A stronger version of Kolmogorov Inequality I came across this question which says that there is a stronger version of the Kolmogorov Inequality for symmetrically distributed random variables. The question is as follows

Let $\xi_1, \ldots, \xi_n   $ be a sequence of independent random variables, each $\xi_i $ has a symmetric distribution. That is $P(\xi_i \in A) = P(\xi_i \in -A) $ for any Borel set $ A \subseteq \mathcal{R}$. Assume that $E \xi_i^{2m} < \infty$, where $m$ is a positive integer. Prove the stronger version of the Kolmogorov inequality:
  $$P\left(\max_{1\le k \le n } |\xi_1+\ldots +\xi_k| \ge t \right) \le \frac{E(\xi_1+\ldots +\xi_n)^{2m}}{t^{2m}}.$$

Apparently, when $m=1$, we go back to the ordinary Kolmogorov Inequality. Any idea on how to prove it for a general $m$?
 A: Let us write $S_{k}=\xi_{1}+\dots \xi_{k}$ an $S=\max_{i}(|S_{i}|)$.  We fix $t$ for the the rest of the proof.
A way to prove Kolmogorov's inequality is to consider a stopping time variable, here the random variable $\tau:\Omega \mapsto \{0,1,2,\dots,n\}$  defined by $\tau(\omega)=k$ where $k$ is the smallest integer such as $|S_{k}|\geqslant t$, or $\tau(\omega)=0$ if such a $k$ doesn't exist. A slight refinement of that method gives your inequality.
We have 
\begin{equation*}
\mathbf{P}(S\geqslant t)=\sum_{k} \mathbf{P}(S_{k}\geqslant t, \tau =k) \leqslant \sum_{k} \frac{1}{t^{2m}}\int_{\tau=k} S_{k}^{2m}
\end{equation*}
We'd like to change the $S_{k}$'s into $S_{n}$'s in the last term, so that it simply becomes $t^{-2m}\mathbf{E}(S_{n}^{2m})$.
By symmetry of the $\xi$'s, for positive uneven integers $i$, $(S_{n}-S_{k})^{i}$ has mean $0$. It is also independent from the variable $S_{k}$ and the event $\tau=k$. Consequently
\begin{equation*}
 \enspace
\int \mathbf{1}_{\tau=k}S_{k}^{2m-i}(S_{n}-S_{k})^{i}=0,
\end{equation*}
and developping in the integral
\begin{align*}
\int_{\tau=k} S_{n}^{2m} = & \int \mathbf{1}_{\tau=k}(S_{k}+S_{n}-S_{k})^{2m} \\
= & \int \mathbf{1}_{\tau=k} \left(S_{k}^{2m}+ \binom{2m}{2}S_{k}^{2m-2}(S_{n}-S_{k})^{2}+\dots +(S_{n}-S_{k})^{2m}\right) \\
\geqslant & \int_{\tau=k} S_{k}^{2m}.
\end{align*}
(The dropped terms are positive).
So we can indeed change the $S_{k}$'s into $S_{n}$'s and the inequality is proved.
