Prove/disprove the matrix convergence in probability

Question

Problem:

Given $\alpha\in\mathbb{R}^+$, $N\in \mathbb{Z}^+$ and $D\in \{N, N+1, N+2, \cdots\}$, a random matrix $\mathbf{A}$ is generated by the following steps:

$(1)$ Randomly select $N$ numbers from $\{1,2,\cdots,D\}$ to form a sequence $p=\{p_i\}_{i=1}^N$.

$(2)$ Then calculate $\mathbf{A}=[a_{ij}]_{N\times N}$, where $a_{ij}=e^{-\alpha |p_i - p_j|}$.

Please prove or disprove the following proposition:

$\mathbf{A}$ converges to $\mathbf{I}_N$ in probability, i.e., for any $\epsilon>0$ and choice of norm $\|\cdot\|$, there is: $$ \mathbb{P}[\|\mathbf{A}-\mathbf{I}_N\|\geq\epsilon]\to0~~(D\rightarrow \infty). $$

---

My Efforts:

I am confused by how to start.

I may know that the diagonal elements of $\mathbf{A}$ will be all ones, since $|p_i-p_i|=0$.

And I may know that all elements of $\mathbf{A}$ are in $[0,1]$ and $\mathbf{A}$ is symmetric.

Intuitively, I guess that when $D$ increases, the absolute distances between each two $p_i$s may become larger and larger, so $a_{ij}$ is expected to be smaller and smaller.

I also write the following Python program for numerical validation:

import numpy as np
import random
from scipy import spatial
alpha = 1
N = 10
I = np.eye(N)
for D in range(N, 10000):
    MSE = 0.0
    for i in range(100):
        p = np.array(random.sample(range(1, D + 1), N)).reshape(N, 1)
        A = np.exp(-alpha * spatial.distance.cdist(p, p))
        MSE += np.sum((A - I) ** 2.0)
    MSE /= (100 * N * N)
    print(MSE)

I can see that when $D$ increases, the mean squared error between $\mathbf{A}$ and $\mathbf{I}_N$ converges to zero.

0.027683220252563596
0.02508590350202309
0.02317795057344325
...
0.0001934704436327538
0.00032059290537374806
0.0003270223508894337
...
5.786435956425624e-05
1.1065792791574203e-05
5.786469182583059e-05

How to prove/disprove the proposition by exactly analysing the process of $D\rightarrow \infty$?

Answer 1

The (original, now edited out of the question) claim that $A=I_N$ whp is false.

For any $j\neq k$, the $(j,k)$^th entry of $I_N$ is $0$. But the $(j,k)$^th entry of $A$ is $e^{-\alpha|p_j-p_k|}$. Thus \begin{align*} \mathbb{P}[{A=I_N}]&\leq\mathbb{P}[{0=e^{-\alpha|p_j-p_k|}}] \\ &=\mathbb{P}[{\infty=\alpha|p_j-p_k|}] \\ &=0 \end{align*} since $\alpha$ is a constant and $|p_j-p_k|$ is a.s. finite.

With that said, $A$ converges to $I_N$ in probability: for any $\epsilon$ and choice of norm $\|\cdot\|$, the value $$\mathbb{P}[\|A-I_N\|\geq\epsilon]\to0$$

To see this, note that all norms are equivalent on the finite-dimensional vector space of $N\times N$ matrices; I will choose the $\infty\to1$ norm $$\|M\|=\sum_{j,k}{|M_{j,k}|}$$

Then \begin{align*} \|A-I_N\|&=\sum_{j,k}{|(A-I_N)_{j,k}|} \\ &=\sum_{j<k}{2e^{-\alpha|p_j-p_k|}} \tag{1} \end{align*}

Now, pick your favorite function $f(D)$ that is $\omega(1)$ (i.e., grows without bound) but $o(D)$ (for example, $f(D)=\sqrt{D}$). Since $p_j$ and $p_k$ are (assumed) pairwise independent, $$\mathbb{P}[{|p_j-p_k|\leq f(D)}]\leq\frac{f(D)}{D}\to0$$ Taking the complement, $$2e^{-\alpha|p_j-p_k|}\leq2e^{-\alpha f(D)}\to0$$ on an event of probability arbitrarily close to $1$.

The value (1) is the sum of $\binom{N}{2}$-many such terms; the latter is finite and independent of $D$, so if each term converges to $0$ whp, the sum must converge to $0$ as well.