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?

• 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
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.