0
$\begingroup$

Assume that for every $n\geq 1 $ we are given a real random variable $X_n$ such that $(X_n-n)/\sqrt n$ follows the standard normal distribution. Furthermore, assume that the $X_n$ are independent. Fix any $z\geq 0 $. Can we estimate the quantity $$ \frac{1}{n}\#\{1\leq i\leq n :X^{(i+1)}-X^{(i)}>z\},$$ where $X^{(m)} $ is the $m$-th smallest number among all $\{X_n:n\geq 1 \}$. I am thinking it should be tending to $\mathrm e ^{-z}$ but would like to have a proof.

(Edited $X_i\to X^{(i)}$.)

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ Something is off in your formulation. Note that $X_n$ normal of mean $n$ and variance $n$ follows your assumptions. Now for $i$ even, the sequence $X_{i+1}-X_i$ is i.i.d., and $P(X_{i+1}-X_i>z)\to 1/2$ for any $z$. So by the LLN, the expression you wrote is at least $1/4$, whp, regardless of $z$. In fact, it converges to $1/2$. $\endgroup$
    – ofer zeitouni
    Commented May 8, 2021 at 16:41
  • 1
    $\begingroup$ I am not sure about the limiting constant, but it seems that a first and second moment computation will get you the answer $\endgroup$
    – Nicolas Rivera
    Commented May 8, 2021 at 16:46
  • $\begingroup$ @oferzeitouni Indeed, this is a very good point! I think I was probably having in mind to ask for $X^{(I+1)}-X^{(i)} > z $ instead. Where $X^{(i)} $ is the I-th smallest value. $\endgroup$
    – Dr. Pi
    Commented May 8, 2021 at 16:52
  • 1
    $\begingroup$ @Dr. Pi If you want an answer, probably you should edit the question to reflect what you mean. I believe the formulation you just wrote makes no sense either, since the average spacing will be of order $1/\sqrt{n}$ then. The man of course changes nothing. $\endgroup$
    – ofer zeitouni
    Commented May 8, 2021 at 17:54
  • $\begingroup$ thank you for the comment ofer. I edited-please let me know if that is ok and I am happy to edit again. I am very sorry for the confusion! Actually I am a bit confused about the averaged spacing: shouldnt it be 1 instead of $1/\sqrt n $? Sorry if this is wrong! $\endgroup$
    – Dr. Pi
    Commented May 8, 2021 at 20:08

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .