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The parameter estimation errors are a = a a and = b What are the b b mean parameter estimation errors, the variance of the parameter estimation errors, and the covariance between the parameter estimation errors 1 The mean value of a is E a = E b n = 0 2 The mean value of is E = E b 0 = 0 b b 3 The covariance of a is E a2 = E (b + n)2

1 2 3 1 2 3 1 2 3 1 2 3

Value 1,000 pF /001 F 1,500 pF /0015 F 2,000 pF /002 F 2,200 pF /0022 F 4,700 pF /0047 F 5,000 pF /005 F 5,600 pF /0056 F 6,800 pF /0068 F 01 015 02 022 033

5 The covariance between a and is b E a b E (a a)(b b) = E (b + n)(b b) 2 = E b nb 2 = b =

1 1 1 1 1 1 1 1 1 1 1 1

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We won t be able to make any of these elements vanish without making a whole row vanish, giving us the equation 0x + 0y + 0z = 0 which is utterly useless However, the above matrix tells us that x+y+z=1 An infinite number of ordered triples (x,y,z) satisfy this equation Our original three-bythree linear system is actually one equation stated three different ways It s redundant, so a single solution does not exist

Based on the results of Steps 3 5 and eqn (423), the correlation coe cient between a and is a = b 2 b 2 b

Type Ceramic / Mylar Ceramic / Mylar Ceramic / Mylar Ceramic / Mylar Ceramic / Mylar Ceramic / Mylar Ceramic / Mylar Ceramic / Mylar Ceramic / Mylar Mylar Mylar Mylar Mylar Mylar Mylar Mylar Mylar Mylar Mylar Mylar Mylar Mylar Mylar Mylar Mylar

These worked-out solutions do not necessarily represent the only way a problem can be figured out If you think you can solve a particular problem in a quicker or better way than you see here, by all means go ahead! But always check your work to be sure your alternative answer is correct

If we de ne P = cov([ , ), then based on the above analysis a b] P= which checks with eqn (424) Analysis similar to that of Example 49 will have utility in Part II for the initialization of navigation systems Related to this example, the case where a is a vector and the case where y is a nonlinear function of a are considered in Exercise 411

21

For the vector random variable x Rn , the notation x N ( , P) is used to indicate that x has the multivariate Gaussian or Normal density function described by px (X) = 1 (2 )n det(P) 1 exp (X ) P 1 (X ) 2

1 The 0th power of j is equal to 1 We know that j 2 = 1, so according to the difference of powers law, j 0 = j 2 2 = j 2/j 2 = 1/( 1) =1 2 We determined in this chapter that j 2 = 1, j 4 = 1, j 6 = 1, j 8 = 1, and so on for increasing even-integer powers of j Based on this knowledge, we can use the negative powers rule to determine the following facts: j 2 = 1/(j 2) = 1/( 1) = 1 j 4 = 1/(j 4) = 1/1 =1

Code 102 152 202 222 472 502 562 682 103 203 223 333 473 503 563 683 104 204 224 334 474 564 105 205

Again, the density of the vector Normal random variable is completely described by its expected value Rn and its covariance matrix P Rn n

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