132B_1_HW2_Solutions

132B_1_HW2_Solutions - EE1'32B Professor Izhak Rubin...

Info icon This preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
Image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: EE1'32B. - Professor Izhak Rubin Solution for Homework No. 2 Set-No. 4 Problem 1 Let F‘x(s) denote the moment generating function of the distribution function for X Fm) .—. ma“ ) 1 m __ _ 2 = / e‘sz-J‘21—3e 202(3 g) dz —m ' fa. = j“ 21 a —E1;§[=' 202 a’s)=+u’]: -9; Id 3 ’2 = j; 21 e Tizllr-(ss- 05)] +2M’s- «and: V x175 l [r-(#- d 8)]2 e _263 dz. 1| ”Ix- [O 'L': h I 2" “In, 5.2 h Using the moment generating function, the mean and the variance of X is calculated as follows: E(X) = -'li_rg zFxh) —- 2 3_a2‘2 -lim-d-e ‘2( p ) ll 1-».0 d3 _ —- 2 03:2 = - lim --(2p- -20 3):? ;( #1- ) 1—0 :: - ling —-(2p-—- ---2ar2 s)Fx(s). Since iim._.o fix“) 2 0, we have ELK) =: p 2 d2 1'7}le =- lim—F‘vfs‘l I {13 A : [WE—J: Ee-‘Ml‘ld: 0 A on 2 Lunch” Eeudx-i-fo e"”-g-e_”dz CI )1 on )1 _. _ {l-Ilrd / _ -(»\+:3= _ c z + 3 dz [m2 a 2 _ A l 1 “ 2 A-s Jul-5 ' Problem.2.4 . d - Em - glad—fins) ~_ 1' d) l +- l _, iflds? A-s A+s = —:imi[ 1 +—lL—] ”-02 0—3)? 0+5)“ = 0. a . d: - E(X') = 'lflgfiqu) :—od33'2 A—s A+s : limi[ 2 + 2 ] 3—02 (iv—s)“ (Ad-.1)a _ 3. _. A3 = Vor(X). Problem 3 Set q =1- p. Problem3.1 Fo'rm=1.2,---,m-landn=2,3,---. P(X=m,Y=n) = P(Y=n|X=m)-P(X=m) = P(Y—X=n—m|X=m)-P(X=m). Set 2 = Y—X. Then, the random variable Z is the number of trials until the second head comes up after the first. head occurs. Since a. geometric distribution has the memorylesa property, Z is governed by a geometric distribution and-is independent of X. Therefore, we have P(X=m,Y=n) =I P(Z=n—m}-P(X=m) {1- elq"'"‘" '(1- WP“ (l-q)°q“". for n =2,3,---. Problem 3.2 P(X=m} fl-qiom”. PH” 2 n) = Z“ -—qr}:'qr""'2 m=l = ('1 v 1H1 ‘ 4024"":- Problem 3.3 Since P[Y = 1) = 0. the conditional probability that X = m given Y for :1 =2.3.---. = n is defined onl: (1 41’4“" {'1 -1}(1- new I n-li for m=1,2.---,n—l. Otherwise, PfX=me=nJ = 0. Note that given Y = n. the random variable X has a. discrete uniform distribution in {1.2. - - -, n -1}. ' Problem 4 Recall the memoryleas property:r of an exponential distribution. - Problem 4.1 For an exponentially distributed random variable X with parameter A, P(X > t) = e'“. P X > 50. 'x > 30 Hit .1» 50) For > 30) e—LS -—_ €419 -O.G = e = 0.549. Est: nit: POI: >—5iJ—iux $36: PIX > 20}. Problem 4.2 P(X>t+le>t) 11 -II re The solution above shows that P(X :> t+ : | X > r} = P(X > a), (i.e.. the random variable X has the rnernorylese property}. P roblem 5-(Spragins 4.1} I) The can! cncoding of states. malogoug to that in Figure 421:. is: WW Ring- Gull Tilt- Lina in: Cat. in: in: in: 0 0 o o o 1 I a o 1 0 2 o 1 o 1 3 o 1 1 0 4 l o o 1 s 1 o 1 0 6 1 1 o l 7 1 1 1 ______ __ __ _ . "-7; _____________________ No‘m 1mi I 0H 7.4 t: ' T V “1.4 4—7.5 .6 new Listening . E. 1' T A few comments may clarify the state diam "Tn-o possible sequences benedi- ately following the calling party picking up thephone (taking it ofl‘ book) are shown: either thedial toneorthe userlistening mymmefirnfollowedbythemhamepencfingonhow quickly the caller puts the phone to his or her on). The same tinging signal has been assumed to be observed by both the cslling party and the calledpatt-ty.1 All conversations have been assumed to be initiated by the celled party talking first (eg. saying 'Hello"): although telephone company equipment does" not require this to hcppen, it is normal pro- cedure. Three possible results for an attempted cell are shown. a busy signal followed by the calling party hanging up. no answer followed by the ceiling party. banging tip. 'or. ”suc- cessful completion of a call. If a call is successfully completed, the two partiee are assumedmbepoumenoughthutheynevuanemptmnlkfimnlnneouslyifimhermneh always assumed to be talking dtning the conversation. to there are no periods with both silent. Solutions which do not involve these simplifying assumptions will be more com- I tytskestbepboneofl’hoohndstartslisteningwotethedlutone). Thedinltooeister- minntedbythesurtofdialingmotshown-indetnfl). Thisladstotheotherphonerinfing. with tetminstion of ringing when it is answered. Thebn'ef conversation described then ensues, with the eallingpartyhanging upsoonafieneeognizing thmheeelled party hes hunam Calling Party 0,—— Vain mm, W, Vain I Rx. 3.4} (1.410.: 1.5 u 5.: 7 .s ' ,4 _ ( )(H) () (.33 Ma) (7104) cured Party ma, .4 . 1. .4 .4 (3 1 (2.4) (H) ( 515”“! 0.5) a 1 a ) Prob-Kern figSPra-gin: 4.2) Although this pro-Elem can be wind readily by use of equations in Section 4 1'1. «nth appmpriate interpretation 'a'f-te'rms, it is intended L10 be solved di‘ractly. It sheuld he sokvable Immgdtately after treat-in; the material in Section 4.3. See the félltmt'tng firs-urn which illustrate: wavefurma seen by transmitter and retain:- af the data. message. {the figure is not drawn to sca‘laJ Aka the figure. Indicates. the receiver completes reception of the data manage {plus overhead) at a.- time equal to t, plus the time. (9M + na-JJR. req'n' 24+;§§3+ 3“ + 0.001;: 2 + 11020 + 0.019;: 0'.14117(secancls}.. This yields an e‘fl'ecté—ive data. rat-£4: of R. z lUODlDJilI-T = 7034 (bitafseaond}. ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern