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arrow_back Prove Kolmogorov's inequality

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Prove Kolmogorov's inequality

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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
(106,844 points)

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