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If we have iid observations $\mathbf {(X_1,Y_1),(X_2,Y_2),\dots}$ from bivariate distribution $G$ supported on unit disc $\mathbf {[(x,y): 0 \le (x^2,y^2) \le 1]}$. Suppose that distribution has a continuous density $\mathbf {g(x,y)}$ with $\mathbf{g(0,0) \gt 0}$.

Let $\mathbf {D_i=(X_i^2+Y_i^2)^{1/2}}$, for $\mathbf{i=1,2,\dots}$

Let $\mathbf{D_{(1,n)} = min(D_1,D_2,\dots,D_n)}$.

So can someone please help me find real constants $\mathbf {a_n}$ and $\mathbf{b_n\gt0}$ such that $\mathbf {b_n(D_{(1,n)}-a_n)}$ converges to a non-degenerate distribution?

Actually, previously, I had done a question where I had to find limiting distribution of $\mathbf{{(Y_n-a_n)\over b_n}}$. Where $\mathbf {Y_n={\sum X_n\over n}}$ and $\mathbf {X_j;j\ge1}$ were independent $\mathbf {Bernoulli ({1\over j^{1/2}})}$. In that it was easy using $\mathbf{Lyapunov CLT}$. Since $\mathbf{X_j-{1\over j^{1/2}}}$ is bounded for all $\mathbf{j\ge 1}$. Suitable choices of $\mathbf {a_n}$ and $\mathbf{b_n}$ and proving conditions on them showed that this converged to $\mathbf {N(0,1)}$ which is non-degenerate distriubution.

Will this question be similar?

Michael Hardy
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user87455
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1 Answers1

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Tools:

  • When $r\to0$, $P[D_1\leqslant r]\sim\pi g(0,0)r^2$.
  • For every $n$, $P[D_{(1:n)}\geqslant r]=P[D_1\geqslant r]^n=(1-P[D_1\geqslant r])^n$.
  • For every $(c_n)$ such that $c_n\to c$, $\left(1-c_n/n\right)^n\to\mathrm e^{-c}$ when $n\to\infty$.

Thus, $P[D_{(1:n)}\geqslant c/\sqrt{n}]\to\exp(-\pi g(0,0)c^2)$ when $n\to\infty$.

Let $Z$ denote a standard exponential random variable, then $P[Z\geqslant z]=\mathrm e^{-z}$ for every $z\geqslant0$ hence $P[D_{(1:n)}\geqslant c/\sqrt{n}]\to P[Z\geqslant\pi g(0,0)c^2]$, that is, $\sqrt{n}D_{(1:n)}\to\sqrt{Z/\pi g(0,0)}$ in distribution. Equivalently, $\sqrt{n}D_{(1:n)}$ converges in distribution to the Rayleigh distribution with parameter $\sigma=1/\sqrt{2\pi g(0,0)}$.

Did
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  • Thank you. But we constructed the whole proof assuming r to be close to 0 (so that we were able to take g(0,0) almost constant there). But how does this cover the case when r is not small enough? – user87455 Aug 21 '13 at 08:11
  • Obviously $D_{(1:n)}\to0$ almost surely hence, to get a nondegenerate limit, one should consider events $[D_{(1:n)}\geqslant r_n]$ with $r_n\to0$. What the post shows is that the good scaling is $r_n$ of the order of $1/\sqrt{n}$. If $r_n\sqrt{n}\to\infty$, $P[D_{(1:n)}\geqslant r_n]\to0$. – Did Aug 21 '13 at 08:23
  • oh off course. Sorry, as it is ${c\over n^{1/2}}$ also goes to zero anyway. Thank you once again – user87455 Aug 21 '13 at 09:33