Integrating the inverse-Wishart density

Question

It is alleged in this question and in the Wikipedia article and elsewhere that the density function for the inverse-Wishart distribution with $n$ degrees of freedom on $p\times p$ positive-definite symmetric matrices is $$ f(X) = c\times (\det X)^{-(n+p+1)/2} \exp\left( -\frac{1}{2} \operatorname{tr} \left(\Sigma X^{-1}\right) \right)$$ where $\Sigma$ is the scale matrix and $c$ the normalizing constant.

With respect to what measure is this the density? One could guess that it means $$ \Pr( X\in A) = \int_A f(X) \prod_{i,j} dx_{ij} $$ where $x_{ij}$ is the $ij$ entry of the matrix $x,$ i.e. it's with respect to Lebesgue measure on a $p^2$-dimensional space. But then it's really a $p(p-1)/2$-dimensional space, since $x$ is symmetric, and that complicates the question. And that's something of a wild guess. The bounds of integration would be messy since the support of the density is the space of symmetric positive-definite matrices.

Answer 1

In complete analogy to the (non-singular) Wishart distribution with $n$ degrees of freedom, for $n \in \mathbb N_{\geq p},$ the function $f$ can be interpreted as a joint probability density of the $p(p+1)/2$ distinct entries of $X$, e.g. of $\mathrm{vech}(X)$, w.r.t. the corresponding Lebesgue measure defined by $$ \mathrm d\hspace{0.05em} X = \prod_{1\leq j \leq i \leq p} \mathrm d\hspace{0.05em} x_{ij}. $$