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Y µ } v v Á v P t Z v P µ v P Z µ U D Z o v P U y µ } v P , µ v P • Z W l l Æ À X } P l l í ô í î X ì ï ñ õ ïL } d o } ( } v v v } v W µ v / v i µ Ç ~ / U } v v µ , } u © v } v W W Z u Ç W } À Y Y Y Y Y Y Y Y Y Y Y X Y X X X Y Y Y Y X X X Y3 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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/ ª H µ ( Å Ç Å ª0y0 2 y0m² y ) !!Xw § 2 µ Þ ® µ H @ Å Ä Á A É Á Â L Æ Å È Ã 2 µ µ Ê Á À Ã Á Ð P 2 P § ÿC » ` h ¯ Ø Ì Ñ Z t o Í h } ¼ g " w G O Fig 1 t Ô b } ^ t { c ® V s ` w h t ¼ g " † j Ø !Ie E(X) = µ As Hays notes, the idea of the expectation of a random variable began with probability theory in games of chance Gamblers wanted to know their expected longrun winnings (or losings) if they played a game repeatedly This term has been retained in
Title AEG2 RFA Questions and Answers Created Date AMZ w ¼ g " w P q ` o ¦ µ Â Æ Ä % µ Â ï è µ ï(SUS316L) ;U = (Y − µ Y) − b(X−µx), where b is a fixed constant that I choose It follows that U2 = (Y − µ Y) 2 − 2b(X−µx)(Y − µ Y) b 2(X−µ x) 2 Hint Do not expand this expression Keep the terms grouped as they are (a) (4 points) Derive the expected value of U E(U) = E(Y − µ Y) − b(X−µx) = E(Y) − µ Y
Suppose that µ(x) < ∞ and f X × 0,1 → C is a function such that f(·,y) is measurable for each y and f(x,·) is continuous for each x Then for every > 0 there exists a measurable set E ⊂ X with µ(E) < and f(·,y) converges to f(·,0) uniformly on Ec as y → 0 The solution is a modification of the proof of Egoroff's TheoremD v v P E µ o E Á } l W E } u o Ì } v U Z P µ o Ì } v X / v } } > v v P U & o o î ì î í Y µ l Z W d v v P v Á } l Ç P v v1 random v ec tor with mean µ y and v ar iance
X and Y, ie corr(X,Y) = 1 ⇐⇒ Y = aX b for some constants a and b The correlation is 0 if X and Y are independent, but a correlation of 0 does not imply that X and Y are independent 33 Conditional Expectation and Conditional Variance Throughout this section, we will assume for simplicity that X and Y are discrete random variablesÁ Â Ã Å Æ Ç È É Ê Ë Ì Í Î Ï Ð Ñ Ò Ó Ô Õ J Ö × t Ø Ù Ú Û z E o µ h 0 1?M h } ¼ g " ;
Ï ´ É µ â Ç ¿ Ä 5& ) 1 5 Í Æ ¹ Ç ¿ «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 coefficientsD Z } v } o u W Á } E r v µ u } } µ E = í r v µ u
E −(ln(x)−µ)2 2σ2, if x ≥ 0;Y } u µ } } } v } v W '^ r ï ñ& r ì ò í íZ Á Á Á X Æ } u µ X } u W P n ï ^/E y Dd } Á ¡ y t } l } v ð r } l ô rd Z / v o y } v t ^ o o t r î í î ïS Z s ` e ° r Z Â Á t i d ^ s r Î Ê Á r > R ¤ Î B ¥ r X ° ² Z c Q ¤ ¿ Þ q y Î Ó O ¥ ³ r ß ± Ô Þ s ½ Æ Æ £ Å ¢ v Q A _ A ¤ Â Â ¤ ¨ ñ ç ¢ ñ µ å Á Õ ã è ¥ J Â ½ < Ì ¤ ® é Õ ¥ /¥ Ì v ½ u y b
Title Microsoft Word VLAB Meeting Minutes (draft) Author zku Created Date PMTitle WW_AllReports__1530hrsxlsm Author SJablons Created Date PMG y R29 å12 D5 Ô yReceived ⃝c 18 The Society of Materials Science, Japan ∗ Y q » y ý Á G ¶ y ¶ æ ß y ý Á ¢
And µ 2 = EX 2 θ Hint Note that X = e Y and X 2 = e 2Y where Y ∼ N(θ, σ 2) and use the momentgenerating function of Y (b) Suppose that X 1,,X n is an iid sample from the Lognormal(θ, σ 2) distribution of size n Find the method of moments estimates of θ and σ 2 Hint evaluate µ 2 /µ 2 and find aTitle Microsoft Word EVV FAQs as of 116docx Author Tim Created Date 11/6/ PMTitle Microsoft Word Section 502A Bidding Registration Application effective before 21 Upset Sale FOR INDIVIDUAL Author sberry Created Date
B j Ï r C Ç ¯ µ Ä Þ y Ñ v Á d Ì , d } ½ ð ¬ 0 ¾ P Ð _ ¨ P ¤ â / J v ¤ ´ ^ r y ð y µ ¤ à ¥ â v ~ / J v ¤ z T ç U ð ¬ 0 v ç E ¤ à ¥ r Æ Â È = b q O u O P Ð _ u(e) the variance of Y 4 Let Y be a random variable having mean µ and suppose that E(Y −µ)4 ≤ 2 Use this information to determine a good upper bound to P(Y −µ ≥ 10) 5 Let U and V be independent random variables, each uniformly distributed on 0,1 Set X = U V and Y = U − V Determine whether or not X and Y are^ È y j y Ì LQGRZV Õ è ¯ µ å ñ ç ¤ Ç ³ Ñ y E » Ó Æ ¤ ¥ E ,7 ³ Û Æ è n q U d } 9 Û r z ® Þ y
Title Microsoft PowerPoint EQC Essentials Aging and the Body Author OR Created Date PMDistributions 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, σWZKs/ Z^ h^/E' ïZ W Zdz v / v P v P Á Z ,, y Z v P Z } µ P Z l v P &&^ l U v U h, U t o o W } À µ u À À ,, y W/ Á Z À
Title Microsoft Word Questions and Answers for Early Intervention COVID19 _1_ Author Jen&Tony Created Date PMHZ hK&> E D E ' D Ed U W ZdD Ed K&d, /Ed Z/KZ d Z o } µ } ( } Z Y µ Ì À o o l } µ v Ç Ç Á Ç Z Ç v ó u UW ( ^ vX, and let Y b e a q !
Math 541 Statistical Theory II Methods of Evaluating Estimators Instructor Songfeng Zheng Let X1;X2;¢¢¢; be n iid random variables, ie, a random sample from f(xjµ), where µ is unknown An estimator of µ is a function of (only) the n random variables, ie, a statistic ^µ= r(X 1;¢¢¢;)There are several method to obtain an estimator for µ, such as the MLE,1 ra ndom v ector with mean µ x and v aria nce co v ar iance ma trix !Þ ) ½ w Ô ù w ¤ = Á ` 0 p V b { Þ » ¿ ½ Ý ï Ä Ç ó Ú ¨ å µ Ê Ä t ó Ú ¨ å µ 7 0 p V b { µ æ Ü » Ó Û Å ç » Ó µ æ Ü » Ó " ¾ V ¾ ü Z " ¾ V ¾ ü Z × ¾ V z Á ` r j D z Á ` r j D ± 6 #*" 3"$564 º º " ¾ V ¾
>> W> E^ W o u i µ Á Z µ } À Title Microsoft Word EVV Billing Workflows All Plans All Solutions FINAL Author kenoch1 Created DateTitle TNGVGP List and Tables_Finalxlsx Author BeFernan Created Date 12/9/ PMTitle Microsoft Word EVV Toolkit 721 df edits v2 Author Jennd Created Date 532 PM
è Ã u y y ' y a S Ñ b ® y _ L ?Y µ v Ç v Y µ v Ç v Y µ v Ç u ò î ï ì í ¨ ï ì U ì ì ì ^ À v h v v l ^ Á h v µ u } Ç ' v À o o } v ò î î ð ò ¨ ñ U ì ì ì0, if x < 0 This is derived via computing d dx F(x) for where Θ(x) denotes the cdf of N(0,1) Observing that E(X) = E(eY) and E(X2) = E(e2Y) are simply the moment generating function (MGF) M Y (s) = E(esY) of Y ∼ N(µ,σ2) evaluated at s = 1 and s = 2 respectively yields E(X) = eµσ 2 2 E(X2) = e2µ
^^ v } Á u v Y µ o _ ~> o ~ ( o L ^ v Y µ o _ ~ } À W ' ñ WZ /Z/ WK/Ed Z E } À u ì í U î ì í õ E Á D u 9Wff`Y fa =`ai Kag ^/ Á } v v } v ( u } ( } P Ç UIs imp orta n t b ecause it tells us w e can a lw a y s pr etend the mea n eq uals ze ro when calculat ing co v aria nce ma trices 6Let X b e a p !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
Á¨ ¸É¦®´ ´ Á¦¸¥ ºÉ° µ¤ »¨ ®¤µ¥Á® » µ¥°£·¦´  ¤¸ µ¥ µ¥¡¦®¤ ¦³ µ Á¡·É¤Ã£ µ µ¥ µ¥°´ ¦ ´ r°Á ¡·¡´ r »¨ µ¥ µ¥ ¥» Á¨¸¥ 妻 Ê µ¥ µ¥³ Á ¸¥¦ µª «r µ¥For µ,and S2 is an unbiased statisticfor σ2 in a random sample 3 METHODS OFESTIMATION Let Y1,Y2,···Yn denote a random sample from a parent population characterized by the parameters θ1,θ2,···θk It is assumed that the random variable Y has an associated density function f( ·;θ1,θ2,···θk) 31 Method of Moments 311DW>Z À Á } ( yZ µ W W P î } ( í ð y ì ì ì í í ð ð U y ì ì ì í í ñ î U y ì ì ì í í ò ð U y ì ì ì í í ò õ U y ì ì ì í í ó ï U
Title Microsoft Word Supplemental Q&A1 Author MariaG Created Date 7/9/21 PM1 lv wkh qxpehu ri v\perov lq wkh wdujhw rxwsxw 6 l lv wkh lwk v\pero lq wdujhw rxwsxw 7 ohqjwk ri lqsxw )luvw fuhdwh rxwsxw wdeoh)ru l 1S s k h l D j r} g ± H g k r Ü ` g o ¯ Ý k ¯ Þ z < 0 h n o r ±} Ý g s k k ° s h g o ß k r g h k n ~ g 0 g } f a > H E ' B 1 m ~ g 0 k l h r o µ à á â ã ä å æ ç è é è â ê ë J ê ì è í L î â ã ê Ô ç à ï Ú q z
U ÿ Â o M h f w D ó Q y ` o S X A U K { y æ µ « Ñ « » w x v Ø ´ p K ^ * w C ± p U ô X s ¢ æ ï p C ± p U ô M £ { f w w w æ µ « Ñ « » x Õ Ý ¥ ì ñ £ B ¤ í ô £ q E ~ º ü { ~ B ÷ º J % w ) U { h ³ æ ¶ y ® ¤ Ï Ã ï µ t , n X g ¶ O y g ¶ OY µ o Ç D v P u v , } Á } u } v / v ( o µ v Y µ o Ç M u } v Y µ o Ç o W À v P l v v Z W l l Á Á Á X ( o l X } u l Z } } l v l î í ô ò í î ì ò î ó v Z u ÇY ö À i !
µ µ µ µ µ µ1971 !0" È ³ µ Â Ü ¼ Ø C z S þ q j Ý Ç á L ` o S b { ¢ Ä è ý Ä a ^ tuuuuuuuuuuuuuuuuuuuuuuu y Í ë ;ã Õ ¥  ø µ p o å á Á È ú2 ãh { h h h h h h h < 6 k < 6 l µ q o 8 ã á à ~ Ù Ü ¾ û å ¶ Ô ã ) á ½ ú c ò µ µ µ ¹ Ì Í Ù Æ ù Á Ã Ù Ü ú º µ ¯ k l å µ µ µ µ r o à å á 7 þ Ú Æ ù ¶ Ô Ê á b à ¾ Ú Æ µ
Title Microsoft Word SBPS Phase 6 Questionnaire FINAL Update with NegTst Author bonne330 Created Date 8/5/21 AM
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