hw9 - SECTIflN 6.4 Expected Value and Variance it By...

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Unformatted text preview: SECTIflN 6.4 Expected Value and Variance it. By Theorem-2 the expected number MSW for ra Bernoidlimlale 151 rap; In the present problem wehm 11: 10 and 113135. Therefinetbe expected numberofeuocemee (Le, appeareneeeofa head} is 1i)-{1f2J= 5- e. Teammaeemeeeiee, maptthet p: ISLE. .Thua the expected member at beanie ie Iii-0.626. II= -p1e IE! we saw that the variance of the number of aeooesees in n. Bernoulli trials is ape. Here 11- _ 10 . il=1ffi and 9: 51%. Therefore thevarianee is Eoflfl. “all?! need to do is copy the proof 0! Theorem 7.1'eplacing some of two events with some of 11 events. u . gets only slightly meaeier. We will use Summation notation. Note that by the distributive Jaw we (I: leer tiled-22 135a,. i-1 igiejsn WM?(: .1 £111“) 4131.11)“ ' ' that tone-we Iflom algebra and linearity of eJ-ipectation 'Ilt V(ZX1)=EE[}:XE +2 2 Mg) — (:EHL'JY. i—l 1511:3511: I } I'l- ' - = -2 - tumour; . 'X " Ell-X1) 2 5 XEEX?) + IEEECHELXI- If) E I lgicjgnI I mum that l: ‘ ' . t 1 die' int- by Theorem 5 we have EILXiXJ-J = ELX1J.II'3(XJJ s are pairwise JO , 11. n __ '62 = n Vtx'l}' (12" ’2 Because til? Wt . I51 .. -Lrn .I\_. _.I'l9fil1 HEELS-tn- : Barn pimle thla by doing tonne algebra on the definition, using the facts [Theorem 3} that the expectation sum [or difl'ereneeJ is the etun {or difierenee} oi" the expectations and that the expectation of a constant ' a random variable equals that constant timee the empeotation of the random variable: II‘Z‘WKX_YJ=EIE{X— Elxll' {F- ETD):E(XY-1’-'E[XJ-X'E{YJ+E{X}'EG’D = Eon/1'— E:YJ-‘E1X)— Epo- at) + Eon Em= EKXY)— Eon-1.1m and Y ai'eindependent, thmbyThmrmnfitheeehsttmItamemethesemeaotheir dilferenoelsll. use the reeult ofExarelee 38. It' is easy to ace that ELK]: 7 and Efl’J- fl '1' {the Example 4J. 'Ib find ation of KY, we construct the foflowhig table to Show the value of 2i{i +j) for the 3B equally-likely 3 (i la the row label, 1 the column label): 1 2 3 4. 5 e 1 4 e e 10 12 it 2 12 1e 2e .24 as '32 3 24. en ea 42 as 54 4 411 411 55 e4 12 en 5 to 111 so a] 1m 1111 e 134 911.1113 1-211 132 144 :Value of X? is therefore the aura of these entries divided by Bfi,namelyI—I197:l.fflfi — 329,26. (minimise 329,!6— 1'_ 1:35/e~ ea WWW: Wwflta a ' .jh n._'.1- . _ “EWMOEWIJImnm-ita - ta” ' ' I. qhnf 01:— 'I‘hmphfgfilln for. ”L and 5° 0“— Whfll we reach $123 ofthrf ““72 {usmthflmurrence ' Him us an ' ' pl'Oceflum! - Emphmt formula £91- the answer or it will give us a biggie Erwin]: 3 W $9 obtain an aacplinit formula for the ammu- ufl =_ —n + (In—1 6) = -n. + (—(H 41} + tin-2} = "{H + {“1 ' 1)] +Iflfl-3 —.- -—[n+, [n -— 1}) + (—(n'— 2) +G-n—a] = —{fi + K“ ‘_' 1] + [11" 2)] +“n73 =_.{n+(n—1)+[n—2)+---+lfi-{Tl—’13))"FEW-'1' s—[n+[n—-1]+(n—-'2]+'-'+1)‘lffl-n d) “11. = -3I+2Gw-1 == —-3 + Eli—3 + Zfln_-3:I -—— '3 + 2(fl3} + 40...;3 .- = —3 + 2931+ 4(—3 + 2am} = 4 + 2(4) + lit-3) + Saw-'3 - = —3 +2(—3) + 4l— 3) +' a(-3 + an“) = _3 + 2(—3) + sq—s) + 8[-3) + 15.3“ _ = -—3{1+2+4+---+2""1}+ 2%.... =-—3{2“ - 1) +2"{-1) = 4“” + a e)- on = (n+ nee-3-1 = (n + man—2 ' 5 {n + 1mm - l) (n - flan—4 =—- {n + 1}nl‘;n —- Dan-3 5 (n. + Unfit a my: # 21m 43) - - - (n — [a — 2)) “H = cn+ lhnln r W — ZKH- 3') -- =(n+1]1-2=I2(n+1)1 -2+fl.n_ f) an = Emu—1 = 2n[2{n -— ljunflg] =_21{n{n —- lflaflra = 22bit: — 1]] {fin — mama] = 23' (uh: —_- 1)[n -— 2)] and = 2530”. —- 1}[n .—- 2)(n -4 3] - r-{n — GI. - ”hm... =2“n[n—1){n—-2}(n—-3)---1«In ' = 3 -'2"n1 g) an=n-—1—a.n_1- . 3:01-11 *(fl—3}+Gn—2 zfirl-(Eflf1‘1}‘ufl-3 =(n—1)_—(n*2)+{[n—-3-1}—an_3]={fl—1}—{n—_-2)+[n—-3)—an-a = (n. g 11— [n —-2}+---+[—-1}"_1(n — n)+{—1)"a._1_fl '—-1 .— n =EL§LEL+(._1)".7 23. This is identical to Exercise 27, one level deeper. a] Let an be the nmnber of ways ire-climb n. stairs. In'nrder to climb is stairs, at hereon must either start with a step of one stair and then climb a — 1 stairs {and this can be done in a.._1 ways) or else start with a. stepoftwu stairs andthsn climb 11— 2 stairs (andthiacsn be'dons in «3.3 ways} orelsestartwith eaten nfthreestairaandthen climb n— 3 stairs {and thisean bedmrein a..- 3 ways). Ramthis anslysisweean immediately write dawn the recurranflla relation, 1raliqi for all :12 3: [13:13»: +ma_2.+a3_3. _ b) The initial conditions are so = 1, a1: 1. and a: = 2,-sinee there is one way tnelimb lie-stairs (do nothing), clearly only one way to climb one stair, and ties ways to climb tam stairs (one step twice or two steps at ones). Note that the reenrrenee relation is the same as that for Ermreise 25. a) Each term in our sequence :{an} is the sum of the previous three terms, so the sequence begins :13 = 1, a1=i., a; =2, s3=4, a4=7, n3=13, s3=24, {37:44, (13:31. Thusapersan canelirnbaflightuis stairs in 31 wags uncler the restrictions in this problem. “ted the characteristic equationsnd findits mats. Usingthiswewrgir; mm 6 '1 BE in the initial conditions to obtain a when: of linear equating; the arbitrary constants in the general militiamen and finally we as dflfifl uniqueanswer- ' - . _ ' (Hr: —2r+1——.D 3:1’1 a)r2—r-fi=fl ‘-"="2=3 . fln=flllfl+fl2nlnzcx1+am un=_w1(—3l“+”93" 4=a1 3=e1+fl£a ' 1~-a:1+33,a s=—2n1+3a2 _ {31:4 =_3 ale-sis mar-1W5 .' a3=4—-3s' M=(3’{5}[_-2}n+|:12f5)3“ EJ r2._1=fl i"=—1, 1 ' -fl-n— Ci“ —1"' ’1: n b}r3~?r+lfl=fl' r=2,a' s 3:111:35”! HIE 1}+ _fla=a12"+'cu35“ 2=fll+fls 1=hl+5flg fl1=3 marl—1 flfl=3,2n_5n “Jr'a—flr+$=fl 1-:2’4 1a. This is e. third degree reeuneeee relatlfln. The chemeterietla equetienie r3 - Be“ + 1n e e -_.— a: By the rational roe-t test, the pomihle rational roots are :Ei,:I:2,:i:4. We find that r = 2 is eroot. Dividing r- 1 into 1-3-52: +1251—B, wefind that ra-flr?’+12r—-B = (1—52)?” fi4r+4ji Rhine-pectin wisetar the-rest, obtaining r3 — iii"2 4- 12:- 8= {r - 2313. Hence the only root is 2 with multiplicity 3, so the general salutian is Ilby Theorem 4} an = £112" + m2“ +3351“ 'i .Thfind time eoeifltientst we plug 111 the initial annditiout 'wh: au=a1 4t=n1=2a1+2u3+2a3 J I ' 88¥m=4fil+flm+lfiag. Solving- this system of equations, We get m = *5, as = 1(2, and on: 13:2. There-hare theansweria e,,==es-ene-{efe}-2fl+_[1sefi,12)-2II=~5-2“+n-5I“41+13e3~e"-1. S. a} BytheEinominl Theorem {the third line afTable 1) weget arm= C(3, it) for In=0,1,2 3, and theather eoefliclents are all 0. Alternatively, we amid just multiply out this finite polynomial end note the nonzero coefficients: 013:1, {12:3, e4=3, egzl. b) Thieie likepert {a}. Flrsthe need to feetor out —1 and mitethiess —[l —3.1:}3. Thenbythe Bmornisi Theorem {theeeoandlineafTahle 1:1 weget a“ _='—{313, n)[— 3)“ far n= [1,-1, 2, 3, and the othereoefiicients are all 0 Alternatively, we can Id (by hand or with Maple} just multiply out this finite polynomial and note ' the nonzero eeeneieeeee' as = 41, e, = e, n: = —27, e3 :27. a] This problem requires a. eemhineeen of the results of the sixth and seventh identities in Title 1. The oneflieientaffl 132“, endtheoddoaefieienteareallfl. _ . a) We time that I=,I‘(l— 1:)3= z“ 2°10 C(In+2, 23st“— — 23;” U(n+2,2)1'“+2= Siamese“. Therefore nn=0[n,2}=n{n—1}f2forn>23nd flg=dl=fl (Actually,sinee 0(1), 2): C(1,2)=fl,mreellydnnlt - needtomsheeepeeialstatementfinrn<2J e]Thelesttermgivesus,fi-om‘1hblel, anal)“. Wenmdtoadjustthisforn= Dendn=1beeeuseo£the firsttwaterms. ThueaD=—1+3°=U', and e1=1+31= -4 .Q We split this into two parts and proceed as in pert (d): (1+1e)3 {1 + E‘s—Ts 2(5) ”“5"“ + “)3“ + a" Zi—ilwme 2, etc“ __Z(—1}fIC{e+e, me" + Z[—1)“C(n+2, met-ii - n-fl . = 2(1—1 )“0{n+2 2)s“+Z(—1)”'3I5'(n-1 3J3“ ’ 'Natethat n and 11—3 heme opposite paritim. Thalefare an =(— 3,1]""C[n+2 2) +(—1)“"30(n—1. 2)= 71]“ (Chat-2, 2) (III?! — 1 ,2]]= {—1]“3n for vi 2* 3 and in, =[— l)"6’{n+ 2, 2): (—1}”[n+2}(n+l);’2- it < 3. This answer can be confirmed using the series command in Maple. ' . Ehahey here' is to renal] the algebraic identity 1 —illa= [l- n}[l +z+n3). Therefore the given function Ehe renaitten as $[1— 3:)[11'1 — 1:31, which can then be split into zfll -— 2:3] plus —:t3j{1—.1:3]. From __ 1 mines; that 1311—35] = 1+:a +:i:fl I+239+- .Therefine 3/(1 —1:3]I =r+14 +27+21"+-- :33 I'll $3}=-:I:3— 3:5— 4:3—— 9:11— .Thusweseethsrt in, is [i when it ieamultlpleoffl,1t1el. THE 1 greater then a mulfipleef 3, audit is —1 when n is 2 greater theha. raultiple of 3 Onenen tiasenswerwithfidhple. . ‘Ihble 1 we knew that eIII = 1 + e + maffl! +£3r‘r3! +-- .11: tenses that ' ’ ' (3:9)“ {Ea' )3 s! + _31 + - read off the coefficients of the generating function for e359—1.'First, clearly nn= Ill. Second, it is odd. Finally, when n is even, we have a2", = Emj'mi. e331=1+3m2+ 1.4.3:, 1501' i= 1.2... “n. be the expnnent of 3 taken from the ith fisctsr in forming s term :1," in the- ' Thusel+eg+r -+c.,-=_fi. Thenoeifieientofmfiisthereforethemunberofwuystosolvethis tips with megstive integers which, from Section 5 h is" 16(11 + 6— 1,6) = C(n + h ,6). Its value. of depends on 11.. 3|]. :1) Multiplisstinn distributes over addition, even when we are talking about infinite sums, so the generating function is just 261$]. , In) What used to be the eoefiicient of :5“ is now the eoefleient of m1 and similarly for the other terms. T113 way that happened is that the whole series get multiplied by 3. Therefore the generating function for $11“ series is 113(3). In symbols, oon+a1x2+agms+---:n:{nu+e1:r+a2:nz+w)=39|{x)! c} The terms involving an and 1:1 are missing; Gfim)— an — 131:1; = G232 + (134:3 +- --. Here1 however 1"“ went at to betheesefieientofs“, not :1:2 (sndsindlsrlyfisrtheotherpowers) sowemustthrawinanflfiflf factor. Thus the answer is a aTIE-1:13) — an — sis}. is. ‘ cl) Thls is just like part (12') except that we slide the powers deem. Thus the answer is [G{r:]— au— mill-1': .E e) Fhllowing the' hint, we differentiate G(:c]= minus“ to obtain G'{a:] = Ziunsnnhl. By it change of variable this beeemes Eff—tit+ 1]nn+1z“=e1 + 24.1532: + 31133.3 +- -, which is the generating fimetlon for premiselj.r the sequence we are given. Thus G'[:::) is the genereting function for this sequence. t‘) Ifwe looket Theorem 1, it isnot hardtossethst thesequenes shown here is preciselytheeoeflisients of 9(5) - GTE) . La- 6(a) = EEO cite". Then :tG[n]= I 22:0 ath1= —E::Ie ohm" [by changing the heme Iof the variable from it to i: +1)I and flee): Erintmflz_ -— E"; flk— 39:". Thus r I -— 0(3) — 533(2) - 213th}= Zeta?“ —— gable" —- Z fiflk_21‘k = nu + ale: — 1301: + E 2'" as" . ' ' e=u -k=1 11:: ' _ e=2 . ' 4 — 12:1:2 1—2x‘1“?m=_ 1—2.t‘ me of the given recurrence Deletion, the initial conditions, Table 1, and sigehrh. Since the left-hand side - equaled-in factors as G[z'](1 — 23}[1 + e}, we have Sh?) = (4 — 1233),! {(1 +'3}{1 — 2:]3}. At this point _ Just use partial Emotions to break up the denominator.__3stfing ‘4—1st3 _A+B+ G" [1+n]{1—2n)“~_ 1+2: 1 —2m (1—23)“ -. 'ms through by the common denominator, and equating eosflieients,_ we find that A_=' —8}9, B = C = 233. Thus git): 1-:f_:_,_138f; u—E’Efl—B=E(—S —{—:J"+:—3 -2"+-(s+112")t* =1: +st+ 1]. ‘I‘hersfcu'e flllt— — {— 3f9][—-1)h + (38(9)? + (2f3){.t + ink. Incidentally, it would lie-wise to ' answers either with a computer algebra. package, or by computing the next term of the sequenoo the recurrenos end the formula {here 43— _. 24-botl1 ways). 4-“- 3133’ Mi“ ' (4x2) (—uzurszfl—Efil H?" * 1W 11. -(27L— 1} =( llnl‘jfifr 5.42541} 2-‘4-6-(25'31'. -_— (”WIT ' We! 275i = [- ‘1)“; n'lzfl “all“ =E— 1)“(25")1t=(2:)T—11F (Theaters 2), with -4J:) in plaoe of s: and u = -1;’2, we have b) By the attended Binomial Theorem . (i—dmi‘l’z—E(5 1;? his)“ 25; 4)“? WEZGDIH' nFU n=ll 50. 8.} Since all 4" bhse-fourstringsoflengthn tellintooneafthefour estegeries coumedbjr on, hm (:5,st d",DbfiDl.lflly dnzntflhmfl—bn——q5. Nextie‘t’sseehowsstringofmious typesofiength n+1 eenbe ohtainedfi'omastringoflength-nbysddiogonsdigit.Togetestringoflengthn+lwithsnesennumber offlsandnnevennumherofls wecantekeestriugofleugthnwiththesssemepsritiesendeppende r2me3(thustheresr92can suchstringsofthis‘type},orwemntakssstringoflengthnwithaneven number ofOesndsnodd numberoi'ls end appendel (thus thereare b5 such strings oi’thistype),orwe can take a stung oflength n. with an odd number :1st and uneven number of Is and append all (thus there ares-,5 suchstringsoithistym). Thereforeweheves55+1- 2a,,+b,5 +_c55. Inthessmewsywsfiudthst b5“ = 25» +fln+dm which equals 315—65 +4“ after substituting the identity with which we began this solution. Similarly, c5+1=2c5+a5+d5=cn— —EI,,_+4“. ' _ b)ThestflugsoflengthiereU 1, 2, 2:11:13. So'seasyafiz m: =1' and a]: u. {Netsthetfliseu cvenuumber.)lnfactweesusleosuythet.a5=1{theemptystr‘hrg)endbn=cg=da.=fl. c) We apply the recurrences frompert(e) twice; _.;5— 22+1+1=s a5_=2-6+4+4:20 52:1‘1+4=4. 'sa=4+is—4=16 m=1—1+4=4 m=4+lfi—4=16 s5=1s—- s— 4 4: 2 s5= 54— flfl—lfi— ls=12 11] Before proceeding as the problem asks, we note a shortcut. By symmetry, iifl must he the some as c... Substituting this into our recurrences, we find immediately that ii“: cn— — Lin—1 tor n I:- 1. Therefiore on: 2on4 +104. This resurrenee with the initial condition ul— — 2 can easily be solved by the methods 'ofeither this section or Section 7. 2 to give 055— — 2"“1 + 4“ 1 But let's proceed as instructed. 5., _ Let Aim] Bit]. and 0(3) be the desired generating Emotions. Than 554(5) _ 23:65H5w1 = _.E,.=10n—11“ and 8111111le for B and C', so we have A{z)—sB{er]—1:G(m)—21‘=A{3J 2mg: —Zt_1z —Ec5_1"$ —Z:s5_ 1: = _ =1. ' 'I'I-Il ' n-l I B(s}I—zB(m]+sC(s}= Zine“ —Zb5_ 1s“ +£43.41" “:1 11:1 . n-l n n 3 =bo+24 a: =D+m§4zs =1_4¢- llV'lDllBljl' G sstisflmthesameequsflon. Therefore oursystemofthree equetmns (suppressing the arguments B,_and (7)13 _' . . (1—2s)A—:tB'—r(7=1 - ' I: [1_—:t)B+zO=1_4$ wB+(1— {10: a: 1143' 415:. Wanting the third “I‘m“ In part (I!) from the Hand film the semfid equation immediatelyfl 53qu W8 that 3:11.33ch than plugging that. -.- film into the first uquniion yields I (1 —3$]A.— Ew- '1 ' —1 far A aim US 1— 4:1: ’ 3(3) = fl_ (1—33“: - 4:1 Now that we Im'ow the génamting fumfifmnsz.J we can recover the wafiicients. For B and C (using Table 1] = (L- We rewrite ALI} using paztial mimmadlatelygetnooefflcientofldfil forall n2 1,with bg=cg fiastmaa I ,' 1 . A(m}— 1 1:2 ,1 sowehava an=§-2“_-|-i.4"=2“-1+4"*1 for n31,w1than=§+%+i=1. ...
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