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M 8k ê ö b *Ë #Õ q _ > E q 3 b N 4 _ 6õ M 2 '¼ b M 4 5 G M 2 L 0)X '¨ M ê ö b *Ë #Õ q _ > E q 3 b N 4 _ 6õ M 2 b M 4. HOMEWORK 4 SOLUTIONS Exercise 1 Let X, Y be two random variables on (Ω, F, P) Let A⊂F be a subσalgebra The random variables X and Y are said to be independent conditionally on A is for every nonnegative measurable. F E 9 så x E c±Ï E 9± , ¼ FFF F 9 å s 9±Ï E 9± , X± 9 ´å x ±Ï F 9 å 9 %± 9 ı a±Ú s Úk 9 , 9å N ?Ø È ¼ N x c E ´ å 5Æ så x È È µ µ µ cÚ± 9 ı 5± x 9 ´å x ±Ï E 9 cÏ E 5.

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This is part of a long question im doing I have just come from e^y = e^x c to get to y = ln(e^x c) so dont want to go backwards, (e both sides) I recall (ln) rules are a bit wierd so im sorta stuck I final got y = x ln(c), But i dono if i can ln(e^x c) = ln(e^x) ln(c) Can some1 check this?. E^x = 271 This article is compiled by Rahul and reviewed by GeeksforGeeks team Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above Attention reader!. 3 " ¥ f & ò E ¥ d ö V 1 $ Æ ¥ H ¸" ð ¦ µ ¥ ¤ ¹ ù ¤!.

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E Ê ¯ ~ ª » ° > a J = ¨ ü = É s £ Ä ¬ Ê ¯ ± Ù ¶ î É Â ò / ¯ A s í ¥ Í º ² Ù ´ è ï ) ÷ ï æ ) 8 Ë L r s> è ß C Ë L r s> è F · ò ð ÿ ´ ¸ Þ r Ç Õ ¯ Ó Ç ¢ ï * ù ù ä @ # , 3 ø. To Z and Y we obtain EE(ZjY)= E(Z) which is the same as EE(g(X)jY)= E(g(X)) 2 This property may seem to be more general statement than (i) in Theorem 1 The proof above shows that in fact these are equivalent statements 3 E(XYjY)=YE(XjY) Proof E(XYjY =y)=E(yXjY =y)=yE(XjY =y)(because y is a constant) Hence, E(XYjY)=. è Â" Å S E Ð E ® Ú X æ$"#$%&’()*,/#0% 4 ³ B ñ µ ¥ C `!!.

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