Rate of convergence of (random) minimum Euclidean distance

Question

Suppose two random variables $X$ and $Y$ are independent and both distributed uniformly on $[0,1]$. I am interested in the vectors $Z=(X, Y)$.

Suppose I have $N$ independent realizations of $Z$: $Z_{i}=(X_i, Y_i)$, $i=1, \ldots, N$.

I know the following univariate results: $N\cdot\min\limits_{i=1, \ldots, N}(1-X_i)$ converges in distribution to a non-degenerate distribution and $N\cdot\min\limits_{i=1, \ldots, N}(1-Y_i)$ converges in distribution to a non-degenerate distribution.

But now I am interested in considering something like for bivariate $Z$. In other words, I am interested in getting the rate of convergence of $$\min\limits_{i=1, \ldots, N}\|(1,1)-(X_i,Y_i)\|,$$ where $\|\cdot\|$ stands for the Euclidean distance. I am mostly interested in the rate of convergence $a(N)$ such that $$a(N)\cdot\min\limits_{i=1, \ldots, N}\|(1,1)-(X_i,Y_i)\|$$ converges in distribution to a non-degenerate limit (or some bounds on this $a(N)$), not in the limiting distribution itself.

How do I proceed?

Answer 1

1. If $X$ is $U[0,1]$, so is $1-X$.
2. In light of the above and the independence of $X_i,Y_i$, $(1,1)-(X_i,Y_i)$ has the same distribution as $(X_i,Y_i)$.
3. Let $D_i = \|(X_i,Y_i)\|=\sqrt{X_i^2+Y_i^2}$. Then for any $x>0$, $$P(D_i >x) = P(\sqrt{X_i^2+Y_i^2} > x).$$ If $x<1$, this is equal to $1-\frac\pi 4 x^2$ (draw the picture to see why).
4. By independence, $$P(a_n \min_{i=1}^n D_i >\alpha) =P(\min_{i=1}^n D_i >\alpha/a_n)=P(D_1>\alpha/a_n)^n=(*)$$ Suppose $a_n\to\infty$, then $\alpha/a_n<1$, and so from 3. we obtain

$$(*) = (1-\frac \pi4 \frac{\alpha}{a_n})^n.$$ 5. Letting $a_n=\frac{\pi}{4} n$, it follows that the RHS in 4. tends to $e^{-\alpha}$.

Conclusion. The answer to your question is $a_n=\frac{\pi}{4}n$, and the limit in this case is exponential with parameter $1$ (of course, we can choose any other positive constant instead of $\frac \pi 4$, resulting in a different rate).

Answer 2

I would start with a distribution of a distance from $(1,1)$. Once you know that, calculating distribution of a minimum distance out of $n$ realizations is simple given distribution of order statistics.

To determine the distribution of a distance from $(1,1)$, $(X,Y)$ is a point in unit square and since both $X$ and $Y$ are uniform, the point has uniform density over the square.

What is the distribution of the distance? Let me make a heuristic argument.

Locus of points at distance $d\in[0,1]$ from $(1,1)$ is quarter of a circle, that is, should have density $2\pi d/4=\frac{\pi d}{2}$.

Locus of points at distance $d\in(1,\sqrt{2}]$ is subset of a circle and little bit of trigonometry gives its density equal to $d\left[\frac{\pi}{2}-2\arcsin{(\frac{\sqrt{d^{2}-1}}{d})}\right]$.

Just as a sanity check, integral over the density should be equal to unity and my Mathematica claims this to be the case.

To calculate the distribution of the minimum, you need cdf, which is $\frac{\pi x^{2}}{4}$ for $x\in[0,1]$ and $\frac{\pi x^{2}}{4}+\sqrt{(x^{2}-1)}-x^2\arccsc{\frac{x}{\sqrt{(x^{2}-1)}}}$ for $x\in(1,\sqrt{2}]$. Figure of pdf for $n\in\{1,2,3,10\}$ is below.