In Bayesian statistics, does variance parameter actually mean 1/variance (i.e. precision)?

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?

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I think people use both, but precision is more common. It would help if you linked to the paper in question. – John Salvatier Apr 26 '12 at 20:23
@JohnSalvatier: jds-online.com/file_download/353/JDS-746.pdf (equation 2.11) – user1061210 Apr 26 '12 at 20:32
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. – Dilip Sarwate Apr 26 '12 at 20:57
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. – guest Apr 26 '12 at 22:37
@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. – user1061210 Apr 27 '12 at 0:38
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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.