# Kolmogorov's inequality

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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.Fact|date=May 2007

tatement of the inequality

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

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

where "S""k" = "X"1 + ... + "X""k".

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 $\left(Z_i\right)_\left\{i=0\right\}^n$ as follows. Let $Z_0 = 0$, and:for all $i$.Then $\left(Z_i\right)_\left\{i=0\right\}^n$ is a also a martingale. Since $ext\left\{E\right\} \left[S_\left\{i\right\}\right] = ext\left\{E\right\} \left[S_\left\{i-1\right\}\right]$ for all $i$ and $ext\left\{E\right\} \left[ ext\left\{E\right\} \left[X|Y\right] = ext\left\{E\right\} \left[X\right]$ by the law of total expectation,:The same is true for $\left(Z_i\right)_\left\{i=0\right\}^n$. Thus:by Chebyshev's inequality.

ee also

* Chebyshev's inequality
* Doob's martingale inequality
* Landau-Kolmogorov inequality
* Markov's inequality

References

* (Theorem 22.4)
*

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