# arrow_back Prove Kolmogorov's inequality

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Prove Kolmogorov's inequality

For $k=1,2, \ldots, n$, let $A_{k}$ be the event that $\left|S_{k}\right| \geq \lambda$ but $\left|S_{i}\right|<\lambda$ for all $i=1,2, \ldots, k-1$. Note that the events $A_{1}, A_{2}, \ldots, A_{n}$ are disjoint, and
$\bigcup_{k=1}^{n} A_{k}=\left\{\max _{1 \leq k \leq n}\left|S_{k}\right| \geq \lambda\right\}$
Let $I_{A}$ be the indicator function of event $A$. Since $A_{1}, A_{2}, \ldots, A_{n}$ are disjoint, we have
$0 \leq \sum_{k=1}^{n} I_{A_{k}} \leq 1$
Hence, we obtain

$\sum_{k=1}^{n} \operatorname{Var}\left[X_{k}\right]=E\left[S_{n}^{2}\right] \geq \sum_{k=1}^{n} E\left[S_{n}^{2} I_{A_{k}}\right]$
After replacing $S_{n}^{2}$ by $S_{k}^{2}+2 S_{k}\left(S_{n}-S_{k}\right)+\left(S_{n}-S_{k}\right)^{2}$, we get
\begin{aligned} \sum_{k=1}^{n} \operatorname{Var}\left[X_{k}\right] & \geq \sum_{k=1}^{n} E\left[\left(S_{k}^{2}+2 S_{k}\left(S_{n}-S_{k}\right)+\left(S_{n}-S_{k}\right)^{2}\right) I_{A_{k}}\right] \\ & \geq \sum_{k=1}^{n} E\left[\left(S_{k}^{2}+2 S_{k}\left(S_{n}-S_{k}\right)\right) I_{A_{k}}\right] \\ &=\sum_{k=1}^{n} E\left[S_{k}^{2} I_{A_{k}}\right]+2 \sum_{k=1}^{n} E\left[S_{n}-S_{k}\right] E\left[S_{k} I_{A_{k}}\right] \\ &=\sum_{k=1}^{n} E\left[S_{k}^{2} I_{A_{k}}\right] \\ & \geq \lambda^{2} \sum_{k=1}^{n} E\left[I_{A_{k}}\right] \\ &=\lambda^{2} \sum_{k=1}^{n} \operatorname{Pr}\left(A_{k}\right) \end{aligned}

\begin{aligned} &=\lambda^{2} \operatorname{Pr}\left(\bigcup_{k=1}^{n} A_{k}\right) \\ &=\lambda^{2} \operatorname{Pr}\left(\max _{1 \leq k \leq n}\left|S_{k}\right| \geq \lambda\right) \end{aligned}
where in the third line, we have used the assumption that $S_{n}-S_{k}$ is independent of $S_{k} I_{A_{k}}$
by Platinum
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