习题二

1.

• \[EX_n=E(A\cos\lambda n+B\sin\lambda n)=EAE\cos\lambda n+EBE\sin\lambda n=0\]
• \[Cov(X_n,X_m)=EX_nX_m=E(A\cos\lambda n+B\sin\lambda n)(A\cos\lambda m+B\sin\lambda m)=E( A^2\cos\lambda n\cos\lambda m+B^2\sin\lambda n\sin\lambda m + AB\cos\lambda n\sin\lambda m+AB\sin\lambda n\cos\lambda m )=\cos(\lambda(n-m))\]

2.

• \[EZ_n=0,VarZ_n=1\Rightarrow EZ_n^2=1\]
• \[EX_n=E\sum\limits_{i=o}^r\alpha_iZ_{n-i}=\sum\limits_{i=o}^r\alpha_iEZ_{n-i}=0\]
• 若\(|m-n|>r\),则\(Cov(X_n,X_m)=E\sum\limits_{some\;i\neq j}\alpha_i\alpha_j Z_{n-i}Z_{n-j}=0\)因为\(Z_{n-i},Z_{n-j}\)独立且\(EZ_{n-i}=EZ_{n-j}=0\)
• 若\(s=|m-n|\leq r\),则不妨设\(m\geq n\),从而有\(Cov(X_n,X_m)=E\sum\limits_{i=0}^{r-s+1}\alpha_{r-i}Z_{m-r+i}\alpha_{r-s-i}Z_{n-(r-s)+i}=\sum\limits_{i=0}^{r-s+1}\alpha_{r-i}\alpha_{r-s-i}\)

3.

• 两侧求均值有\(EX_n=\alpha EX_{n-1}+EZ_n\Rightarrow EX_n=0\)
• 两侧求方差有\(VarX_n=\alpha^2VarX_{n-1} + VarZ_n,\Rightarrow VarX_n=\frac{1}{1-\alpha^2}\),则\(m=0\)时满足所需证的等式
• 不妨设\(m>0,\)不然有\(Cov(X_n,X_{n+m})=Cov(X_{n+m},X_n)\overset{i=n+m,j=-m>0}{=}Cov(X_i,X_{i+j})\)
• 根据递推关系有 \(\(X_{n+m}=\alpha^mX_n+\alpha^{m-1}Z_{n+1}+\cdots+Z_{n+m}\)\),从而有 \(\(Cov(X_n,X_{n+m})=EX_nX_{n+m}=E\{ \alpha^mX_n^2+\alpha^{m-1}X_nZ_{n+1}+\cdots +X_nZ_{n+m}\}=\alpha^m EX_n^2=\frac{\alpha^m}{1-\alpha^2}\)\).得证

4.

• \[X(t)\sim \mathcal{N}(0,\cos^2(\theta t)+\sin^2(\theta t))=\mathcal{N}(0,1),\forall t\in R\]

5.

• \[E\xi_i=\frac{1}{2},Var\xi_i=\frac{1}{4},E\xi_i^2=\frac{1}{2}\]
• \[EV_n=\sum\limits_{i=1}^\infty\frac{1}{2^{i+1}}=\frac{1}{2}\]
• \[EV^2_n=E(\sum\limits_{i=1}^\infty\frac{1}{2^i})^2=\lim\limits_{m\to\infty}\{ \sum\limits_{i=1}^m\frac{E\xi_{n+i}^2}{4^i}+\sum\limits_{i=1}^{m-1}\sum\limits_{j=i+1}^m\frac{E\xi_{n+i}E\xi_{n+j}}{2^i2^j}\}=\frac{3}{4}\]
• \[VarV_n=EV^2_n-(EV_n)^2=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}\]
• 从而\(m=0\)时,\(Cov(V_n,V_{n+m})=VarV_n=\frac{1}{2}\)
• 当\(m\neq 0\),不妨取\(m>0\)(理由类似第三题),此时有
  • \[EV_nV_{n+m}=E\sum\limits_{i=1}^\infty\frac{\xi_{n+i}}{2^i}\sum\limits_{i=1}^\infty\frac{\xi_{nm+i}}{2^i}=(\sum\limits_{i=1}^m\frac{\xi_{n+i}}{2^i}+\frac{1}{2^m}\sum\limits_{i=1}^\infty\frac{\xi_{n+m+i}}{2^i})\sum\limits_{i=1}^\infty\frac{\xi_{n+m+i}}{2^i}=\frac{1}{2^m}\frac{3}{4}+\frac{1}{4}(1-\frac{1}{2^m})=\frac{1}{2^{m+1}}+\frac{1}{4}\]
  • 故有\(Cov(V_n,V_{n+m})=\frac{1}{2^{m+1}}\),对\(m=0\)也符合
• 综上\(Cov(V_n,V_{n+m})=\frac{1}{2^{m+1}}\)
• \[(\xi_{n_1+n},\xi_{n_2+n},\cdots,\xi_{n_s+n})\overset{d}{=}(\xi_{n_1},\xi_{n_2},\cdots,\xi_{n_s})\]

从而

\[(V_{n_1+n},V_{n_2+n},\cdots,V_{n_s+n})\overset{d}{=}(V_{n_1},V_{n_2},\cdots,V_{n_s})\]

6.

• \[p(1_{\xi_k\leq t}=1)=t,p(1_{\xi_k\leq t}=0)=1-t\]
• \[EX(t)=E\frac{1}{n}\sum\limits_{k=1}^n1_{\xi_k\leq t}=\frac{1}{n}\sum\limits_{k=1}^nE1_{\xi_k\leq 1}=\frac{1}{n}nt=t \implies EX(t)=t\]

• \(Cov(X(s),X(t))=EX(s)X(t)-EX(s)EX(t)\)

  • 其中
• \[EX(s)X(t)=E\frac{1}{n^2}\sum\limits_{i=1}^n1_{\xi_i\leq s}\sum\limits_{j=1}^n1_{\xi_j\leq t}=\frac{1}{n}E( \sum\limits_{i=1}^n1_{\xi_i\leq s}1_{\xi_i\leq t} +\sum\limits_{i\neq j}1_{\xi_i\leq s}1_{\xi_j\leq t})=\frac{1}{n^2}( n\min\{s,t\} + (n^2-n)st )= \frac{\min\{s,t\}+(n-1)st}{n}\]
• \(\implies Cov(X(s),X(t))=\frac{\min\{s,t\}+(n-1)st}{n}-st=\frac{\min\{s,t\}-st}{n}\)