Assume the following linear discrete system:

$x_k = Fx_{k-1} + w_{k-1}$ where $w_{k} \sim N(0, Q)$

$y_k = Hx_k + v_{k}$ where $v_{k} \sim N(0, R)$

One way to prove that the Kalman filter is optimal is to show that it minimises the following cost function:

$J_m = \frac{1}{2}(y_k - Hx_k)^TR^{-1}(y_k - Hx_k) + \frac{1}{2}(x_k - Fx_{k-1})^TP_{k|k-1}^{-1}(x_k - Fx_{k-1})$

Why do we use $P_{k|k-1}$ instead of Q as the covariance in the second part of $J_m$? By the way, $P_{k|k-1}$ is the covariance of estimation

  • $\begingroup$ Hi: Assuming that $M$ is the variance of the state variable $x_{k}$ and that $Q$ is the variance of $w$, , then $M$ is included because $Q$ is the variance of $w$ so the formula for $M$ will have $Q$ in it. This is because at any update , $M$ is a function of the variance of the state ( often referred to as $C$ but notations vary ) and $Q$. I'm not sure if this answers your question. It's best to explain what $M$ and $Q$ are because the notation is often not consistent. $\endgroup$ – mlofton Sep 8 '18 at 17:34
  • $\begingroup$ If we know $x_{k-1}$, then isn't the variance of $x_k$ just Q instead of M? $\endgroup$ – Vykta Wakandigara Sep 8 '18 at 17:39
  • $\begingroup$ Hi: Since this is a KF, then you never actually know $x_{k}$. It's a state that has a variance associated with it at each step $k$. It's best to read up on how the kalman filter update is derived. Then you will see what I mean more clearly. $\endgroup$ – mlofton Sep 9 '18 at 19:32
  • $\begingroup$ I have made the appropriate edits $\endgroup$ – Vykta Wakandigara Sep 10 '18 at 18:34

Because you are using Bayes' theorem (and conditional independence):

$$ p(x_k \mid y_{1:k}) \propto p(y_k \mid x_k) p(x_k \mid y_{1:k-1}). $$ Note that, regarding the observation density: $$ p(y_k \mid x_k) \propto \exp\left[-\frac{1}{2}(y_k - Hx_k)^TR^{-1}(y_k - Hx_k) \right]. $$ Also, the "prior" for the state, or the state prediction density is $$ p(x_k \mid y_{1:k-1}) \propto \exp\left[-\frac{1}{2}(x_k - Fx_{k-1})^TP_{k \mid k-1}^{-1}(x_k - Fx_{k-1}) \right]. $$

You do not use $Q$ because that is the covariance matrix for the state transition distribution $p(x_k \mid x_{k-1})$. However, you would use $Q$ in the calculation of $P_{k \mid k-1}$ because that state prediction density can be written in terms of your outdated filtering distribution, and the state transition distribution: $$ p(x_k \mid y_{1:k-1}) = \int p(x_k \mid x_{k-1})p(x_{k-1} \mid y_{1:k-1}) dx_{k-1}. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.