马尔可夫不等式, 切比雪夫不等式, 大数定律

防盗https://www.cnblogs.com/setdong/p/17325420.html

1. Markov’s inequality

Theorem: X X X 为非负随机变量, 且 E [ X ] < ∞ \mathbb{E}[X]<\infty E[X]<. 那么对于任意 t > 0 t>0 t>0
P [ X ≥ t E [ X ] ] ≤ 1 t \mathbb{P}[X\geq t\mathbb{E}[X]]\leq\frac{1}{t} P[XtE[X]]t1
或对于任意的 ϵ > 0 \epsilon>0 ϵ>0
P [ X ≥ ϵ ] ≤ E [ X ] ϵ \mathbb{P}[X\geq\epsilon]\leq\frac{\mathbb{E}[X]}{\epsilon} P[Xϵ]ϵE[X]
证明, 以上两式等价, 这里证第一个:

由定义有
P [ X ≥ t E [ X ] ] = ∑ x : x ≥ t E [ X ] P [ X = x ] \mathbb{P}[X\geq t\mathbb{E}[X]]=\sum_{x:x\geq t\mathbb{E}[X]}\mathbb{P}[X=x] P[XtE[X]]=x:xtE[X]P[X=x]
对于 x ≥ t E [ X ] x\geq t\mathbb{E}[X] xtE[X] 部分, 有 x t E [ X ] ≥ 1 \frac{x}{t\mathbb{E}[X]}\geq 1 tE[X]x1, 所以
P [ X ≥ t E [ X ] ] ≤ ∑ x : x ≥ t E [ X ] P [ X = x ] x t E [ X ] \mathbb{P}[X\geq t\mathbb{E}[X]]\leq\sum_{x:x\geq t\mathbb{E}[X]}\mathbb{P}[X=x]\frac{x}{t\mathbb{E}[X]} P[XtE[X]]x:xtE[X]P[X=x]tE[X]x
对于 x < t E [ X ] x < t\mathbb{E}[X] x<tE[X] 部分, 有 P [ X = x ] x t E [ X ] ≥ 0 \mathbb{P}[X=x]\frac{x}{t\mathbb{E}[X]}\geq 0 P[X=x]tE[X]x0 , 所以加上这些非负部分, 有:
P [ X ≥ t E [ X ] ] ≤ ∑ x P [ X = x ] x t E [ X ] = E [ X ] t E [ X ] = 1 t \mathbb{P}[X\geq t\mathbb{E}[X]]\leq\sum_{x}\mathbb{P}[X=x]\frac{x}{t\mathbb{E}[X]}=\frac{\mathbb{E}[X]}{t\mathbb{E}[X]}=\frac{1}{t} P[XtE[X]]xP[X=x]tE[X]x=tE[X]E[X]=t1

2. Chebyshev’s inequality

Theorem: X X X 为随机变量, 且 Var [ X ] < ∞ \text{Var}[X]<\infty Var[X]< (方差). 那么对于任意 t > 0 t>0 t>0
P [ ∣ X − E [ X ] ∣ ≥ t Var [ X ] 1 2 ] ≤ 1 t 2 \mathbb{P}[|X-\mathbb{E}[X]|\geq t\text{Var}[X]^{\frac{1}{2}}]\leq\frac{1}{t^2} P[XE[X]tVar[X]21]t21
或对于任意 ϵ > 0 \epsilon>0 ϵ>0
P [ ∣ X − E [ X ] ∣ ≥ ϵ ] ≤ Var [ X ] ϵ 2 \mathbb{P}[|X-\mathbb{E}[X]|\geq \epsilon]\leq\frac{\text{Var}[X]}{\epsilon^2} P[XE[X]ϵ]ϵ2Var[X]
证明:

首先有
P [ ∣ X − E [ X ] ∣ ≥ t Var [ X ] 1 2 ] = P [ ( X − E [ X ] ) 2 ≥ t 2 Var [ X ] ] \mathbb{P}[|X-\mathbb{E}[X]|\geq t\text{Var}[X]^{\frac{1}{2}}]=\mathbb{P}[(X-\mathbb{E}[X])^2\geq t^2\text{Var}[X]] P[XE[X]tVar[X]21]=P[(XE[X])2t2Var[X]]
然后对上式右侧应用 Markov’s inequality, 其中 X X X ( X − E [ X ] ) 2 (X-\mathbb{E}[X])^2 (XE[X])2, ϵ \epsilon ϵ t 2 Var [ X ] 2 t^2\text{Var}[X]^2 t2Var[X]2
P [ ( X − E [ X ] ) 2 ≥ t 2 Var [ X ] ] ≤ E [ ( X − E [ X ] ) 2 ] t 2 Var [ X ] = 1 t 2 \mathbb{P}[(X-\mathbb{E}[X])^2\geq t^2\text{Var}[X]]\leq \frac{\mathbb{E}[(X-\mathbb{E}[X])^2]}{t^2\text{Var}[X]}=\frac{1}{t^2} P[(XE[X])2t2Var[X]]t2Var[X]E[(XE[X])2]=t21
其中 E [ ( X − E [ X ] ) 2 ] = Var [ X ] \mathbb{E}[(X-\mathbb{E}[X])^2]=\text{Var}[X] E[(XE[X])2]=Var[X] 为方差的定义.

3. Weak law of large numbers

Theorem: ( X n ) n ∈ N (X_n)_{n\in\mathbb{N}} (Xn)nN 是一列独立的随机变量, 它们有相同的期望 μ \mu μ 和方差 σ 2 > ∞ \sigma^2>\infty σ2>. X ˉ n = 1 n ∑ i = 1 n X i \bar{X}_n=\frac{1}{n}\sum_{i=1}^n X_i Xˉn=n1i=1nXi 为均值. 那么对于任意的 ϵ > 0 \epsilon>0 ϵ>0
lim ⁡ n → ∞ P [ ∣ X ˉ n − μ ∣ ≥ ϵ ] = 0 \lim_{n\rightarrow\infty}\mathbb{P}[|\bar{X}_n-\mu|\geq\epsilon]=0 nlimP[Xˉnμϵ]=0
证明:

已知变量是独立的, 根据期望和方差的特性有,
E [ X ˉ n ] = ∑ i = 1 n μ n = μ Var [ X ˉ n ] = ∑ i = 1 n σ 2 n 2 = σ 2 n \mathbb{E}[\bar{X}_n]=\sum_{i=1}^n \frac{\mu}{n}=\mu\\ \text{Var}[\bar{X}_n]=\sum_{i=1}^n \frac{\sigma^2}{n^2}=\frac{\sigma^2}{n} E[Xˉn]=i=1nnμ=μVar[Xˉn]=i=1nn2σ2=nσ2
对上式使用 Chebyshev’s inequality, 其中 t = ϵ / Var [ X ˉ n ] 1 2 t=\epsilon/\text{Var}[\bar{X}_n]^{\frac{1}{2}} t=ϵ/Var[Xˉn]21, 有
P [ ∣ X ˉ n − μ ∣ ≥ ϵ ] ≤ σ 2 n ϵ 2 \mathbb{P}[|\bar{X}_n-\mu|\geq\epsilon]\leq\frac{\sigma^2}{n\epsilon^2} P[Xˉnμϵ]nϵ2σ2
所以当 n n n 趋于无穷时, 概率趋于 0 0 0. 可以理解为样本均值依概率收敛于期望值.
防盗https://www.cnblogs.com/setdong/p/17325420.html

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