ch_14 - 14-1 14—2 W It suffices to show that [see...

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Unformatted text preview: 14-1 14—2 W It suffices to show that [see (14-41)] H(A 'BlBj) = H(AIBj) Since AiBj k a j AinBj = and 12minj Inj) =P(AiIBj) {a} k # j (14-40)y1e1ds “(A oBIBj) = —i§k P(AinIBj) log P(A1BkIBj) If a <8, then ¢'(u) >¢'(B) because ¢'(0) ‘ ¢'(B) = 103(8/a) >0. Hence, b b+c I ¢'(0)dfl > I ¢'(a)du c >0 a a+c This yields p1+p2 p2 ¢(p1-sz) — ¢(pl) = J ¢'(a)da < I ¢'(a)du = ¢(p2) P1 0 Similarly ¢(p1-+e) - ¢(p1) - ¢(92) +¢(p2— 6) 4 p1 p1+6 p2 ¢'(a)da - I ¢'(a)da > o p2-€ ==a=msF___.__________________________________________________‘ 203 14-3 14—4 Applying the identity H(A1 0A2) - MAI) + H(A2IA1) (1) to the partitions All-A, Az-B - C and Al-A - B, A2 -C, we obtain the first line. The second line follows from the first [see (1)]. The third line is a consequence of the first two. M It follows if we apply the identity 1(A1.A2) ' MAI) + H(A2) - H(A1 ~A2) to the partitions A1 -A - B, A2 -C. 204 14-5 (a) From (14-53) and since (see Prob. I(A,B o C) = MA) + 1105 -c) - H(A o B -C) I(A.c) - um) + 3(a) - H(A - C) 14-4) 1m. -3 -C) - B(A -C) - 11(A onlc) - H(A|C) we conclude with (14-49) that I(A,B o0) -I(A -C) - male) + male) - H(A eBIC) If B -C is observed, then the resulting prediction in the un- certainty of A equals I(A, B -C). But, if B -C is observed, then C is observed, hence, the reduction in the uncertainty of A is at least I(A,C). Hence I(A,B - C) 3 I(A,C‘ with equality only if I(A,BIC) - 0, i.e., if in the subsequence of trials in which C occurred, knowledge of the occurrence of B gives no information about A. 205 14-6 14-7 14-8 The partition 11(A3) has eight elements with respective probabili tics 3 2 2 2 2 2 2 3 P 3? q’P Q9P Qapq 9Pq opq sq Hence 3 2 2 2 3 3 3 3 2 H(A ) = - p 108? - 3!: qlogp q - 3m logpq - q logq 2 2 2 2 " - 313(1) +2pq+q )logp - 3q(p +qu+q )logq = - 3plogp - 3qlogq 8 3H(A) The density of the RV g - x + :1 equals fx(w-a). Hence, H(§+a) = - J fx(w-a) 10g fx(w-a)dw a .. I fx(x) log fx(x)dx = 110:) The joint density of the RVs x and z =x+y equals fxy(x,z-x). Hence [see (14—30)] H(E|§) = — J J fxy(x,z—x) log fxy(x,z-x)4'x(x)dxdz —¢) —m G) on = — J J fxy(x,y) log fxy(x,y}éx(x)dxdy " “Zip —m -w W The RVs :5 and y take the values x iff x=x . = + i and y3 respectively {Ab-15 xi )7:] i and y ==yJ (assumption). Hence, {f=x1+yj}=(§=xi}n{z=yj] ~ This shows that Az =Ax - By. Furthermore, since the RVs-x and y are independent, the events {;§=xi} and {z=yj} are also independent. This shows that the partitions AK and B are independent and [see (14-44) and Prob . 14-1] Y 206 H(AzIAx) = H(Ax -AyIAx) = amyle) = Buy) From this it follows that H(§I§) - HQ!) because [see (14—88) "and (14 -41) ] ugly - H<AZIAX) 14-9 As we see from (14-80) HQ) 8 lna where we assume that aaNG. The RV y takes the values 0,6,...,(N—1)6 with probability UN. The conditional density of § assuming y-k6 is uniform in the interval (k6,k6 +6). Hence, k6+6 H(xly=k6) a -J k6 f(x|y-k6)£n f(x|y=k6)dx = in 6 And as in (14-41) N H(x|y) =2 H(x|y=k6)P{y=k6} - 1:16 - - k'O I. ~ ' Finally [see (14-95)] Hag!) = H(§) - 8635'!) = lna- 2:16 W. 14-10 If yi-g(x1), yj-g(xj) and yi-Iy:l then xi=xj. Hence, Pi i=3 p.Lj - : pi = P{§=xi} and 207 14-11 14-12 From Prob. 10-10 it follows with g(x) - x that 1105,15) 1- 11(3). And since [see (14-103)] 1105,35) - H(§|§) + H05) we conclude that H(x|x) = 0. From Prob. 14—3 it follows that H(y,x|x) - nu -AxIAx) - aux. Ax) + no.ny -Ax) - ~ - y - H(AyIAx) - mzlf) becauseA -A -A and H(A -A ) -H(::,::) -0. x x x x x .. WW E{x}-0 E{x2}-S E{y}-0 2 " -2k 2 -20 2 Hzn} - 2 2 E{§n_ L7 E{§nzn} - Ehfn} = 5 k-O (a) From (14-95) ,(14-84) , and (15—86) with 1111 -5, 1122 =20/3, and ulz =5 11(3) - 2n¢101re HQ) s lnMOne/3 Hng) = lnlOnt/fi- I(x,y) = 9412 (b) The process y(t) is the output of the system L(z) =..—1._._:1 £0 1 ‘005 z = l with input 35“. Since 13(3) = HQ) and [see (12A-1) 3117 I ln|L(ej¢)|d¢ ¥ 2112.0 = 0 -1r (14-133)y1e1ds fi(y) = {1(35) = 11(5) - £m’101re. 208 14—13 6 .. 1 H(§)-H(§)--%J£n§dx-£n2 4 And as in Prob. 14-12 with 2.0 I 5, 171(y) - I-l(x) + in 5 - in 10 14—14 Given that f(x) - 0 for |xl>l and E{x) - 0.3, find f(x). With g(x) = x, (14-143) yields f(x) - Ae""‘ where 1 A A I 9"" dx - -—(e" - fl) - 1 .1 A 1 A I -1 xe"‘* dx - 4:70" - e"‘) - 35—0.," - e“) - 0.31 Solving, we obtain A :2 0.425, A z - l ---- gu————_———-—-——-n—--—--—-—--—_--—-——-———-——_---—----————-———_-—--—————-- 14—15 {(x) - Ar“ for l<x<5 and 0 otherwise, 5 5 A J- 9"" dx - 0.31 A I xe-de - 3 3-7- 1 , 60 209 14—16 From (14-151) with xk=k, 310(k) = gl(k) = k, k=l, ..., 6 ( ) {o k-1,3,5 Ac“ k-l,3,5 k l k-2,4,6 k 641"": k-2,4,6 Since p1 + p, + p5 = 0.5 and E{x} = 4.44, we conclude with z = e42 and w = e'A2 that A(2+23+25) = Aw(zz+z4zs) A(z +3z3+525) + Aw(222+4z4+6z°) = 4.44 This yields A a: 0.0437, 2 = l/w g 1.468 14—17 (a) The transformation y = 3x is one-to-one, hence, H(y) = H(x) (b) From (14-113) with g(x) = 3x: 111(2) = HQ) + in 3 l l - l4 ?, P(ll) = —— P{neither 7 nor 1]) = —— 14—18 (a) For fair dice, P(7) = 18, 18 l l l 1 l4 l4 HA=- —— —— — —— —— =. () [6ln6+18£n18+18£n18) 0655 (b) From (14-10) with n=100 and N=3: nT g enHWg 2.79 x 1028 na 2 N“ g 5.16 x 1047 210 14 —19 The process xn is WSS with entropy rate fi(x). Show that, if n 8“ =k§0 r“-k fik then lim—l—lflw ,...,w) =fi(x) +£n|l I (i) n+1 ~o -n 0 11+» Proof. The RVs genuyn are linear transformations of the RVs xo,...,xn and the transformation matrix equals 9.0 0 ... 0 2.1 9.0 ... 0 in gin-10¢. 0 Since the determinant of this transformation equals [2. (14—115) yields I n+1 o I H(yo.-u.yn) = 11(§°,...§n) + (n+1)£nl£ol Dividing by (n+1) we obtain (31) as n-W’. 14—20 As in Example 14-19, f(p) = A e-Ap. To find A, we usc the k—n curve of Fig. 14-16. This yields A = - 1.23 f(p) = 0.51 e1'23p 211 14-21 As in Example 14-22,!)k = A e-‘Xk. To find A, we use the w-r. curve of Fig. 14-17. This yields (see also Jayncs) w = 1.449 A = - 0.371 pl 92 D3 94 P5 P6 0.051. 0.079 0.114 0.165 0.240 0.348 14-22 The unknown density is normal as in (14-157) Where 4 1 1 A = 1 4 -— lo 2 +' 2 + 56 "‘23 “‘23 m23 1 m23 4 The moment x323 = E{x2x3} must be such as to maximize A. This yields 11123 = 0.25. 14-23 6AMMOM 212 Illa II/I to NO 213 14-24 If 25“ = 0, then J-tn - 000 and g“ = 1 iff §n consists of one 0 or no zeros. The probability of one and only one zero equals 382(1..s) [see (3-I3)]; the probability of no zeros equals 83. Hence, M2“ - 1|35n - 0} - 382(1-8) + 83 Thus, the redundantly coded channel of Example 14-29 is symmetrical as in (14-191) with probability of error 81- 82. 'W 14-25 If the received information is always wrong, then Pfyn=1l§n=0} = B - 1, hence C - 1—r(8) = 1 214 ...
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This homework help was uploaded on 04/09/2008 for the course ENGR, STAT 320, 262, taught by Professor Harris during the Spring '08 term at Purdue.

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ch_14 - 14-1 14—2 W It suffices to show that [see...

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