# A generalization of Kolmogorov's maximal inequality

Suppose $X_1, X_2, \ldots$ are independent random variables with $E(X_n) = 0$ for all $n$. For $\varepsilon > 0$, define $A_n = \left\{\max\limits_{1 \leqslant k \leqslant n} |S_k| \geqslant \varepsilon\right\}$ for all $n$, where $S_n = \sum\limits_{k = 1}^n X_k$ for all $n$. Prove that for any real $r \geqslant 1$,$$\varepsilon^r P(A_n) \leqslant E(|S_n|^r I_{A_n}) \leqslant E(|S_n|^r). \quad \forall n \geqslant 1$$

This is from the problem set of my homework and it appears to be a really nice generalization of Kolmogorov's inequality. However, when I was trying to imitate the standard proof of Kolmogorov's inequality and the proof given by Sergio for cases in which $r$ is an even integer, there occurred the problem that binomial expansion cannot be properly applied to general cases in which $r$ is an arbitrary real. I wonder if the step that uses binomial expansion in the proof of Kolmogorov's inequality is correct for this generalization or if there is a proof that avoids this particular step. Thanks!

First, define $\mathscr{F}_n = \sigma(X_1, \cdots, X_n)$ and $f_n(x_1, x_2, \cdots, x_n) = (x_1, x_1 + x_2, \cdots, x_1 + \cdots + x_n)$ for all $n$. Since $f_n$ is bijective and $f_n$ and $f_n^{-1}$ are both Borel-measurable, then$$\sigma(S_1, \cdots, S_n) = \sigma((S_1, \cdots, S_n)) = \sigma(f_n(X_1, \cdots, X_n)) = \sigma((X_1, \cdots, X_n)) = \mathscr{F}_n.$$
Then for all $n$, because $X_1, \cdots, X_{n + 1}$ are independent, $X_{n + 1}$ and $\mathscr{F}_n$ are independent. Therefore,\begin{align*} E(S_{n + 1} \mid \mathscr{F}_n) &= E(S_n \mid \mathscr{F}_n) + E(X_{n + 1} \mid \mathscr{F}_n)\\ &= S_n + E(X_{n + 1}) = S_n, \quad \text{a.s.} \end{align*} so $\{S_n\}$ is a martingale with respect to $\{\mathscr{F}_n\}$. Note that the function $y = |x|^r$ is convex on $\mathbb{R}$ since $r \geqslant 1$, so $\{|S_n|^r\}$ is a submartingale with respect to $\{\mathscr{F}_n\}$.
Now, define a stopping time $\tau = \min\{n \geqslant 1 \mid |S_n| \geqslant \varepsilon\}$ where $\min \varnothing \mathrel{:=} +\infty$, and $B_n = \{\tau = n\}$ for all $n$, then $B_n \in \sigma(S_1, \cdots, S_n) = \mathscr{F}_n$. Note that $A_n = \bigcup\limits_{k = 1}^n B_k$, therefore,\begin{align*} E(|S_n|^r) &\geqslant E(|S_n|^r I_{A_n}) = \int_{A_n} |S_n|^r \,\mathrm{d}P\\ &= \sum_{k = 1}^n \int_{B_k} |S_n|^r \,\mathrm{d}P = \sum_{k = 1}^n \int_{B_k} E(|S_n|^r \mid \mathscr{F}_k) \,\mathrm{d}P\\ &\geqslant \sum_{k = 1}^n \int_{B_k} |S_k|^r \,\mathrm{d}P \geqslant \sum_{k = 1}^n \int_{B_k} \varepsilon^r \,\mathrm{d}P\\ &= \varepsilon^r \sum_{k = 1}^n P(B_k) = \varepsilon^r P(A_n). \end{align*}