Limit of some random variable sequence: $\lim_{n\rightarrow\infty}\dfrac{\max_{1\le i\le n} Y_i}{n}$

Question

Suppose that we have some infinite sequence of random variables $\{Y_i\}$. They may or may not be independent.

I wonder whether there is some existing theory or method which would deal with $\lim_{n\rightarrow\infty}\dfrac{\max_{1\le i\le n} Y_i}{n}$.

Answer 1

Without making any assumptions whatsoever, I cannot imagine that much can be said. Maybe what you're looking for is extreme value theory, which treats similar questions under some assumptions of independence. Also related is the theory of large deviations.

Answer 2

In some cases this will at least converge in distribution. Suppose $Y_i$ were independent with CDF $F_Y$. Define $M_n = \frac{\max_{1\le i\le n} Y_i}n$. The event that $M_n \le x$ is equivalent to $Y_1 \le xn, \ldots, Y_n \le xn$, and so the CDF of $M_n$ is $F_{M_n}(x) = F_Y(xn)^n$.

If we take, for example, a random variable with CDF $$ F_Y(x) = \begin{cases} 0, & x < 1 \\ 1-1/x, & x \ge 1 \end{cases} $$ The limit $$ \lim_{n\to\infty} F_{M_n}(x) = \lim_{n\to\infty} F_Y(xn)^n = \lim_{n\to\infty} \left(1-\frac{1}{xn}\right)^n = e^{-1/x}, x \ge 0 $$ and so $M_n$ converges in distribution to a random variable with CDF $e^{-1/x}, x\ge 0$. For other random variables $Y$ the limiting distribution depends on the shape of the tail, but the first steps for turning it into the limit CDF will be the same.

Answer 3

If the $\{Y_i\}$ is a sequence of independent RVs distributed with common cdf $F_Y(\cdot)$ then, the cdf of $M_n:=\displaystyle \frac{\max_{1\le i\le n}Y_i}{n}$ is given by $F_{M_n}(y)=(F_Y(yn))^n$ as shown by @pokes. If $\{Y_n\}$ is a Martingale or submartingale sequence with some finite moment of order $p\ge 1$, then by Doob's inequality $$P(|M_n|\ge \lambda)\le \frac{\mathbb{E}(|Y_n|^p)}{n^p\lambda^p}$$ So as $n\to\infty$ the sequence $M_n$ converges to $0$ in probability.