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

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