In the context of particle filtering. We assume a standard state space model where k is time and i is the particle index.Note: $w_k^i$ are normalised weights. Assume I have a set $\{x_k^i, w_k^i\}_{i=1\cdots N_p}$, what does it mean to compute the "emprirical covariance matrix" of $\{x_k^i, w_k^i\}_{i=1\cdots N_p}$. Take the covariance matrix to be called $S_k$. If it helps,this leads on to the computation of the white transform to have matrix $D_kD^T_k=S_k$.

I have a feeling that $D$ is the covariance matrix of the white transformed of a data matrix $X$ i.e. $Y=W_hX$, $D=Cov(Y)$ and $W_h=E^T$, where $E$ is the matrix of eigenvectors of covariance matrix $S_k$. See here https://theclevermachine.wordpress.com/2013/03/30/the-statistical-whitening-transform/

Does it mean to compute the covariance matrix, $S_k$ where it's elements are $(S_k)_{ij}= cov(x_k^i,x_k^j)$??? Or are the weights somehow incorporated into the covariance?

This appears in the paper https://pdfs.semanticscholar.org/7f85/23f6a2d6a3a4c029416ebf800ccab1abac6d.pdf.


1 Answer 1


Is $k$ time? Are the weights normalized? You didn't define $D$. It's probably $$ \widehat{\Sigma} = \sum_{i=1}^{N_p}w^i_k x^i_k (x_k^i)^T. $$

  • $\begingroup$ thanks for your answer. Your answer has to be correct. In the case where I only have one dimension to deal with, I think the $\hat{\Sigma}$ is just a scalar by your formula. And so $D$ is a diagonal matrix with $\sqrt{\sum{w_k^ix_k^i}}$ in the diagonals. $\endgroup$ Commented Feb 28, 2017 at 21:35
  • $\begingroup$ I was confused because I considered $x^i_k$ as a random variable, when really it is a realisation of a random variable, and hence $\hat{cov(x_k,x_k)}=sample covariance(x_k^i)$ $\endgroup$ Commented Feb 28, 2017 at 21:42
  • 1
    $\begingroup$ @tintinthong right, we rarely have knowledge of true parameters. Also, this formula only works if your weights are normalized. $\endgroup$
    – Taylor
    Commented Feb 28, 2017 at 21:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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