F08 Sol

F08 Sol - Atlanta 1 / 2 1 / 6 1 / 3 Chicago 1 / 3 1 / 3 1 /...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
IE 336: Operations Research - Stochastic Models Exam #1 Solutions Fall 2008 1. (a) Z c 3 Z 0 ye - 3 x dxdy = Z c 3 y | - 1 3 e - 3 x | 0 dy = 1 6 | y 2 | c 3 = c 2 - 3 6 = 1 Therefore, c 2 = 9 < = > c = 3 (b) f x ( x ) = Z y ye - 3 x dy = e - 3 x 1 2 | y 2 | 3 3 = 3 e - 3 x f y ( y ) = Z x ye - 3 x dx = y Z 0 e - 3 x dx = y 3 f ( x, y ) = f x ( x ) f y ( y ) Therefore, X and Y are independent. (c) Since X and Y are independent, E ( X | y ) = E ( X ), E ( Y | x ) = E ( Y ) and E ( XY ) = E ( X ) E ( Y ) . Here, the random variable X follows an exponential distribution. That is, f x ( x ) = λe - λx = 3 e - 3 x Hence, E(X) = 1 λ = 1 3 . In case of the random variable Y, E ( Y | x ) = E ( Y ) = Z 3 3 y 2 3 dy = 3 - 3 3 . Finally, E ( XY ) = E ( X ) E ( Y ) = 1 3 3 - 3 3 ! = 1 - 3 9 ! . 2. P = 1 2 3 1 0 . 1 0 . 5 0 . 4 2 0 . 3 0 . 1 0 . 6 3 0 . 1 0 . 1 0 . 8 , P (2) = 1 2 3 1 0 . 20 0 . 14 0 . 66 2 0 . 12 0 . 22 0 . 66 3 0 . 12 0 . 14 0 . 74 1
Background image of page 1

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

View Full DocumentRight Arrow Icon
(a) P ( X 1 = 1 | X 0 = 2) = 0 . 3 P ( X 6 = 2 | X 5 = 1 ,X 4 = 3) = P ( X 6 = 2 | X 5 = 1) = P 12 = 0 . 5 (b) P (2) 32 = 0 . 14 (c) P 13 P 32 P 21 P 11 P 11 P 12 = 0 . 00006 (d) e (2) 31 = P (1) 31 + P (2) 31 = 0 . 1 + 0 . 12 = 0 . 22 (e) 2 P ( X k = 3) = P ( X k = 1) 1 2 P ( X k = 1) = P ( X k = 2) P ( X k = 1) + P ( X k = 2) + P ( X k = 3) = 1 Therefore, P ( X k = 1) = 1 2 , P ( X k = 2) = 1 4 and P ( X k = 3) = 1 4 . P ( X k +2 = 3) = 3 X i =1 P ( X k +2 = 3 | X k = i ) P ( X k = i ) = 0 . 66 × 1 2 + 0 . 66 × 1 4 + 0 . 74 × 1 4 = 0 . 68 3. (a) The diagram is omitted. P = Atlanta Chicago Detroit
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Atlanta 1 / 2 1 / 6 1 / 3 Chicago 1 / 3 1 / 3 1 / 3 Detroit 1 / 3 2 / 3 P (2) = Atlanta Chicago Detroit Atlanta . 417 . . 139 . 444 Chicago . 389 . 167 . 444 Detroit . 389 . 056 . 056 * Dont have to show the 2nd step transition matrix. (b) Since it is not possible to have a trip from Detroit to Chicago, the probability is 0. (c) f (2) AA = P (2) AA-f (1) AA P (1) AA = . 417-. 25 = . 167 (d) e (3) DC = P (1) DC + P (2) DC + P (3) DC = 0 + 0 . 056 + 0 . 083 = . 139 2...
View Full Document

Page1 / 2

F08 Sol - Atlanta 1 / 2 1 / 6 1 / 3 Chicago 1 / 3 1 / 3 1 /...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online