solutions2_chapter2

solutions2_chapter2 - 2-4 Solutions Manual for Statistical...

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Unformatted text preview: 2-4 Solutions Manual for Statistical Inference 2.9 From the probability integral transformation, Theorem 2.1.10, we know that if M56) = F9568), then FAX) ~ uniform(0, 1). Therefore, for the given pdf, calculate 2.10 a. b. 2.11 a. b. 0 ifccgl “($)=Fm(x)={(x—1)2/4 if1<x<3 1 if3§ac We prove part b), which is equivalent to part a). Let Ag 2 {w : Fm(a:) S y}. Since Fm is nondecreasing, Ay is a half infinite interval, either open, say (~oo,xy), or closed, say (—oo,my]. If Ag is closed, then FM?!) = 130’ S y) = P(Fm(X) S y) = P(X E A,) = 111051;) S y‘ The last inequality is true because my 6 Ag, and Fm(x) _<_ y for every x E Ay. If Ay is open, then Fy(y) = 1’0” S y) = P(F1(X) S y) = P(X E Ag), as before. But now we have P(X 6 A,,) = P(X 6 (— oo,m,,))=1$ir1;1P(X e (—oo,m]), Use the Axiom of Continuity, Exercise 1.12, and this equals limmTy Fx(x) S y. The last inequality is true since Film) 3 y for every 3: 6 Ag, that is, for every x < my. Thus, Fy (y) S y for every y. To get strict inequality for some y, let y be a value that is “jumped over” by Fx. That is, let y be such that, for some my, limFX(ac) < y < Fx(my). am! For such a y, Ag = (—00,36,,), and Fy(y) = lirany FX(x) < y. -12 Using integration by parts with u = II) and d1} = xercc then °° 1 ,sz 1 _m2 °° °° *2 1 EX22/ (132—6 2 d3: — —-a:e 2 +/ e 2 dzc 25—(27r):1. —OO _00 27r : 27r -00 Using example 2.1.7 let Y = X 2. Then Therefore, EY /00 y eLzldy —-1 [ 2y%e;zfl°°+/w ’71 :2Kdy] —1 (V2) 1 : r: — y e : 7T : , 0 V2719 0 M271" :1 This was obtained using integration by parts with u = 211% and d1) = %e 2 and the fact the fy(y) integrates to 1. Y = [XI where ~00 < :c < 00. Therefore 0 < y < 00. Then Fyly) = P(YSy) = P(lesm P(—ysX = P($Sy)-P(XS—y) = Fx( |/\ 1/) II S, I Q? I S Second Edition 2-5 Therefore , d 1 —y 1 —y 2 v3 fly) dy v(y )= fx(y)+fx( 9) fire 2 + 2W6 2 7Te 2 EY=/Oooy\/§c;2udy=\/§/Oooe’[email protected][HB—"lgol = 3; whereu:—. I / flew—131w? w] m1 This was done using integration by part with u = y and dv 2 ye 2 ’fldy. Then Var( (Y—) — 1 — 9— 2.12 We have tanx = y/d, therefore tan 1(y/d) = ac and Hd—y tan 1(y/d)= md 1dy— — dzt. Thus, Thus, 2 1 fi/(y) = a ——-——1+(y/d)2, 0 < y < 00. This is the Cauchy distribution restricted to (0, 00), and the mean is infinite. 2.13 P(X = k) =(1 —p)k 12+]? (1—— 19,) k “—1 2. ..Therefore, EX 2 Zk[(1—p)kp+p’°(1—p)l = (1—17)}; [Zk(1—p)k‘1+2kpk-1 16:1 16:1 16:1 _ _ i 1 -_ 1—2p+2p2 “ (1 plplpz <1—p)2l “ pu—p) ' 2.14 /oo(1‘FX(w))dx == /OOP(X>ac)dgc 0 0 : /0 / fX<y)dydw = / MW = foo ny( yd?!) = EX, 0 where the last equality follows from changing the order of integration. 2.15 Assume without loss of generality that X g Y. Then X V Y = Y and X /\ Y 2: X. Thus X + Y :- (X /\ Y) + (X V Y). Taking expectations E[X+Y]:E{(XAY)+(XVY)]=E(XAY)+E(XVY). Therefore E(X V Y) : EX + EY ~ E(X /\ Y). 2.16 From Exercise 2.14, -—ae“)‘t (1 — a)?“ a 1— a ET 2/ aehM+ 1 — a 6““ dt : = g 0 [ ( ) ] A M 0 A M 2—6 Solutions Manual for Statistical Inference 2.17 a. fom 3.70%.): = m3 32‘; s m = (am 2 .794. b. The function is symmetric about zero, therefore m = 0 as long as the integral is finite. 1 0° 1 1 1 0° 1 7r 7r — 2— ~ . =— — — =1. 7r/_001+$2d$ ”tan (I) 7r (2+2) This is the Cauchy pdf. 2.18 EIX — (1| 2 foo In: — a]f(;r)d3: :. film —(x —ya)f(9c)dzr + £3061: ~ a)f(a:)dac. Then, '—00 d a 00 se EElX—a|:/_oof(x)dx—/a f(;c)dm £0. The solution to this equation is a = median. This is a minimum since dZ/da2E|X —a| : 2f (a) > 0. 2.19 %E(X — (1)2 Edd /00 (ac — a)2fX(:c)dx = /00 —(a: — a)2fX(x)dx _00 ~00 da ll /00 —2(:r — a)fX(a:)da: = —2 [/oo: fo(x)dx -—- (if—C: fX (m)da:] -00 1.. = ~2[EX — (1,]. Therefore if adEE(X —a)2 = 0 then —2[EX —a] = 0 which implies that EX 2 a. If EX 2 a then dezE(X at)? : 2[EX a] : 2[a a] = 0. EX 2 a is a minimum since dZ/da2E(X — (1)2 2 2 > 0. The assumptions that are needed are the ones listed in Theorem 2.4.3. 2.20 From Example 1.5.4, if X = number of children until the first daughter, then P(X = k) = (1 —P)k‘1p, where p = probability of a daughter. Thus X is a geometric random variable, and EX = Elsa—my“ 1p = p— d_(1..p)k = —pd— [SQ—Mk4] k=1 k=1 p p k=0 d 1 1 = — — ——1 = —. poll) [1) i p Therefore, if p = % ,the expected number of children is two. 2.21 Since g(m) is monotone 00 00 ‘— d — 00 E900 = / g<x>fx<x>dx = / yfx(9 WW 1<y>dy= / mom/2m: where the second equality follows from the change of variable y 2 g(x), :1: : g_1(y) and div = 3%9"1(y)dy- 2.22 a. Using integration by parts with u = a: and do : {re—”02m2 we obtain that 00 2 ‘2 2 2 2 00 2 2 / are—“3 /fi d3: =—/ 6—95 #3 dm. 0 2 0 The integral can be evaluated using the argument on pages 104-105 (see 3.3.14) or by trans— forming to a gamma kernel (use 3/ 2 «A2/fi2). Therefore, fooo (3‘9“2/[32 d3: = fifi / 2 and hence the function integrates to 1. 2.24 a. 2.25 a. 2.26 a. Second Edition 2—7 . EX 2 2fi/fi EX2 :’ 352/2 VarX : 52 [3—4] . 2.23 a. 71' Use Theorem 2.1.8 with A0 = {0}, A1 = (»1,0) and A2 = (0,1). Then g1(x) = 1:2 on A1 and 92(30) 2 11:2 on A2. Then 1 _ fY(y):§y 1/22 0<y<1. .EY : fol yfy(y)dy = — = fol yzfy(y)dy 2% VarYz %_ (2)2 = 7143' EX: [01 27am“ 101:1: M 30 axadx— — “(1:411 2: : #1: EX2 : 01x2a17“ 1dac 2 fol axa+1dx = “:1: 0 3 fl 2 VarX ~ a—+2 _ (m): = 2+2—><2+T>7 g _ 1 _ 1 n(n2=+1) n+__1 VarX : (22+1)((:2n+1)l _ (%1)2_ _ 2222 +63n+1_ n 2+inil : 222131 . EX 2 fozxg (a: -— 1)2d;c = gf02(:c3 — 21:2 +a:)d;c _ 1 EXQ: 0:m2%(27~31)2d:c=—3—f2( 4—2a:3+:r2)dx:§. VarX= ——12:: Y = -X and g 1(y) = -y- Thus fy(y) = fx(g‘1(y))lzfgg‘1(y)| = fx(—y)| - 1| = fx(y) for every y. Uilw . To show that M X(t) is symmetric about 0 we must show that M X (0 + e) 2 M X(0 — e) for all e > 0. oo 0 oo MX(0 + 6) = / e(O+e)me(x)dsc 2 / eewfx(ac)dx +/ eewa(x)dx _00 ~00 0 oo 0 oo = / e€<~$>fX(—2)d2 + / e€<-$>fx(~12)d2 r. / e-wfxmmx 0 ~ ~00 00 00 = / 6(0—6)mfx($)d112 : MX (0 — 6). There are many examples; here are three. The standard normal pdf (Example 2.1.9) is symmetric about a a: 0 because (0 —~ 6)2 =(0 + e)2. The Cauchy pdf (Example 2.2.4) is symmetric about a = 0 because (0 ~ 6)2= (0 + (3)2 . The uniform(0,1) pdf (Example 2.1.4) is symmetric about a = 1/2 because f((1/2)+6) =f((1/2) 7) 2 {a 1130 < 6 <1. if 1 < e < 2oo /00 f(a:)d;r = /000 f(a + e)de (change variable, 6 r: a: —— a) = /oof(a—E)d€ (f(a+e)=f(a—e)foralle>0) 0 (Change variable, a: = a — 6) ll \ Q T: :3, &. H 2-8 Solutions Manual for Statistical Inference /~:Of(:r)dzc+/:Of(w)dx=/_:f(x)dmZ /_;f(x)d$:/lmf(m)dx=1/2, Since it must be that Therefore, a is a median. c. EX ~ (1 = — a) 2]: (a: — a) = /X (ac — we) )dx + fmf (m — a>f<x>dx ‘55 “00 = / (—e)f(a—6)de+ ef(a+e)de 0 0 With a change of variable, 6 = a — x in the first integral, and e = a: — a in the second integral we obtain that EX —~ a : E(X —— a) = —/ 6f(a — e)de +/ ef(a — €)d€ (f(a + e) : f(a — e) for all e > 0) 0 0 = 0. (two integrals are same) Therefore, EX = a. d. If a > 6 > 0, f(a _ e) : e_(a_e) > e_(“+6) : f(a -|— 6). Therefore, f (1:) is not symmetric about a > 0. If —6 < a S 0, f(a — e) : 0 < 6'01“) 2 f(a + 6). Therefore, f (at) is not symmetric about a g 0, either. e. The median of X = log2 < 1 = EX. 2.27 a. The standard normal pdf. b. The uniform on the interval (0,1). c. For the case when the mode is unique. Let a be the point of symmetry and b be the mode. Let assume that a is not the mode and without loss of generality that a = b+e > b for e > 0. Since b is the mode then f(b) > f(b+€) Z f(b+26) which implies that f(a — e) > flat) 2 f(a+ e) which contradict the fact the f (51:) is symmetric. Thus a is the mode, For the case when the mode is not unique, there must exist an interval (371,322) such that f(x) has the same value in the whole interval, i.e, f(:c) is flat in this interval and for all I) E (171, $2), b is a mode. Let assume that a §Z (x1, 332), thus a is not a mode. Let also assume without loss of generality that a z (b + e) > b. Since b is a mode and a 2 (b + 6) ¢ (931,272) then f(b) > f(b + e) 2 f (b + 26) which contradict the fact the f (.76) is symmetric. Thus a 6 ($1,172) and is a mode. (1. f(m) is decreasing for a: 2 0, with f(0) > f(:c) > fly) for all 0 < a: < y. Thus f(a:) is unimodal and 0 is the mode. Second Edition 2—9 2.28 a. #3 = /’00 (1r — a)3f(:1:)d:c = /a (x — a)3f($)dac + /Oo(x ~ a)3f(m)da: 0 00 = / y3f(y + (Ody +/ y3f(y + a)dy (change variable y = as — a) _00 0 ll [)0 —y3f(—y +a)dy + /00 y3f(y +60% 0 0 = 0- (f(—y+a)=f(y+a)) b. For f(x) = 6—37, 111 : 112 z 1, therefore a3 = 113. #3 z / (a: _ ”36—17013; :: / (303 —~ 312 + 3m — 1)e“xdx 0 0 : r(4)~3r(3)+3r(2)~r(1) 3!—3><2!+3><1—1 :2. H c. Each distribution has 111 = 0, therefore we must calculate 112 = EX 2 and 114 = EX 4. (i)f($)=\/——,T€_$2/2a M2=17 114:3, (1423. (ii) f($)=% 1<x<1 112—5 “4:1, 6,42%. (iii) f($)=%e lml ,—-oo<:z:<oo, 11222, 114224, 044:6. As a graph will show, (iii) is most peaked, (i) is next, and (ii) is least peaked. 2.29 a. For the binomial EX<X—1) = Zw(x—1>(:)p$<1—p)"-x 93:2 : ’I’L( _1)2n ”—2 w—21_ n—z‘ n 10 El m )1) ( 1)) 71—2 = n(7”L—1)192y:0(TL;2)19y(1mp)" 2 y=n(n—1)p2, ‘ where we use the identity 33(96 ~ 1)(Z )— — n(n — 1X”; 2), substitute y = :1: — 2 and recognize that the new sum is equal to 1. Similarly, for the Poisson 00 —A x 00 «A y EX(X—1)=Zm(x—1)ex')‘ 21226; :12, 36:2 ' y=0 where we substitute y z a: —- 2. b. Var(X) = E[X(X —— 1)] + EX — (EXP. For the binomial Var(X) = n(n — 1)p2 + m9 “("192 = np(1 — 19). For the Poisson Var(X) = A2 + A — A2 = A. (a+b—) 1 n TL _ 1 (a+b— 1) CL :Zyya _1_ a(Z )———_ (n+a+b—) )1 _ 1) (a + 1) (1) ((n—1)+(a+1)+b—1) y+a y=1 (y— 1)+(a+ 1) 2-10 Solutions Manual for Statistical Inference " a n—1 (32-1) ;n(y*1)+(a+1)<yw1>_(ln—-W_lj (y~l)+(a+1) ll = $(afi—1) i: a + 1 (n — 1) _(a+;:l_1) (”+1121“) <3 —1>+<a +1> y -1 (“31311335 nu __ a + 1 (n — 1) (“iii—1) na h ' 7W ‘ 7 *7 ( n (j+?a+1) ) a + b since the last summation is 1 being the sum over all possible values of a beta-binomial(n — 1,a + 1, b). EY[ (Y — 1)]— — W is calculated similar to EY but using the identity y(y ~ 1)(Z )— — n(n — 1)(:::) and adding 2 instead of 1 to the parameter a. The sum over all possible values of a beta-binomial(n — 2, a + 2, b) will appear in the calculation. Therefore nab(n + a + b) Var(Y) = E[Y(Y — 1)] + EY (EY)2= m 2.30 a E(e ::) = 1‘06 625$;de Lam‘s 2 16m_ 11 = _1_(etc_1)‘ ct ct at b E(e X2) foe}? 2x 8md$— — C2372 (Ctetc — etc + 1). (integration—by—parts) c. _ t a ( Vfi 0° < )/fi E837 : /OO _1_ew— a €t$d$+/ ___e—m— a etmdx ( ) 2B a 23 06 00 : Baa/,3 1 ex(%+t) _ea/fl 1 e—m(-}§~t) 2n (2+1) m 25 <3 - t) a eat ‘3 - < t < . 1—6212” W W It 01. E(etX) = 22:0ew(7"+: 1Wu —p):1——’"p 22°: 0 (”2 1) ((1 —p)et) . NOW use the fact a: ’I‘ that 2:020 (T+:)1 ((—110)e°) (1p)et) = 1 for (1 —p)et < 1, since this is just the T sum ofthis pmf to get E(et X): (;—(:———_ (1_ p)? > , t < —log(1—p). 2 31 Since the mgf 1s defined as MX(t ) X, we necessarily have MX (0): E60 = 1. But t/(l — t) is 0 at t— — 0 therefore it cannot beE an emgf H 2.32 d d 11—142 t X , as“) t20 I E (log(Mx(t)) t=0 : d§w$(f)) t=0 = ET 2 EX (s1nce MX(0) :2 E60 2 1) d_2 _ 1 MM Mz<t>MZ<t> -~ [MM]? 3125“) 2:0 ‘ dt (M2051) i=0 [M20012 2:0 2 2 ~——————1 -EX21—(EX) = VarX. 2.33 a. MX(t) =Zg°=oem e?” :. e—A 2:0 1L A)”_e—161et : 6121.1)~ $1 (1 e — _ EX : me(t)lt=0 = meMt 1Met t=0 — A. C. 2.35 a. Second Edition 2— 1 1 EX2 2 $2Mx(t)‘ 0 2‘ Maw-1)Aet+AeteA<e‘-1>‘ 0 = A2 + A. t: t: VarX = EX2 — (EX)2 = A2 + A _- A2 = A. 00 OO MmCt) = Z emml —P)$ = 192 ((1 *p)et)$ 11:0 w: 1 p 1—(1 —p)et 1-(1 —p)et ( ) d ~10 t EX = —Mm(t) — ( (1 W) dt no (1 — (1 —p>et>2 i=0 2 20(1 — p) 1—p p2 10 d2 EX2 = 71mm <11? t=0 2 (1-(1 “104) (10(1 —p)et) +p(1 — p)6t2 (1—(1 WM“) (1 vp)et _ (1 — <1 — p)et>4 t=0 _ 103(1 —p) + MO —;0)2 _ 20(1 —p) + 2(1 — p)2 _ p4 _ p2 1 — 2 1 — 2 1 — 2 1— VarX : p( 10);:2 ( p) _( p210) 2 p210 M050?) 2 ff; 6” 21W e_(x_”)2/2”2dx 2 21m ff; e-($2_2”$_2"2m+“2)/2‘72dx. Now com- plete the square in the numerator by writing 2:2 —- 2px — 202t$+u2 :- 332 ~ 2(M + 0203: i (u + 0202 + M2 = (w — (M + 02t))2 - (M + 0202 + M2 = (:1: ~ (M + 02t))2 — [2u02t + (0202]. Then we have Mm(t) : e[2wzt+(gzt)2]/202 “317;” [fem e—;}§($“(M+02t))2dx = €Mt+#. 2 2 EX = %M$(t)lt20 = (n+02t)e“t+‘7 t ”LO 2 p. EX 2 = £52m) L0 = <u+a2tfeut+azt2/2+a2eut+a2t/2{H = M2 + 02. VarX 2: #2 + a2 — M2 2 0'2. 00 1 1 2 2 EX; : / 31:7~ e_( 0‘5””) / d3; (f1 is lognormal with u = 0, 02 2 1) 0 27111) = —1— 00 ey(’"_1)e_y2/2eydy (substitute y : logx dy : (1/m)d:z:) x/27r _oo 5 1 00 2 1 00 2 2 2 = __ e—y /2+ryd : _/ e—(y ~2ry+r )/26r /2d V 271‘ \/—(x) y V 27? —00 y 2 er /2. H 2—12 Solutions Manual for Statistical Inference b. °O °° 1 2 . / $Tf1(x) sin(27r10g 3:)d3? = / 36’" 6—00“) ” sin(27rlogm)d;r 0 , 0 27m: /00 (“7")? 1 e‘wi'flz/2 sin(27r + 27r7")dy : e ————-— ‘ -00 V27? y (substitute y = log a:, dy : (1/x)d:c) m I = / 1 e("z_y2)/2 sin(27ry)dy _00 271' (sin(a + 2717") : sin(a) if 7" = 0,1, 2, . . .) z 0, because e(’"2“yz)/2 sin(27ry) : —e(T2”(”y)2)/2 sin(27r(—y)); the integrand is an odd function so the negative integralcancels the positive one. 2.36 First7 it can be shown that lim tam—(1035902 = 00 IK—>OO by using l’Hopital’s rule to show t’ — lo :1: 2 hm us; :1, :B—>oo t3? and, hence, lim tx ~ (logaz)2 : lim tm : oo. IE—‘>00 III—>00 Then for any k > 0, there is a constant c such that 00 1 2 2 0° 1 oo / _etme( ng) / d2: 2 0/ ~—dx = clogxlk : 00‘ k: 5” k 36 Hence M9605) does not exist. 2.37 a. The graph looks very similar to Figure 2.3.2 except that fl is symmetric around 0 (since it is standard normal). ‘ b. The functions look like t2 / 2 — it is impossible to see any difference. c. The mgf of fl is €K1(t). The mgf of f2 is 6K2“). d. Make the transformation y 2 em to get the densities in Example 2.3.10. a. % fox 6“}‘tdt = 6—”. Verify d m—At _d 1—At' (ml/0 e dtl_d$[ A6 b. 719%wa e”)‘tdt = fooo fie">‘tdt= [00° —te_’\tdt= —F)E2) = —>\1 . Verify d 0° _M _ d 1 _ 1 dA 0 e dt‘dM‘ A2 c 2217f: m—lgdzr —;12— Verify d 1 1 d 1 1 d 1 1 --— — 2—- ———- 2— —1 -— 2—— dt U1. $2de dt< xi) dt< +t) t2 d 00 1 00 d 1 d _ 002, ‘3d'~* —2°°_ 1 V ~f Cl 32‘]; de: 1 dt (W) III—fl (.E—t) .L‘a*-(£IZ—t) ‘1 —(1—:t—)g.er1y d 00 ”2 I _ _1‘OO _ d 1 _ 1 a 1 (m—t) (113—de t) 1l_dt1—t_(1_t)2 (-2.13 “43935 Use ‘Hne Known mo ' T meld“: 360 mej'rlc 0U8+Ffbbk+lbhw O‘F The From awe/MEX) 31C YN GQOWTV’C (10)) The“ EY Z“ 2 ”1007331“: .1», Kzl "10 This {mpkes iKflquK": L. ’31 4331 Repacnmj 10 by Pp m fins 34VQS Z KPK~13 4...... K2: 0 ‘10)2 Plujjiflj “These vaMas baaK {n+0 43%, QX‘PFQ/SSYGW {”01” EX we, 5611’ LL’Y K [Oi 0» OUSCVmDa FMOQOM MIL/[0,6 whosc POAAQQ/ \S- ”Ukfi A0“ flegmfldc .‘Vveegc’yVS. , Po 0° ,4 a Mfiflb “F gszUfi) + gag) : Z K: MK» 5”“ gr (KB: 1* Fly-D J” J (a) X : :D’Fx‘r‘fl = i O “ 3533] J:‘ 9.:0 [2J9] Assume EX:L exu‘sfs (is FI’nfie). 77/133 ‘imPHes EX QXISTS. Now no‘fe ‘WIOLT EGG-201 = E ( Xl*23X+&Z) = E><9~~—2a(E><)+ 212*. Thus C742: ECX“8)Z :- “Z EX 4' 23 (file) aha! H” (S Clear “H041— ...é—l EO<~3)Z‘ :2 2 > O Thus a: EX {5 The unigue Value 0F 3 . . ’2, which mimmYZeS ECX-fi) . We only assumph'on we, needed, {3 ‘fi‘af Exz axis‘k. Inc X {S confinuouf wT/‘h pd? 7&9) I This means ‘fha‘f f%%1£<(9<)0[x< 00 ., a)! Ike, Original/4P, Jr‘s—woke?" ’51 JR) , _ X 0 ‘3L [2’22 (ALI/ERNGTE SOLUTlON] 0 133” ,3, "15%: 0291 M * “mg: , 00 “’7’ z gag—’RKA «3; M “M 7 P n, X’fl? ’ <9 ’7 king We .47 J? Reca)’ I PM“) (”(3)3—{5 sow" (q) 50%er 3W W W 600 «v w) PKoBLEM 02-22 FL Valpamlx X 3 ('P “a" 451i“ f {352% “6%“? :ffirz J” Y’s/“3%? fiifl =1 %L®%Q"€"% 2 :33; W2) . L 04" 7/ if“; my ; and; Ax 5' [£47 in \l , ‘8‘ fi K 3 437' problem 13?? WSW/ks ‘ X IMAM, a, Aeflwaxto bsmwvboi JkfiJ’LwE/LW Lax ' x: 0; z , ‘ , MB ;L”1"”§wce»p¥‘ 3 o 402; (‘>' M .‘vcf’egéf )(po .0 + t ’5 I” {v (“:43 (040363 ; L, P; “X" 4 .- e X rem!» x’o >4 >('”c("’°3>(€ [1103) : F ]r fl—r'jf‘” M Z. ‘10 (‘— [1430105 0 JC 3 (a) New, *KM E (mix\5L1’c+(t~p)\(60p)§x >8 ALL Sm f F020 :\ 3% N Nag 6M6 P) PC“ («103) 0.3 (0ng as (7)103" (Nth/”3 “705) {)mHm 2. 3? MSIMQ mm P -/ (up) for £1 fir 2’0 {(1103 ’0 F /‘ '6’?wa ya 4 ”EM Lu. 240 CQWWWM) z:- E % (“2) W“? = 2% I: we (:2) W‘Www] I (x M .. E0 (22) Pk (I‘P) ~lc : (M~<y)(:’x> gal? 76:mhorl(HQ/x476+C. Lab P=(. 0: 0+4, '- C10 Ix ‘ ~ -. E) (“7%) P“-P)‘"" = (M (3?) S "’ W“ Mme . D ...
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What students are saying

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    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.

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

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    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.

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    Dana University of Pennsylvania ‘17, Course Hero Intern

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    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.

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    Jill Tulane University ‘16, Course Hero Intern