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assignment8_solutions

# assignment8_solutions - EE 505 B Fall 2011 Assignment 8...

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EE 505 B Fall 2011 Assignment 8 Solutions For full credit, you must show all of the steps or reasoning that you used to get the answer. 1. (a) The definition of the region can be written as B = { ( x 1 , x 2 ) : x 1 > 0, - x 1 < x 2 < x 1 } . The region is shown in Figure 1 . x 2 = - x 1 x 2 = x 1 x 1 x 2 Figure 1: The shaded part is the region B = { ( x 1 , x 2 ) : x 1 > 0, | x 2 | < x 1 } . (b) integraldisplay - integraldisplay - f X 1 X 2 ( x 1 , x 2 ) dx 2 dx 1 = 1 integraldisplay 0 integraldisplay x 1 - x 1 ce - x 1 dx 2 dx 1 = 1 2 c integraldisplay 0 x 1 e - x 1 dx 1 = 1 Integrating by parts with u = x 1 and dv = e - x 1 dx 1 , gives us 2 c parenleftbigg - x 1 e - x 1 vextendsingle vextendsingle 0 + integraldisplay 0 e - x 1 dx 1 parenrightbigg = 1 2 c bracketleftbig e - x 1 bracketrightbig 0 = 1 therefore, c = 1 2 1

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(c) The value of probability density function for X 1 at a value x 1 is found by integrating the joint probability density function over all values of x 2 for that value of x 1 . Mathematically, this idea is expressed as f X 1 ( x 1 )= integraldisplay - f X 1 X 2 ( x 1 , x 2 ) dx 2 For x 1 0, the integrand is 0, so f X 1 ( x 1 )= 0. For x 1 > 0, f X 1 ( x 1 )= integraldisplay x 1 - x 1 1 2 e - x 1 dx 2 = x 1 e - x 1 x 1 x 2 x 1 - x 1 x 1 Therefore, f x 1 ( x 1 )= braceleftBigg 0 x 1 0 x 1 e - x 1 x 1 > 0 This function is drawn in Figure 2 . x 1 0 f X 1 ( x 1 ) 1 e - 1 Figure 2: A graph of the function f X 1 ( x 1 ) . Similarly, f X 2 ( x 2 )= integraldisplay - f X 1 X 2 ( x 1 , x 2 ) dx 1 For x 2 < 0, f X 2 ( x 2 )= integraldisplay - x 2 1 2 e - x 1 dx 1 = 1 2 bracketleftbig e - x 1 bracketrightbig - x 2 = 1 2 e x 2 x 1 x 2 x 2 - x 2 2
For x 2 0, f X 2 ( x 2 )= integraldisplay x 2 1 2 e - x 1 dx 1 = 1 2 bracketleftbig e - x 1 bracketrightbig x 2 = 1 2 e - x 2 x 1 x 2 x 2 x 2 Therefore, f x 2 ( x 2 )= 1 2 e -| x 2 | This function is drawn in Figure 3 . x 2 0 f X 2 ( x 2 ) 1 2 Figure 3: A graph of the function f X 2 ( x 2 ) . (d) The joint cumulative distribution function is found by integrating the joint probability density function. F X 1 X 2 ( x 1 , x 2 )= integraldisplay x 1 - integraldisplay x 2 - f X 1 X 2 ( α 1 , α 2 ) d α 2 d α 1 For { x 1 0 } or { x 1 > 0 and x 2 ≤ - x 1 } , there is no overlap between the region of integration and the nonzero portion of the pdf, so the integral is 0. ( x 1 , x 2 ) x 1 x 2 F X 1 X 2 ( x 1 , x 2 )= 0 3

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For { x 1 > 0 } and {- x 1 < x 2 < 0 } , the overlap between the region of integra- tion and the nonzero portion of the pdf is shown in the figure. ( x 1 , x 2 ) α 1 - α 1 - x 2 x 1 x 2 F X 1 X 2 ( x 1 , x 2 )= integraldisplay x 1 - x 2 integraldisplay x 2 - α 1 1 2 e - α 1 d α 2 d α 1 = 1 2 integraldisplay x 1 - x 2 e - α 1 ( x 2 + α 1 ) d α 1 = 1 2 integraldisplay x 1 - x 2 x 2 e - α 1 + α 1 e - α 1 d α 1 = 1 2 parenleftBig x 2 bracketleftbig e - α 1 bracketrightbig - x 2 x 1 + bracketleftbig e - α 1 ( α 1 + 1 ) bracketrightbig - x 2 x 1 parenrightBig = 1
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assignment8_solutions - EE 505 B Fall 2011 Assignment 8...

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