台交大随机过程 Lecture 4

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

这次回顾第四讲，介绍了线性时不变系统，脉冲响应函数以及卷积，对应视频9-11。

有随机输入的系统

输入：$x(t)$，其中$\{x(t), t \in \mathcal{I}\}$定义在$(S, \mathcal{F}, P)$

算子：$\boldsymbol{T}$

输出：$\boldsymbol{y}(t)=\boldsymbol{T}(\{\boldsymbol{x}(s), s \in \mathcal{I}\}, t)=\boldsymbol{T}_{t}(\{\boldsymbol{x}(s), s \in \mathcal{I}\})$

系统分类

确定性系统：$\boldsymbol{T}_{t}\left(x^{\mathcal{I}}, \zeta\right)=\boldsymbol{T}_{t}\left(x^{\mathcal{I}}\right)$

随机系统：存在$\zeta_{1} \neq \zeta_{2}$，使得$\boldsymbol{T}_{t}\left(x^{\mathcal{I}}, \zeta_{1}\right) \neq \boldsymbol{T}_{t}\left(x^{\mathcal{I}}, \zeta_{2}\right)$

无记忆系统

系统是无记忆的，如果$T_{t}\left(x^{\mathcal{I}}, \zeta\right)=T\left(x^{\mathcal{I}}, \zeta\right)$。

引理

如果无记忆系统的输入$x(t)$是SSS，那么其输出$y(t)$也是SSS

证明：

考虑概率

$$P\{\zeta \in S: \boldsymbol{y}(t+c, \zeta) \in A\} =P\{\zeta \in S: \boldsymbol{T}(\boldsymbol{x}(t+c, \zeta), \zeta) \in A\}$$

定义

$$B \triangleq\{x \in \Re: \boldsymbol{T}(x, \zeta) \in A\}$$

那么

$$\begin{aligned} P\{\zeta \in S: \boldsymbol{y}(t+c, \zeta) \in A\} &=P\{\zeta \in S: \boldsymbol{x}(t+c, \zeta) \in B\}\\ &=P\{\zeta \in S: \boldsymbol{x}(t, \zeta) \in B\}\\ &= P\{\zeta \in S: \boldsymbol{T}(\boldsymbol{x}(t, \zeta), \zeta) \in A\}\\ &= P\{\zeta \in S: \boldsymbol{y}(t, \zeta) \in A\} \end{aligned}$$

推论

• 如果输入$x(t)$是$n$阶平稳，那么输出$y(t)$是$n$阶平稳
• 如果输入$x(t)$是区间平稳，那么输出$y(t)$也是区间平稳
• 但是，如果输入$x(t)$是WSS，输出$y(t)$不一定是WSS

最后一点的反例如下：

考虑$y(t)=x^2(t)$，其中$\boldsymbol{x}(t)=\boldsymbol{a} \cos (\omega t)+\boldsymbol b \sin (\omega t)$，$a,b$独立$0$均值。

计算相关函数：

$$\begin{aligned} E\left[\boldsymbol{y}\left(t_{1}\right) \boldsymbol{y}\left(t_{2}\right)\right]=& E\left[\left(\boldsymbol{a} \cos \left(\omega t_{1}\right)+\boldsymbol{b} \sin \left(\omega t_{1}\right)\right)^{2}\left(\boldsymbol{a} \cos \left(\omega t_{2}\right)+\boldsymbol{b} \sin \left(\omega t_{2}\right)\right)^{2}\right] \\ =& \frac{E\left[\boldsymbol{a}^{4}\right]+E^{2}\left[\boldsymbol{a}^{2}\right]}{2}+\frac{E\left[\boldsymbol{a}^{4}\right]-E^{2}\left[\boldsymbol{a}^{2}\right]}{2} \cos \left(2 \omega t_{1}\right) \cos \left(2 \omega t_{2}\right) +E^{2}\left[\boldsymbol{a}^{2}\right] \sin \left(2 \omega t_{1}\right) \sin \left(2 \omega t_{2}\right) \end{aligned}$$

只有当$E\left[\boldsymbol{a}^{4}\right]=3 E^{2}\left[\boldsymbol{a}^{2}\right]$时，$y(t)$才是WSS。

例子：Arcsine law

定义

$$\boldsymbol{y}(t)=\left\{\begin{array}{ll} +1, & \text { if } \boldsymbol{x}(t)>0 \\ -1, & \text { if } \boldsymbol{x}(t)<0 \end{array}\right.$$

其中$x(t)$是$0$均值高斯平稳过程。计算$y(t)$的均值和自相关函数。

解：

显然

$$E[\boldsymbol{y}(t)]=0$$

对于$R_{y y}\left(t_{1}, t_{2}\right)$，注意到

$$\boldsymbol{y}\left(t_{1}\right) \boldsymbol{y}\left(t_{2}\right)=\left\{\begin{array}{l} +1, \text { if } \boldsymbol{x}\left(t_{1}\right) \boldsymbol{x}\left(t_{2}\right)>0 \\ -1, \text { if } \boldsymbol{x}\left(t_{1}\right) \boldsymbol{x}\left(t_{2}\right)<0 \end{array}\right.$$

现在计算$x(t_1),x(t_2)$的联合概率分布，首先显然是联合高斯分布，期望为$0$，协方差矩阵为

$$\Sigma=\left[\begin{array}{cc} R_{x x}(0) & R_{x x}\left(t_{1}-t_{2}\right) \\ R_{x x}\left(t_{2}-t_{1}\right) & R_{x x}(0) \end{array}\right]=R_{x x}(0)\left[\begin{array}{cc} 1 & \rho \\ \rho & 1 \end{array}\right]$$

其中$\rho=R_{x x}\left(t_{1}-t_{2}\right) / R_{x x}(0)$。

所以

$$\begin{aligned} \operatorname{Pr}\left[\boldsymbol{x}\left(t_{1}\right) \boldsymbol{x}\left(t_{2}\right)>0\right] &=2 \int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{2 \pi|\Sigma|^{1 / 2}} \exp \left\{-\frac{1}{2}[x, y] \Sigma^{-1}\left[\begin{array}{l} x \\ y \end{array}\right]\right\} d x d y \\ &=2 \int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{2 \pi\left(1-\rho^{2}\right)^{1 / 2} R_{x x}(0)} \exp \left\{-\frac{x^{2}-2 \rho x y+y^{2}}{2\left(1-\rho^{2}\right) R_{x x}(0)}\right\} d x d y \\ &=\frac{1}{2}+\frac{1}{\pi} \arcsin (\rho) \end{aligned}$$

具体推导见课件81-82。

所以

$$\begin{aligned} E\left[\boldsymbol{y}\left(t_{1}\right) \boldsymbol{y}\left(t_{2}\right)\right] &=\operatorname{Pr}\left[\boldsymbol{x}\left(t_{1}\right) \boldsymbol{x}\left(t_{2}\right)>0\right] - \operatorname{Pr}\left[\boldsymbol{x}\left(t_{1}\right) \boldsymbol{x}\left(t_{2}\right)<0\right] \\ &= 2\operatorname{Pr}\left[\boldsymbol{x}\left(t_{1}\right) \boldsymbol{x}\left(t_{2}\right)>0\right]-1\\ &=\frac{2}{\pi} \arcsin (\rho)\\ &=\frac{2}{\pi} \arcsin \left(\frac{R_{x x}\left(t_{1}-t_{2}\right)}{R_{x x}(0)}\right) \end{aligned}$$

定理（Bussgang）

假设系统是无记忆的，如果输入为$0$均值，高斯，那么互相关函数$R_{x y}(\tau)$和$R_{xx}(\tau)$成正比。

证明：

$$\begin{aligned} R_{x y}\left(t_{1}, t_{2}\right) &=E\left[\boldsymbol{x}\left(t_{1}\right) \cdot \boldsymbol{T}\left(\boldsymbol{x}\left(t_{2}\right)\right)\right] \\ &=E\left[E\left[x_{1} \cdot \boldsymbol{T}\left(x_{2}\right) | \boldsymbol{x}\left(t_{1}\right)=x_{1}, \boldsymbol{x}\left(t_{2}\right)=x_{2}\right]\right] \\ &=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty} x_{1} y d P_{\boldsymbol{y} | \boldsymbol{x}}\left(y | x_{2}\right)\right) \frac{1}{2 \pi|\Sigma|^{1 / 2}} \exp \left\{-\frac{1}{2}\left[x_{1}, x_{2}\right] \Sigma^{-1}\left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right]\right\} d x_{1} d x_{2} \end{aligned}$$

协方差矩阵为

$$\Sigma=\left[\begin{array}{cc} R_{x x}(0) & R_{x x}\left(t_{1}-t_{2}\right) \\ R_{x x}\left(t_{2}-t_{1}\right) & R_{x x}(0) \end{array}\right]=R_{x x}(0)\left[\begin{array}{cc} 1 & \rho \\ \rho & 1 \end{array}\right]$$

其中$\rho=R_{x x}\left(t_{1}-t_{2}\right) / R_{x x}(0)$。

令

$$g\left(x_{2}\right)=E\left[\boldsymbol{T}\left(x_{2}\right)\right]$$

那么

$$\begin{aligned} R_{x y}\left(t_{1}, t_{2}\right) &=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{x_{1} \cdot g\left(x_{2}\right)}{2 \pi R_{x x}(0)\left(1-\rho^{2}\right)^{1 / 2}} \exp \left\{-\frac{x_{1}^{2}-2 \rho x_{1} x_{2}+x_{2}^{2}}{2 R_{x x}(0)\left(1-\rho^{2}\right)}\right\} d x_{1} d x_{2} \\ &=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{x_{1} \cdot g\left(x_{2}\right)}{2 \pi R_{x x}(0)\left(1-\rho^{2}\right)^{1 / 2}} \exp \left\{-\frac{(x_{1}-\rho x_2)^{2}}{2 R_{x x}(0)\left(1-\rho^{2}\right)}\right\}\exp \left\{-\frac{x_{2}^{2}}{2 R_{x x}(0)}\right\} d x_{1} d x_{2} \\ &=\rho \int_{-\infty}^{\infty} \frac{x_{2} \cdot g\left(x_{2}\right)}{\sqrt{2 \pi R_{x x}(0)} }\exp \left\{-\frac{x_{2}^{2}}{2 R_{x x}(0)}\right\} d x_{2} \\ &=\rho E\left[E\left[x_{2} \cdot g\left(x_{2}\right) | x(t)=x_{2}\right]\right]\\ &=R_{x x}\left(t_{1}-t_{2}\right) \frac{E[x(t) \cdot g(x(t))]}{R_{x x}(0)} \end{aligned}$$

如果$g’(.)$存在，以及

$$\lim _{x \rightarrow \infty} g(x) \exp \left\{-\frac{x^{2}}{2 R_{x x}(0)}\right\}=\lim _{x \rightarrow-\infty} g(x) \exp \left\{-\frac{x^{2}}{2 R_{x x}(0)}\right\}<\infty$$

那么

$$\frac{E[x(t) \cdot g(x(t))]}{R_{x x}(0)} = E\left[g^{\prime}(\boldsymbol{x}(t))\right]$$

证明：

$$\begin{aligned} & \frac{1}{R_{x x}(0)} \int_{-\infty}^{\infty} \frac{x \cdot g(x)}{\sqrt{2 \pi R_{x x}(0)}} \exp \left\{-\frac{x^{2}}{2 R_{x x}(0)}\right\} d x \\ &=-\left.\frac{g(x)}{\sqrt{2 \pi R_{x x}(0)}} \exp \left\{-\frac{x^{2}}{2 R_{x x}(0)}\right\}\right|_{-\infty} ^{\infty}+\int_{-\infty}^{\infty} \frac{g^{\prime}(x)}{\sqrt{2 \pi R_{x x}(0)}} \exp \left\{-\frac{x^{2}}{2 R_{x x}(0)}\right\} d x\\ &=\int_{-\infty}^{\infty} \frac{g^{\prime}(x)}{\sqrt{2 \pi R_{x x}(0)}} \exp \left\{-\frac{x^{2}}{2 R_{x x}(0)}\right\} d x \end{aligned}$$

Dirac Delta函数

定义Dirac Delta函数$\delta(t)$为满足如下条件的函数：

$$\delta(t)=\delta(-t)=\left\{\begin{array}{ll} \infty, & t=0 ; \\ 0, & t \neq 0 \end{array} \text { and } \int_{-\infty}^{\infty} \delta(t) d t=\int_{-\infty}^{\infty} \delta(-t) d t=1\right.$$

Dirac Delta函数满足如下性质：

$$g(t)=\int_{-\infty}^{\infty} g(\tau) \delta(t-\tau) d \tau=\int_{-\infty}^{\infty} g(\tau) \delta(\tau-t) d \tau$$

线性系统

如果系统满足

$$T\left(\alpha x_1 +\beta x_2\right) =\alpha T(x_1)+\beta T(x_2)$$

那么系统$T$被称为线性系统。

线性系统可以表示为卷积的形式，为了证明这点，首先考虑如下引理：

引理

如果一阶可微函数$f$满足

$$f(\vec{x})+f(\vec{u})=f(\vec{x}+\vec{u})$$

那么

$$f(\vec{x})=\sum_{i}\left(\left.\frac{\partial f(\vec{x})}{\partial x_{i}}\right|_{\vec{x}=0}\right) x_{i}$$

证明：

$$\begin{aligned} \frac{\partial f(\vec{x})}{\partial x_{i}} &=\lim _{\delta \downarrow 0} \frac{f\left(\cdots, x_{i-1}, x_{i}+\delta, x_{i+1}, \cdots\right)-f\left(\cdots, x_{i-1}, x_{i}, x_{i+1}, \cdots\right)}{\delta} \\ &=\lim _{\delta \downarrow 0} \frac{\left[f\left(\cdots, x_{i-1}, x_{i}, x_{i+1}, \cdots\right)+f(\cdots, 0, \delta, 0, \cdots)\right]-f\left(\cdots, x_{i-1}, x_{i}, x_{i+1}, \cdots\right)}{\delta} \\ &=\lim _{\delta \downarrow 0} \frac{f(\cdots, 0, \delta, 0, \cdots)}{\delta} \\ &=\text {constant } \end{aligned}$$

所以

$$f(\vec{x})=\sum_{i}\left(\left.\frac{\partial f(\vec{x})}{\partial x_{i}}\right|_{\vec{x}=0}\right) x_{i}+C$$

注意到线性系统显然有

$$f(\vec 0) = 0$$

这说明

$$C=0$$

即

$$f(\vec{x})=\sum_{i}\left(\left.\frac{\partial f(\vec{x})}{\partial x_{i}}\right|_{\vec{x}=0}\right) x_{i}$$

卷积

线性系统可以表示为卷积的形式。

连续系统：

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

离散系统：

$$\boldsymbol{y}(t)=\sum_{n=-\infty}^{\infty} \boldsymbol{h}(n ; t) \boldsymbol{x}(t-n)$$

脉冲响应：

$$h(t-s;t)= T(\delta(t-s))$$

以离散情形的例子证明该结论：

定义线性系统

$$\boldsymbol{T}_{t}(\{\boldsymbol{x}(s), s \in \mathcal{I}\})=T_{t}(\{\boldsymbol{x}(s), s \in\{t, t-1\}\})=T_{t}(\boldsymbol{x}(t), \boldsymbol{x}(t-1))$$

那么由引理可得

$$\begin{aligned} \boldsymbol{y}(t) &=\left(\left.\frac{\partial T_{t}\left(x_{1}, x_{2}\right)}{\partial x_{1}}\right|_{x_{1}=x_{2}=0}\right) \boldsymbol{x}(t)+\left(\left.\frac{\partial T_{t}\left(x_{1}, x_{2}\right)}{\partial x_{2}}\right|_{x_{1}=x_{2}=0}\right) x(t-1) \\ &=\sum_{n=0}^{1} h(n ; t) \boldsymbol{x}(t-n) \end{aligned}$$

同理可得

$$\boldsymbol{y}(t)=\sum_{n=-\infty}^{\infty} \boldsymbol{h}(n ; t) \boldsymbol{x}(t-n)$$

时不变系统

系统被称为时不变，如果$\boldsymbol{T}_{t}\left(x^{\mathcal{I}}, \zeta\right)=\boldsymbol{T}\left(x^{\mathcal{I}}, \zeta\right)$。

含义：$x(t-s)$的输出为$y(t-s)$

此时

$$\boldsymbol{h}(\tau ; t)=\boldsymbol{h}(\tau)$$

时变系统

考虑

$$\boldsymbol{h}(\tau ; t)=\delta(\tau) e^{j \omega_{0} t}$$

那么

$$\boldsymbol{y}(t)=\int_{-\infty}^{\infty} \delta(\tau) e^{j \omega_{0} t} \boldsymbol{x}(t-\tau) d \tau=\boldsymbol{x}(t) e^{j \omega_{0} t}$$

显然该系统是线性的，但是如果输入为$x(t-s)$，那么输出为

$$x(t-s)e^{j \omega_{0} t} \neq y(t-s)$$

所以是时变的。

基本定理和定理9.2

对于任意线性系统，我们有

$$\begin{aligned} \eta_{y}(t) &=E\left[\boldsymbol{h}(\tau ; t) * \eta_{x}(t)\right]\\ R_{xy}(t_1,t_2)&=E\left[\boldsymbol{h}^{*}\left(\tau ; t_{2}\right) * R_{x x}\left(t_{1}, t_{2}\right)\right]\\ R_{yy}(t_1,t_2)&=E\left[\boldsymbol{h}^{*}\left(\tau ; t_{2}\right) * \boldsymbol{h}\left(\tau ; t_{1}\right) * R_{x x}\left(t_{1}, t_{2}\right)\right] \end{aligned}$$

证明：

$$\begin{aligned} E[\boldsymbol{y}(t)] &=\int_{-\infty}^{\infty} E[\boldsymbol{h}(\tau ; t) \boldsymbol{x}(t-\tau)] d \tau\\ &=\int_{-\infty}^{\infty} E[\boldsymbol{h}(\tau ; t)] E[\boldsymbol{x}(t-\tau)] d \tau \\ &=\int_{-\infty}^{\infty} E[\boldsymbol{h}(\tau ; t)] \eta_{x}(t-\tau) d \tau \\ R_{x y}\left(t_{1}, t_{2}\right) &=E\left[\boldsymbol{x}\left(t_{1}\right) \boldsymbol{y}^{*}\left(t_{2}\right)\right]\\ &=E\left[\boldsymbol{x}\left(t_{1}\right) \int_{-\infty}^{\infty} \boldsymbol{h}^{*}\left(\tau ; t_{2}\right) \boldsymbol{x}^{*}\left(t_{2}-\tau\right) d \tau\right] \\ &=\int_{-\infty}^{\infty} E\left[\boldsymbol{h}^{*}\left(\tau ; t_{2}\right)\right] E\left[\boldsymbol{x}\left(t_{1}\right) \boldsymbol{x}^{*}\left(t_{2}-\tau\right)\right] d \tau \\ &=\int_{-\infty}^{\infty} E\left[\boldsymbol{h}^{*}\left(\tau ; t_{2}\right)\right] R_{x x}\left(t_{1}, t_{2}-\tau\right) d \tau \\ &=E\left[\int_{-\infty}^{\infty} \boldsymbol{h}^{*}\left(\tau ; t_{2}\right) R_{x x}\left(t_{1}, t_{2}-\tau\right) d \tau\right]\\ R_{y y}\left(t_{1}, t_{2}\right) &=E\left[\boldsymbol{y}\left(t_{1}\right) \boldsymbol{y}^{*}\left(t_{2}\right)\right]\\ &=E\left[\int_{-\infty}^{\infty} \boldsymbol{h}\left(\tau ; t_{1}\right) \boldsymbol{x}\left(t_{1}-\tau\right) d \tau \int_{-\infty}^{\infty} \boldsymbol{h}^{*}\left(s ; t_{2}\right) \boldsymbol{x}^{*}\left(t_{2}-s\right) d s\right] \\ &=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} E\left[\boldsymbol{h}^{*}\left(s ; t_{2}\right) \boldsymbol{h}\left(\tau ; t_{1}\right)\right] E\left[\boldsymbol{x}\left(t_{1}-\tau\right) \boldsymbol{x}^{*}\left(t_{2}-s\right)\right] d \tau d s \\ &=E\left[\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \boldsymbol{h}^{*}\left(s ; t_{2}\right) \boldsymbol{h}\left(\tau ; t_{1}\right) R_{x x}\left(t_{1}-\tau, t_{2}-s\right) d \tau d s\right] \end{aligned}$$

推论

对于任意线性系统

$$\begin{aligned} C_{xy}(t_1,t_2)&=E\left[\boldsymbol{h}^{*}\left(\tau ; t_{2}\right) * C_{x x}\left(t_{1}, t_{2}\right)\right]\\ C_{yy}(t_1,t_2)&=E\left[\boldsymbol{h}^{*}\left(\tau ; \boldsymbol{t}_{2}\right) * \boldsymbol{h}\left(\tau ; \boldsymbol{t}_{1}\right) * C_{x y}\left(t_{1}, t_{2}\right)\right] \end{aligned}$$

例1：微分算子

微分算子是线性时不变系统，其输出为输入的导数，即

$$\eta_{y}(t)=\frac{\partial \eta_{x}(t)}{\partial t}$$

注意微分系统可以表示为

$$\boldsymbol{y}(t)=\lim _{\epsilon \to 0} \int_{-\infty}^{\infty} \boldsymbol{h}_{\epsilon}(\tau) \boldsymbol{x}(t-\tau) d \tau$$

其中

$$h_{\epsilon}(\tau)=\frac{\delta(\tau+\epsilon)-\delta(\tau)}{\epsilon}$$

那么根据定理9.2，我们有

$$\begin{aligned} R_{xy}(t_1,t_2) &=E\left[\boldsymbol{h}^{*}\left(\tau ; t_{2}\right) * R_{x x}\left(t_{1}, t_{2}\right)\right]\\ &= \lim _{\epsilon \to 0} \int_{-\infty}^{\infty} \boldsymbol{h}_{\epsilon}(\tau) R_{x x}\left(t_{1}, t_{2} -\tau\right) d \tau\\ &= \frac{\partial R_{x x}\left(t_{1}, t_{2}\right)}{\partial t_{2}} \end{aligned}$$

同理可得

$$R_{y y}\left(t_{1}, t_{2}\right)=\frac{\partial R_{x y}\left(t_{1}, t_{2}\right)}{\partial t_{1}}=\frac{\partial^{2} R_{x x}\left(t_{1}, t_{2}\right)}{\partial t_{1} \partial t_{2}}$$

利用推论得到

$$C_{x y}\left(t_{1}, t_{2}\right)=\frac{\partial C_{x x}\left(t_{1}, t_{2}\right)}{\partial t_{2}}$$

以及

$$C_{y y}\left(t_{1}, t_{2}\right)=\frac{\partial C_{x y}\left(t_{1}, t_{2}\right)}{\partial t_{1}}=\frac{\partial^{2} C_{x x}\left(t_{1}, t_{2}\right)}{\partial t_{1} \partial t_{2}}$$

例2：微分方程

微分方程的形式如下：

$$a_{n} \boldsymbol{y}^{(n)}(t)+\cdots+a_{0} \boldsymbol{y}(t)=\boldsymbol{x}(t)$$

这里假设初值为$0$，$y(t)$唯一。

那么根据基本定理：

$$a_{n} \eta_{y}^{(n)}(t)+\cdots+a_{0} \eta_{y}(t)=\eta_{x}(t) \text { with } \eta_{y}(0)=\cdots=\eta_{y}^{(n-1)}(0)=0$$

将系统理解为输入$y(t)$，输出$x(t)$，那么根据定理9.2可得

$$\begin{array}{l} a_{n} \frac{\partial^{n} R_{x y}\left(t_{1}, t_{2}\right)}{\partial t_{2}^{n}}+\cdots+a_{0} R_{x y}\left(t_{1}, t_{2}\right)=R_{x x}\left(t_{1}, t_{2}\right) \text { with } R_{x y}\left(t_{1}, 0\right)=\cdots=\frac{\partial^{n-1} R_{x y}\left(t_{1}, 0\right)}{\partial t_{2}^{n-1}}=0 \\ a_{n} \frac{\partial^{n} R_{y y}\left(t_{1}, t_{2}\right)}{\partial t_{1}^{n}}+\cdots+a_{0} R_{y y}\left(t_{1}, t_{2}\right)=R_{x y}\left(t_{1}, t_{2}\right) \text { with } R_{y y}\left(0, t_{2}\right)=\cdots=\frac{\partial^{n-1} R_{y y}\left(0, t_{2}\right)}{\partial t_{1}^{n-1}}=0 \end{array}$$

定理9-2的推广