Expected value for amount of nondominated points in a set of n 2d points,

Question

We have a set of $n$ 2-d points ${(x_1​,y_1),(x_2​,y_2​),…,(x_n​,y_n​)}$ where $x$ and $y$ are sampled from independent normal distributions. A point $(x_a, y_a)$ is said to dominate another $(x_b, y_b)$ if $x_a>x_b$ and $y_a>y_b$. What is the expected value the number of nondominated points in the set?

Answer 1

Let $Z_k$ be the indicator function of the event that the $k$-th point is nondominated, then $Z=Z_1+\dots+Z_n$ is the total number of nondominated points, and thus by additivity and symmetry \begin{align*} \mathbb{E}Z=n \mathbb{E}(Z_n)=n\mathbb{P}(Z_n=1) =n\mathbb{P}(\text{for each $k=1,2,\dots,n-1$ we have }x_k<x_n\text{ or }y_k<y_n) \end{align*} Let us condition on $(x_n,y_n)=(x,y)$. Then \begin{align*} &\mathbb{P}(\forall k\in[1,n]\ x_k<x_n\text{ or }y_k<y_n\mid x_n=x,y_n=y) =\prod_{k=1}^{n-1} \left[1-\mathbb{P}(x_n>x)\mathbb{P}(y_n>y)\right] \\ &=\left[1-\frac1{\sqrt{2\pi}}\int_x^{\infty} e^{-u^2/2}du \frac1{\sqrt{2\pi}}\cdot\int_y^{\infty} e^{-u^2/2}du\right]^{n-1}. \end{align*} Recalling that $x$ and $y$ itself are normal, by taking the expectation we get $$ \mathbb{E}(Z)=\frac{n}{2\pi} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \left[1-\frac1{2\pi}\int_x^{\infty} e^{-u^2/2}du \cdot\int_y^{\infty} e^{-u^2/2}du\right]^{n-1} e^{-x^2/2}e^{-y^2/2} dxdy. $$ By changing the variable $z=1-\Phi(x)$ and $w=1-\Phi(y)$ where $\Phi(t)$ is the CDF of Normal, we get $$ \mathbb{E}(Z)=n\int_0^1\int_0^1 \left[1-zw\right]^{n-1} dz dw=\int_0^1 \frac{(1-w)^n-1}{w} dw=\frac1{n!}\left[n+1\atop 2\right] $$ where the quantity on the right is the Stirling number.