proce’ 3.35 sone AC SNDOM PROCESS 6 ERG OERAGE ime average of a random process {X (f)} is defined as MBLE AVERAGE The Ensemble Average of a Random process {X (#} is the geNSE! e of the random variable X at time f. Ensemble Average = E[X (a) ERGODIC RANDOM PROCESS A Random Process {X (f)} is called ergodic if all its ensemble averages equals appropriate time averages. Q MEAN ERGODIC PROCESS expected valu {X ()} is said to be mean ergodic lim Ramee. Tey ocean H iT ™ isan X (dt O MEAN ERGODIC THEOREM 1 3 Let {X ()} be a random process with constant mean p and let Xp beits Time ewe Then {X (f)} is mean - ergodic if Var Xp =0. aye Q CORRELATION ERGODIC PROCESS TI the pro, he Stationary process {X (0} is said to be correlation ergodic if cess {Y (1) is mean ergodic where Y(t) = X@X(t+A). EVO! = i \ Wi Ta hen Yr is the Time Average of Y (1).PROBABILITY AND RANDOM prog, 3 Ess; 3.36 ¢ SOLVED PROBLEMS ON ERGODIC RANDOM PROCESS ¢ Example |: Let {X ()} be a WSS process with Zero Mean ang i Ut Correlation function Rx @et a where T is a consta n Find the mean and variance of the time average of {x ) t, (0, T). OVer Solution : [A.U. Noy 05) Given the Auto Correlation function Ry (t) = 1 — i Since {X (O} isa WSS process with mean zero, we get E[X(@] = 0 ae We know that time average is given by w — 1 Xr = ap J Xa. eT Since the interval is (0, T), we have iT 1 - eh X (f) dt (1) 0 To find (i) E[ Xr ] (i Var | Xt ] @E(XT] = Ty By definition,E [ Xt ] = elt f xo a| 0 1 ah = 7f Ex lat it = 0 [By (A) 0 B| Xr | 20 ae bo 1 2T (ij) = Var{ Xp] = ap S Ryy (1) Cxx (1) at a Q) rol where Cx x (t) is the Auto covariance function. B UNIT3ocESSES. 3.37 ee 00 that Cxyx@ = Rex @O-EIXOIEIX G+ 9] Ryxx (t) — [By (A)] ' TOD cubstituting (3) in (2), we ty es 1 var[ Xt] = WIS Rxx Pat -2T il ¥ f (Rx (t))? at -T i 2 ae al (i = iH) dt [. Rxx ( is an even function] : 2¢ (1 -4) ai fF. n@, 2), 1= 4) Ml SIy aie iy Soren! co: a ns Ww q\t a nly als. —— Ss A — 2 lim var{ Xr] =3 + Toe F 2a 2 REMARK : Since fees var [Xt] = 3 79% from the Mean Ergodic Theorem, we conclude that {X (#)} is not mean ergodic. Example 2: Consider the process X(N=Acosor+B sin @f, where A and B are random variables with E (A) = 0 =E (B) and E (AB) = 0. Prove that {X ()} is mean ergodic. Solution : [A.U. June 05] Given X(f1)=Acosor+B sinof nee) To prove that {X (#)} is mean ergodic, we have to show that Witlarcrs: Seer moe X, = E[X(] (Xp is the Time Average) pee) Consider F(X()] = E{[Acos@s +B sin we] [Using (1)] = cos@tE(A)+ sinwt E(B) [A & Barer.v’s] eabesiiecamee sssneogeasaneonsuctntseens}sanaoearsznasetagnstseussaanajagugguinamatiaanansnassonssa AODPROBABILITY AND RANDOM PROCES 3.38 or (0) + sino! (0) [Given B(A) = B(B)= 9) SJ = co ceo) by: ra ses f X()at Time Average XT art t = 55 f (Aces ot + Bsin wf) dt (From (1) ai F =54 ff acoser dr + f pan ra a) -T eT, Since sin wf is an odd function, a f sinwt dt = 0 (5) oT 1 i Xt = oF awl cos wf dt (cos wt is even) ft - f fener = 2f fer i790 | T = afin ik AL jr =heay A 5 aa [ sin 0=0] 6) =—_ A 2. Sonprgae (1) lim 3 — tim Asin (oT, rie = im Asin(ot) aad = A tim sin(oT) Result Oll|>a 7” Lt sind aioe a 00 0 0, of From (3) & (8, [." Maximum Value of sin @= 1 and =0) , We lim Bet Mwo OT * 9=ELX@, Hence {X (1)} is mean ergodic, UNITSConsider the random process (X (1) with x (= A cos (A' 1+ 4), where 4 is a uniformly distributed se : sot random variable in (1 1), Prove that (X ()} is correlation ergodic. IA.U. Apr. "08, Nov. °05, Dec '04] gattn n that Given to = Acasa +4), @ yi) = XOXU+A). iQ) ; . sobs for X (4) in (2), we get y(t) = Acos (A2t+ ) A cos [A2(t +2.) +4] = A2cos (A+) cos (A7t + A2A + 6) as Thus (3) becomes, ¥ (0 = 5 (eos (pert Her Af) +cos(A2t+o + A2t+ A2A + 6)] [. _- cos A cos B= } [os (A~B) + 20s (A+B) 2 AE [20s (-A?A) + 00s (2A21 + 2+ A?A)] i - [eos (A2A) + cos (2A2¢+ 2+ A2A)] (4) [+ cos (8) = cos 8] To prove that {X (#)} is correlation ergodic, we have to show that {Y ()} is mean-ergodic, where {Y (1)} is given by (4). 4 ENO) = i arf Yoa TT Tofind B [Y (#] A2 ELY (| = e[S [cos (A?A) + cos (2A2/+ 2 + any} [From (4)] Ad * 7 {E [00s (A2A)] + B [cos (2A% + 24 + A2A)}} nae - 2 { (AYA) + f (q) cos(2A% + 29 + A2A) a| wi) L E (constant) = constant | and so E {cos (A? A)] = co (PROBABILITY AND RANDOM oa 3.40 ; a fac * : ; Since is uniformly distributed : * rs) z n), fo) = m—-(-—m) 2m’ a in (8), we get, 2 aol = cos (A2A) + lim = [sin (2A2T + 2 + A2A) 2 T308T lim ie = T?0 — sin (-2A2T + 26 + A2A) 2 a , = x cos (A2A) +0 +0 [using (ne 0| ee A2 rim Y 7 = 7 00s (Ah). 9) From (7) and (9), we have tim = Ae EON yn Vrs 7) 8 (A?A). Hence the process {X (1)} is correlation ergodic. NOTE : Note that Ergodicity is a weaker condition than stationary. i.e., All Ergodic processes need not be stationarity. But all stationary processes are ergodic. Example 4: Consider the random process X (1) = 10 cos (100 ¢ + 6), where is uniformly distributed in the interval (-1, nm). Show that {X ()} is correlation ergodic, Solution : JA.U, Dec. "09, June 07] Taking A = 10 in the previous problem, we get the answer, Example 5; Prove that the random processes X (0 =A cos (wt + 0) where Aand o are constants and ‘O’is uniformly distributed variable in (0, 21) is correlation Ergodic, Solution : IA.U. Apr, 106, | Given X() =Acos (ot +0) | ee seneeennnnecneneneccngnnnnnen eeennnee Seapee3.42 PROBABILITY AND RANDOY Ta YQ) =XO°XU+) ma Substituting (1) in (2) we Bet = ¥ (0 = Acos (wt +8) Acos {o(¢+2) +9} = A2 cos (wt + 8) cos (af +2 +8) 2 =* feos (1 + 8 -(0f + 2 +8)) on ))) E cos Acos B= [cos (A - +08 (Asa = cos (ph +fl- phon -B) +cos(@!+0+ of +h +6)| 2 yo=4[ cos (02) + 60s (201 +28 + 2)] «.() To prove that {X (/)} is correlation ergodic, we have to show that {Y(d)} is mean-ergodic. een 7 E[Y(@] = \tloo 2T jf Y@dr -T To find E [Y (9] 2 E(Y @)) =E {i [ cos oa + cos (2@1 + 20 +wi)}} 2 = + Eteas0n)+ [eos 2or ct 20+ 02))} A2 eS = {cose f1@) 0s 201 +28 + wh) 8} (4) ~o Since 6 1s uniformly distributed in (0, 2) we have 1 1 >= 0X0 f(®) = 2n-0 2m oy p(*), 0, Otherwise Substituting (5) in (4), we get A2 Qn ae E[Y ()] = F[ower+j Qn 69S Quer +26 +n) a0| BEUNIT 3 esrnnnnn Slprocesses oi a) 2 Bey cose +5 2 le : ao . x cosoA + G_ 7 esin (20 +d) —sin (2of ab oni} 2 A 2 cos or [.; sin (42 + @) = sin 6] A2 «BLY @] = 7 008 © (6) els Now ‘Gaerage Y (1) dt T, A2 — j F[ cosa + cos wt +20 + 0A)| dt N A2 T li =4T {cos @A dt + i) cos (2mt + 20 + wd) dt aT, “1 sin (2@t +20 + on] 20 T A2 “fi ewon ys sin (2@T +20 +@A) | 2T cos mA + Xe 20 sin (-2oT + 20 + 2 Y= Mw? TS 2 COSMA +o T sin (2oT +20 +@A) —sin (-20T + 20 +0) (2) As T ~ o in (7), we get, ccenatnannennnegnqnnnnnenne en sssneseesnssanerel NETPROBABILITY AND RANDOM PROCES, SEs = ae Him Atos ia mn ARON se BOT sin (207 + 20 +A) woe tte jos hee am 4 Using ,!™ Ha lim — AC ‘i ao : tr. coon +0 Fe ne T+ §o7 0 ny. ee p08 Dh joa we ag) From (7) and (8) we get lm — EV O]= toe YT Hence the process {X (} is correlat Show that the random process' ion ergodic. es X (f) = cos (t + >) where 4 is Example 6 : : a random yariable uniformly distributed in (0, 2n) is correlation ergodic. Solution : Put A> 1,@—> | in the previous example. QEXERCISES 0 A. Review Questions: Define Time Average of a random process. Define Ensemble Average of a random process. Define an ergodic random process. State the mean ergodic theorem. Define an correlation ergodic random process. B. Problems: 1. A random process X (#) has sample values x, ( x4 (0) =-1, Xs (t) = -3, x6 () =—S. Find the Mean and process. Is the process ergodic in the mean. 2, Let X (t) =A cos +B sin ¢ where A, B are rando E (A) = E(B) = 0 and E (AB) = 0, Show that {X ()} is mean 3, Let {X ()} be a WSS process with zero Mean find the Mean a Ans. 9,3 a = 5) Oa Variance of the m variables with ergodic: Variance of the time average of {X (#)} over (0, 1).