GubnerSolutions2007July13

GubnerSolutions2007July13 - Solutions Manual for...

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Unformatted text preview: Solutions Manual for Probability and Random Processes for Electrical and Computer Engineers John A. Gubner University of WisconsinMadison File Generated July 13, 2007 CHAPTER 1 Problem Solutions 1. = { 1 , 2 , 3 , 4 , 5 , 6 } . 2. = { , 1 , 2 ,..., 24 , 25 } . 3. = [ , ) . RTT > 10 ms is given by the event ( 10 , ) . 4. (a) = { ( x , y ) IR 2 : x 2 + y 2 100 } . (b) { ( x , y ) IR 2 : 4 x 2 + y 2 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 B x y-1 B-1-1 x y x y x y 3 x y 3 H 3 J 3 C 1 1 2 Chapter 1 Problem Solutions x y x y 3 3 3 3 H 3 J 3 U = M 3 U H 3 J 3 N 3 = x y 2 2 M N 3 = 2 M 2 x y 4 3 4 M N 3 4 3 7. (a) [ 1 , 4 ] parenleftBig [ , 2 ] [ 3 , 5 ] parenrightBig = parenleftBig [ 1 , 4 ] [ , 2 ] parenrightBig parenleftBig [ 1 , 4 ] [ 3 , 5 ] parenrightBig = [ 1 , 2 ] [ 3 , 4 ] . (b) parenleftBig [ , 1 ] [ 2 , 3 ] parenrightBig c = [ , 1 ] c [ 2 , 3 ] c = bracketleftBig (- , ) ( 1 , ) bracketrightBig bracketleftBig (- , 2 ) ( 3 , ) bracketrightBig = parenleftbigg (- , ) bracketleftBig (- , 2 ) ( 3 , ) bracketrightBig parenrightbigg parenleftbigg ( 1 , ) bracketleftBig (- , 2 ) ( 3 , ) bracketrightBig parenrightbigg = (- , ) ( 1 , 2 ) ( 3 , ) . (c) intersectiondisplay n = 1 (-1 n , 1 n ) = { } . (d) intersectiondisplay n = 1 [ , 3 + 1 2 n ) = [ , 3 ] . (e) uniondisplay n = 1 [ 5 , 7-1 3 n ] = [ 5 , 7 ) . (f) uniondisplay n = 1 [ , n ] = [ , ) . Chapter 1 Problem Solutions 3 8. We first let C A and show that for all B , ( A B ) C = A ( B C ) . Write A ( B C ) = ( A B ) ( A C ) , by the distributive law , = ( A B ) C , since C A A C = C . For the second part of the problem, suppose ( A B ) C = A ( B C ) . We must show that C A . Let C . Then ( A B ) C . But then A ( B C ) , which implies A . 9. Let I : = { : A B } . We must show that A I = A B . : Let A I . Then A and I . Therefore, B , and then A B . : Let A B . Then A and B . We must show that I too. In other words, we must show that A B . But we already have B . 10. The function f : (- , ) [ , ) with f ( x ) = x 3 is not well defined because not all values of f ( x ) lie in the claimed co-domain [ , ) . 11. (a) The function will be invertible if Y = [-1 , 1 ] ....
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GubnerSolutions2007July13 - Solutions Manual for...

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