Say, $\psi | Y, {Y}^{*}, {\sigma}^{2}$~$N(\mu, {\Lambda}^{-1})$ this is a bivariate normal

Does the ${\Lambda}^{-1}$ actually mean precision or variance. I was told in some Bayesian papers, the second parameter actually means precision. I didn't think so, but maybe someone else has read more paper would know?

The paper I am reading www.jds-online.com/file_download/353/JDS-746.pdf

  • $\begingroup$ I think people use both, but precision is more common. It would help if you linked to the paper in question. $\endgroup$ – John Salvatier Apr 26 '12 at 20:23
  • $\begingroup$ @JohnSalvatier: jds-online.com/file_download/353/JDS-746.pdf (equation 2.11) $\endgroup$ – user1061210 Apr 26 '12 at 20:32
  • $\begingroup$ In some Bayesian circles, the notation $N(\mu, b)$ does mean a normal random variable with mean $\mu$ and variance $b^{-1}$. Since the inverse of the correlation matrix (I have seen $\Sigma$ used much more often than $\Lambda$ in this context) occurs inside the exponent in the multivariate normal distribution, I would not be surprised if someone says that this usage extends to the mutivariate case as well, with $\Lambda$ denoting the correlation matrix. See the discussion following this question and the answer which summarizes the comments. $\endgroup$ – Dilip Sarwate Apr 26 '12 at 20:57
  • 1
    $\begingroup$ The BUGS language parameterizes its Normal distribution in terms of mean and precision - both for univariate and multivariate versions. But it would be unusual to use this notation in a paper, at least without very clearly noting that one was doing so. $\endgroup$ – guest Apr 26 '12 at 22:37
  • $\begingroup$ @DilipSarwate I have trouble with the VarCov matrix (See answer section). if I invert it's a very small number, like 3.146047e-14 if I don't it's a very big number, like 3.243281e+13. $\endgroup$ – user1061210 Apr 27 '12 at 0:38

It's more conventional to write $N(\mu, \sigma)$ or $N(\mu, \sigma^2)$ rather than $N(\mu, \sigma^{-1})$. $\sigma^2$ denotes the variance.

  • $\begingroup$ So, this paper meant precision matrix, as I thought. Then, why am I getting very small numbers? like -5.292398e-08 2.105710e-08 from bivariate normal. $\endgroup$ – user1061210 Apr 27 '12 at 0:40
  • $\begingroup$ Precision is different from variance but related. The prior is put on precision. The data with the prior leads to a posterior on precision and from that we derive the inference on the varinace. $\endgroup$ – Michael Chernick May 27 '12 at 5:00

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.