assignment8_solutions

# X 1 by differentiating f x 1 x 2 1 x 1 with respect

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x 1 ) by differentiating F x 1 |{ X 2 > 1 } ( x 1 ) with respect to x 1 : f x 1 |{ X 2 > 1 } ( x 1 )= d dx 1 F x 1 |{ X 2 > 1 } ( x 1 )= braceleftBigg 0 x 1 1 ( x 1 - 1 ) e - ( x 1 - 1 ) x 1 > 1 This function is plotted in Figure 6 . 10

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x 1 0 f X 1 |{ X 2 > 1 } ( x 1 ) 1 2 e - 1 Figure 6: A graph of the function f X 1 |{ X 2 > 1 } ( x 1 ) . (k) y 1 = g 1 ( x 1 , x 2 )= x 1 + x 2 y 2 = g 2 ( x 1 , x 2 )= x 1 - x 2 and x 1 = h 1 ( y 1 , y 2 )= y 1 + y 2 2 x 2 = h 2 ( y 1 , y 2 )= y 1 - y 2 2 Differentiating g 1 ( x 1 , x 2 ) and g 2 ( x 1 , x 2 ) looks easy, so we use the form of the Jacobian determinant of J ( x 1 , x 2 )= det bracketleftBigg g 1 x 1 g 1 x 2 g 2 x 1 g 2 x 2 bracketrightBigg = det bracketleftbigg 1 1 1 - 1 bracketrightbigg = - 2 therefore | J ( x 1 , x 2 ) | = 2. Therefore, f Y 1 Y 2 ( y 1 , y 2 )= 1 2 e - x 1 2 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle x 1 = y 1 + y 2 2 x 2 = y 1 - y 2 2 = 1 4 e - y 1 + y 2 2 For this expression to be complete, we need the ranges for y 1 and y 2 . x 1 and x 2 are related by x 2 > - x 1 and x 2 < x 1 . Substituting the expressions for x 1 = y 1 + y 2 2 and x 2 = y 1 - y 2 2 , we can find the ranges of y 1 and y 2 . x 2 > - x 1 x 2 < x 1 y 1 - y 2 2 > - parenleftbigg y 1 + y 2 2 parenrightbigg y 1 - y 2 2 < y 1 + y 2 2 y 1 > - y 1 - y 2 < y 2 y 1 > 0 y 2 > 0 11
Putting everything together, f Y 1 Y 2 ( y 1 , y 2 )= braceleftBigg 1 4 e - y 1 + y 2 2 y 1 > 0 and y 2 > 0 0 otherwise Y 1 and Y 2 are independent random variables because we can write the joint probability density function as the product of the margin probability density functions. The marginal probability density functions are f Y 1 ( y 1 )= integraldisplay - f Y 1 Y 2 ( y 1 , y 2 ) dy 2 = integraldisplay 0 1 4 e - y 1 2 e - y 2 2 dy 2 = 1 4 e - y 1 2 bracketleftBig 2 e - y 2 2 bracketrightBig 0 = 1 2 e - y 1 2 for y 1 > 0, and f Y 2 ( y 2 )= integraldisplay - f Y 1 Y 2 ( y 1 , y 2 ) dy 1 = integraldisplay 0 1 4 e - y 1 2 e - y 2 2 dy 1 = 1 4 e - y 2 2 bracketleftBig 2 e - y 1 2 bracketrightBig 0 = 1 2 e - y 2 2 for y 2 > 0. Therefore, f Y 1 ( y 1 ) f Y 2 ( y 2 )= parenleftbigg 1 2 e - y 1 2 parenrightbigg parenleftbigg 1 2 e - y 2 2 parenrightbigg = 1 4 e - y 1 + y 2 2 = f Y 1 Y 2 ( y 1 , y 2 ) 12

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2. (a) Using the definition for conditional probability density functions, f XY ( x , y )= f X | Y ( x ) f Y ( y ) = braceleftBigg e - ( x - y ) x y and 0 y 1 0 otherwise (b) We use the principle of total probability to find f X ( x ) : f X ( x )= integraldisplay - f XY ( x , y ) dy For x < 0, f X ( x )= 0 x y x 1 1 For 0 x 1, f X ( x )= integraldisplay x 0 e - ( x - y ) dy = e - x [ e y ] x 0 = e - x ( e x - 1 ) = 1 - e - x x y x x 1 1 For x > 1, f X ( x )= integraldisplay 1 0 e - ( x - y ) dy = e - x [ e y ] 1 0 = e - x parenleftBig e 1 - 1 parenrightBig = e - ( x - 1 ) - e - x x y x 1 1 Therefore, f X ( x )= 0 x < 0 1 - e - x 0 x 1 e - ( x - 1 ) - e - x x > 1 13
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• Spring '12
• MUNK
• Probability theory, probability density function, X1

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