Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.