依概率收敛问题的证明
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/08/03 22:52:24
依概率收敛问题的证明
![](http://img.wesiedu.com/upload/e/9d/e9dccb85a0d3178ee317dc405490166f.jpg)
![](http://img.wesiedu.com/upload/e/9d/e9dccb85a0d3178ee317dc405490166f.jpg)
![依概率收敛问题的证明](/uploads/image/z/16273015-7-5.jpg?t=%E4%BE%9D%E6%A6%82%E7%8E%87%E6%94%B6%E6%95%9B%E9%97%AE%E9%A2%98%E7%9A%84%E8%AF%81%E6%98%8E%26nbsp%3B)
令Y=1/n*∑{i=1,n}X{i}²,考虑到X{i}独立同分布,则
E(Y)=1/n*∑{i=1,n}E(X{i}²)= E(X{i}²)=D(X{i})+ E(X{i})²=σ²
D(Y)=1/n²*∑{i=1,n}D(X{i}²)=1/n* D(X{i}²)=1/n*{ E(X{i}⁴)-[ E(X{i}²)]²}
∵E(X{i}⁴0,有
0≤P{|Y-E(Y)|≥ε}≤D(Y)/ε²,即
0≤P{|Y-σ²|≥ε}≤1/n*{ E(X{i}⁴)-[ E(X{i}²)]²}/ε²
≤1/n*(M-σ⁴)/ε²
而lim{n→∞}1/n*(M-σ⁴)/ε²=0
由夹逼定理可知,lim{n→∞}P{|Y-σ²|≥ε}=0
即Y=1/n*∑{i=1,n}X{i}²依概率收敛于σ²
再问:![](http://img.wesiedu.com/upload/1/db/1dba842f6fe4d8b0130a6bb67dfbad89.jpg)
E(Y)=1/n*∑{i=1,n}E(X{i}²)= E(X{i}²)=D(X{i})+ E(X{i})²=σ²
D(Y)=1/n²*∑{i=1,n}D(X{i}²)=1/n* D(X{i}²)=1/n*{ E(X{i}⁴)-[ E(X{i}²)]²}
∵E(X{i}⁴0,有
0≤P{|Y-E(Y)|≥ε}≤D(Y)/ε²,即
0≤P{|Y-σ²|≥ε}≤1/n*{ E(X{i}⁴)-[ E(X{i}²)]²}/ε²
≤1/n*(M-σ⁴)/ε²
而lim{n→∞}1/n*(M-σ⁴)/ε²=0
由夹逼定理可知,lim{n→∞}P{|Y-σ²|≥ε}=0
即Y=1/n*∑{i=1,n}X{i}²依概率收敛于σ²
再问:
![](http://img.wesiedu.com/upload/1/db/1dba842f6fe4d8b0130a6bb67dfbad89.jpg)