i.e., if and only if x 2 f −1(Bn) for all n. But this says that x 2 ¥ n=1 f −1(Bn). 16. If B = S i{bi} and C = S i{ci}, put a2i := bi and a2i−1 := ci. Then A = S i ai = B[C is countable. 17. Since each Ci is countable, we can write Ci = S j ci j . It then follows that B := ¥[ i=1 Ci = ¥[ i=1 ¥[ j=1{ci j} is a doubly indexed sequence and is therefore countable as shown in the text. 18. Let A = S m{am} be a countable set, and let B  A. We must show that B is countable. If B = ?, we’re done by definition. Otherwise, there is at least one element of B in A, say ak. Then put bn := an if an 2 B, and put bn := ak if an /2 B. Then S n{bn} = B and we see that B is countable. 19. 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. 20. Suppose A is countable and B is uncountable. Wemust show that A[B is uncountable. We prove this by contradiction. 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. 21. MATLAB. OMITTED. 22. MATLAB. Intuitive explanation: Using only the numbers 1,2,3,4,5,6, consider how many ways there are to write the following numbers: 2 = 1+1 1 way, 1/36 = 0.0278 3 = 1+2 = 2+1 2 ways, 2/36 = 0.0556 4 = 1+3 = 2+2 = 3+1 3 ways, 3/36 = 0.0833 5 = 1+4 = 2+3 = 3+2 = 4+1 4 ways, 4/36 = 0.1111 6 = 1+5 = 2+4 = 3+3 = 4+2 = 5+1 5 ways, 5/36 = 0.1389 7 = 1+6 = 2+5 = 3+4 = 4+3 = 5+2 = 6+1 6 ways, 6/36 = 0.1667 8 = 2+6 = 3+5 = 4+4 = 5+3 = 6+2 5 ways, 5/36 = 0.1389 9 = 3+6 = 4+5 = 5+4 = 6+3 4 ways, 4/36 = 0.1111 10 = 4+6 = 5+5 = 6+4 3 ways, 3/36 = 0.0833 11 = 5+6 = 6+5 2 ways, 2/36 = 0.0556 12 = 6+6 1 way, 1/36 = 0.0278 36 ways, 36/36 = 1 23. Take W := {1, . . . ,26} and put P(A) := |A| |W| = |A| 26 . Chapter 1 Problem Solutions 5 The event that a vowel is chosen is V = {1,5,9,15,21}, and P(V) = |V|/26 = 5/26. 24. Let W := {(i, j) : 1  i, j  26 and i 6= j}. For A  W, put P(A) := |A|/|W|. The event that a vowel is chosen followed by a consonant is Bvc =  (i, j) 2 W : i = 1,5,9,15, or 21 and j 2 {1, . . . ,26}{1,5,9,15,21} . Similarly, the event that a consonant is followed by a vowel is Bcv =  (i, j) 2 W : i 2 {1, . . . ,26}{1,5,9,15,21} and j = 1,5,9,15, or 21 . We need to compute P(Bvc [Bcv) = |Bvc|+|Bcv| |W| = 5 ·(26−5)+(26−5) ·5 650 = 21 65  0.323. The event that two vowels are chosen is Bvv =  (i, j) 2 W : i, j 2 {1,5,9,15,21} with i 6= j , and P(Bvv) = |Bvv|/|W| = 20/650 = 2/65  .031. 25. MATLAB. The code for simulating the drawing of a face card is % Simulation of Drawing a Face Card % n = 10000; % Number of draws. X = ceil(52*rand(1,n)); faces = (41 = X & X = 52); nfaces = sum(faces); fprintf(’There were %g face cards in %g draws.n’,nfaces,n) 26. Since 9 pm to 7 am is 10 hours, take W :=[0,10]. The probability that the baby wakes up during a time interval 0  t1 t2  10 is P([t1, t2]) := Z t2 t1 1 10 dw. Hence, P([2,10]c) = P([0,2]) = R 2 0 1/10dw = 1/5. 27. Starting with the equations SN = 1+z+z2+···+zN−2+zN−1 zSN = z+z2+···+zN−2+zN−1+zN, subtract the second line from the first. Canceling common terms leaves SN −zSN = 1−zN, or SN(1−z) = 1−zN. If z 6= 1, we can divide both sides by 1−z to get SN = (1−zN)/(1−z). 6 Chapter 1 Problem Solutions 28. Let x = p(1). Then p(2) = 2p(1) = 2x, p(3) = 2p(2) = 22x, p(4) = 2p(3) = 23x, p(5) = 24x, and p(6) = 25x. In general, p(w) = 2w−1x and we can write 1 = 6å w=1 p(w) = 6å w=1 2w−1x = x 5å w=0 2w = 1−26 1−2 x = 63x. Hence, x = 1/63, and p(w) = 2w−1/63 for w = 1, . . . ,6. 29. (a) By inclusion–exclusion, P(A[B) = P(A)+P(B)−P(AB), which can be rearranged as P(AB) = P(A)+P(B)−P(A[B). (b) Since P(A) = P(AB)+P(ABc), P(ABc) = P(A)−P(AB) = P(A[B)−P(B), by part (a). (c) Since B and ABc are disjoint, P(B[(ABc)) = P(B)+P(ABc) = P(A[B), by part (b). (d) By De Morgan’s law, P(Ac Bc) = P([A[B]c) = 1−P(A[B). 30. We must check the four axioms of a probability measure. First, P(?) = l P1(?)+(1−l )P2(?) = l ·0+(1−l ) ·0 = 0. Second, P(A) = l P1(A)+(1−l )P2(A)  l ·0+(1−l ) ·0 = 0. Third, P  ¥[ n=1 An  = l P1  ¥[ n=1 An  +(1−l )P2  ¥[ n=1 An  = l ¥å n=1 P1(An)+(1−l ) ¥å n=1 P2(An) = ¥å n=1 [l P1(An)+(1−l )P2(An)] = ¥å n=1 P(An). Fourth, P(W) =l P1(W)+(1−l )P2(W) =l +(1−l ) = 1. 31. First, since w0 /2 ?, μ(?) = 0. Second, by definition, μ(A)  0. Third, for disjoint An, suppose w0 2 S n An. Then w0 2 Am for some m, and w0 /2 An for n 6= m. Then μ(Am)=1 and μ(An)=0 for n 6=m. Hence, μ S n An  =1 and ån μ(An)=μ(Am)= 1. A similar analysis shows that if w0 /2 S n An then μ S n An  and ån μ(An) are both zero. Finally, since w0 2 W, μ(W) = 1. Chapter 1 Problem Solutions 7 32. Starting with the assumption that for any two disjoint events A and B, P(A [ B) = P(A)+P(B), we have that for N = 2, P  N[ n=1 An  = Nå n=1 P(An). () Now we must show that if () holds for any N  2, then () holds for N +1. Write P N[+1 n=1 An  = P  N[ n=1 An  [AN+1  = P  N[ n=1 An  +P(AN+1), additivity for two events, = Nå n=1 P(An)+P(AN+1), by (), = N+1 å n=1 P(An). 33. Since An := Fn Fc n−1 ···Fc 1  Fn, it is easy to see that N[ n=1 An  N[ n=1 Fn. The hard part is to show the reverse inclusion . Suppose w 2 SNn =1 Fn. Then w 2 Fn for some n in the range 1, . . . ,N. However, w may belong to Fn for several values of n since the Fn may not be disjoint. Let k := min{n : w 2 Fn and 1  n  N}. In other words, 1  k  N and w 2 Fk, but w /2 Fn for n k; in symbols, w 2 Fk Fc k−1 ···Fc 1 =: Ak. Hence, w 2 Ak  SNn =1 An. The proof that S¥n =1 An  S¥n =1 Fn is similar except that k := min{n : w 2 Fn and n  1}. 34. For arbitrary events Fn, let An be as in the preceding problem. We can then write P  ¥[ n=1 Fn  = P  ¥[ n=1 An  = ¥å n=1 P(An), since the An are disjoint, = lim N!¥ Nå n=1 P(An), by def. of infinite sum, = lim N!¥ P  N[ n=1 An  = lim N!¥ P  N[ n=1 Fn  . 8 Chapter 1 Problem Solutions 35. For arbitrary events Gn, put Fn := Gc n. Then P  ¥ n=1 Gn  = 1−P  ¥[ n=1 Fn  , by De Morgan’s law, = 1− lim N!¥ P  N[ n=1 Fn  , by the preceding problem, = 1− lim N!¥  1−P  N n=1 Gn  , by De Morgan’s law, = lim N!¥ P  N n=1 Gn  . 36. By the inclusion–exclusion formula, P(A[B) = P(A)+P(B)−P(AB)  P(A)+P(B). This establishes the union bound for N = 2. Now suppose the union bound holds for some N  2. We must show it holds for N +1. Write P N[+1 n=1 Fn  = P  N[ n=1 Fn  [FN+1   P  N[ n=1 Fn  +P(FN+1), by the union bound for two events,  Nå n=1 P(Fn)+P(FN+1), by the union bound for N events, = N+1 å n=1 P(Fn). 37. To establish the union bound for a countable sequence of events, we proceed as follows. Let An := Fn Fc n−1 ···Fc 1  Fn be disjoint with S¥n =1 An = S¥n =1 Fn. Then P  ¥[ n=1 Fn  = P  ¥[ n=1 An  = ¥å n=1 P(An), since the An are disjoint,  ¥å n=1 P(Fn), since An  Fn. 38. Following the hint, we put Gn := S¥k =n Bk so that we can write P  ¥ n=1 ¥[ k=n Bk  = P  ¥ n=1 Gn  = lim N!¥ P  N n=1 Gn  , limit property of P, Chapter 1 Problem Solutions 9 = lim N!¥ P(GN), since Gn  Gn+1, = lim N!¥ P  ¥[ k=N Bk  , definition of GN,  lim N!¥ ¥å k=N P(Bk), union bound. This last limit must be zero since å¥k =1 P(Bk) ¥. 39