Given $X_i, i \geq 1$ iid random variables, with mean zero and variance one, I would like to show that $$ M_n = \max_{1 \leq k \leq n}\left\{\frac{|X_k|}{\sqrt n}\right\} \xrightarrow{d} 0$$ as $n \to \infty$.
I know that $P(M_n \leq x) = \big(P(|X_1| \leq x \sqrt n)\big)^{n}$. However, as $n \to \infty$, it may not be necessary that this goes to one for positive $x$.
I tried using Chebyshev's inequality after finding the distribution functions, but the factors coming from the variance and the scaling factor cancel out, so the right hand side is not meaningful there.
I would like to see if I can use characteristic functions to do this (find the characteristic function of $M_n$, and show that it goes to $1$ everywhere). But I was unable to proceed this way as well.
Best Answer
Note that $$ \begin{align} P\left(\frac{\max_{1\leq i\leq n}|X_i|}{\sqrt n} > \epsilon \right)&= 1-P\left(\max_{1\leq i\leq n}|X_i| \leq \epsilon \sqrt n \right)\\ &= 1-P\left(|X_1| \leq \epsilon \sqrt n \right)^n \\ &= 1-P\left(|X_1|^2 \leq \epsilon ^2 n \right)^n \end{align}$$
For any integrable $X$, $\lim_{x\to \infty} xP(X> x )=0$. Indeed $$\begin{align} xP(X> x ) = x\int \mathbb 1_{X> x}(w) dP(w) &\leq x\int \mathbb 1_{X> x}(w)\frac{X(w)}{x} dP(w)\\ &=\int \mathbb 1_{X> x}(w) X(w) dP(w) \end{align}$$ and $\displaystyle \lim_{x\to \infty} \int \mathbb 1_{X> x}(w) X(w) dP(w)=0$ by dominated convergence.
Here, this implies $P\left(|X_1|^2 > \epsilon ^2 n \right)=o\left( \frac{1}{n}\right)$, hence $$ \begin{align} P\left(\frac{\max_{1\leq i\leq n}|X_i|}{\sqrt n} > \epsilon \right)&= 1-\left(1+o\left( \frac{1}{n}\right) \right)^n\\ &= 1-\exp\left(n\ln \left(1+o\left( \frac{1}{n}\right) \right) \right)\\ &= 1-\exp\left(o\left( 1\right) \right)\\ &= o\left(1\right) \end{align}$$
This proves convergence in probability to $0$, hence convergence in distribution to $0$.