This exam is a take home exam. The time for this exam is 2 hours. The exam must be taken in a continuous 2 hours slot beginning when you start reading the first problem. The completed exam, must be turned in to DH 2017 (prof. Aazhangs office in Duncan Hall) by 5pm on Monday 30th of October. This exam is a closed books, notes exam. You are only allowed to use two cheat sheets 8.5 by 11 inches, and two sided (you may write whatever you think is necessary on these cheat sheets, as in definitions, formulas, Fourier transform formulas, densities formula, etc.)(ex. two sheets of notes: one for this exam and the one from Exam 1). No other references are allowed (example: class notes, homework, homework solutions, books or the internet). Please show your work for more rational grading. All problems are weighted equally. Please, write down the time you start taking the exam and the time you finish it. This exam is covered by the Rice Honor Code. Please write and sign the pledge bellow. (Pledge: On my honor, I have neither given nor received any unauthorized aid on this examination). Exam date: ……………….. Begin time: ………… Finish time: ……….. PLEDGE: NAME: …………………………………………………… GOOD LUCK !!! 1 ELEC 533 (Behnaam Aazhang ): Midterm Examination 2017 Problem 1 Let W0,W1, be independent, identically distributed Gaussian random variables with mean zero and variance 1. Let X1 = 0 and Xn =rXn1 +Wn n0, (1) a) Show that Xn as n + converges in distribution to a random variable X. What is the distribution of where 0 < r < 1. the random variable X? b) It is not that hard to show that (please do not try to show that and just believe me) P{|Xn Xn1|2}P{Wn 2}0.02n1. (2) Then show that for X in part (a) we have (It turns out that this inequality is true for any X.) P{|Xn X|1}+P{|Xn1 X|1}P{|Xn Xn1|2}n1. (3) Knowing that, then does Xn as n + converge in Probability to our random variable X from part a)? Solution1 a) X0 =W0 X1 = rX0 + W1 is Gaussian (0, 1 + r2) X2 = rX1 + W2 is Gaussian (0, 1 + r2 + r4) So on Xn = rXn1 + Wn is Gaussian (0, ni=0 r2i) CDF of X converges to Gaussian (0, 1 ) n 1r2 b) For the second part P{|Xn X| 1}+P{|Xn1 X| 1} P{|Xn X| 1}orP{|Xn1 X| 1} P{|Xn Xn1| 2} 0.02 n 1. As n + P{|Xn X| 1} does not converge to 0 so then Xn does not converge in probability to X. Page 2 of 5 ELEC 533 (Behnaam Aazhang ): Midterm Examination 2017 Problem 1 Problem 2 Suppose we have a random process Xt with t T . We take three samples of the process at t = 1, 2, and 3 X1 seconds to form a three dimensional vector X = X2. Assume that X is a Gaussian random vector with X3 1 mean = 1 and covariance matrix denoted by CX or |X = 1 with some || 1/2. 1 0 0 01 a) We wish to form an estimate of X3 of the form X3 = aX2 +b where a and b are some constants. Find the value of a and b that minimize the mean squared error (MSE) given X2 = x2. What is the corresponding MSE value? b) Find the minimum mean square estimator of X3 given X1 = x1 and X2 = x2. Find the corresponding value of MSE and compare that with the value in Part (a). Solution2 a) X3= E[X3] + Cov(X3,X2)(X2-E[X2]) = (x2 1) V ar(X2) Var(X3)=Cov(X3,X2)Cov(X3,X2) =12 b) Similarly X3= 0 1 1 x11 1 x21 V ar(X2) 1 10 Var(X3)=1 0 1 =(112)1 2 2 Page 3 of 5 X(1) = t1 Xt2 N ,11 0 Rx(t1,t1) Rx(t1,t2) 1 et1t2 ELEC 533 (Behnaam Aazhang ): Midterm Examination 2017 Problem 2 Problem 3 Suppose {Xt;t R} is a Gauss-Markov process, meaning that Xt is a zero-mean wide sense stationary (WSS) Gaussian process with autocorrelation RX (t, s) = e|ts| . 1. Show that if you sample Xt at three different points of time t1, t2, t3, then E[Xt3 |Xt1 , Xt2 ] depends only on Xt2, and not on Xt1. 2. Show that Xt1 , Xt2 , and Xt3 form a Markov chain; that is, Xt3 conditioned on Xt2 is independent of Xt1 . Solution3 a) We will assume that: t1 < t2 < t3 Rx(t, s) = e|ts| = E[XtXs] because the mean is zero. Therefore we can write the following: Rx(t1, t2) = Rx(t2, t1) = et1t2 and similar we can write: Rx(t2, t3) = et2t3 , Rx(t1, t3) = et1t3 . X 0 11 = Rx(t2, t1) Rx(t2, t2) = et1t2 et1t2 1 1 1 1 e2(t1t2) det(11) = 1 e X(2) = Xt3 N(0,22) with 22 = Rx(t3,t3) = 1. 1 1 11 = et1t2 2(t1t2) X 1,2=E t1 Therefore, Xt3 |Xt1 , Xt2 N ( , ), where : = + 1(X(1)) After the terms are replaced we obtain: 22 21 11 12 =Xt2et2t3 = 1 e2(t2t3) Xt2 (Xt3) = e t2t3 et1t3 t3 21 11 = 1 Because Xt3 , Xt1 , Xt2 are Gaussian, than Xt3 given Xt1 , Xt2 is Gaussian with mean and variance , which are both not function of t1 and depend only on t2, therefore E[Xt3 |Xt1 , Xt2 ] depends only on Xt2 and not on Xt1 . b) We will see what is the form of E[Xt3 |Xt2 ]. Rx(t2,t2) Rx(t2,t3) 1 et2t3 t2t3 = Rx(t3,t2) Rx(t3,t3) = et2t3 1 ,11 =1=22 and12 =e =21. Page 4 of 5 ELEC 533 (Behnaam Aazhang ): Midterm Examination 2017 Problem 3 We can conclude that Xt3 |Xt2 N (, ) where both the mean and variance are independent of t1: =0+ 1(X 0) 21 11 t2 = 1 22 21 11 12 Therefore,f(Xt3|Xt2)isindependentofXt1 andsoE[Xt3|Xt2]isindependentoft1 andsuchXt1,Xt2,Xt3 form a Markov Chain. Page 5 of 5

Related posts: