0
$\begingroup$

What does this expression mean?

Normal distribution with condition

I am reading a research paper and found the following expression (Eq.28 in the paper below).

enter image description here

It means a Gaussian distribution, but the mean component seems conditional probability-like expression $\it{\bf{s}}_t | \it{\bf{m}}_{b, t, m}^{(j)}$. I have never seen this expression before and cannot find any info about it.

The variables $\it{\bf{s}}_t$ and $\it{\bf{m}}_{b, t, m}^{(j)}$ are both vectors and $\bf{\Sigma}_{b}$ is a covariance matrix.

Does anybody have an idea of what this expression means?

Original paper where the expression is.

The original paper can be found here: https://eprints.soton.ac.uk/437941/1/08340823.pdf

Thanks in advance.

$\endgroup$
3
  • $\begingroup$ The $m$ is the vector of means and $\Sigma$ the covariance of the multivariate normal distribution. The $s$ is in the place of the variable of the density function. So, $\mathcal{N}(x,m,\Sigma)=(2\pi)^{-k/2}|\Sigma|^{-1/2}\exp(-2^{-1}(x-m)^T\Sigma^{-1}(x-m))$, where $k$ is the dimension of the vectors $x,m$ and $\Sigma$ is $k\times k$. $\endgroup$ – plop Aug 4 '20 at 10:36
  • $\begingroup$ Thank you very much, that is exactly what I was looking for. $\endgroup$ – salto Aug 4 '20 at 13:52
  • $\begingroup$ @plop, can you write that as an answer? We discourage writing answers in the comments. Thank you! $\endgroup$ – D.W. Aug 4 '20 at 17:00
1
$\begingroup$

$$\mathcal{N}(x|m,\Sigma)=(2\pi)^{-k/2}|\Sigma|^{-1/2}\exp\left(-2^{-1}(x-m)^T\Sigma^{-1}(x-m)\right)$$

where $x,m$ are vectors of dimension $k$, $\Sigma$ is a $k\times k$ matrix, $|\Sigma|$ is its determinant. The vector $m$ represents the mean of the multivariate normal, and $\Sigma$ the covariance matrix, which must be positive definite.

$\endgroup$

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.