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ECE302HW4Soln_SP10

ECE302HW4Soln_SP10 - ECE 302 Homework 4 Prof Saul Gelfand...

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Unformatted text preview: ECE 302: Homework 4 Prof. Saul Gelfand Spring 2010 Due March 5, 2010 Problem 1: 3.42 in text Let X denote the number of heads, T denote the number of tails, and Y = X ~ T denote the difference between number of heads and number of tails. Since in n tosses, we have T : n — X ., then Y 2 2X — n, and E[Yl : 2E[X] e n, and Va7“[Y] = 4Var[X]. Since we have that X is a binomial random variable, E [X ] : p77, and Var[X] = 71])(1 — p), so that Em and Va7‘[Y] 271]) ~71 2 71(21) ~ 1) 47'L])(1 — p). Problem 2: 3.43 in text Note that this problem of the hacker attempting to choose a correct password out of 2m passwords is identical to the previous homework problem of the professor trying to unlock his door using 77, = 27” keys. Recall from that problem that pXCIC) = i = 2—}; for 1 S i, S 2"“. a. By the hint, we have that p; is the probability of guessing the correct password on attempt '27. On attempt 7;, we will have tried 2% 1 passwords, leaving 2""‘—/z', to try, of which 1 is correct. Therefore, W 1 V pl- 7 2,71% and 1 pX(’zC) = (1 ~p1)(1 7p?) . . . (1 —p7»_1)p7; : E77 for 1 S i 3 2"“'. From this PMF, we now compute 2m 2m , 1 2’” ~ it PM > M = Z we : 5— : 2 1i:k+1 2':1;+1 . 73X7(i) : ‘ J 27m > X > A” pX('I,|X > ls) : { étx>k1 Z'hk elee# b. Using the result of (a) 00 277‘! MMX>k ijme>k) = ijfluX>m ::—oo i:k:+1 2771 . = Z i:}€+1 2m _ k 2171 : 2m M k. 2 I i: l+1 2 C' Wye have found E [X IX > [If], so before commuting E [X]7 We find MMXsm; k P[X S k] : _P[X > [3]: 2m 7W) —1 < 1 < L : PX< If S k) : PMS/g] 1 _ Y #1 0» else [XIX < ID] : Zipfiflz g A) k? 1 k w+u I ? Z Z (k+1)§ I 2 21:1 We now have that E [X ] is related to the conditional expectations as follows: E[X] : E[X|X > Is] ]P[X > k] + E[X|X g IslP[X : k7] _ (22m+k+1) 2771—1- . (m1) I; ~< M31323 (1+ 2’”) Problem 3 This problen'121s now identical to the previous homework prob— lem of the professm trying to unlock his door using 71 m 2"" keys but placing 111c01rect ke3s back on the kc3 ring. Recall horn that pIoblein that pX(1) = (n 1y 1(21) for z 2 1. n a. From the hint, we have that p is the probability of the hacker guessing correctly on any given try (all atternps are now indis— tinguishable since he forgets what passwords he has tried). Since there are 2"” passwords and one which is correct, we have 1) 2 —L 21717 and . 1 14 1 m1) = (17%) 7 :> P[X > k] = 2103(1) 11 M8 A 1—1 I file 3 “as H [\3 :3 ‘ >—‘ H H F‘g-‘fi A l A I ‘9‘_ f c, [\4 r at i v H l—l " 1 2m _ 1'17 Z < 2771 > ' We now find the conditional pinf: M # _ L i’k*1 L ,n pX(z'lX > 1:) :{ P[X>k] # (1 ) , X > A, y M . oo . > 06 , , 1 i—k—l 1 E[XlX > k] = Z 2413me > k) 7, <1 ~ 57;) 57 izioo i:k+1 (letlzi—kfil) H : p; + t: A r—t I slH K/N “Ede || 3: + C M8 ‘ A >-—t t “£1?“ V [\D are In order to evaluate these sums, we now consider now a geometric random variable, Z with 102(1) 2 (1 ~ p)lp. Since Z is geometric, then 00 2(1 amlp: 1 1:0 and E[Z] : Z [(1 — ml]? : p 2) 1:0 We relate these to our sums from above by letting p : 2—17, so that (k+1)§:<1*%>l§l; : (k+1) 0° 1 1 1 1—7}: , §[<1__>_ 12'22m_1 2m 2m 2T1 (kwr 1) +27” — 1 : 27714—11»: ll 5, 5 >< v 3 H We have that Ele = Z 'zipxtzil = 21(1 ~p)lp7 1:700 2:] where p 2 ~17. We showed in the previous problem that 2 00 i 1 e 9 Zifl—pfl): I =2m~1. 1:: p 0.8- Dwights snowballs 1.5‘ Kama) f>< 0.5- leghts good snowballs W 15 05* J 111175 snowballs —1 O 1 X Jim‘s good snowballs 2 1.5— E X w—Xj 0‘5 1 2 0 —2 _2 —1 1 2 0 x All snowballs (from either person) Figure 1: Problem 4 C7! Problem 4 We have the PDF’s fXD(:1:) and fXJ(x) plotted in Figure 1. a. We note that fXD(:1:) is syi'm'i'ietric about 0, therefore E [X13] 2 0; similarly fXJ(r) is symmetric about :13 2 %, therefore E [X J] = 5. Now E12651 : / ngxuem 0 1 1 2 1 = / m2(— + —L‘)da: +/ r2(— — ~1)dz _2 2 0 _ L3 + $4 _ 2 " 6 16 7:2 _ 3 , z 2 :> Wrap] = E[X})] — (E[XD])2 : g E[X3] = / :rgijhfii? T72 3/2 = / :L‘2(1/2 +:z:)d:L‘ +/ 12(3/2 — I)dL 71/2 1/2 [13 $4] 1/2 L3 1,4 3/2 = — + »~ + 1— — -—1 6 4 93:~1/2 2 4 :czl/Z _ 1 + 3 _ 5 _ 24 8 _ 12 x 1 :5 vaJ} z E[X3] — (E[XJ])~2 = 6 b. The probability that Dwight throws a good toss is 1 1 1/2 , 7 P7 [~—2— < XD < E] _ 71/2 fXD(1L)dl — 16, The probability that Jim throws a good toss is P7“ —— < XJ < —- = fXJ(q:)dzz': : — 2 2 . 71/2 We then have the conditional PDF’s fXDW) , '11 < l fXJW) meme) = Prt—sxer 2 , 111147210): 1’Tl~%<XJ%l , IR?! 2 § 0, [Oh—A kib— These functions are plotted in Figure 1 We have that , 1 . 1 , jX($) I EfXD (T) + ifX/(w) This is shown in Figure 1. We then have the following: E[X[ : /:a?f'X(-93)d93 = /_°O GM) + glam) cm- 11 Em: /00562J‘X(513)d$ = [[email protected]<a/~)+§ij<m)>d2: It I E ii .1. | E 3: H | I l | => VaflX] : E[X2]*(E1Xl)2 The average of the variances of X D and X J is 1V [X [+1v [X] 1+ 1 5 #V( [X] ~a7” ~a7’ :7 ~27 L?" . 2 D 2 J 3 12 12 Var[X] is not simply an average of VarlXDl and Var [X J] because variance7 unlike expected value, is not a linear operator. [3 Now7 we are given that a good toss occurs. We find that 1 1 7 PHGlDwight throws] : Pfl—i < XD < 5] : T6 1 1 1 P7”[GIJlIH throws] : PHTE < X] < E] 2 i' 7 Since P7”[G|Ji1n throws] > PT[G|Dwight throws], we have that Jim is more likeiy to have tossed the good si'iow1:)a11. Problem 5 a. Current I z % so that since R > 0, P7’[X/R >1]: P7“[X > R] 2 [:0 @273 exp (—577) d1“ = Q(R/0). b. Power P = 5}; so that Pr [9; > I] = P7~[X > Vi] +Pr[X < mx/E] : 262 (g) Em = E[X/R] : %E[X} : 0 Varm 2 va/R} : Elgva] : f; d. END] 2 EiX2/R] = gm : % (mm + (EiXW) = j;— E[P2] : E[X4/R2] : 1213 [X4] : 3% (since X = 0) 2» vmp] = E[P2] — (E[P])2 = 3% — 2:1 “ ‘4) if 1M, {Mnfl‘iler U; k ”45%04 v-fliL My Miami/L We“ Le. ' \4 / lra X6, CML I XH ‘ 1”“ u :9) < t) K 'W“ I< ),¢:I,~“,I< Men X‘ I? I'n ‘H‘e— 'lfiMA’L» TAK‘ 0W Y6 : 40‘) I“; QM +0 aka Mdlbalfi 0+ 411$ [MWVNL' \C) Y: 400 Y.‘; adc‘sm Mm VanaHL Uflvlng'L FrWWl 5%??ij )JDT [M (L:)W— 1):, N _L ‘ ‘ \ - _\ b 303’ U?’ 1);, K [email protected]*[a+%1) £Fr(a+(1~brl)xkl‘”<)(<mw Ij)‘[t ] .. T; /< wagflw/ég/ 1- UNRfiEJTeX‘ E 3% x W93?) ...
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