Page 1 Least squares solution with equality constraints l Ax v = + , 0 ...
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Page 1 Least squares solution with equality constraints l Ax v = + , 0 ...
Least squares solution with equality constraints l Ax v. = + ,. 0. Cx d. + = 2 (. ) (. ) (. ) 2 (. ) T. T. T. T v Pv k Cx d l Ax P l Ax k Cx d. Φ = +. + = -. -. +. +. 2(. ) ( ) 2. 0. T.
Least squares solution with equality constraints
l = Ax + v ,
Cx + d = 0
Φ = vT Pv + 2k T (Cx + d ) = (l − Ax)T P(l − Ax) + 2k T (Cx + d ) ∂Φ = 2(l − Ax)T P(− A) + 2k T C = 0 ∂x
⇒
AT PAxˆ + C T k = AT Pl
∂Φ = −2(Cx + d )T = 0 ∂k
⇒
Cxˆ + d = 0
xˆ = ( AT PA)−1 AT Pl − ( AT PA) −1 C T k
Cxˆ + d = C ( AT PA) −1 AT Pl − C ( AT PA)−1 C T k + d = 0 k = [C ( AT PA)−1 C T ]−1 C ( AT PA) −1 AT Pl + [C ( AT PA)−1 C T ]−1 d xˆ = ( AT PA)−1 AT Pl − ( AT PA) −1 C T {[C ( AT PA)−1 C T ]−1 C ( AT PA) −1 AT Pl + [C ( AT PA) −1 C T ]−1 d } xˆ = ( AT PA)−1 AT Pl − ( AT PA) −1 C T [C ( AT PA)−1 C T ]−1 C ( AT PA)−1 AT Pl + d
xˆ0 = ( AT PA)−1 AT Pl
unconstrained solution
xˆ = xˆ0 − ( AT PA) −1 C T [C ( AT PA) −1 C T ]−1[Cxˆ0 + d ] Covariance propagation:
l = Ax + v
v ~ (0,σ 2 P −1 ) , Cv = σ 2 P −1
AT PA ≡ N
xˆ = N −1 AT P( Ax + v) − N −1C T [CN −1C T ]−1 CN −1 AT P( Ax + v) + d =
= N −1 AT PAx + N −1 AT Pv − N −1C T (CN −1C T )−1 CN −1 AT PAx + CN −1 AT Pv + d = = x + N −1 AT Pv − N −1C T (CN −1C T ) −1 Cx + CN −1 AT Pv + d = = x + I − N −1C T (CN −1C T )−1 C N −1 AT Pv
C xˆ = {[ I − N −1C T (CN −1C T ) −1 C ]N −1 AT P} Cv {[ I − N −1C T (CN −1C T )−1 C ]N −1 AT P} , Cv = σ 2 P −1 T
1
σ
2
Cxˆ = {[ I − N −1C T (CN −1C T ) −1 C ]N −1 AT P} P −1 {[ I − N −1C T (CN −1C T ) −1 C ]N −1 AT P} =
= {[ I − N −1C T (CN −1C T ) −1 C ]N −1 AT P} P −1 { PAN −1[ I − C T (CN −1C T )−1 CN −1 ]} =
T
= { I − N −1C T (CN −1C T ) −1 C} N −1 AT PP −1 PAN −1 { I − C T (CN −1C T ) −1 CN −1} = = { I − N −1C T (CN −1C T ) −1 C} N −1 { I − C T (CN −1C T ) −1 CN −1} =
= N −1 − 2 N −1C T (CN −1C T )−1 CN −1 + N −1C T (CN −1C T ) −1 CN −1C T (CN −1C T ) −1 CN −1 = = N −1 − N −1C T (CN −1C T ) −1 CN −1 Cxˆ = σ 2 N −1 − N −1C T (CN −1C T ) −1 CN −1
σˆ 2 =
vˆT Pvˆ (l − Axˆ )T P(l − Axˆ ) = f f
f =degrees of freedom = number of observations – number of unknowns + number of constraints
Cxˆ = σ 2 N −1 − N −1C T (CN −1C T ) −1 CN −1
----------------------------------------------------------------------------------------Otherwise: C x0 = σ 2 ( AT PA) −1 ≡ σ 2 N −1 x = x0 − N −1C T [CN −1C T ]−1[Cx0 + d ] = [ I − N −1C T (CN −1C T ) −1 C ]x0 − N −1C T (CN −1C T )−1 d C x = [ I − N −1C T (CN −1C T )−1 C ]Cx0 [ I − N −1C T (CN −1C T ) −1 C ]T 1
σ
2
Cx = [ I − N −1C T (CN −1C T ) −1 C ]N −1[ I − C T (CN −1C T )−1 CN −1 ] =
= N −1 − 2 N −1C T (CN −1C T )−1 CN −1 + N −1C T (CN −1C T ) −1 CN −1C T (CN −1C T ) −1 CN −1 = = N −1 − 2 N −1C T (CN −1C T ) −1 CN −1 + N −1C T (CN −1C T ) −1 CN −1 = Cx = σ 2 N −1 − N −1C T (CN −1C T ) −1 CN −1
Use of matrix identity:
( A + C T KC ) −1 = A−1 − A−1C T ( K −1 + CA−1C T ) −1 CA−1
