ee302hw2soln_Fa11

# ee302hw2soln_Fa11 - 4 z {3 . é: , _ w ‘ y 2 jyx 553\$ 7...

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Unformatted text preview: 4 z {3 . é: , _ w ‘ y 2 jyx 553\$ 7 ‘ “M 0%» i (a J é w/V/(inﬁ 0’1“ V. \ *T uev;«,;£«,\/W“~Q , ; X) a J a»:{/~{..r<./ Problem a. The error process described is Bernoulli. Let K be the number of errors out of n operations. We have then HK=M=<\$pW=MWE fork=1,...,n. We have that type 1 errors occur with probability ap. Thus, type 1 errors also form a Bernoulli process with parameter op. If we let K1 be the number of type 1 errors out of 71 operations, then Pm = m = (am — am“, for k1 21,...,n. Similar to above. Let K2 be the number of type 2 errors out of 71 operations. n We = a = <<1~ a>p>k<1 -<1— comm-'2 for k221,...,n. We ﬁrst rewrite the joint probability P[K1=k1ﬂK2=k2] : =Pm=am=a+awmzm+m. 1 As mentioned in the hint, we will have that if we are given that K = k: 2 k1 + k2 errors occur in 71 operations, then each of the k errors are either type 1 (with probability a) or type 2 (with probability 1 —oz). Thus, type 1 errors amoung errors are Bernoulli trials. We then have that: P[K = kllK = k1 + k2] = (k1 2; k2) ak1(1 — or)“, and P[K1=k1ﬂK2=k:2] _ k1 + k2 [131 _~ [432 n kl—l-kz __ n—k1~k2 — ( k1 )a <1 a) Mb 12 <1 p) Problem 3 a. We have that the pmf pX(k) = P(X 2 k1) = P(ﬁrst k — 1 guesses are wrong)P(k—th guess is correct). When the key is replaced in the pocket, we will have that the probability of picking the right key will be 1/11 on each try, since we are picking uniformly random from n keys on each attempt. Therefore. 1 ’H 1 (71 ~ 1)’H Z 1 — — —— : _ fork=1,2,3,.... b. When the attempted keys are removed from the pocket7 we will have diminishing probability of choosing incorrectly as more and more keys are chosen. In other words, there remains to be a single correct key in the pocket, but more and more keys are taken away. On the ﬁrst attempt, we have that there are 71 keys, of which n —— 1 are incorrect. in the event that the ﬁrst key chosen is incorrect, there now remain n — 1 keys in the pocket, of which n — 2 are incorrect. Therefore: *6 >< Si N :2 N H 35 || /-'\ “:9 (Z:>---<Z:E:::§> (Mi—11> W‘";J2ﬂgf>' | l 3|“ fork=1,2,...,n. Problem 52% 3. Since [5000 fX(x)d:v = 1, we have 5 / keﬂdx = k:(e — 8—5) = 1 —1 1 k: = —-— :’ <e—e-5> b. 3 3 6—:1: P[1 3X < 3] 2/ fx(\$)d\$ :2 / (e_€_5)dsc 1 1 _ 6—1 _€—3 — (6—6—5) 5 5 8—2: 1 1 _ e 1 _ 6—5 _ (€- 6‘5) c. X3<52>X<51/3. 3 1/3 51/3 6% 6—1 — #51/3 PX <5=PX 5 2 —~—-—d =———- [ ] [ < 1 (e—ees) “3 (e—e-5> d. :1: e_)\ 07 17 _<_ —1 é/ : (:::::), "1< in S 5 _00 (e W e ) 1‘ it > 5 The PDF and CDF are plotted in Figure 1. 3 pdf CDF FX(:I:) Figure 1: Problem 3(d) Problem a 5 a. Since Z:_OO pX(£l7i)d:E = 1, we have P11£X<51 = px(2)+px(3)+px(4)+px(5) _ 32 1 2+32 1 3'3214+32 1 5_15 _ 127 2 127 2 '127 2 127 2 ‘127 C. P[X3<5] :2 P[X<51/3] 32 1 #1 112 _ 1 =— - — ”” _ :— pX( 1)+PX(0)+PX( ) 127 (2) +1274r 127 (2) 127 ll 0.8 Q7 0.6' 0.5 px<x> 0.4' 2:] PMF pX CDF FX Figure 2: Problem 4(d) _§:£1”_§2_4_1\$ " I _1127 2 _‘127 2 ’ 331: for x 6 [—1,5]. FX(x) = 0 for a: < —1 and FX(:1:) = 1 for :13 > 5. The PMF and CDF are plotted in Figure 2. Problem 5 a. True. _ PM D B Q C] _ PM]P]B]P[C] _ Pmmnq_ Pwnq _ Hmﬂm ‘PW‘ b. True. P]S ﬂ (0] 2 PM] = 0 = P[S]P[(D] = O P]SﬂA] 2 PM]: P[S]PM] = 1 PM] 2 PM] PW) H A] :2 PM] = O = P[(D]PM] = 0 PMoSﬂmzszzoszPumzo 0. True. PM ﬂ B] 2 PM] = 0, but PM]P[B] > 0 Since PM] > O and P[B] > 0. Therefore PM D B] 74 M]P]B]. 5 True. P[An]P[An_1|An]P[An_2|An m An_1]..,P[A1|An m . . . 0 A2] PM” o An_1]P[An_2|An m An_1]...P[A1|An n . . . n A2] PM” (i An_1n An_2] . . . P[A1|An m . . . ﬂ A2] 2 PM” o An_1 HA,” 0 . . . 0 A1]. ll False. By the binomial theorem: 1=QH%1—MV :(;)wu—pww. Therefore, (@pwvmrr=1—u—mn 19:1 False. fo>—-1F<) <1Pm<xl #ﬂX—rl X 0 — dag X33 “we dm _ mm — 0 True. P[a<X<b]=P[a§X<b]=P[a<X<b]=P[a<XSb]:/bfx(x)d\$. True. By deﬁnition, pX ((130) = P[X :2 x0]. True. By deﬁnition, P[a g X S b] 2: Emidmwpxwi). False. Strictly speaking, we have P[a<X§b] : FX(b)—FX(a) and P[a£X§b] = P[X=a]+FX(b)—Fx(a). If X is a discrete RV. at X = a, then P[X : a] # 0. Therefore, in general, P[a g X g 1)] 7E FX(b) — FX(a). ...
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## This note was uploaded on 02/14/2012 for the course ECE 302 taught by Professor Gelfand during the Spring '08 term at Purdue University-West Lafayette.

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ee302hw2soln_Fa11 - 4 z {3 . é: , _ w ‘ y 2 jyx 553\$ 7...

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