E x x e xx x x e xx e x e x e xx e xx where the last

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E  X  X  E XX X X  E XX E X E X  E XX    E XX  where the last expression can be useful as a computational device. 71
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Sometimes we need to use the variance matrix of a row vector. Naturally, we will arrange the variances and covariances in the same way as when X is a column vector. For computation, it is easiest to just note that if X is a row vector then we define Var X Var X . For now, we take X to be a column vector. 72
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Why does it make sense to define Var X as we have? Consider linear combinations of X that are random variables. Let a be an m 1 vector and let Y a X a 1 X 1 ... a m X m . Then we already know that Y E Y a E X a X j 1 m a j j . 73
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Now we can compute Var Y : Var Y E  Y Y 2 E  a X a X 2 E  a X X  2 E  a X X  a X X  E a X X  X X a a E  X X  X X a a Var X a . 74
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So we have shown that, given the definition of Var X , Var a X a Var X a We can write this out using summations as Var j 1 m a j X j j 1 m a j 2 j 2 2 i 1 m 1 j i 1 m a i a j ij With a 1,1,. ..,1 , Var j 1 m X j j 1 m j 2 2 i 1 m 1 j i 1 m ij 75
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If we have two RVs X and Y , Var aX bY a 2 Var X b 2 Var Y 2 abCov X , Y Var X Y Var X Var Y 2 Cov X , Y Var X Y Var X Var Y 2 Cov X , Y If X and Y are uncorrelated, Var X Y Var X Y X 2 Y 2 76
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We already noted that Var X is always a symmetric matrix. It has another important property. Because Var a X 0 for any m 1 vector a Y a X is a random variable, and so its variance must be nonnegative – we must have a Var X  a 0 This means that Var X is always a positive semi-definite (psd) matrix. If strict inequality holds for all a 0 then the matrix is positive definite (pd). 77
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If a Var X  a 0for a 0 then Var a X 0 for a nontrivial linear combination of X , which means that a X c (with probability one), where c is a constant. Consequently, at least one of the elements of X is an exact linear function of the other elements of X .If a 1 0, X 1
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E X X E XX X X E XX E X E X E XX E XX where the last...

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