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Unformatted text preview: 16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Lecture 4 Last time: Left off with characteristic function. 4. Prove ( ) = i ( ) where X = X 1 + X 2 + ... + X ( X i independent) t t x x n Let S X 1 + X + ... X where the X i are independent. = 2 n ... t jtS = E e jt X 1 + X 2 + + X n ) ( ) = E e ( s jtX 1 E e jtX jtX n = E e 2 ... E e t = ( ) X i This is the main reason why use of the characteristic function is convenient. This would also follow from the more devious reasoning of the density function for the sum of n independent random variables being the n th order convolution of the individual density functions and the knowledge that convolution in the direct variable domain becomes multiplication in the transform domain. 5. MacLaurin series expansion of t ( ) Because f(x) is non-negative and f ( ) x dx = 1 (or, even better, f ( ) x dx = 1 ), it follows that f ( ) x dx = 1 converges so that f(x) is Fourier transformable. Thus the characteristic function t ( ) exists for all distributions and the inverse relation t ( ) is analytic for all ( ) f ( x ) holds for all distributions. This implies that t real values of t. Then it can be expanded in a power series, which converges for all finite values of t . 1 2 2 n n ( t ) = (0) + ( ) ( ) t + 1 ( ) ( ) t + ... + 1 ( ) ( ) t + ... 2! n ! jtx d x , t ) ( ) = f ( x e (0) = 1 Page 1 of 6 1 16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde d t n jtx f x n ( ) = ( )( jx ) e dx n dt n n n n 0 ) ( ) ( ) = j x n f ( x d x = j X n n n ( ) X t + ... + 1 ( ) X t + ......
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- Fall '04