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2 ANALYTIC FUNCTIONS 3 Sequences going to z 0 are mapped to sequences going to w 0 Example 23 Many functions have obvious limits For example lim z!2 z2 = 4 and lim z!2 (z2 2)=(z3 1) = 6=9 Here is an example where the limit doesn’t exist because di erent sequences give di. ^ dKZ U W E ,^, > E ' Z >K< ^, ZD õ í ì ì í ì õ ð ì õ X l Z o P u o X } u u v o o í í D Z i u í í ð ð D t Z ,K^W/d > Wsd >d D Z r í ó ñ U , Z/ , h hW ,z z E ' Z D /E U E Z '>/d /E D Z E/d/E ^ E ,z. · Respostas dolivrogeometriaanaliticaalfredosteinbruchepaulowinterle 1 28 Problemas Propostos 1 Determinar a extremidade do segmento que representa o vetor v = (2, −5), sa bendo que sua origem ´e o ponto A(−1, 3).
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F § ï Â Ó è µ æ æ µ ¤ y Ú ï ³ ã ï A ¨ å D Ô ~ G N ¢ M ~ A N ¢ ¤ y Ú ï ³ ã ï70 · A ¨ D § U ò j ' w ¨ A x p à f ª. Basically E X Y = E E X Y Y = E Y E X Y The first step is the iterated rule of conditional expectation For the second, use the fact that given Y, Y is like a constant However if you are looking for the usage of rigorous definition of conditional expectation, the solution by Davide Giraudo is the one to go for Share. H&&/ /K WK^d > ~ í ì l í î í dZ ^ KZ >E Z/K E µ u } / v ( À } > Z u } W } o } v o } }.
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1 Answer1 One wants to show that E ( T ∣ X, Z) = T with T = E ( Y ∣ X) This holds true in full generality since (i) the random variable T is σ ( X) measurable by definition hence T is σ ( X, Z) measurable, and (ii) E ( U ∣ X, Z) = U for every σ ( X, Z) measurable random variable U. E Z v © Z } v Z ^ Z o Z µ Z X À Z Ì Z ^ P X > ï ( } Z µ À o v P µ o o µ À } o v µ o v P µ v ( } µ Z X. E } ó r í ì Á } l v P Ç E } z î / z d Z Ç } / v } v E } î r ï Á } l v P Ç z E } î / o v E } v l E } î / o E } v l E } ò.
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