Expected number of Pareto-optimal points for $d$-dimensional points

Question

I was wondering if it is possible to generalize any of the solutions previous question for the 2-dimensional case to derive the size of the set of the Pareto optimal points for an arbitrary dimension $d>2$.

Here is where I start my attempt (I'm not a mathematician): if I naïvely attempt to generalize the reasoning to a finite set of $d$-dimensional points arbitrarily in $\mathbb{R}^d$, I guess that I can always normalize a finite set of points to be enclosed within $[0,1]^d$, so this assumption from the previous solution should not be too restrictive. Thus, this boils down to evaluate the following expression: $$P(I_i=1)=\int_0^1\dots\int_0^1 \left(1-\prod_{i=1}^d x_i\right) dx_1\cdots dx_d$$

But then, I stopped, as I never had to compute an integral with an arbitrarily large number of variables in my life. Perhaps there is a more straightforward solution, but I fear I would trivialise the approach too much (I sense that the trivial consideration for $H_n^d$ is not necessarily the right solution).

Answer 1

In view of the linked thread, an exponent is missing and we have $$ P(I_i=1)=\int_0^1\dots\int_0^1 \left(1-\prod_{i=1}^d x_i\right)^{n-1} dx_1\cdots dx_d=:I_{n,d}. $$ One has, for $a\in (0,1)$, \begin{align} \int_0^1(1-at)^{n-1}dt&=\left[-\frac{(1-ta)^n}{an}\right]_{t=0}^{t=1}\\ &=\frac{1-(1-a)^n}{an}\\ &=\frac 1n\sum_{j=0}^{n-1}(1-a)^j\\ &=\frac 1n\sum_{k=1}^{n}(1-a)^{k-1} \end{align} hence letting the role of $a$ played by $x_1\dots x_{d-1}$ in $I_{n,d}$ where $x_1,\dots,x_{d-1}$ are fixed, we find that
$$ I_{n,d}=\frac 1{n}\sum_{k=1}^nI_{k,d-1}. $$ Since $I_{k,1}=\frac 1k$, we recover Dinesh's result. Note that $$ I_{n,3}=\frac 1n\sum_{k=1}^nI_{k,2}=\frac 1n\sum_{k=1}^n\frac 1k\sum_{i=1}^kI_{i,1}, $$ hence for $d=3$, denoting $H_k=\sum_{\ell=1}^k \frac 1\ell$, the wanted expectation is $$ \sum_{k=1}^n\frac 1k\sum_{i=1}^k\frac 1i= \sum_{k=1}^n\frac 1k H_{k}. $$ I have not looked whether a nice general formula exists for arbitrary $d$ but here is the idea.

Answer 2

There is a well-known general result: $$I=\int_{(0,1)^{d}}f\left(\prod_{k=1}^{d}x_k\right)dV=\int_{0}^{1}f(u)\frac{1}{(d-1)!}\log^{d-1}\left(\frac{1}{u}\right)du$$ note: $dV=dx_1\wedge dx_2\wedge\cdots\wedge dx_d$.

See this question for more information (about probability aspect) and other derivations.

Proof using multivariable calculus:
Change variable $\forall l>0,u_l=\prod_{k=1}^l x_k\implies \begin{cases}x_l=\frac{u_l}{u_{l-1}}, l\ge1\\ u_0 = 1\end{cases}$ . The Jacobian is: $$J_{x\mapsto u}=\left(\frac{\partial}{\partial u_j}x_i\right)_{i,j}=\left(\frac{\partial}{\partial u_j}\frac{u_i}{u_{i-1}}\right)_{i,j}=\left([j=i]u_{i-1}^{-1}-[j=i-1]u_iu_{i-1}^{-2}\right)_{i,j}$$ $J_{x\mapsto u}$ is triangular, hence its determinant is product of main diagonal elements, we have $$|\det(J_{x\mapsto u})|=\prod_{k=1}^du_{k-1}^{-1}=\prod_{k=1}^{d-1}u_k^{-1}$$ The new bounds is obtained by solving $0<\frac{u_k}{u_{k-1}}<1, k=1,..,d$. This set of inequalities is equivalent to $0<u_d<u_{d-1}<...<u_1<1$, thus our bounds are $u_{k+1}<u_k<1,k=1,...,{d-1}$ and $0<u_d<1$, for convenience, rename $u_d$ as $u$
Our integral is $$I=\int_{(0,1)^{d}}f\left(\prod_{k=1}^{d}x_k\right)dV=\int_{0}^1 f(u)du\int_u^1\frac{du_{d-1}}{u_{d-1}}\int_{u_{d-1}}^1\frac{du_{d-2}}{u_{d-2}}\ldots\int_{u_2}^1\frac{du_1}{u_1}$$ Note that this integral must be calculated from right to left, inductively: $$I=\int_{0}^{1}f(u)\frac{1}{(d-1)!}\log^{d-1}\left(\frac{1}{u}\right)du$$ [end proof]
To solve OP's problem, let $f(x)=(1-x)^{n-1}$ $$I_{n,d}:=\int_{(0,1)^{d}}\left(1-\prod_{k=1}^{d}x_k\right)^{n-1}dV=\int_{0}^{1}(1-u)^{n-1}\frac{1}{(d-1)!}\log^{d-1}\left(\frac{1}{u}\right)du$$ $$\implies \sum_{d\ge1}I_{n,d}x^{d-1}=\int_{0}^{1}u^{-x}(1-u)^n du=\frac{\Gamma(1-x)\Gamma(n)}{\Gamma(n+1-x)}$$ This generating function can be used to calculate $I_{n,d}$