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Unformatted text preview: AMS 311 Joe Mitchell Practice Exam 3 Solution Notes 1. (a). f Y ( y ) = integraldisplay  f ( x,y ) dx = braceleftbigg integraltext y (2 x + 2 y ) dx = 3 y 2 if 0 < y < 1 otherwise (b). P ( X > . 1  Y = 0 . 5) = integraldisplay . 1 f X  Y ( x  . 5) dx = integraldisplay . 1 f ( x, . 5) f Y (0 . 5) dx = integraldisplay . 5 . 1 2 x + 2(0 . 5) 3(0 . 5) 2 dx = 64 75 , where we have used the answer to part (a) to obtain f Y (0 . 5) = 3(0 . 5) 2 . (c). E ( X  Y = 0 . 5) = integraldisplay  x f X  Y ( x  . 5) dx = integraldisplay . 5 x 2 x + 2(0 . 5) 3(0 . 5) 2 dx = 5 18 2. We are told that Y is exponential with mean 120 minutes. (Note, since Y is the lifetime in MINUTES, its mean is 120, NOT 2.) Thus, Y is exponential( 1 120 ). Let X be the number of minutes after 5:00pm when Joe enters the room. Then, given the Y = y , X is Uniform(0 ,y ); thus, E ( X  Y = y ) = y/ 2, for any y > 0. (a). We can compute E ( X ) by conditioning on Y : E ( X ) = integraldisplay  E ( X  Y = y ) f Y ( y ) dy = integraldisplay y 2 1 120 e y 120 dy = 60 Thus, I expect Joe to enter at 60 minutes after 5:00pm, i.e., at time 6:00pm. (b). We want to compute P ( X > 100) (since 6:40pm is 100 minutes after 5:00pm). We condition on the value of Y : P ( X > 100) = integraldisplay  P ( X > 100  Y = y ) f Y ( y ) dy = integraldisplay 100 P ( X > 100  Y = y ) 1 120 e y 120 dy + integraldisplay 100 P ( X > 100  Y = y ) 1 120 e y 120 dy = integraldisplay 100 1 120 e y 120 dy + integraldisplay 100 y 100 y 1 120 e y 120 dy where we have used the fact that P ( X > 100  Y = y ) = braceleftbigg y 100 y if y > 100 if y < 100 Alternatively, both parts can be solved directly, using the joint density (see Example 8.17): f ( x,y ) = f X  Y ( x  y ) f Y ( y ) = braceleftbigg 1 y 1 120 e y 120 < y ,0 < x < y o.w. Then, we can compute (look at a picture of the support set): P ( X > 100) = integraldisplay 100 integraldisplay x 1 y 1 120 e y 120 dydx and E ( X ) = integraldisplay integraldisplay x x 1 y 1 120 e y 120 dydx 3. (a). Let p i = P ( she makes it to the summit  she starts i feet above the cliff). Our goal is to find p 1 . We condition on the first step she takes, which takes her up by 1 foot with probability 0.3 and down by 1 foot with probability 0.7. Thus, we get a system of 3 equations in 3 unknowns: p 1 = (0 . 7) 0 + (0 . 3) p 2 p 2 = (0 . 7) p 1 + (0 . 3) p 3 p 3 = (0 . 7) p 2 + (0 . 3) 1 1 Now, solve the system of three equations in three unknowns; the final answer is p 1 . (b). Let t i = E (number of steps she takes before she is done  she starts i feet above the cliff). Condition on the result of the first step: t 1 = (0 . 7)(1 + 0) + (0 . 3)(1 + t 2 ) t 2 = (0 . 7)(1 + t 1 ) + (0 . 3)(1 + t 3 ) t 3 = (0 . 7)(1 + t 2 ) + (0 . 3)(1 + 0) Now, solve the system of three equations in three unknowns; the final answer is t 1 ....
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This note was uploaded on 12/06/2011 for the course AMS 311 taught by Professor Tucker,a during the Fall '08 term at SUNY Stony Brook.
 Fall '08
 Tucker,A

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