Part a
Notice $P(X_i \le t) = F(t),i = 1,\cdots,n$P(Xi​≤t)=F(t),i=1,⋯,n
$E[\hat{F}_n(t)] = E \left[ \frac{1}{n}\sum_{i=1}^n I(X_i \le t) \right] = \frac{1}{n}\sum_{i=1}^nE \left[ I(X_i \le t) \right] = \frac{1}{n}\sum_{i=1}^nP(X_i<t) = F(t)$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
$P(\hat{F}_n(t)=k/n) = P \left( \frac{1}{n}\sum_{i=1}^n I(X_i \le t) = q \right) = P \left( \sum_{i=1}^n I(X_i \le t) = k \right) \\ = C_n^k[F(t)]^{k}[1-F(t)]^{n-k},k=1,\cdots,n$P(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 $n\hat{F}_n(t) \sim Binom(n,F(t))$nF^n​(t)∼Binom(n,F(t)).

Part c
Calculate
$E[\hat{F}_n(t)] = \frac{1}{n}E[n\hat{F}_n(t)] = F(t) \\ Var(\hat{F}_n(t)) = \frac{1}{n^2}Var(n\hat{F}_n(t)) = \frac{F(t)[1-F(t)]}{n}$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,
$\frac{\hat{F}_n(t) - F(t)}{\sqrt{\frac{F(t)[1-F(t)]}{n}}} \to_d N(0,1) \Rightarrow \sqrt{n}[\hat{F}_n(t) - F(t)] \to_d N(0,F(t)[1-F(t)])$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定理说的是当$n \to \infty$n→∞时，$\hat F_n$F^n​收敛到$F$F。上面的题目提到了，对于给定的$t$t，$\hat{F}_n(t)$F^n​(t)会收敛到$F(t)$F(t)，因为
$Var(\hat{F}_n(t)) = \frac{F(t)[1-F(t)]}{n} \to 0,\ as \ n\to \infty$Var(F^n​(t))=nF(t)[1−F(t)]​→0, as n→∞

因此$\hat{F}_n(t) - F(t) \to_{L_2} 0$F^n​(t)−F(t)→L2​​0。一个从分析出发的证明会显得更简单，经验分布的性质说明
$|\hat{F}_n(x_j) - F(x_j)| \le \frac{1}{n},\forall j = 1,\cdots,n-1$∣F^n​(xj​)−F(xj​)∣≤n1​,∀j=1,⋯,n−1

下面估计
$\sup|\hat{F}_n(x) - F(x)| \le \max|\hat{F}_n(x_j) - F(x_j)| + \frac{1}{n}$sup∣F^n​(x)−F(x)∣≤max∣F^n​(xj​)−F(xj​)∣+n1​

当$n \to \infty$n→∞时，$\max|\hat{F}_n(x_j) - F(x_j)| \to 0$max∣F^n​(xj​)−F(xj​)∣→0，因此$\hat F_n \to_{L_{\infty}} F$F^n​→L∞​​F。