台交大随机过程 Lecture 6

课程主页：https://ocw.nctu.edu.tw/course_detail.php?bgid=8&gid=0&nid=558

老师主页：http://shannon.cm.nctu.edu.tw/

课程视频：https://www.bilibili.com/video/BV1sW411U7xK?from=search&seid=2411740640027894887，

https://www.youtube.com/watch?v=IsQSWVbAKy0&list=PLj6E8qlqmkFvw7Rt63yBqai2HmPKF0V0J

这次回顾第六讲，介绍了均方误差下的等式，对应视频13-14。

几个重要定理

定理9-5

如果WSS过程$x(t)$的自相关函数$R_{x x}(\tau)$满足$R_{x x}\left(\tau_{1}\right)=R_{x x}(0)$，对于某个$\tau_1\neq 0$，那么$R_{xx}(\tau)$的周期为$\tau_1$。

证明：

利用Cauchy-Schwartz不等式

$$\left.\left|E\left[\left(\boldsymbol{x}\left(t+\tau+\tau_{1}\right)-\boldsymbol{x}(t+\tau)\right) \boldsymbol{x}^{*}(t)\right]\right|^{2} \leq\left. E\left[\mid \boldsymbol{x}\left(t+\tau+\tau_{1}\right)-\boldsymbol{x}(t+\tau)\right)\right|^{2}\right] E\left[|\boldsymbol{x}(t)|^{2}\right]$$

这等价于

$$\left|R_{x x}\left(\tau+\tau_{1}\right)-R_{x x}(\tau)\right|^{2} \leq\left(2 R_{x x}(0)-R_{x x}\left(\tau_{1}\right)-R_{x x}\left(-\tau_{1}\right)\right) R_{x x}(0)$$

注意$R_{x x}(0)$为实数，所以

$$R_{x x}\left(-\tau_{1}\right)=R_{x x}^{*}\left(\tau_{1}\right)=R_{x x}^{*}\left(0\right)=R_{x x}(0)$$

从而

$$2 R_{x x}(0)-R_{x x}\left(\tau_{1}\right)-R_{x x}\left(-\tau_{1}\right) =0$$

因此

$$R_{x x}\left(\tau+\tau_{1}\right)=R_{x x}(\tau)$$

推论

如果WSS过程$x(t)$的自相关函数$R_{x x}(\tau)$在原点连续，那么它在每处都连续。

证明：

利用Cauchy-Schwartz不等式

$$\left.\left|E\left[\left(\boldsymbol{x}\left(t+\tau+\tau_{1}\right)-\boldsymbol{x}(t+\tau)\right) \boldsymbol{x}^{*}(t)\right]\right|^{2} \leq\left. E\left[\mid \boldsymbol{x}\left(t+\tau+\tau_{1}\right)-\boldsymbol{x}(t+\tau)\right)\right|^{2}\right] E\left[|\boldsymbol{x}(t)|^{2}\right]$$

这等价于

$$\left|R_{x x}\left(\tau+\tau_{1}\right)-R_{x x}(\tau)\right|^{2} \leq\left(2 R_{x x}(0)-R_{x x}\left(\tau_{1}\right)-R_{x x}\left(-\tau_{1}\right)\right) R_{x x}(0)$$

因此

$$\lim _{\left|\tau_{1}\right| \downarrow 0} R_{x x}\left(\tau+\tau_{1}\right)=R_{x x}(\tau)$$

引理：交叉谱的上界

对于任意$a,b$，

$$\left|\int_{a}^{b} S_{x y}(\omega) d \omega\right|^{2} \leq\left(\int_{a}^{b} S_{x x}(\omega) d \omega\right)\left(\int_{a}^{b} S_{y y}(\omega) d \omega\right)$$

证明：

令$z(t),w(t)$是WSS输入$x(t),y(t)$通过滤波器$H(\omega)=1 \cdot 1\{a<\omega<b\}$的输出，那么根据Cauchy-Schwartz不等式，

$$\begin{aligned} \left|E^{2}\left[z(t) \boldsymbol{w}^{*}(t)\right]\right| & \leq E^{2}\left[\left|\boldsymbol{z}(t) \boldsymbol{w}^{*}(t)\right|\right] \\ &\leq E\left[|\boldsymbol{z}(t)|^{2}\right] E\left[\left|\boldsymbol{w}^{*}(t)\right|^{2}\right] \\ &=R_{z z}(0) R_{w w}(0) \\ &=\frac{1}{2 \pi}\left(\int_{-\infty}^{\infty} S_{x x}(\omega)|H(\omega)|^{2} d \omega\right) \frac{1}{2 \pi}\left(\int_{-\infty}^{b} S_{y y}(\omega)|H(\omega)|^{2} d \omega\right) \\ &=\frac{1}{4 \pi^{2}}\left(\int_{a}^{b} S_{x x}(\omega) d \omega\right)\left(\int_{a}^{b} S_{y y}(\omega) d \omega\right) \end{aligned}$$

注意到

$$\begin{aligned} E\left[z(t) \boldsymbol{w}^{*}(t)\right] &=E\left[\int_{-\infty}^{\infty} h(\tau) \boldsymbol{x}(t-\tau) d \tau \cdot \int_{-\infty}^{\infty} h^{*}(s) \boldsymbol{y}^{*}(t-s) d s\right] \\ &=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} h(\tau) h^{*}(s) R_{x y}(s-\tau) d \tau d s \\ &=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} h(\tau) h^{*}(s) \frac{1}{2 \pi}\left(\int_{-\infty}^{\infty} S_{x y}(\omega) e^{j \omega(s-\tau)} d \omega\right) d \tau d s \\ &=\frac{1}{2 \pi} \int_{-\infty}^{\infty} S_{x y}(\omega)|H(\omega)|^{2} d \omega \\ &=\frac{1}{2 \pi} \int_{a}^{b} S_{x y}(\omega) d \omega \end{aligned}$$

所以结论得证。

均方误差意义下的等式

均方（MS）意义相等

两个过程$\{\boldsymbol{x}(t), t \in \mathcal{I}\},\{\boldsymbol{y}(t), t \in \mathcal{I}\}$在均方意义下相等， 当且仅当

$$E\left[|\boldsymbol{x}(t)-\boldsymbol{y}(t)|^{2}\right]=0,\forall t\in \mathcal I$$

定义：均方周期

过程$x(t)$被称为均方周期，如果对每个$t$

$$E\left[|\boldsymbol{x}(t+T)-\boldsymbol{x}(t)|^{2}\right]=0$$

定理：均方周期性和自相关函数的关系

过程$x(t)$是MS周期，当且仅当自相关函数是双重周期的，即对每个整数$m,n$，都有

$$R_{x x}\left(t_{1}+m T, t_{2}+n T\right)=R_{x x}\left(t_{1}, t_{2}\right)$$

证明：

$\Rightarrow$：

利用Cauchy-Schwartz不等式

$$\begin{aligned} \left|E\left\{\boldsymbol{x}\left(t_{1}\right) \cdot\left(\boldsymbol{x}\left(t_{2}+T\right)-\boldsymbol{x}\left(t_{2}\right)\right)^{*}\right\}\right| & \leq E\left\{\left|\boldsymbol{x}\left(t_{1}\right) \cdot\left(\boldsymbol{x}\left(t_{2}+T\right)-\boldsymbol{x}\left(t_{2}\right)\right)^{*}\right|\right\} \\ & \leq E^{1 / 2}\left[\left|\boldsymbol{x}\left(t_{1}\right)\right|^{2}\right] E^{1 / 2}\left[\left|\boldsymbol{x}\left(t_{2}+T\right)-\boldsymbol{x}\left(t_{2}\right)\right|^{2}\right] \\ &=0 \end{aligned}$$

这推出

$$R_{x x}\left(t_{1}, t_{2}+T\right)=R_{x x}\left(t_{1}, t_{2}\right)$$

重复上述过程即可证明结论。

$\Leftarrow$：

由定义可得

$$R_{x x}(t+T, t+T)=R_{x x}(t+T, t)=R_{x x}(t, t+T)=R_{x x}(t, t)$$

因此

$$\begin{aligned} &E\left[|\boldsymbol{x}(t+T)-\boldsymbol{x}(t)|^{2}\right] \\ =&R_{x x}(t+T, t+T)-R_{x x}(t+T, t)-R_{x x}(t, t+T)+R_{x x}(t, t) \\ =&0 \end{aligned}$$

定义：均方连续

过程$\boldsymbol{x}(t)$被称为均方连续，如果对每个$t$，都有

$$\lim _{\epsilon \downarrow 0} E\left[|\boldsymbol{x}(t+\epsilon)-\boldsymbol{x}(t)|^{2}\right]=0$$

因为

$$E\left[|\boldsymbol{x}(t+\epsilon)-\boldsymbol{x}(t)|^{2}\right]=R_{x x}(t+\epsilon, t+\epsilon)-R_{x x}(t+\epsilon, t)-R_{x x}(t, t+\epsilon)+R_{x x}(t, t)$$

所以MS连续当且仅当自相关函数连续。

对于WSS过程，我们有

$$E\left[|\boldsymbol{x}(t+T)-\boldsymbol{x}(t)|^{2}\right]=2 \cdot \operatorname{Re}\left\{R_{x x}(0)-R_{x x}(T)\right\}$$

定义：均方可微

过程$\boldsymbol{x}(t)$被称为均方可微，如果对每个$t$，都有

$$\lim _{\epsilon \downarrow 0} E\left[\left|\frac{x(t+\epsilon)-x(t)}{\epsilon}-\left(\lim _{\gamma \downarrow 0} \frac{x(t+\gamma)-x(t)}{\gamma}\right)\right|^{2}\right]=0$$

因为

$$\begin{aligned} &E\left[\left|\frac{x(t+\epsilon)-x(t)}{\epsilon}-x^{\prime}(t)\right|^{2}\right] \\ =&\frac{R_{x x}(t+\epsilon, t+\epsilon)-R_{x x}(t+\epsilon, t)-R_{x x}(t, t+\epsilon)+R_{x x}(t, t)}{\epsilon^{2}} \\ &-\frac{R_{x x^{\prime}}(t+\epsilon, t)-R_{x x^{\prime}}(t, t)}{\epsilon} -\frac{R_{x^{\prime} x}(t, t+\epsilon)-R_{x^{\prime} x}(t, t)}{\epsilon}+R_{x^{\prime} x^{\prime}}(t, t) \end{aligned}$$

因为

$$\begin{aligned} R_{x x^{\prime}}\left(t_{1}, t_{2}\right)&=\frac{\partial R_{x x}\left(t_{1}, t_{2}\right)}{\partial t_{2}}\\ R_{x^{\prime} x^{\prime}}\left(t_{1}, t_{2}\right)&=\frac{\partial R_{x x^{\prime}}\left(t_{1}, t_{2}\right)}{\partial t_{1}}=\frac{\partial^{2} R_{x^{\prime} x^{\prime}}\left(t_{1}, t_{2}\right)}{\partial t_{1} \partial t_{2}} \end{aligned}$$

从而$x(t)$均方可微当且仅当$\partial^{2} R_{x x}\left(t_{1}, t_{2}\right) /\left.\partial t_{1} \partial t_{2}\right|_{t_{1}=t_{2}}$存在。

定义：均方（黎曼）可积

过程$\boldsymbol{x}(t)$被称为均方（黎曼）可积，如果对任意$a,b$都，都有

$$\lim _{\Delta \downarrow 0} E\left[\left|\int_{a}^{b} x(t) d t-\sum_{i=0}^{\lfloor(b-a) / \Delta\rfloor} x(a+i \Delta) \times \Delta\right|^{2}\right]=0$$

证明：

令

$$\boldsymbol{y}(t)=\int_{t-(b-a)}^{t} \boldsymbol{x}(s) d s=\int_{-\infty}^{\infty} h(\tau) \boldsymbol{x}(t-\tau) d \tau$$

其中

$$h(\tau)=1,0 \leq \tau<(b-a)$$

那么

$$\begin{aligned} &E\left[\left|\boldsymbol{y}(b)-\Delta \sum_{i=0}^{\lfloor(b-a) / \Delta\rfloor} \boldsymbol{x}(a+i \Delta)\right|^{2}\right]\\ =&R_{y y}(b, b)+\Delta^{2} \sum_{i=0}^{\lfloor(b-a) / \Delta\rfloor} \sum_{k=0}^{\lfloor(b-a) / \Delta\rfloor} R_{x x}(a+i \Delta, a+k \Delta)\\ &-\Delta \sum_{i=0}^{\lfloor(b-a) / \Delta\rfloor} R_{x y}(a+i \Delta, b)-\Delta \sum_{i=0}^{\lfloor(b-a) / \Delta\rfloor} R_{y x}(b, a+i \Delta) \end{aligned}$$

所以$x(t)$是均方（黎曼）可积，当且仅当$\int_{a}^{b} \int_{a}^{b} R_{x x}\left(t_{1}, t_{2}\right) d t_{1} d t_{2}$黎曼可积。