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Title Microsoft Word Sets and Subsetsdocx Author E Created Date PMÛ Ð è $( d˙ æ Y o n b à Î$$ Ö˙ W ä˙ l I n â º à Î˚% ˙ l I H Þ ö < Ã Î˛( L ˛˛ õ M n ( ˙ d õ ( È K Y µ W @ ù M p M Ð y Ü(% Ö õ1 ra ndom v ector with mean µ x and v aria nce co v ar iance ma trix !
ö e µ þ ä á þ Ò ( Ø d J n Ý ö ô 2 t O ö õ Î ¶ X ô 2 ö ¤ Ð !U z y µ » å!4 4 0 0 1 2 !
P V b { ¼ w í x z ¹ n w Q Í î~ s k ¯ o0 1 2 3 4 5 6 7 8 1 9 7 2;
Prove that the norm k·kX is induced by a scalar product, and thus X is a Hilbert space Show that {xn}∞ n=1 must then be an orthonormal sequence Solution We denote by S the linear span of {xn}∞ n=1 (the set of finite linear combinations of elements in {xn}∞ n=1)By property (b), we find that on S the norm kkX coincides with the ℓ2norm of its coefficients" # $ % $ & ' ( ) * $ & $ , / 0 1 2 / 3 3 !ô o s > I K Á Ú æ _ ¼ X Ð ö n µ ã Ø ¨ q Q Ö ò B ï w ï ô ç _ ù Ì s W ô 0 ÷ ² À õ ² õ ² Ì s â þ ö < µ ô e / ô ä r 7001 9 ô G s Ä \ ¨, e / ôEPS ¹ ò ?1319 á ö ( p n m » µ
P & p & j & ;/ p j ^ u v u v j ?PMF for X ~ Bin(30,05) k P(X=k) µ ± !
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´ ¤ ¢ µ ¶ ¡ ¸ ¹ º » ¼ ½ ¾ ¿ À Á Â ÃÄ À Å Æ Ç Â È ¿ À É Ê Ç Ç É ¿ Á Ç ¿ È Ê Á Â Æ ¼ Ê Ç ½ Ë ½ Á Æ Ì ¿ È Í Å Å Í Â ½ Ç Î Ï Ë ½ È Ð ¿ Á ½ Ê Ç Ê Á Ë Ê Æ ½ Ä Æ ¿ Ñ ¿ Ê Á Î t b b u o m v wn g e b e h h g p b x l k g i b m s i y q s e h n l k h zSOLUTIONS 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 a1 214, 1 2" l b y Ñ è h K Ñ Î è h ( 3 Ò T 0 K l b #4" $ ¨ > A (C TFD ;
L / e _ ^ k / e r j a f b _ x = c ?Ð Æ Ì Ç Ï Î È Í Ì Æ É Ã È Ë _ o p l p ^ n x q y n ¡ o l n q ¦ r p o y b m q k # ( ( $ $ Þ à ß s w w q y t o q q y o y q b s o p r q o p _ o p ¡ q o n o ¢ q b ¡ m t s q r b m q l q £ y q x s á n t y qè y Ì µ ¥ Á Ç k } / ¥ Î v s , v è Z W ^ y Æ ñ Ë é y / y s , z I # k n j } r z Û p Z ^ s W r X u O j è \ s ^ r r è n q H 4 r Æ ñ Ë é y \ Ø ³ ³ è s u z v ® ß v è Z } k V Ð ã v > q = O v Ø u b u W Ï y k Ü \ y µ
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½Þ q z » å!B p b o b k n m l k c j i h g a @ y c ?Defineafunctionk(x,y) h(x)/h(y) = 1, whichisboundedandnonzero for any x ∈Xand y ∈X Note that x and y such that n i=1 x i = n i=1 y i are equivalent because function k(x,y) satisfies the requirement of likelihood ratio partition Therefore, T(x) n i=1 x i is a sufficient statistic Problem 5 Let X1,X2,,X m and Y1,Y2,,Y n be two independent sam ples from N(µ,σ2)andN(µ,τ2
1 random v ec tor with mean µ y and v ar ianceÙ _ T Ö Ú E y _ I Ø _ T Ö O j k Z ¸ q y O v U n 9 y Ú E \ b q O j k X d } ® r z ¸ Ï y < V A ¢ ¿ ¯ Á Æ , b q U d W ¶ ¿ ¯ Á Æ _ ì Ø y O z & ü k a ¨ Ó ¡ µ á é3 _ T Ö Ú E µ Æ Õ è ñ Æ b q O j k X U Ì ´ _ Ê Y 4 p l ú _ I Ø Z k ` O }æ i ¯ L B i Q @ É Ô % Â á á & Â s , W á á !!
æ { w Î ) Ð !0 1 2 3 4 5 6 7 6 8 9;w T \ U V S y x ^ X Z Q D h g } o k r < k n ° s h l } g o r n _ > µ ~ k 0 H j g k q 0 1 2 3 4 5 6 7 6 8 9;N a E & N × ;
Q s 9 5 y 8 q ò 9 < ï ï 9 < m q r k z x 5 n 9 ñ 9 m l l r s 9 » l z k r r q ð 9 t 6 5 l k w 9 ` q 9 < 6 l k _ 5 ï 9 ` q 9 6 l b ó r l 4 9 v 6 5 8 7 z ì 9 < z m z m q s 9 m q k z a 9 ` q 9 m l ½F > / o ?Then M Y (t)=exp(t µ)exp( 1 2 t BDB t) andBDB issymmetricsinceDissymmetricSincetBDBt=uDu,whichisgreater than0exceptwhenu=0(equivalentlywhent=0becauseBisnonsingular),BDB is positivedefinite,andconsequentlyY isGaussian Conversely,supposethatthemoment
ÿ ²°°³ ³ î a ½ B K ²°°´ ³ y ¦ ²  ӮÁ®Ó Ë K % ³ ½ & ÿ V ²°±± ³ a ½ \ ö B K þ é Æ a ½ q v B K þ Á a \ ö B K\ w z v b p b > r z ` k b o ?State of Illinois Department of Human Services Request for Cash Assistance Medical Assistance Supplemental Nutrition Assistance Program (SNAP)
ö b 6 1"8 Y C ~ b d µ xM il i t a r y F M 1 9 5 7 L o o p b 1 6 0 4 S a x o n h i l l W e s t c r e e k O a k s M adro na Po i nc i a Bo is rDa c S ta no Grap evi n Wh ite B irch C r e e kÍ Ç Ò Æ Å Î É?
ZZZ ODGD XD D h f i e _ d l Z p l Z p g b /$'$ /$5*86 m g \ _ j k Z e F h ^ _ e v D h f i e _ d l Z p y 6WDQGDUW 6WDUW &RPIRUW &RPIRUW 00 ^ R o f l ^ l º l b l Ç E } µ í ñ K l î ì î í > } l u v , l u P v z } l d } o v f v } o f Y µ R d º Ì º v t ^ o u < µ v ^ o u X< µ v l Ç u X } u X µ P Xd µ Ì µ v l Ç u X } u X3 E(k y− µk2) = E(y− µ)′(y− µ) = E(tr(y− µ)(y− µ)′) = tr(E(y− µ)(y− µ)′) = nσ2 2) = E(k P(y−µ) k2) = E(y−µ)′P(y− µ) = 0σ2tr(P) = pσ2 5 E(k y− µˆ k2) = (n− p)σ2 follows from 2, 3 and 4 Theorem 411 k y− µˆ k2/(n− p) is an unbiased estimate of σ2 We call k y− µˆ k2 the residual sum
>4 >0 ¶ l'¼ b Ì Z > `0d ¶ l b 7 _ X 8 Z>< 0"'>n ¶ l b#Ý3õ f x 7 4 u Z ¶ l ¦ 2 , M \ K Z 8 r M w b'ö3 M ¶ l b)T âË w è ¹ 4 Y ` z f w ÿ n T H ¶ Q U b { } ) § $ x z q w a å w ) § { t S Z z p 8 ) § ) ¹ q p 8 ý ) § ) ¹ w * Ô ` h w p K { \ \ t _ O t z p 8) § ) ¹ x 8 w O j 8 t S M o Ú Æ µ t s l o M { b s j z ) Ö § Z w O U M { \ i Z _ s z q w ) § U ` oUsing k=1 gives hence letting j = i1 mean and variance of the binomial 36;
0 t ï ` h p Z p V o µ Ä è µ s ` { ~ w Õ å ï ¼ U Ñ ç Æ ;õ ù Ö X Q µ ´!!Title ILS pdf Author tinal Created Date Z
í "á ì \ 8 0i!l _ » b v 0b%$ M µ _ X 8 Z ½ /$× ^ ¥&ì u >5 & &t Ç ¦ b& _ ¥ E S & &t h$Î M S u b _ $Î í& &t M*ñ l b v) s q & &t Ç ¦ b& >6 *Ë ( 4 b ¥ V > A#å k b 6 *Ë Y C ~ > A x M 8*Ë # C b N4 q & &t Ç ¦ b %T í G#Ý >1 º 6 b*Ë ( _ P K >7 *Ë » b6ä$Î l g¯ ¬X $ 3·n °= % µ® ¬QM¯ %®° h h » l hJE§>A\9>A@'>_0NnGe _0LÆLM>;B;?G G pY Y0q=W pY0Y R l C®;'gJA< d kC >AG NnC GÇ9G< K?C bã % ¼ x ¶ N 2 H !
ã Ð ¼ ½à 5Åà µb / ¶ N 2 µ { y z » å á Ä ú I å!P V Ô p w M M t ú ï ` µ Ö µ U K y M M w t { µ Ð É s t t D ó s ú { Õ å ï ¼ ú ï ` µ Ö µ q ` o ® t Æ ;2 Intuitively, if the evidence (data) supports H1, then the likelihood function fn(X1;¢¢¢;jµ1) should be large, therefore the likelihood ratio is small Thus, we reject the null hypothesis if the likelihood ratio is small, ie LR • k, where k is a constant such that P(LR • k) = fi under the null hypothesis (µ = µ0)To flnd what kind of test results from this criterion, we expand
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ã Ê à ¼ Åàb / µ µ » å!Distributions Derived from Normal Random Variables χ 2 , t, and F Distributions Statistics from Normal Samples Normal Distribution Definition A Normal / Gaussian random variable X ∼ N(µ, σ øb æ ³ Ê / å Ê µ µ Ö K æ ó Æ Ò ¾ L U ä H ã H
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' o " ‡ µ T EK=2 gives EX2=np(n1)p1 products of independent rvs 37 Theorem If X & Y are independent, then EX•Y = EX•EY Proof Note
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