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lecture04 - 16.322 Stochastic Estimation and Control Fall...

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16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Lecture 4 Last time: Left off with characteristic function. 4. Prove φ ( ) = Π φ 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 ) φ ( ) = E e ( s jtX 1 E e jtX jtX n = E e 2 ... E e t = φ ( ) 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 φ 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 φ t ( ) exists for all distributions and the inverse relation φ t ( ) is analytic for all ( ) f ( x ) holds for all distributions. This implies that φ t real values of t. Then it can be expanded in a power series, which converges for all finite values of t . 1 2 0 2 n n 0 0 φ ( t ) = φ (0) + φ ( ) ( ) t + 1 φ ( ) ( ) t + ... + 1 φ ( ) ( ) t + ... 2! n ! jtx dx , φ t ) ( ) = f ( x e φ (0) = 1 −∞ Page 1 of 6
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1 16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde d φ t n jtx f x n ( ) = ( )( jx ) e dx n dt −∞ n n n n 0 ) φ ( ) ( ) = j x n f ( x dx = j X −∞ n n n ( ) X t + ... + 1 ( ) X t + ... φ ( t ) = + jXt + 1 j 2 2 2 j 2! n !
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