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
Unformatted text preview: IE 336: Operations Research  Stochastic Models Homework Set #3 Solutions Fall 2008 1. From HW#1 we know that f X ( x ) = 2 e x 2 e 2 x and f Y ( y ) = 2 e 2 y For E(X): E ( X ) = Z x xf X ( x ) dx = Z x ( 2 e x 2 e 2 x ) dx = Z 2 xe x dx Z 2 xe 2 x dx = 2 " e x ( 1) 2 ( 1 x 1) #  2 " e 2 x ( 2) 2 ( 2 x 1) # = 2 e + 1 2 4 e + 1 4 e E ( X ) = 1 . 5 Note: It may be tempting to think that the interval for E(X) is from y to infinity, but you are asked for the E(X) (which is independent of y) so the interval is from 0 to infinity. An alternative point of view would be to look the problem as if you did not know the joint distribution, but were given the marginal of X (which is the pdf of X, independent of Y) and the interval 0 x . If you were asked for E(X  y) then the interval would be from y to infinity. This distinction is the focus of an alternative solution. Alternative solution: E ( X  y ) = e y R y xe x dx = ( y + 1) and f ( y ) = 2 e 2 y Then, E ( X ) = Z y E ( X  y ) f ( y ) dy = Z ( y + 1)2 e 2 y dy = ye 2 y + 0 . 5 e 2 y + 0 . 5 e 2 y + 1 = 1 . 5 1 For E(Y  x): E ( Y  x ) = Z y yf ( y  x ) dy * = Z x y e y 1 e x dy = 1 1 e x Z x ye y dy = 1 1...
View Full
Document
 Spring '08
 Leyla,O
 Operations Research

Click to edit the document details