习题八

1

样本平均: \(\mu=E X(t)=E X_0 E(-1)^{N(t)}=0\)

时间平均 \(X_T=\frac{1}{T} \int_0^T X_0(-1)^{N(t)} d t=0\) (对称)

于是 \(\tau = \lim\limits_{T\to\infty} X_T=0\)

故 \(\tau=\mu\).从而满足均值遍历性。

2

\[ \begin{aligned} & \mu=E X(t)=e^{-\alpha t / 2} E B\left(e^{\alpha t}\right)=0 . \\ & X_T=\frac{1}{T} \int_0^{\top} e^{-\alpha t/2} B\left(e^{\alpha t}\right) d t=0 \implies \tau = 0 \\ & \Rightarrow \tau=\mu . \end{aligned} \]

3

\[ E X_n=\begin{cases} 0 & n=0 \\ \lambda E x_{n-1} & n \geq 1 \end{cases} \]

\(\Rightarrow \mu=E X_n=0\)

$$ X_n=Z_n+\lambda Z_{n-1}+\cdots+\lambda^n Z_0 $$ \(r_X(k)=E X_0 X_k=E Z_0\left(Z_n+\cdots+\lambda^k Z_0\right)=\lambda^k E Z_0^2=\frac{\lambda^k \sigma^2}{1-\lambda^2}\)

\(\lim\limits_{k \rightarrow \infty} r_X(k)=0=\mu^2\)

故 \(X\) 满足均值遍历性

4

显然 \(X_n\) 平稳

\(\mu_X=E X_n=E\left(\frac{\xi_n+\cdots+\xi_{n-k}}{k+1}\right)=\mu\).

取 \(l>2k\) 充分大,则有

\[r_X(k) = E\left(\frac{\xi_k+\cdots+\xi_{0}}{k+1}\right)\left(\frac{\xi_{k+l}+\cdots+\xi_l}{k+1}\right)=E\xi_k E\xi_{k+l}=\mu^2\]

于是

\[\lim\limits_{l\to\infty}r_X(l)=\mu_X^2=\mu^2\]

从而满足均值遍历性

5

数分知识. \(x_n \to a \implies \frac{1}{n} \sum\limits_{i=1}^n x_i \to a\).

\(X\) 满足均值遍历性 \(\implies r_x(k) \rightarrow \mu^2 \implies \frac{1}{n} \sum\limits_{k=1}^n r_x(k)=\mu^2\)

反过来,因为

\(\frac{1}{n^2} \sum\limits_{k=1}^n\left(r_{X}({k})-\mu^2\right) \leq \frac{1}{n^2} \sum\limits_{k=1}^{n} k\left(r_X(k)-\mu^2\right) \leq \frac{1}{{n}^2} \sum\limits_{{k}=1}^{{n}} {n}\left(r_{X}({k})-\mu^2\right)\)

两边同时取极限得到左右两边均为 0 ,所以

\[\lim _{{n} \rightarrow \infty} \frac{1}{{n}^2} \sum_{{k}=1}^{{n}} {k}\left(r_{{X}}({k})-\mu^2\right)=0\]

于是

\[\frac{1}{n^2}\sum\limits_{k=1}^n (n-k)(r_X(k)-\mu^2)\to 0\]

从而满足均值遍历性