# Find limit in law of $Y_n=\frac{n}{\max\{X_1, \ldots, X_n\}}$

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.

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

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?
• 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 $Y_n$? May 26, 2021 at 12:34
• 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 May 26, 2021 at 15:55
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 $$1-e^{-\frac{y}{\pi}}$$.