proof e xy e x e y

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proof e xy e x e y This shows why independence of X and Y implies that E XY E X E Y The converse does not necessarily hold that is we can come up with examples of random

X E X and Y E Y and k be a positive integer 1 The kth moment of X is de ned as E Xk If k 1 it equals the expectation 2 The kth central moment of X is de ned as E X X k I 1x i y i approaches the expectation E XY For example if X is height and Y is weight E XY is the average of height weight We are interested in E XY because it is used for calculating

proof e xy e x e y

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Basically E XY E E XY Y E YE 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 Theorem 1 We have 1 For any two random variables X and Y E X 2 For any real number a E aX aE X 3 For any real number a var aX a2var X Y E X E Y Proof For 1

Here s the proof Cov X Y E X X Y Y E XY XY X Y X Y E XY XE Y E X Y X Y E XY X Y Covariance can be positive zero or negative Positive indicates that there s an For random variables X and Y we have jE XY j q E X2 q E Y2 Equality holds if and only if P X cY 1 for some constant c Proof For any real k we have E kX Y 2 0 This implies

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From the proof X Y 1 iff X E X X Y E Y Y equality with probability 1 i e iff X E X is a linear function of Y E Y In general X Y is a measure of how closely X or estimated by a 1 If X and Y are independent r v s then E XjY E X Proof As we know X and Y are independent if and only if fX Y x y fX x fY y or equiva lently fXjY xjy fX x But then

Let X colon Omega to mathbb R k and Y colon Omega to mathbb R k be random variables in some probability space Omega mathcal F mathbb P and let alpha Conditional Expectation as a Function of a Random Variable Remember that the conditional expectation of X given that Y y is given by E X Y y xi RXxiPX Y xi y Note that

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proof e xy e x e y - In general we have mathbb E XY mathbb E mathbb E Y mid X X which tells that expectation of XY can be computed by first averaging Y over given