Consider $ S^n,$ the space of Schur-convex, simply connected, closed topological $n-$manifolds as subsets of the unit $(n+1)-$cube, which include $p=(0,0,\cdot\cdot\cdot,0)$ and $q=(1,1,\cdot\cdot\cdot,1).$ Consider the set of all Schur-convexity preserving maps from $S^n$ to itself.
Define a class of $S^n,$ denoted $\zeta(\mathbf X),$ in the following way:
Let $ L^n_+$ be the set of all $n$-dimensional nonnegative random vectors $\mathbf X = (X_1, X_2,\cdot\cdot\cdot,X_n)^⊤$ with finite and positive marginal expectations, and let $\mathbf Ψ^{(n)}$ be the class of all measurable functions from $\Bbb R^n_+$ to $[0, 1].$ Then $\zeta(\mathbf X)$ of the random vector $\mathbf X$ with joint CDF $F$ is:
$$\zeta(\mathbf X)=\bigg\{\bigg(\int \psi(\mathbf x)dF(\mathbf x), \int \frac{x_1\psi(\mathbf x)}{E(X_1)}dF(\mathbf x),\cdot\cdot\cdot,\int \frac{x_n\psi(\mathbf x)}{E(X_n)}dF(\mathbf x):\psi \in \mathbf Ψ^{(n)}\bigg)\bigg\}, $$
$$ =\bigg\{\bigg(E\psi(\mathbf X), \frac{E(X_1\psi(\mathbf X))}{E(X_1)},\cdot\cdot\cdot,\frac{E(X_n\psi(\mathbf X))}{E(X_n)}:\psi \in \mathbf Ψ^{(n)}\bigg)\bigg\}. $$
Let me provide a concrete example. I have a 3D graph of $\zeta$ for a bivariate II Pareto distribution with parameters $(\mu_1,\mu_2,\sigma_1,\sigma_2)=(0,0,1,1)$ and $\alpha=9$ so I know the qualitative nature of the shape of any given surface given by any distribution. Here is a picture to give some intuition about the definition:
I was interested in obtaining the volume for $\zeta(\mathbf X).$ I calculated it to be:
$$\mathrm{Vol}(\zeta(\mathbf X))= \frac{1}{(n+1)!}E\big(|\det(Q)|\big). $$
where $Q$ is a $(n+1)\times(n+1)$ matrix whose $i$th row is $(1, \mathbf{ \bar X}^{(i)}),$ $i=1,2,\cdot\cdot\cdot , n+1.$
Here I'll use a normalized version of $\mathbf X$ denoted by $\mathbf{\bar X}$ such that $\bar X_i= X_i/E(X_i),$ $i=1,2,\cdot\cdot\cdot, n.$ Consider $n+1$ $\mathrm{iid}$ $n$-dim. random vectors $\mathbf{ \bar X}^{(1)},\cdot\cdot\cdot, \mathbf{ \bar X}^{(n+1)} $ each with the same distribution as $\mathbf{\bar X}.$
I made the following observation:
$$\lim_{n \to \infty} \frac{\mathrm{Vol}(\zeta(\mathbf X))}{\mathrm{Vol}(H^{n+1})}= \lim_{n \to \infty} \mathrm{Vol}(\zeta(\mathbf X))=0$$
where $H^{n+1}$ is the unit $(n+1)$-cube.
How do we get precise asymptotics for the following?
$$G(x)=\sum_{n \in \Bbb N}^x \mathrm{Vol}(\zeta(\mathbf X)). $$
Let $\mathbf X$ be normally distributed.