I want some help about the initial conditions for the derivative of a Kalman filter. (Differentiating the filtering equations necessary for the calculation of the gradient of the log-likelihood function based on Kalman filtering: $$\frac {\partial \hat{x}_{0|0}}{\partial \theta_l}\text{ and }\frac {\partial P_{0|0}}{\partial \theta_l},\quad l =1,...,p,$$ and $p$ is the number of parameters.) In the Kalman filter it is familiar that $\hat{x}_{0|0}$ and $P_{0|0}$ contain a scalar values, for example the initial conditions of Kalman filter for the estimation of a nonlinear state space model with 3 variables are $\hat{x}_{0|0}=[1, 0, 1]^T$ and $$P_{0|0}= \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{pmatrix}.$$ Are the initial conditions caculated by simply performing the derivative of $\hat{x}_{0|0}$ and $P_{0|0}$ with respect to $\theta_l$, which leads us to that all components of both $\displaystyle\frac {\partial \hat{x}_{0|0}}{\partial \theta_l}$ and $\displaystyle\frac {\partial P_{0|0}}{\partial \theta_l}$ are zeros?

Or should I initialize $\displaystyle\frac {\partial \hat{x}_{0|0}}{\partial \theta_l}$ and $\displaystyle\frac {\partial P_{0|0}}{\partial \theta_l}$ like in Kalman filter by a vector $\displaystyle\frac {\partial \hat{x}_{0|0} }{\partial \theta_l}=[.,.,.]^T$ and a matrix $$\frac {\partial P_{0|0}}{\partial \theta_l}= \begin{pmatrix} . & . & .\\ . & . & .\\ . & . & .\\ \end{pmatrix}\ ?$$


We initialize with the derivative of $\frac {\partial \hat{x}_{0|0}}{\partial \theta_l}$ and $\frac {\partial P_{0|0}}{\partial \theta_l}$ wrt. the parameter. So, indeed, if the initial state distribution does not depend on the parameters, we initialize with zeros.

  • 2
    $\begingroup$ Could you clarify the wording a bit: you probably meant "with the derivative of $\hat{x}_{0 \mid 0}$ and $P_{0\mid 0}$ wrt.the parameter" ("the derivative of $\frac{\partial \cdot }{\partial \theta}$ wrt the parameter" as written in the answer would be the second derivative of $\cdot$ wrt the parameter...) $\endgroup$ Jan 13 '18 at 18:17

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