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Fundamentals of Probability and Statistics for Engineers

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Â344 <strong>Fundamentals</strong> <strong>of</strong> <strong>Probability</strong> <strong>and</strong> <strong>Statistics</strong> <strong>for</strong> <strong>Engineers</strong><strong>and</strong> (11.23) with the Cramer–Rao lower bounds defined in Section 9.2.2. Inorder to evaluate these lower bounds, a probability distribution <strong>of</strong> Y must bemade available. Without this knowledge, however, we can still show, in Theorem11.2, that the least squares technique leads to linear unbiased minimum-varianceestimators <strong>for</strong> <strong>and</strong> ; that is, among all unbiased estimators which are linearin Y , least-square estimators have minimum variance.Theorem 11.2: let r<strong>and</strong>om variable Y be defined by Equation (11.4). Givenasample (x 1 ,Y 1 ), (x 2 ,Y 2 ), . . . , (x n ,Y n ) <strong>of</strong> Y with its associated x values, leastgivenby Equation (11.17) are minimum variancesquare estimators ^A <strong>and</strong> ^Blinear unbiased estimators<strong>for</strong> <strong>and</strong> , respectively.Pro<strong>of</strong> <strong>of</strong> Theorem 11.2: the pro<strong>of</strong> <strong>of</strong> this important theorem is sketchedbelow with use <strong>of</strong> vector–matrix notation.Consider a linear unbiased estimator <strong>of</strong> the <strong>for</strong>mQ * ˆ‰…C T C† 1 C T ‡ GŠY:…11:24†We thus wish to prove that G ˆ 0 if Q * is to be minimum variance.The unbiasedness requirement leads to, in view <strong>of</strong> Equation (11.19),GC ˆ 0:…11:25†Consider now the covariance matrixUpon using Equations (11.19), (11.24), <strong>and</strong> (11.25) <strong>and</strong> exp<strong>and</strong>ing the covariance,we haveNow, in order to minimize the variances associated with the components <strong>of</strong> ,we must minimize each diagonal element <strong>of</strong> GG T . Since the iith diagonalelement <strong>of</strong> GG T is given bywhere g ij is the ijth element <strong>of</strong> G, we must have<strong>and</strong> we obtaincovfQ * gˆEf…Q * q†…Q * q† T g: …11:26†covfQ * gˆ 2 ‰…C T C† 1 ‡ GG T Š:…GG T † ii ˆ Xnjˆ1g 2 ij ;g ij ˆ 0; <strong>for</strong> all i <strong>and</strong> j:Q *G ˆ 0:…11:27†TLFeBOOK

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