practice-exam3-sol

practice-exam3-sol - AMS 311 Joe Mitchell Practice Exam 3...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 &lt; y &lt; 1 otherwise (b). P ( X &gt; . 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 &gt; 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 &gt; 100) (since 6:40pm is 100 minutes after 5:00pm). We condition on the value of Y : P ( X &gt; 100) = integraldisplay - P ( X &gt; 100 | Y = y ) f Y ( y ) dy = integraldisplay 100 P ( X &gt; 100 | Y = y ) 1 120 e- y 120 dy + integraldisplay 100 P ( X &gt; 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 &gt; 100 | Y = y ) = braceleftbigg y- 100 y if y &gt; 100 if y &lt; 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 &lt; y ,0 &lt; x &lt; y o.w. Then, we can compute (look at a picture of the support set): P ( X &gt; 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 ....
View Full Document

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.

Page1 / 7

practice-exam3-sol - AMS 311 Joe Mitchell Practice Exam 3...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online