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×ÍÆš gC Ùf ˆV gC ú ÿÿÿ ü ÿÿÿ hC ú ü ÿÿÿ ð ð ÷ 4 5 4 5 ú % Þq?@ r @Þq?@ % r @ rù?. 2 9 1 8 7 6 5 4 3 2 1 0 < > 6 a 6 ?. @ a b c > d e f g h i @ j k < d h @ j e h g l m@ i i @ g a g.

á ¶ Å Ä · Ñ Ñ · Ý ¹ Ò Ø Ù Í Ï µ ¹ · Ò ¹ Ó Ó ´ Ô ¹ µ Ç ¼ ¹ Í ¸ Ç 5 h = > 4 0 9 2, 1?. ç 0 û ö ¯ 5 ö N » Ð ° 5 ö Á ¥ k z 4 j Î T a m k í z Ð ° 5 ö P ³ ý é!. A unique characterization of µ n terms of x and y is obtained as µ = 8 < arctan y x;.

SOLUTIONS OF SELECTED PROBLEMS Problem 36, p 63 If µ(E n) < ∞ and χ E n → f in L1, then f is ae equal to a characteristic function of a measurable set Solution By Corollary 232, there esists a subsequence χ. ` _ m h I æ w » É U F î Z Ö U 9 ` X o 1 q Q C 4 O w \ q ß Q h 0 p ( b s â ^ Q t h Ï o ª » p ^ _ s Z Ö ± t Ç Z p V Ý ï Â Æ ï µ Ù ¼ Q 0 ° a Å " V. "@ ö ì Û å Û å 1Ï#ã ´4{ ú ·4 · b ¿ Ý î*Y c>* ) é L í ä Ø l R E s t p F ï ä Ø í ä Ø l R E s t p 6 y ä s s ;.

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N d & Ç k Q ä ò Y z Ö À * ® Q I k C b H Ø Q · = h ) ( Q Ø & J u z H q V H 4 þ * Z e b 4 ¥ u z B Y z Ö À C * Ø A Q 9 ÷ = É ÷ N d NJ T Ø Ø f Õ ú ) N d NJ N d NJ Ð > PP ª 7 d u z Î ó 7 d u z Ø Ø N d ú NJ P B È ú k à ¼ C. Nsuch that f n!f Then by monotone convergence theorem and the de nition of integral for simple functions, we have (E) = Z E fd = Z E (limf n)d = lim Z E f nd = 0 Then we may apply RadonNikodym theorem to see that f = d d and see that g 2L1( ) if and only if R X jgjd < 1which is equivalent to R X jgjfd = R X jgjd d d < 1 Since fis. N Z@U N\@f ª@a ¬@ M À@ N Ä@ 0 ‚Æ@ M HA D LA M wA 1 ‡{A 1 ‡ B 0 ƒ‰B 0 Ø C `äC M DD 0 HDU 6gEU > E M ÛE M ßE N ãE C åE 0 ‚ÿE M F 0 1F 0 =¶F 0 ÈóF 0 >»G 0 ÊùG 0 ´ÃH 0 ´wI 0 aJ i FŒJ 0 ´ÒJ 0 ´†K 0 9L 0 ÜsL ÆOM 0 2 N 0 üGN 0 CO 0 }O 0 _ P 0 aP 0 k kQ 0 Ç ÖR 0 Ç T 0 çdV 0 6KW 0 W 0 ‰X 1 ‡ÃX.

H p > p n o v 0 Æ Ð > > h q > Ã Ý Æ p n o w 0 Æ Å ä Î Æ Ð > y f _ l c b _ > g l l m t _ r g m l > a g r w À x ö > À x ö Â Ê ¨ Æ ¥ ¬ æ k K r 7 J >. J º = ì à ¬ r ® = q ë œ Ö ' q l Æ é Ž q ‡ â 1 € v Ï þ Ò ï N Ö L À ¨ ¤ € D ¼ ð b 3 t í á d ’ þ ³ Ä ó É ò > Ý { l 7 1 ¿ ± & ¬ ä Û / b ¤ ¢ n !. D } u Ç D ' } µ > X r ^ } µ v Æ h } ( î X / v } v v } v Á Z v l > µ u } À o } ( Z o } ( D } ( } Æ D } u Ç } À ¨ ïD µ Z } } v l > µ u P v o o.

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If n = 1 or n = 2, the volume of a rectangle is its length or area, respectively We also consider the empty set to be a rectangle with µ(∅) = 0 We denote the collection of all ndimensional rectangles by R(Rn), or R when n is understood, and then R → µ(R) defines a map µ R(Rn) → 0,∞). × Õ é ³ ¢ Ç r y Á ñ z Ò ¯ } ¤ Á Ç Å Á « r y Á ñ z F d W I ³ Ö y N < ê y j r Ö z ë ½ è O Þ ¤ ² O ê è ¥ s u d } " ½ y ¨ Ö v J n q _ ê è > ` O } Title Microsoft Word , ÞÌëqxK)4óö ' !. Title CS_1pdf Author andremueller Created Date 8/4/16 PM.

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¯ Ê Ò Á ¥ C ª d y 4 q Á ¥ Q T y 4 q ì ½ = ;. 5 é u Y é Ã » = M á Ñ = ¾ ¿ ¾. 3 y y ' 2 ' X Y FigureS13 2 Themomentgeneratingfunctionofc 1X 1 c 2X 2 is Eet(c 1X 1c 2X 2)=Eetc 1X 1Eetc 2X 2=(1−β 1c 1t) −α 1(1−β 2c 2t) −α 2 Ifβ 1c 1 =β 2c 2,thenX 1 X 2 isgammawithα=α 1 α 2 andβ=β ic i 3 M(t)=Eexp( n i=1 c iX i)= n i=1 Eexp(tc iX i)= n i=1 M i(c it) 4 ApplyProblem3withc i=1foralliThus M Y(t)= n i=1 M i(t)= n i=1 expλi(et−1.

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If x ‚ 0 arctan y x ;. If x < 0 The complex conjugate of z = xjy is z⁄ = x¡jy or z⁄ = jzje¡jµ jz⁄j = jzj;. Proposition 4 All sets with µ∗(E) = 0 are µ∗ measurable Proof If µ∗(E) = 0, then for an arbitrary set A⊂Xwe have µ∗(A) ≥µ∗(A\E) = µ∗(A\E) µ∗(A∩E) {z } 0, because A∩E⊂E Let M∗ be the class of all µ∗measurable sets Theorem 5 (Carath´eodory) M∗ is a σalgebra and µ∗ M∗ →0,∞ is a measure Proof We will split the proof into several steps.

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N g p c j j g > J # J < J > P N P N > ù O Æ l _ n _ a > H _ q s n c p > r c a > P R Ó > J ö 2 7 J = w t V L P T x H ¢ û h Ê û h & Â 1 í ß ü Ý ä å M ð @ Å W h ù Ê ¢ û h Ê û h & Â 1 ö 6 0 J ÿ E ð @ Å W h ù Ê. N £ m, AG is an m £ m projection matrix and GA is n £ n In general if P is a projection matrix, then P = P2 implies Py = P(Py) and Pz = z for for all z = Py in the range of P That is, if P is n £ n, P moves any x 2 Rn into V = fPx x 2 Rng (the range of P) and then keeps it at the same place. The sum S =åN j=1 Xj where the number in the sum, N is also a random variable and is independent of the Xj’s The following statement now follows from Theorem 1 Theorem 2 (i) ES=E(X)£E(N)=aE(N) (ii) Var(S)=Var(X)£E(N)E(X)2 £Var(N)=s2E(N)a2Var(N).

51 n × Ò ý µ > 6 Ã i r É Ä á × r É n Ò ý µ n c ¤ w ¤ ô ¾ á _ § 6 Ã w Ö ¨ 52 ¤ / µ ¢ ± i » á f ¤ á ¾ á 53 y / µ ë y i » ¤ w á f m y é n è n y l z Ò ¬ !. ¢ ´08 Ë z2,014 ª £ p x \ \3 ñ D x N Ð s. > 8 6 = 8 7 6 1 9 = 9 < 8 7 6 5 ;.

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Math 594 Solutions 2 Book problems x41 9 Suppose that G acts transitively on the finite set A and let H be a normal subgroup of G with O1Or the distinct orbits of H on A (a) Prove that G permutes the sets O1Or transitively Deduce that all orbits of H on A have the same cardinality Solution For each i there is some ai 2 A such that Oi = HaiNow since H C G;. " # # $ % & ' * % , " ' (* $ % / ' & " 0 1 2 * 3 !. @ 7 A > 6 3 B 8C D / E / t 1 f a b 0 j p e q x c u m gl h m v 1 c l f 0 jb g a / e u d h e t.

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à 4 4 à Ò j r > q o 8 gog g9g ô 8o% g!·fég gug gm8o% 0 gegxg gh ö h 6ä g 'h ß0b3¸fúfù ggwgg a ggwgg)r lfå 8 a Êfþ'8® gcgqge#Ý7 (Ù Ù if÷fþ æ&g gug gmfû _ fÜfúfßf¸gog g9g fÜ =føfçfö g föfÔg g fþ 5 ½ v Ü9×fåfúfù h gog g9g fÜ g g 'f'¢ h. 5 é u Y é à » = M á Ñ = ¾ ¿ ¾. N (d) According to Corollary A on page 309 of the text, the maximum likelihood estimate is a function of a sufficient statistic T In part (b), the maximum likelihood estimate was found to be θˆ MLE = − n P n i=1 log(x i) −1 2.

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