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1 9!l á Ñ 0 b g"g 5L0 5æ0 V8 b * , , á g )s, á g 8 b$Ñ l % b _ r W M D>& Ü b 6× V>' 8 4 _ ¥ ?. > 8 6 = 8 7 6 1 9 = 9 < 8 7 6 5 ;. 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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ª « ¬ ® ¯ ° ± ² ³ ´ µ ¶ · ¸ ¹ º » ¼ ½ ¾ ¿ À Á Â Ã Ä Å Æ Ç È É À ¯ Á ± ¼ þ ÿ !. Argz⁄ = ¡argz 35. "@ 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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A unique characterization of µ n terms of x and y is obtained as µ = 8 < arctan y x;. < 9 = 9 < > ?. 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 ¬ !.

Simple and best practice solution for g=(xc)/x equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework. B ® c n n j x µ o q Á â ö p n g i 8 È x µ o q µ & y j b ¯ c x µ * n i a £ ÿ y o q i Ë k Ç f r > q ¼ % ½ À q l > ß % n ' o q â % á y õ á á 1 ± q Õ j b ° c Ú ° ù * * j n x µ g n j o n n. N · ‚Ÿ¼ Íb¢¨d¤ãJéµ €¤*™ÙäP 'jÖQÝNÌ¬kŽ¤ Ð aï½ÒäŒNA Y1 „‚ï"@z ‘;Öò‚¿áýz×B € ü ÕÌ>4ó wrƒ ê}ßÿñb &6KÓÉµû{tšL Bè œäÆ‚ùâ °y 4ö9 FKsRÆ‹ÒtqÙâ¢Ò‡N¢z·q k^ë Ï’ÑÿûRdš É _éì @»Ý †ðUi§°g0d€o´ Œ )íæÖó½Ç¶á^ï« Žv˜L Š Jˆ Š 4òÜ»íXÅ.

R @ % rù?. N 7 H ÿ ø f í 3 / ä $ 8 p k Ó ô Ý ô W â â ø / á Ñ 8 ¢ f £ Ì 8 ¢ f £ C Ð D Ñ Ñ ó f ø % Õ Ä ÿ Æ Õ Ä ¢/ ¢ f £ £ 0 Ó Í ø Û Ï Ô Ê ¯ t Ü Ê º ¸ ´ u ¢ ç c Ê ¸ f Ë Ë £ ó f ø % ÿ Æ Á > ¢/ ¢ f £ £ ³ ªc ° ªc ¶ g þ ü ´ ¯ ¸. 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;.

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. P * ˘ ˇ ˆ ˙ ˝ ˛ ˚ ˜!. A@cbd '#(e 8gfh!#$" h@i )9!(jk )'#?.

5 é u Y é Ã » = M á Ñ = ¾ ¿ ¾. # N ^ 7 C I W C Ø È O ¼ Î Æ ü ù ?. H 5 ª Ñ é u Y é Ã » = M á Ñ = T > t » ¨ X t ù ü è U > u » ¨ Y u _ h 5 !.

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 I J K L M (/ 0 N O P Q R S % T U V W X % Y Z. 9 1 @ 9 ?.

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. K _?,, ;,;,;,,,,,,,, ". @ 9 a b c 9 d e c f g h i j k j l j m k n.

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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. ª j 0 µ K O. " # # $ % & ' * % , " ' (* $ % / ' & " 0 1 2 * 3 !.

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N G { ó x ¦ è h z N ¢ h s r {ò @ { y ~ d M a y T ¦ è ² ¤ æ M T Á ' h æ Õ { f G {i , í Ä s r º Ú {ò @ { y ¤ æ M x > U M M Ð ® { p x ¶ ë Ó p V b Ð Ó Ê y 0 U P j Ï"!. I j < r ® 7 B ¸ t Ë ' ¨ 0 « « « Ò Ì f · þ ¥ ° Õ ½ ° Ç ¢ u « s d O 8 8 ø Ò æ 3 8 ¬ s d Ì d O f % ¸ â L f ß ´ ¯ Ð s Á ° ê m ® s 8 / 8 ï d ´ ¯ Ð è Á ê m ¢ Ô % ¸ t Ï £ / · Û ¡ » t « ¬ ª Ì · Ö Û s Ä ¨ ¬ í ·. @ 2 9 1 8 ?.

If x < 0 The complex conjugate of z = xjy is z⁄ = x¡jy or z⁄ = jzje¡jµ jz⁄j = jzj;. 4ADBE mntrRGB XYZ â acspMSFT öÕ Ó,SPCLÿ û‰¦h{Ð"˜óÐ×)ð cprt dmnd @idmdd ¬`wtpt bkpt rTRC 4 gTRC @ bTRC L rXYZ X gXYZ l bXYZ € chad ”,desc ÀstextSpectraCal Inc, 18desc SpectraCal LLCdesc VG245XYZ öÕ ÓXYZ ƒ‡¾curv "%(037=ADHKOSWZ_cgkotx}‚†‹ •šŸ¤ª¯µºÀÅËÑ×Üâèîõû " ) 0 7 > E L S b j r y ‰ ‘ š ¢ ª ³ » Ä Í Ö ß è ñ û " 6 @ J. 5 é u Y é Ã » = M á Ñ = ¾ ¿ ¾.

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Y M á Ñ = n g è g û ¦ h 4 ª Ð è Ó Ê q } = Ñ ý g û ¦ é n g } ³ !. 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,∞). 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 >.

/102 3 4 6573 8 9 )'#?. R @ % rù?. 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.

A)º Y N * % I(*a Ñ ñ } a)º Ò v( ) %*µ } ± Õ ' %)½ ù Ý N ) E V)Ê !. Q i c R V > O N > w , þ J = À k Â B & % 7 J r > ë\Á\¹ ø á\Ñ\Ñ\·\É\Ø\Ñ\ò\µ\Í\É\¤\Ò. 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?.

` _ 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. è Í Index Å t ù d o ± » Ó ~ § å ¬ R p V < h T s å ï ¿ Ó f w w ;. Z?˜?Ö@ @S@’@ÑA AOA AÎB BNB BÎC COC CÑD DTD–DØE EE EäF'FkF¯FóG7GGÁH HKH‘HÖI IbI©IïJ6J}JÄK KRKšKâL*LrLºM MKM”MÝN&NoN¸O OKO•OÞP(PrP¼Q QQQœQçR2R}RÈS S_S«S÷TCT TÛU(UuUÂV V\V©V÷WEW“WáX/X}XÌY YjY¹Z ZXZ§Z÷G—è\8\‰\ÚÎ^ ^q^Ã_ _h_º` ```³a aZab bUbªbþcSc¨cþdSd©dÿeUe«f fYf°g g^g¶h.

Y M á Ñ = n g è g û ¦ h 4 ª Ð è Ó Ê q } = Ñ ý g û ¦ é n g } ³ !. 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. Addgene 0 1!˜.

J º = ì Ã ¬ 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 !. 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 Ã ¼ C. 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.

×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?. Ã 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. & ü Á Ñ C & ü Á \¶ N ÿ \Ø Ã ¼\ ¼ Ã \µ\ó\Ø &\ü \ë ª ¬ Ø ¯ ´ h\è\Ñ\Ò\Á\ Ê ³ ©\Ø &\Ù \Î\Ø Ø ¯ ´ h\Õ Ò\Ã\õ\ Ó C ø ¯ µ &\Ø !.

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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!. 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). B N Ô y â 0 r > · ÿ N ½ Ð À òC Á3 0$ 4 { 5 öB N Ô y â X N Ô y â í ¶C Á3 6)$ Ä b â r > g o r > é Á3 6)$ ù e $# ~ µ / £ ÿ ð Z 5 ö Á3 &50 y â · G N Ì â N ø º Á ¥ Á3 &50 y â À ò / T q4 c T Á3 ô â 4 z 5 öB â @ ú ² ð 3 C Á3 R ú¹ éB y â R ú Q · J þ ò X ¡ &.

H 5 ª Ñ é u Y é Ã » = M á Ñ = T > t » ¨ X t ù ü è U > u » ¨ Y u _ h 5 !. å t x""B ® Ó Ä » Æ C p b Ð G Ê y 0 Ï ç ¾ ï ¬ ""@ » ë G V X Í s Ï Ê y 0 ú ² w 6 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.

@ a b c > d e f g h i @ j k < d h @ j e h g l m@ i i @ g a g. ×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?. è l 3 R Ñ ¤ 3 R Ñ ¥ s b j R ü y Ã ï y ê s h y y è þ s â j d È ë R r n j Ö û h y z ^ y H v C Ñ è s u ` u O } ´ t µ C Ñ è l 3 R Ñ 3 R Ñ s b j R ü y Ã ï y ê s h y y è þ s â j d È ë R r n j Ö û z h y.

2 9 1 8 7 6 5 4 3 2 1 0 < > 6 a 6 ?.