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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.

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

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  • $\begingroup$ 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. $\endgroup$ 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 $\hat{cov(x_k,x_k)}=sample covariance(x_k^i)$ $\endgroup$ Feb 28, 2017 at 21:42
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    $\begingroup$ @tintinthong right, we rarely have knowledge of true parameters. Also, this formula only works if your weights are normalized. $\endgroup$
    – Taylor
    Feb 28, 2017 at 21:49

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