UA MATH564 概率论 QE练习 Glivenko–Cantelli定理
UA MATH564 概率論 QE練習 Glivenko–Cantelli定理
- Glivenko–Cantelli定理
Part a
Notice P(Xi≤t)=F(t),i=1,?,nP(X_i \le t) = F(t),i = 1,\cdots,nP(Xi?≤t)=F(t),i=1,?,n
E[F^n(t)]=E[1n∑i=1nI(Xi≤t)]=1n∑i=1nE[I(Xi≤t)]=1n∑i=1nP(Xi<t)=F(t)E[\hat{F}_n(t)] = E \left[ \frac{1}{n}\sum_{i=1}^n I(X_i \le t) \right] = \frac{1}{n}\sum_{i=1}^nE \left[ I(X_i \le t) \right] = \frac{1}{n}\sum_{i=1}^nP(X_i<t) = F(t)E[F^n?(t)]=E[n1?i=1∑n?I(Xi?≤t)]=n1?i=1∑n?E[I(Xi?≤t)]=n1?i=1∑n?P(Xi?<t)=F(t)
Part b
P(F^n(t)=k/n)=P(1n∑i=1nI(Xi≤t)=q)=P(∑i=1nI(Xi≤t)=k)=Cnk[F(t)]k[1?F(t)]n?k,k=1,?,nP(\hat{F}_n(t)=k/n) = P \left( \frac{1}{n}\sum_{i=1}^n I(X_i \le t) = q \right) = P \left( \sum_{i=1}^n I(X_i \le t) = k \right) \\ = C_n^k[F(t)]^{k}[1-F(t)]^{n-k},k=1,\cdots,nP(F^n?(t)=k/n)=P(n1?i=1∑n?I(Xi?≤t)=q)=P(i=1∑n?I(Xi?≤t)=k)=Cnk?[F(t)]k[1?F(t)]n?k,k=1,?,n
This means nF^n(t)~Binom(n,F(t))n\hat{F}_n(t) \sim Binom(n,F(t))nF^n?(t)~Binom(n,F(t)).
Part c
Calculate
E[F^n(t)]=1nE[nF^n(t)]=F(t)Var(F^n(t))=1n2Var(nF^n(t))=F(t)[1?F(t)]nE[\hat{F}_n(t)] = \frac{1}{n}E[n\hat{F}_n(t)] = F(t) \\ Var(\hat{F}_n(t)) = \frac{1}{n^2}Var(n\hat{F}_n(t)) = \frac{F(t)[1-F(t)]}{n}E[F^n?(t)]=n1?E[nF^n?(t)]=F(t)Var(F^n?(t))=n21?Var(nF^n?(t))=nF(t)[1?F(t)]?
By CLT,
F^n(t)?F(t)F(t)[1?F(t)]n→dN(0,1)?n[F^n(t)?F(t)]→dN(0,F(t)[1?F(t)])\frac{\hat{F}_n(t) - F(t)}{\sqrt{\frac{F(t)[1-F(t)]}{n}}} \to_d N(0,1) \Rightarrow \sqrt{n}[\hat{F}_n(t) - F(t)] \to_d N(0,F(t)[1-F(t)])nF(t)[1?F(t)]??F^n?(t)?F(t)?→d?N(0,1)?n?[F^n?(t)?F(t)]→d?N(0,F(t)[1?F(t)])
Glivenko–Cantelli定理
Glivenko–Cantelli定理說的是當n→∞n \to \inftyn→∞時,F^n\hat F_nF^n?收斂到FFF。上面的題目提到了,對于給定的ttt,F^n(t)\hat{F}_n(t)F^n?(t)會收斂到F(t)F(t)F(t),因為
Var(F^n(t))=F(t)[1?F(t)]n→0,asn→∞Var(\hat{F}_n(t)) = \frac{F(t)[1-F(t)]}{n} \to 0,\ as \ n\to \inftyVar(F^n?(t))=nF(t)[1?F(t)]?→0,?as?n→∞
因此F^n(t)?F(t)→L20\hat{F}_n(t) - F(t) \to_{L_2} 0F^n?(t)?F(t)→L2??0。一個從分析出發的證明會顯得更簡單,經驗分布的性質說明
∣F^n(xj)?F(xj)∣≤1n,?j=1,?,n?1|\hat{F}_n(x_j) - F(x_j)| \le \frac{1}{n},\forall j = 1,\cdots,n-1∣F^n?(xj?)?F(xj?)∣≤n1?,?j=1,?,n?1
下面估計
sup?∣F^n(x)?F(x)∣≤max?∣F^n(xj)?F(xj)∣+1n\sup|\hat{F}_n(x) - F(x)| \le \max|\hat{F}_n(x_j) - F(x_j)| + \frac{1}{n}sup∣F^n?(x)?F(x)∣≤max∣F^n?(xj?)?F(xj?)∣+n1?
當n→∞n \to \inftyn→∞時,max?∣F^n(xj)?F(xj)∣→0\max|\hat{F}_n(x_j) - F(x_j)| \to 0max∣F^n?(xj?)?F(xj?)∣→0,因此F^n→L∞F\hat F_n \to_{L_{\infty}} FF^n?→L∞??F。
總結
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