0
$\begingroup$

Let $(X_n)_{n \geq 1}$ be a sequence of i.i.d. random variables with density $$f(x) = \frac{1}{\pi (1+x^2)}$$

Let $$Y_n:= \frac{n}{\max\{X_1, \ldots, X_n\}}$$

I'm asked to find the limit in law of $Y_n$.

Here, I'm missing a bit the plan. Would it make sense to first compute $\mathbb P(Y_n < y)$ for $y \leq 0$ and $y>0$?

Thanks for any hint and comment.

$\endgroup$
4
  • $\begingroup$ Computing $\mathbb P(Y_n \geq y)$ is probably easier. No need to distinguish the cases $y\leq 0$ and $y>0$ $\endgroup$ May 26, 2021 at 6:41
  • $\begingroup$ You have a complete answer here. Think to keyword: Cauchy distribution. $\endgroup$
    – Jean Marie
    May 26, 2021 at 7:52
  • $\begingroup$ No comment on my comment ? $\endgroup$
    – Jean Marie
    May 26, 2021 at 15:55
  • $\begingroup$ Thanks for your comment @JeanMarie - if I would have seen this other contribution before... $\endgroup$
    – Stanisla
    May 27, 2021 at 8:29

2 Answers 2

2
$\begingroup$

Yes you can compute $\mathbb{P}(Y_n> y)$ as follows $$ \mathbb{P}(Y_n> y) = \mathbb{P}\left(\frac{n}{\max\{X_1, ..., X_n\}}> y\right) = \mathbb{P}\left({\max\{X_1, ..., X_n\}}< \frac{n}{y}\right) \\ =\mathbb{P}\left(X_1< \frac{n}{y}, ..., X_n< \frac{n}{y}\right) = \mathbb{P}\left(X_1< \frac{n}{y}\right)^n. $$ Therefore $$ \mathbb{P}(Y_n\leq y)=1-\mathbb{P}(Y_n> y) =1-\left(\frac{1}{\pi}\tan^{-1}{\frac{n}{y}}+\frac{1}{2}\right)^n. $$ Can you take the limit now?

$\endgroup$
3
  • $\begingroup$ Letting $n \to + \infty$, the part in brackets after the 1 goes to 0, so the limit of the whole expression will be 1. Is this already the limit in law of $Y_n$? $\endgroup$
    – Stanisla
    May 26, 2021 at 12:34
  • $\begingroup$ It was because I mistakenly wrote $n/y$ as $y/n$. $Y_n$ has a non-trivial limit. You will see that after taking the limit. Best $\endgroup$ May 26, 2021 at 15:55
  • $\begingroup$ Thanks for the correction and for your help. $\endgroup$
    – Stanisla
    May 27, 2021 at 8:18
1
$\begingroup$

It is an exponential distribution with mean ${\pi}$. The reason is that you can say $$ \mathbb{P}(Y_n> y) = \mathbb{P}\left(\frac{n}{\max\{X_1, ..., X_n\}}> y\right) = \mathbb{P}\left({\max\{X_1, ..., X_n\}}< \frac{n}{y}\right) \\ =\mathbb{P}\left(X_1< \frac{n}{y}, ..., X_n< \frac{n}{y}\right) = \mathbb{P}\left(X_1< \frac{n}{y}\right)^n. $$ Therefore $$ \mathbb{P}(Y_n\leq y)=1-\mathbb{P}(Y_n> y) =1-\left(\frac{1}{\pi}\tan^{-1}{\frac{n}{y}}+\frac{1}{2}\right)^n. $$ Next, we find the limit of the second term for large $n$s. To do so, we define $$ g(n) = \left(\frac{1}{\pi}\tan^{-1}{\frac{n}{y}}+\frac{1}{2}\right)^n. $$ Then, we can find the limit of $\log g(1/x)$ for $x\to 0$ as \begin{align} \lim_{x\to 0} \log g(1/x) = \lim_{x\to 0} \frac{\log \left(\frac{1}{\pi}\tan^{-1}{\frac{1}{xy}}+\frac{1}{2}\right)}{x} &\overset{\text{L'Hôpital}}{=}\lim_{x\to 0} \frac{\frac{\partial}{\partial x}\left(\frac{1}{\pi}\tan^{-1}{\frac{1}{xy}}\right)}{1\times (\frac{1}{\pi}\frac{\pi}{2}+\frac{1}{2})}\\ &=\lim_{x\to 0}\frac{-1}{\pi(x^2 y+ \frac{1}{y})} = -\frac{y}{\pi}. \end{align} Finally, because of continuity of exponential function, we know that $$ \lim_{n\to \infty} g(n)= e^{\lim_{x\to 0}g(1/x)} = e^{-\frac{y}{\pi}}, $$ which shows that $\lim_{n\to \infty} Y_n$ has the CDF of $1-e^{-\frac{y}{\pi}}$.

$\endgroup$
4
  • $\begingroup$ Thanks a lot for your explanation. Your answer helped a lot! $\endgroup$
    – Stanisla
    May 27, 2021 at 8:20
  • $\begingroup$ Have you seen the solution I indicated yesterday here ? $\endgroup$
    – Jean Marie
    May 27, 2021 at 8:34
  • $\begingroup$ @Stanisla You're welcome. $\endgroup$
    – ConvXET
    May 27, 2021 at 8:46
  • $\begingroup$ @JeanMarie Sorry I didn't happen to see it before. But thanks! $\endgroup$
    – ConvXET
    May 27, 2021 at 8:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .