Lecture5 - MOMENT GENERATING FUNCTIONS The natural number...

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Unformatted text preview: MOMENT GENERATING FUNCTIONS The natural number e. Consider the irrational number e = { 2 . 7183 . . . } . This is defined by e = lim( n ) 1 + 1 n n . Consider the series expansion of the expression that is being taken to the limit. The binomial expansion indicates that ( a + b ) n = a n + na n 1 b + n ( n 1) 2! a n 2 b 2 + n ( n 1)( n 2) 3! a n 3 b 2 + . Using this, we get 1 + 1 n n = 1 + n 1 n + n ( n 1) 2! 1 n 2 + n ( n 1)( n 2) 3! 1 n 3 + Taking limits as n of each term of the expansion gives lim( n ) 1 + 1 n n = 1 0! + 1 1! + 1 2! + 1 3! + = e There is also e x = lim( p ) 1 + 1 p px = lim( n ) 1 + x n n ; n = px. Using the binomial expansion in the same way as before, it can be shown that e x = x 0! + x 1! + x 2 2! + x 3 3! + . Also e xt = 1 + xt + x 2 t 2 2! + x 3 t 3 3! + . The moment generating function. Now imagine that x f ( x ) is a random variable, and let us define M ( x, t ) = E ( e xt ) = 1 + tE ( x ) + t 2 2! E ( x 2 ) + t 3 3! E ( x 3 ) + . This is the moment generating function or m.g.f. of x . Each of its terms contains one of the moments of the random variable x taken about the origin. The problem is to extract the relevant moment from this expansion. The method is a follows. To find the r th moment E ( x r ), differentiate M ( x, t ) = E ( e xt ) 1 r times in respect of t . Then, set t to zero and the moment drops out. To demon- strate this consider the following sequence of derivatives in respect of the coecient associated with E ( x r ) within the series expansion of M ( x, t ): d dt t r r ! = rt r 1 r ! d 2 dt 2 t r r ! = r ( r 1) t r 2 r ! . . . d r dt r t r r ! = { r ( r 1) 2 . 1 } t r ! = 1 . . . d k dt k t r r ! = 0 for k > r. Thus it follows that d r dt r M ( x, t ) = d r dt r ( X h =0 t h h ! E ( x h ) ) = E ( x r ) + tE ( x r +1 ) + t 2 2! E ( x r +2 ) + t 3 3!...
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Lecture5 - MOMENT GENERATING FUNCTIONS The natural number...

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