Kind of converse of Kolmogorov maximal inequality Let
$S_n=\zeta_1+...+\zeta_n$ where $\zeta_i$ are independent with $E\zeta_i=0, E\zeta_i^2=\sigma_i^2<\infty, |\zeta_i|<K$.
Then, $P(\max_{1 \leq m  \leq n}|S_m| \leq x) \leq (x+K)^2/\operatorname{var}(S_n)$.
How do we show this?  The hint says that I should use the fact that $S_n^2-\sum_{m \leq n}\sigma_m^2$ is a martingale. But I am totally clueless....
Any hint would be appreciated! Thanks and regards.
 A: For fixed $x>0$ define a stopping time $\tau$
$$\tau := \inf\{k \in \mathbb{N}; |S_k|>x\}.$$
Since $M_n := S_n^2 - \sum_{m=1}^n \sigma_m^2$, $n \in \mathbb{N}$, is a martingale, it follows from the optional stopping theorem that $$\mathbb{E}(M_{n \wedge \tau}) = \mathbb{E}(M_0)=0$$
for any $n \in \mathbb{N}$, i.e.
$$\mathbb{E}(S_{n \wedge \tau}^2) = \mathbb{E} \left( \sum_{m=1}^{n \wedge \tau} \sigma_m^2 \right). \tag{1}$$
As
$$\begin{align*} \mathbb{E} \left( \sum_{m=1}^{n \wedge \tau} \sigma_m^2 \right) &\geq \mathbb{E} \left( 1_{\{\tau > n\}} \sum_{m=1}^{n \wedge \tau} \sigma_m^2 \right) = \mathbb{P}(\tau > n) \sum_{m=1}^n \sigma_m^2 \end{align*}$$
it follows from $(1)$ that
$$\mathbb{E}(S_{n \wedge \tau}^2) \geq \mathbb{P}(\tau > n) \sum_{m=1}^n \sigma_m^2. \tag{2}$$
To estimate the left-hand side we note that $$|S_{n \wedge \tau}|^2 \leq (x+K)^2. \tag{3}$$ If $\omega \in \{\tau > n\}$ this estimate is trivial since $$|S_{n \wedge \tau}(\omega)| \leq x^2$$ by the very definition of $\tau$ for any such $\omega$. For $\omega \in \{\tau \leq n\}$ this follows from $$|S_{n \wedge \tau}(\omega)| = |S_{\tau}(\omega)| \leq \underbrace{|S_{\tau}(\omega)-S_{\tau-1}(\omega)|}_{=|\zeta_{\tau}(\omega)|\leq K} + \underbrace{|S_{\tau-1}(\omega)|}_{\leq x} \leq x+K.$$
Combining $(2)$ and $(3)$ we obtain that
$$ \mathbb{P}(\tau > n) \sum_{m=1}^n \sigma_m^2 \leq (x+K)^2. \tag{4}$$
Noting that
$$\mathbb{P}(\tau > n) = \mathbb{P} \left( \sup_{m \leq n} |S_m| \leq x \right) \quad \text{and} \quad \sum_{m=1}^n \sigma_m^2 = \text{var}(S_n)$$
we obtain the desired inequality.
