93 moment generating functions if x is an m 1 random

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Moment Generating Functions If X is an m 1 random vector, we can define its moment generating function from m to 0, . For m 1 vectors t , X t E exp t X  The MGF is useful only insofar it is finite for t in a neighborhood of zero. But now we need to define the neighborhood using a distance measure in m . This is usually done using Euclidean length. 94
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For an m 1 vector u , its Euclidean length, denote u , is the nonnegative number u ‖ j 1 m u j 2 1/2 The Euclidean distance between two vectors is t u ‖ j 1 m t j u j 2 1/2 95
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An open neighborhood around zero with radius 0 is defined as N 0 u m : u So, for the MGF to be useful, we must assume that there exists 0 such that X t E exp t X  for all t N 0 . 96
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As in the scalar case, we can use the MGF to find moments of the random vector X . In particular X t 0 t X 0 E X X 2 t t 0 t 2 X 0 E XX 97
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We can then compute Var X E XX E X E X If the MGF of X is finite in a neighborhood of zero then it is easy to see that each X j has a finite MGF on an interval about zero. Denote these 1 , 2 ,..., m . We can always write X t E exp t X  E exp j 1 m t j X j E j 1 m exp t j X j by a key property of exp  . 98
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If we now assume X 1 , X 2 ,..., X m are independent then E j 1 m exp t j X j j 1 m E exp t j X j  j 1 m j t j , that is X t X t 1 , t 2 ,..., t m j 1 m j t j . 99
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There is a converse to this statement, too. Namely, if for t j j , j for j 0 we can show X t j 1 m j t j , then X 1 , X 2 ,..., X m are independent. The MGF is particularly convenient for establishing facts about the multivariate normal distribution. 100
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There also is an important relationship between the MGF of a sum and the MGFs of the summands when the summands are independent. Let X 1 , X 2 ,..., X m be independent random variables with MGFs j . Define Y X 1 X 2 ... X m . Y inherits its distribution from that of the random vector X . If the X j are independent, that distribution is often much easier to characterize using the MGF of Y . This is because Y t X 1 X 2 ... X m t j 1 m j t 101
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Proof : We again use a key property of exp  along with independence: Y t E exp tY  E exp t X 1 ... X m  E j 1 m exp tX j j 1 m E exp tX j  j 1 m j t 102
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