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 Wisconsin–Madison File Generated July 13, 2007 CHAPTER 1 Problem Solutions 1. Ω = { 1 , 2 , 3 , 4 , 5 , 6 } . 2. Ω = { , 1 , 2 ,..., 24 , 25 } . 3. Ω = [ , ∞ ) . RTT > 10 ms is given by the event ( 10 , ∞ ) . 4. (a) Ω = { ( 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 ω ∈ C . Then ∈ ( A ∩ B ) ∪ C . But then ∈ A ∩ ( B ∪ C ) , which implies ∈ A . 9. Let I : = { ∈ Ω : ∈ 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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