On One Side Kolmogorov type inequalities

时间：2014-01-21 21:35:24 阅读：342 评论：0 收藏：0 [点我收藏+]

Let $X_1,X_2,\ldots,X_n$ be independent random variables. Denote

S n = \sum i = 1 n X i .

$S_n=\sum_{i=1}^n X_i.$ The well known Kolmogrov inequality can stated as for all

ε≥0

$\varepsilon\ge 0$

P (max 1 \leq j \leq n | S j | \geq ε) \leq V a r ( S n ) ε 2 .

$P\left(\max_{1\le j\le n}|S_j|\ge \varepsilon\right)\le\frac{Var(S_n)}{\varepsilon^2}.$

The one side kolmogrov type ineqalites are stated as for all $\varepsilon\ge0$

P (max 1 \leq j \leq n S j \geq ε) \leq V a r ( S n ) ε 2 + V a r ( S n ) .

$?P\left(\max_{1\le j\le n}S_j\ge \varepsilon\right)\le\frac{Var(S_n)}{\varepsilon^2+Var(S_n)}.$ We will prove this inequality in the following.

Proposition. Let $X$ be a random variable with $Var(X)<\infty$ . Then for all $\varepsilon\ge0$

P (X \geq ε) \leq V a r ( X ) ε 2 + V a r ( X ) .

$?P(X\ge \varepsilon)\le\frac{Var(X)}{\varepsilon^2+Var(X)}.$

Proof. Without loss of generality, we may assume that $E(X)=0$ . Then

ε = E (ε ? X) = E {(ε ? X) I X < ε} + E {(ε ? X) I X \geq ε} \leq E {(ε ? X) I X < ε} .

$\varepsilon=E(\varepsilon - X)=E\{(\varepsilon??- X)I_{X<\varepsilon}\}+E\{(\varepsilon??- X)I_{X\ge \varepsilon}\}\le E\{(\varepsilon??- X)I_{X< \varepsilon}\}.$ By Cauchy-Schwardz‘s inequality, We have

ε 2 \leq [E {(ε ? X) I X \leq ε}] 2 \leq E (ε + X) 2 P (X \leq ε) = [ε 2 + V a r (X)] [1 ? P (X > ε)] .

$\varepsilon^2\le \left[E\{(\varepsilon??- X)I_{X\le \varepsilon}\}\right]^2\le E(\varepsilon? + X)^2P(X\le\varepsilon)=[\varepsilon^2+Var(X)][1-P(X>\varepsilon)].$ Therefor,

P (X \geq ε) \leq V a r ( X ) ε 2 + V a r ( X ) .

$P(X\ge \varepsilon)\le\frac{Var(X)}{\varepsilon^2+Var(X)}.$

Proof of the one side Kolmogorov type inequality. Let $\Lambda=\{\max_{1\lej\le n}S_j?\ge \varepsilon\}$ and $\Lambda_k=\{\max_{1\le j <k}S_j < \varepsilon, S_k\ge\varepsilon\}$ , then $\Lambda =\bigcup_{k=1}^{n}\Lambda_k$ . Without loss of generality, we assume that $E(X_j)=0,j=1,\ldots,n.$ Then by the independence of the random variables,

ε = E [ε ? S n] = E [(ε ? S n) I Λ] + [(ε ? S n) I c Λ] = \sum k = 1 n [(ε ? S n) I Λ k] + [(ε ? S n) I Λ c] = \sum k = 1 n E [{(ε ? S k) ? (S n ? S k)} I Λ k] + [(ε ? S n) I Λ c] = \sum k = 1 n E [(ε ? S k) I Λ k] + \sum k = 1 n [E (S n ? S k) I Λ k] + E [(ε ? S n) I Λ c] = \sum k = 1 n E [(ε ? S k) I Λ k] + E [(ε ? S n) I Λ c] \leq E [(ε ? S n) I Λ c .

$\begin{array}{rcl}\varepsilon&=&E[\varepsilon -S_n]=E[(\varepsilon-S_n)I_{\Lambda}]+[(\varepsilon-S_n)I_{\Lambda}^c]&=&\sum_{k=1}^n[(\varepsilon-S_n)I_{\Lambda_k}]+[(\varepsilon-S_n)I_{\Lambda^c}]=\sum_{k=1}^n E[\{(\varepsilon-S_k) -(S_n-S_k)\}I_{\Lambda_k}]+[(\varepsilon-S_n)I_{\Lambda^c}]&=&\sum_{k=1}^n E[(\varepsilon-S_k)I_{\Lambda_k}]+\sum_{k=1}^n[E(S_n-S_k)I_{\Lambda_k}]+E[(\varepsilon-S_n)I_{\Lambda^c}]&=&\sum_{k=1}^n E[(\varepsilon-S_k)I_{\Lambda_k}]+E[(\varepsilon-S_n)I_{\Lambda^c}]&\le&E[(\varepsilon-S_n)I_{\Lambda^c}.\end{array}$

By Cauchy-Schwardz‘s inequality, we have

ε 2 \leq {E [(ε ? S n) I Λ c]} 2 \leq E [(ε ? S n)] 2 P (I Λ c) = [ε 2 + V a r (S 2 n)] [1 ? P (Λ)] .

$\varepsilon^2\le \{E[(\varepsilon-S_n)I_{\Lambda^c}]\}^2\le E[(\varepsilon-S_n)]^2 P(I_{\Lambda^c})=[\varepsilon^2+Var(S_n^2)][1-P(\Lambda)].$ Therefore,

P (X \geq ε) \leq V a r ( X ) ε 2 + V a r ( X )

$P(X\ge \varepsilon)\le\frac{Var(X)}{\varepsilon^2+Var(X)}$ as the inequality claimed.

The inequality is also true for martingale difference sequence, the proof is samilar.

On One Side Kolmogorov type inequalities

原文：http://www.cnblogs.com/levin2013/p/3528824.html

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