elemprob-fall2010-page47

elemprob-fall2010-page47 - P ( a Y n b ) P ( a Z b ). (We...

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The previous proposition then implies that X + Y is a N ( a + c,b 2 + d 2 ). Similarly, if X and Y are independent Poisson random variables with parameters a and b , resp., then m X + Y ( t ) = m X ( t ) m Y ( t ) = e a ( e t - 1) e b ( e t - 1) = e ( a + b )( e t - 1) , which is the moment generating function of a Poisson with parameter a + b . One problem with the moment generating function is that it might be infinite. One way to get around this, at the cost of considerable work, is to use the characteristic function ϕ X ( t ) = E e itX , where i = - 1. This is always finite, and is the analogue of the Fourier transform. The joint moment generating function of X and Y is m X,Y ( s,t ) = E e sX + tY . If X and Y are independent, then m X,Y ( s,t ) = m X ( s ) m Y ( t ). We will not prove this, but the converse is also true: if m X,Y ( s,t ) = m X ( s ) m Y ( t ) for all s and t , then X and Y are independent. The idea behind proving the central limit theorem is the following. It turns out that if m Y n ( t ) m Z ( t ) for every t , then
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Unformatted text preview: P ( a Y n b ) P ( a Z b ). (We wont prove this.) We are going to let Y n = ( S n-n ) / n . Let W i = ( X i- ) / . Then E W i = 0, Var W i = Var X i 2 = 1, the W i are independent, and S n-n n = n i =1 W i n . So there is no loss of generality in assuming that = 0 and = 1. Then m Y n ( t ) = E e tY n = E e ( t/ n )( S n ) = m S n ( t/ n ) . Since the X i are i.i.d., all the X i have the same moment generating function. Since S n = X 1 + + X n , then m S n ( t ) = m X 1 ( t ) m X n ( t ) = [ m X 1 ( t )] n . If we expand e tX 1 as a power series, m X 1 ( t ) = E e tX 1 = 1 + t E X 1 + t 2 2! E ( X 1 ) 2 + t 3 3! E ( X 1 ) 3 + . 47...
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This note was uploaded on 12/29/2011 for the course MATH 316 taught by Professor Ansan during the Spring '10 term at SUNY Stony Brook.

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