Matthieu Bloch
Thursday, September 22, 2022
For \(\matR_\vecy\succ 0\) decomposed as \(\matR_\vecy=\matL\matD\matL^T\) (\(\matL\) lower triangular) \[ \hat{\vecx}_{f} = \mathcal{L}\left[\matR_{\vecx\vecy}\matL^T\matD^{-1}\right]\matL^{-1}\vecy \] where \[ \hat{\vecx}_{f}\eqdef\left[\begin{array}{c}\hat{x}_{0|0}\\\hat{x}_{1|1}\\\vdots\\\hat{x}_{m|m}\end{array}\right]\quad \matR_{\vecy}\eqdef\left[\matR_y(i,j)\right]\quad \matR_{\vecx\vecy}\eqdef\left[\matR_{xy}(i,j)\right] \] and \(\mathcal{L}[\cdot]\) is the operator that makes a matrix lower triangular.
Let \(\matK_s\) denote the smoothing linear estimator, let \(\matK_f\) denote the causal filtering linear estimator. \[ \matK_s = \matK_f + \mathcal{SU}[\matR_y L^{-T}D^{-1}]L^{-1} \] where \(\mathcal{SU}\) denote the strict upper triangularization operator.
The random variable \(\vece_i\eqdef \vecy_i-\hat{\vecy}_i\) is called the innovation
\(\matR_y\) has not reason to be diagonal, but can we "whiten" it with a linear operation? \[ \textsf{Find } \matA\textsf{ such that } \epsilon=\matA\vect\textsf{ has covariance } \matR_\epsilon=A\matR_y\matA^T=\matD \]
\(\matA\) should be non singular to avoid losing information
Many solutions unless we impose more constraints
Required causal relationship: \(\matA=\matL\), lower triangular, and impose \(\matR_y=\matL\matD\matL^{T}\)
LDL decomposition is unique for positive semi definite matrices