part36

part36 - ECE 863 Analysis of Stochastic Systems Fall 2000...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
1 ECE 863 – Analysis of Stochastic Systems Fall 2000 Solutions for Homework Set #5 Michigan State University Department of Electrical & Computer Engineering Problem 1 5.2 [] [] nn n ii i1 ES E X EX n ==  = µ   ∑∑ () () () n nj kk k1 j1 jk VAR S VAR X COV X , X = =+ The above terms can be organized in an (nxn) matrix as follows: 22 2 2 00 K σρ σ ρσ σ ρσ = ρσ ρ σσ ± ± »» » VAR(S n ) is the sum of all terms inside the above matrix. This leads to: VAR(S n ) = n σ 2 + 2(n-1) ρ σ 2 Problem 2 5.4 a) By Eqn. 5.7, we have () () () ww w ZX Y w e e e −α −β − α+β Φ= Φ =
Background image of page 1

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

View Full DocumentRight Arrow Icon
2 b) The above characteristic function is for a Cauch RV with parameter ( α+β ). Therefore, taking the inverse transform leads to: () Z 2 2 fz z α+ β = α+β + Therefore, Z is also Cauchy. Problem 3 5.8 a) jaX bYw jawX jbwY Z wE e E e E e +  Φ= =  () () XY aw bw Φ b) [] '' ' YX ZX Y w0 11 1 E Z w aw a bw bw b aw jj j === Φ Φ where ' Z w Φ is the derivative of the characteristic function with respect to w. Since () ( ) 01 YY XX bw and aw == Φ = Φ = , then the above lead to: [] [] [] EZ aEX bEY =+ 2" Z w = =−Φ () () () () () () "2 X Y aw a bw 2 aw a bw b aw bw b = =− Φ Φ + Φ Φ Φ [][] 22 aE X bE Y 2abEXEY =+− 2 2 VAR(Z) E[Z ] E[Z] 2 a b EXEY =− −− + () [] () () 2 2 VAR Z E Z E Z a VAR X b VAR Y = +
Background image of page 2
3 Problem 4 5.9 () 1 n 1 jw X j n jwM j n M n wE e E e =   Φ= =  w n jX j n j1 Ee = = n MX n w (w) n  Φ   where 12 XX j (w) (w), j , ,. ..n Φ = Problem 5 5.10 () () () X X 1 kk 1 S k 1 k G z Ez ++ == = …… [] nn n k pz q pz q pz q =+ + + where q=1-p, and the second equality follows from the independence of the i X's. 1 k S k Gz p zq The result states that k S is Binomial with parameters ( 1 k ) and p. This is intuitive since k S is the number of “successes” in ( 1 k ) trials. Problem 6 5.11 () () z1 k S k 1 k e e e α− α− α − 1 k e α+ +α = k X Poisson with rate ( 1 k ).
Background image of page 3

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

View Full DocumentRight Arrow Icon
4 Problem 7 5.12 a) Note first that [] n k k1 ES/N n E X nEX =  == =   , thus [] [] [][] ES EES /N ENEX = Now for the variance, we use: () [] () 2 2
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/29/2009 for the course EE 131A taught by Professor Lorenzelli during the Fall '08 term at UCLA.

Page1 / 9

part36 - ECE 863 Analysis of Stochastic Systems Fall 2000...

This preview shows document pages 1 - 5. Sign up to view the full document.

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