Linear LMS estimation based on multiple observations. Let

Chapter , Problem 22

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Linear LMS estimation based on multiple observations. Let 8 be a random variable with mean J-L and variance a5, and let Xl , ... , Xn be observations of the form Xi =8+Wi, where the observation errors Wi are random variables with mean 0 and variance a;. We assume that the random variables e, WI , ... , Wn are independent. Verify that the linear LMS estimator of e based on X I , . . . , X n is n J-L/a5 + L Xi/a; , i=I 8 = -----::- n --- L l/a; i=O by minimizing over al , ... , an, b the function Solution. We will show that the minimizing values of aI , ... ,an. b are b" = I1la'5 n L Ila; i=O " I/a; aj = n j = 1, . . . , n. L Ila; i=O 453 To this end, it is sufficient to show that the partial derivatives of h, with respect to aI, ... ,an, b, are all equal to 0 when evaluated at ai , ... ,a, b" . (Because the quadratic function h is nonnegative, it can be shown that any point at which its derivatives are zero must be a minimum.) By differentiating h, we obtain oh ob From the expressions for b" and a;, we see that Using this equality and the facts n L b* a: -1 = --. . 11 .=1 E[e] = 11, E[Wi] = 0, it follows that oh ob Using, in addition, the equations E [X, (I1 - 8)] = E [(8 -11 + W. + 11) (11 - 8)] 2 = -ao, E[XiWi] = E [(8 + WdWi] = a;, for all i, = 0. for all i and j with i =J. j, we obtain oh oa. o " .b" = E [x; ( ) 8+ t.a:w. +b-) 1 1 = 0. where the last equality holds in view of the definitions of ai and b* .

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