heavy-tailed maximal inequalites

Finite Variance

If $X_1,\dots ,X_n$ are iid with $\mathbb{E}{X}^2\le K$, then

$$\mathbb{E}\underset{1\le i\le n}{\max }|X_i|\le B\sqrt{n}.$$

There is also a partial converse: If $\mathbb{E}{X}^2=\infty$, then

$$\mathbb{E}\underset{1\le i\le n}{\max }|X_i|\ge K{n}^{\frac{1}{2+ϵ}},\text{for all }ϵ>0.$$

I don’t know where this result originally comes from, but it’s not too hard to prove.

Kolmogorov’s inequality

If $X_1,\dots ,X_n$ are independent and mean-zero, then

$$\mathbb{P}(\underset{1\le k\le n}{\max }|S_k|\ge u)\le \frac{1}{u^2}\sum _{k\le n}\mathbb{E}{X}_k^2,$$

where $S_k={\sum }_{i\le k}{X}_i$.