Proof of Chebyshev Type for Maximal Sum of Random Variables I

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Chebyshev type for maximal sum of random variables I.  Let $\{X_n\}$ be independent random variables such that $\mathscr{E}(X_n)=0$ and $\mathscr{E}(X_n^2)=\sigma^2(X_n)<\infty$ for all $n$, then let $S_n=\sum_{j=1}^nX_j$, we have for every $\varepsilon>0$, $$\mathscr{P}\left\{\underset{1\leq j\leq n}{\max}|S_j|>\varepsilon\right\}\leq\frac{\sigma^2(S_n)}{\varepsilon^2}.$$
See List of Inequalities.

$\bullet$ Proof.
Fix $\varepsilon>0$.  For any $\omega$ in the set $$\Lambda=\left\{\omega:\,\underset{1\leq j\leq n}{\max}|S_j|>\varepsilon\right\},$$we define $$\Lambda_k=\left\{\omega:\,|S_1|\leq\varepsilon,\ldots,|S_{k-1}|\leq\varepsilon,\,|S_k|>\varepsilon\right\}=\left\{\omega:\,\underset{1\leq j\leq k-1}{\max}|S_j|\leq\varepsilon,\,|S_k|>\varepsilon\right\}.$$Then $\Lambda_k$'s are disjoint and $\Lambda=\bigcup_{k=1}^n\Lambda_k$.  It follows that $$\begin{array}{rl}\int_\Lambda S_n^2\,d\mathscr{P}
& =\sum_{k=1}^n\int_{\Lambda_k}S_n^2\,d\mathscr{P} = \sum_{k=1}^n\int_{\Lambda_k}[S_k+(S_n-S_k)]^2\,d\mathscr{P}\\ & =  \sum_{k=1}^n\int_{\Lambda_k}[S_k^2+2S_k(S_n-S_k)+(S_n-S_k)^2]\,d\mathscr{P}. \end{array}$$The middle term in the integral, $$\begin{array}{rl}\int_{\Lambda_k}S_k(S_n-S_k)\,d\mathscr{P}&=\int_\Omega I_{\Lambda_k}S_k(S_n-S_k)\,d\mathscr{P}\\ &=\int_\Omega I_{\Lambda_k}S_k\,d\mathscr{P}\int_\Omega(S_n-S_k)\,d\mathscr{P}\;(\because\,\mbox{independence between }X_j\mbox{'s})\\ & = 0\quad\left(\because\,\mathscr{E}(S_n-S_k)=\sum_{j=k+1}^n\mathscr{E}(X_j)=0\right). \end{array}$$Then $$\begin{array}{rl}\sigma^2(S_n) & =\int_\Omega S_n^2\,d\mathscr{P}\geq\int_\Lambda S_n^2\,d\mathscr{P}\geq\sum_{k=1}^n\int_{\Lambda_k}S_k^2\,d\mathscr{P}\\ & \geq\varepsilon^2\sum_{k=1}^n\int_{\Lambda_k}\,d\mathscr{P}=\varepsilon^2\sum_{k=1}^n\mathscr{P}\{\Lambda_k\}=\varepsilon^2\mathscr{P}\{\Lambda\}\end{array}$$since in $\Lambda_k$, $|S_k|>\varepsilon$.  Thus we have $$\mathscr{P}\{\Lambda\}=\mathscr{P}\left\{\underset{1\leq j\leq n}{\max}|S_j|>\varepsilon\right\}\leq\frac{\sigma^2(S_n)}{\varepsilon^2}.$$

$\Box$