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Can I get some insight of how to solve this problem?

Let $X_1, X_2, X_3, ...$ be i.i.d. copies of uniform random variable on $[0, 1]$. Let $M_n = \text{min}_{1\leq i <j \leq n} |X_i - X_j|$. Show $n^2M_n$ converges to $e^{-x}1_{x\geq 0}$ in distribution.

What I have done is:

$$ P(n^2M_n \leq y) = P(M_n \leq \frac{y}{n^2}) = P(\text{min}_{1\leq i < j \leq n}|X_i - X_j| \leq \frac{y}{n^2}),$$ where $$ \begin{align*} P(\text{min}_{1\leq i < j \leq n}|X_i - X_j| \leq \frac{y}{n^2}) = 1 - P(\text{min}_{1\leq i < j \leq n}|X_i - X_j| > \frac{y}{n^2}) \\ \ = 1 - P(\cap_{1 \leq i < j \leq n} \{ |X_i - X_j| > \frac{y}{n^2} \}) \end{align*} $$

But now I don't know how to proceed since many of $|X_i - X_j|$ are dependent (e.g $|X_1 - X_2|$ and $|X_1 - X_3|$) so we cannot easily expand the last term. I will appreciate any insight and help, or just simply a direction to go.

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  • $\begingroup$ Do you mean that $e^{-x}1_{x\geq 0}$ is the limit distribution of the random variable $n^2M_n$? $\endgroup$ – zoli Dec 3 '15 at 10:26
  • $\begingroup$ yes, thanks. it's fixed now. $\endgroup$ – Coroner_Rex Dec 3 '15 at 10:43
  • $\begingroup$ Not sure, but how about finding out the distribution of a subtraction of two uniforms, then the distribution of a minimum? $\endgroup$ – An old man in the sea. Dec 3 '15 at 10:59
  • $\begingroup$ finding distribution of subtraction of uniform is easy, but how would you go about finding distribution of minimum through? $\endgroup$ – Coroner_Rex Dec 3 '15 at 11:51
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That probability is: $$ \left(1-(n-1)\frac{y}{n^2}\right)^n $$ As take $Y_{ni}$ $i\in\{1,2,\ldots,n\}$ to be uniform on $\left[0,1-(n-1)\frac{y}{n^2}\right]$, and order them, and let $Z_{nk}^*=Y_{nk}^*+(k-1)\frac{y}{n^2}$, then the $Z_{nk}^*$ will have the distribution of $X^*_{1},\ldots,X^*_n$ under the condition that $X_{i+1}^*-X_{i}^*>\frac{y}{n^2}$.

$Y^*_{k}$ denotes an ordered sample.

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  • $\begingroup$ Thanks for your reply, first of all. But order sample is quite an unfamiliar concept to me. Could you explain it a more elementary manner? I believe this problem can be solved in a more elementary way. $\endgroup$ – Coroner_Rex Dec 3 '15 at 10:42
  • $\begingroup$ Could you elaborate a little bit more? Why you add $(k-1)y/n^2$ to $Y_{nk}^*$ and what's the motivation to take $Y_{ni}$? $\endgroup$ – Coroner_Rex Dec 3 '15 at 21:09
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Let $X_{(1)},\dots,X_{(n)}$ be order statistics of $X_1,\dots,X_n$ and let $E_1,\dots,E_{n+1}$ be i.i.d. $\exp(1)$. Then $$n(X_{(i)}-X_{(i-1)})\sim \frac{nE_{i}}{\sum_{k=1}^{n+1}E_k}\overset{a.s.}\longrightarrow E_i$$ Hence $$\min_{i\le n}n^2(X_{(i)}-X_{(i-1)})\overset{d}\longrightarrow \exp(1)$$ as a minimum of $n$ i.i.d. $\exp(1)$ random variables.

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  • $\begingroup$ Thanks for your reply. But order statistics is quite an unfamiliar concept to me. Could you explain it a more elementary manner? I believe this problem can be solved in a more elementary way. $\endgroup$ – Coroner_Rex Dec 3 '15 at 10:41
  • $\begingroup$ @Coroner Order statistics is very relevant here since you care only about differences between successive ordered $X_{(i)}$. I also use the fact that given $P_t={n+1}$, the first $n$ Poisson arrivals are i.i.d. on $[0,t]$ + the fact that interarrival time of Poisson is $\exp$. $\endgroup$ – A.S. Dec 3 '15 at 10:55

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