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 Spring 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 ) = integraldisplay x xf X ( x ) dx = integraldisplay ∞ x ( 2 e- x- 2 e- 2 x ) dx = integraldisplay ∞ 2 xe- x dx- integraldisplay ∞ 2 xe- 2 x dx = 2 bracketleftBigg e- x (- 1) 2 (- 1 x- 1) bracketrightBigg ∞- 2 bracketleftBigg e- 2 x (- 2) 2 (- 2 x- 1) bracketrightBigg ∞ = 2 bracketleftbig e- + 1 bracketrightbig- 2 bracketleftbigg 4 e + 1 4 e bracketrightbigg 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 integraltext ∞ y xe- x dx = ( y + 1) and f ( y ) = 2 e- 2 y Then, E ( X ) = integraldisplay y E ( X | y ) f ( y ) dy = integraldisplay ∞ ( y + 1) 2 e- 2 y dy = bracketleftbig- ye- 2 y + 0 . 5 e- 2 y + 0 . 5- e- 2 y + 1 bracketrightbig ∞ = 1 . 5 1 For E(Y | x): E ( Y | x ) = integraldisplay y yf ( y | x ) dy * = integraldisplay x y e- y 1- e- x dy = 1 1- e- x integraldisplay x ye- y dy =...
View Full Document
This note was uploaded on 03/29/2010 for the course IE 383 taught by Professor Leyla,o during the Spring '08 term at Purdue.
- Spring '08
- Operations Research