# Confusion over a likelihood function involving a covariance matrix

I am working through a paper on global optimization using kriging. I can't tell if a term in one of the equations describes the determinant of a matrix, or what. We have $n$ observtions $y=(y_{1} y_{2} \dots y_{n})'$ at $n$ points $(x_{1} x_{2} \dots x_{n})$. These are related by a $n \times n$ matrix, $R$, whose (i,j) elements are given by,

$Cov(Y(x_{i}),Y(x_{j})) = \exp\bigl(-\sum \theta \vert x_{i}-x_{j} \vert^{p} \bigr)$.

We'd like to estimate the parameters $\theta$, $p$, and also the mean vector and variance matrix of the distribution, $Y=(Y(x_{1}) Y(x_{2}) \dots Y(x_{n}))'$. This is where I have an issue--the likelihood function is given by,

$\dfrac{1}{(2\pi)^{n/2}(\sigma^{2})^{n/2}\vert R \vert^{1/2}} exp\biggl(\dfrac{-(y-\mu)'R^{-1}(y-\mu)}{2\sigma^{2}}\biggr)$,

I understand that the numerator in the exponential term evaluates to a scalar, so that the exponential evaluates to a scalar, but what about the $\vert R \vert^{1/2}$ in the denominator of the first term? Is that the square root of the determinant? The paper doesn't mention anything about determinants and I'm unsure how to proceed. Thanks so much!

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Yes, it is the square root of the determinant. The expression is the density of a multivariate normal distribution. –  Stefan Hansen Nov 23 '12 at 19:42
Thanks so much! You're a champ! –  cjohnson318 Nov 23 '12 at 21:44