hw3 - solutions - f08

hw3 - solutions - f08 - IE 336: Operations Research -...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the 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

Page1 / 4

hw3 - solutions - f08 - IE 336: Operations Research -...

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

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