Kolmogorov's inequality

In probability theory, Kolmogorov's inequality is a so-called "maximal inequality" that gives a bound on the probability that the partial sums of a finite collection of independent random variables exceed some specified bound. The inequality is named after the Russian mathematician Andrey Kolmogorov.

Statement of the inequality
Let X1, ..., Xn : &Omega; &rarr; R be independent random variables defined on a common probability space (&Omega;, F, Pr), with expected value E[Xk] = 0 and variance Var[Xk] &lt; +&infin; for k = 1, ..., n. Then, for each &lambda; &gt; 0,


 * $$\Pr \left(\max_{1\leq k\leq n} | S_k |\geq\lambda\right)\leq \frac{1}{\lambda^2} \operatorname{Var} [S_n] \equiv \frac{1}{\lambda^2}\sum_{k=1}^n \operatorname{Var}[X_k], $$

where Sk = X1 + ... + Xk.

Proof
The following argument is due to Kareem Amin and employs discrete martingales. As argued in the discussion of Doob's martingale inequality, the sequence $$S_1, S_2, \dots, S_n$$ is a martingale. Without loss of generality, we can assume that $$S_0 = 0$$ and $$S_i \geq 0$$ for all $$i$$. Define $$(Z_i)_{i=0}^n$$ as follows. Let $$Z_0 = 0$$, and
 * $$Z_{i+1} = \left\{ \begin{array}{ll}

S_{i+1} & \text{ if } \displaystyle \max_{1 \leq j \leq i} S_j < \lambda \\ Z_i & \text{ otherwise} \end{array} \right. $$ for all $$i$$. Then $$(Z_i)_{i=0}^n$$ is a also a martingale. Since $$\text{E}[S_{i}] = \text{E}[S_{i-1}]$$ for all $$i$$ and $$\text{E}[\text{E}[X|Y]] = \text{E}[X]$$ by the law of total expectation,
 * $$\begin{align}

\sum_{i=1}^n \text{E}[ (S_i - S_{i-1})^2] &= \sum_{i=1}^n \text{E}[ S_i^2 - 2 S_i S_{i-1} + S_{i-1}^2 ] \\ &= \sum_{i=1}^n \text{E}\left[ S_i^2 - 2 \text{E}[ S_i S_{i-1} | S_{i-1} ] + \text{E}[S_{i-1}^2 | S_{i-1}] \right] \\ &= \sum_{i=1}^n \text{E}\left[ S_i^2 - 2 \text{E}[ S^2_{i-1} | S_{i-1} ] + \text{E}[S_{i-1}^2 | S_{i-1}] \right] \\ &= \text{E}[S_n^2] - \text{E}[S_0^2] = \text{E}[S_n^2]. \end{align} $$ The same is true for $$(Z_i)_{i=0}^n$$. Thus
 * $$\begin{align}

\text{Pr}\left( \max_{1 \leq i \leq n} S_i \geq \lambda\right) &= \text{Pr}[Z_n \geq \lambda] \\ &\leq \frac{1}{\lambda^2} \text{E}[Z_n^2] =\frac{1}{\lambda^2} \sum_{i=1}^n \text{E}[(Z_i - Z_{i-1})^2] \\ &\leq \frac{1}{\lambda^2} \sum_{i=1}^n \text{E}[(S_i - S_{i-1})^2] =\frac{1}{\lambda^2} \text{E}[S_n^2] = \frac{1}{\lambda^2} \text{Var}[S_n]. \end{align} $$ by Chebyshev's inequality.