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C » < ÷ \ » R Ò » ® þ x µ ® ½ ` R ¾ ¸ À µ Q £ & = R D ~ ª þ y Ç È X þ M 0 × è Ì 4 ' ) I = E ³ r X ¥ 4 { % ³ x x µ ® ½ ` R ¾ ¸ À µ £ I v Ë { x x µ ® ½ ` R ¾ ¸ À µ b m G Q ù B Ï x C ~ * N b O * N G Q ù O * %$ g V l & ÏLet ˜ C (f,µ,ν) = e f x µ x ν w µ w ν y µ ′ y ν ′ B k,r,t (mod B f 1 k,r,t), where B f 1 k,r,t is the twosided ideal generated by e f 1 a) The set {e f x µ x ν w µ w ν y µ ′ y ν ′ d (t) dx κ d (mod B f 1 k,r,t) (t,d,κ d) ∈ δ (f,µ ′,ν ′)} is a basis of ˜ C (f,µ,ν) b) As right B k,r,tmodules, ˜ C (fCase where n = 2, and Σ is diagonal, ie, x = x1 x2 µ = µ1 µ2 Σ = σ2 1 0 0 σ2 2 As we showed in the last section, p(x;µ,Σ) = 1 2πσ1σ2 exp − 1 2σ2 1 (x1 −µ1) 2 − 1 2σ2 2 (x2 −µ2) 2 (4) Now, let's consider the level set consisting of all points where p(x;µ,Σ) = c for some constant c ∈ R In particular, consider

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¾ Á » µ Ä Þ ¸ y Ù û = à Ñ è Ä ¡  t r Î ¢ Õ ê ¶ Å ñ Æ r M ë ê ñ à ñ ½ È è z Ï y Q v ^ q O d } ° ¦ ñ Ç ä ´ y A @ Ù v 2 d C } W z v ö 2(0 z ^ y ÿ b O ¤ Ö v u O è » r 2 " r X Q 9 ( ?µ θσ 2 /2 1 = EX = Ee Y = M Y (1) = e µ 2θ2σ 2 2 = EX 2 = Ee 2Y = M Y (2) = e (b) First, note that µ 2 σ 2 2 /(µ 1) = e It follows that a methodofmoments estimate for σ 2 is σˆ 2 = ln(ˆµ 2 /µˆ 2 1) where µˆ 1 = 1 n X i n i =1 µˆ 2 =U Å ½ s Ú è ñ ® v ï Å ½ v < à Å é Ñ b t u è O b j } t u y C N > y ;

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