Fall 2010 HW11 - University of Illinois Fall 2010 ECE 313...

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

View Full Document Right Arrow Icon
University of Illinois Fall 2010 ECE 313: Problem Set 11: Solutions Joint pdfs and functions of two random variables 1. [Joint densities] (a) f X,Y ( u,v ) = ± A (1 - ( u - v )) , u v A (1 + ( u - v )) , u < v Z 1 0 Z u 0 A (1 - u + v ) dvdu + Z 1 0 Z v 0 A (1 + u - v ) dudv = 1 A = 3 / 2 (b) The support of f X is the interval [0 , 1] . For 0 u 1 , f X ( u ) = Z f ( ) dv = Z u 0 A (1 - u + v ) dv + Z 1 u A ((1 + u - v ) dv Therefore, f X ( u ) = ± - 3 u 2 2 + 3 u 2 + 3 4 0 < u < 1 0 else. Since f X,Y ( ) = f X,Y ( v,u ) , or in other words, ( X,Y ) has the same pdf as ( Y,X ) , the pdfs f Y and f X are the same. (c) By symmetry, P { X > Y } = 1 / 2 . (d) First, use the pdf of X to compute P { X 1 2 } = R 1 0 . 5 f X ( u ) du = 0 . 5 . Also, P { X + Y < 1 ,X > 1 / 2 } is the integral of the joint pdf over the shaded region: u 1 0.5 1 v So P { X + Y < 1 1 / 2 } = R 1 0 . 5 R 1 - u 0 3 2 (1 - u + v ) dvdu = 3 32 . Therefore, P ( X + Y < 1 | X > 1 / 2) = P { X + Y < 1 1 / 2 } P { X > 1 / 2 } = 3 / 16 . 2. [Functions of random variables] The variable Z takes values in the positive reals, and for a 0 , F Z ( a ) = P { Z a } is equal to the integral of the joint pdf over the shaded region: a v u a/2
Background image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.
  • Fall '08
  • Milenkovic,O
  • Probability distribution, Probability theory, Exponential distribution, Cumulative distribution function, shaded region

{[ snackBarMessage ]}

Page1 / 3

Fall 2010 HW11 - University of Illinois Fall 2010 ECE 313...

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