I aim to understand whether the estimation error in a standard Kalman filter with "active noise" can be zero (at some time).

More precisely, a state-space model is given by \begin{align} x_{t+1}&= Fx_t + G w_t\\ y_t&= Hx_t + v_t \end{align} where $w_t,v_t$ are pairwise independent and are i.i.d. according to $w_t\sim N(0,W)$ and $v_t\sim N(0,V)$. Denote the estimation error covariance as: $$\Sigma_{t|t-1} = \mathbf{cov}(x_t-\mathbb{E}[x_t|y^{t-1}]).$$

It is clear that $\Sigma_{t|t-1}$ is positive semidefinite.

  1. If we assume $W,V\succ0$, can we show that $\Sigma_{t|t-1}$ is positive definite at all times? in other words, there will always be an estimation error. (I even think that $W\succ0$ is sufficient)
  2. If the correlation $\mathbb{E}[w_tv_t^T]\neq0$, does it change the answer?

1 Answer 1


In general, $w$ and $v$ are assumed to be uncorrelated, so that doesn't change the answer. The covariance of the Kalman filter (KF) can be written concisely as follows:

$$ \Sigma_{k}^{-1} = \bigr( F_k \Sigma_{k-1} F_k^T + W \bigr)^{-1} + H^T_k V^{-1} H_k $$

where this form is can be computed using the Woodbury Matrix Identity with the most popular form of the KF equations. This form isn't necessary but is the most simple for answering your question.

So, if we assume that $W \succ 0$ and $V \succ 0$ are positive definite, then the covariance matrix as computed by the KF will also be positive definite. This is because of the following properties:

  1. The inverse of a positive definite matrix is positive definite, so if $A$ is positive definite, then $A^{-1}$ is also positive definite.
  2. If $A$ is $n \times n$ positive definite matrix, and $B$ is an $m \times n$ matrix with rank $m$, then $C = BAB^T$ is a positive definite matrix.
  3. If $A$ is an $n \times n$ positive definite matrix and $B$ is an $n \times n$ positive semi-definite matrix, then $A + B$ is positive definite.

This means that both terms in the previous equation are positive definite; thus, the sum of the terms is also positive definite. Therefore, as long as $W$ and $V$ are positive definite, then the covariance matrix will also be positive definite.

  • $\begingroup$ thanks. I do not think that the answer is valid when there is correlation since the prediction term in your formula is not true anymore. Also, in the uncorrelated case, the sufficiency of $V\succ0$ is transparent, but what about $W\succ 0,V\succeq0$? $\endgroup$
    – Morad
    Apr 4, 2021 at 23:53
  • $\begingroup$ Right. The KF assumes the process and observation noise are uncorrelated, so if somehow, there is a correlation, then the KF is not valid. You only need either $W$ or $V$ to be positive definite while the other is positive semi-definite. Properties 1 and 2 hold for positive semi-definite matrices as well (i.e., you can replace the word positive definite with positive semi-definite in 1 and 2) except you don't need rank $m$ on the $B$ matrix. $\endgroup$
    – Ralff
    Apr 5, 2021 at 0:21
  • $\begingroup$ KF is valid when there is correlation, the only thing that changes is the prediction update which is incorrect in your formula. For the sufficiency of $W\succ0$, how does your formula changes when $V$ is not invertible? $\endgroup$
    – Morad
    Apr 6, 2021 at 1:31
  • $\begingroup$ @Morad Can you please share a reference where $w$ and $v$ are correlated? The KF assumes those random variables are not correlated... but I’d be interested if you’ve seen something different in literature. If $V$ is not invertible, then you can’t use the form I provided, but you can instead use the standard form. $\endgroup$
    – Ralff
    Apr 6, 2021 at 1:36
  • $\begingroup$ 1. How can we use the standard form to show that the covariance is PD 2. The standard textbook by Kailath et al. "Linear estimation" presents the general form. $\endgroup$
    – Morad
    Apr 6, 2021 at 2:14

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