自相关(ACF)与偏自相关(PACF)（1）

§1.The Autocorrelation Function for AR(1) Process
Consider the AR(1) process:
$$y_t=a_0+a_1{y}_{t-1}+{\epsilon }_t$$

where ${\epsilon }_t$ is a white noise process. And suppose the process starts at period 0 with value $y_0$.
if $|a_1|<1$,as $t\to \infty$,we have:
$$\underset{t\to \infty }{\lim }{y}_t=\frac{a_0}{1-a_1}+\sum _{i=0}^{\infty }{a}_1^i{\epsilon }_{t-i}$$

we have for sufficiently large enough t,
$$E(y_t)=\frac{a_0}{1-a_1}\equiv \mu$$

$$\begin{array}{l}E(y_t-\mu {)}^2=E[({\epsilon }_t+a_1{\epsilon }_{t-1}+a_1^2{\epsilon }_{t-2}+\cdots {)}^2]\\ =E({\epsilon }_t^2+a_1^2{\epsilon }_{t-1}^2+a_1^4{\epsilon }_{t-2}^2+\cdots )\\ ={\sigma }^2+a_1^2{\sigma }^2+a_1^4{\sigma }^2+\cdots ={\sigma }^2(1+a_1^2+a_1^4+\cdots )\\ ={\sigma }^2\cdot \frac{1}{1-a_1^2}\end{array}$$

$$\begin{array}{l}E[(y_t-\mu )(y_{t-s}-\mu )]\\ =E[({\epsilon }_t+a_1{\epsilon }_{t-1}+\cdots a_1^s{\epsilon }_{t-s}+a_1^{s+1}{\epsilon }_{t-s-1}+\cdots )({\epsilon }_{t-s}+a_1{\epsilon }_{t-s-1}+\cdots )]\\ =a_1^s{\sigma }^2+a_1^{s+2}{\sigma }^2+a_1^{s+4}{\sigma }^2+\cdots \\ ={\sigma }^2(a_1^s)(1+a_1^2+a_1^4+\cdots )\\ ={\sigma }^2(a_1^s)\cdot \frac{1}{1-a_1^2}\end{array}$$

then we have:
$$\begin{array}{l}{\gamma }_0={\sigma }^2/(1-a_1^2)\\ {\gamma }_1={\sigma }^2(a_1^s)/(1-a_1^2)\end{array}$$

And forming the autocorrelations by dividing each ${\gamma }_s$ by ${\gamma }_0$, we find that:
$$\begin{array}{l}{\rho }_0=1\\ {\rho }_s=a_1^s\end{array}$$

$$\{\begin{array}{l}0<a_1<1时，自相关系数直接收敛到0\\ -1<a_1<0时，自相关系数震荡收敛到0\end{array}$$

§2.The Autocorrelation Function for AR(2) Process
Consider the AR(2) model：
$$y_t=a_1{y}_{t-1}+a_2{y}_{t-2}+{\epsilon }_t$$

If the roots of $1-a_{1}L-a_2{L}^2=0$ are outside the unit circle, the process is stationary.
thus,we have:
$$\begin{array}{l}E(y_t)=\mu \\ E(y_t)=a_{1}E(y_{t-1})+a_{2}E(y_{t-2})+E({\epsilon }_t)\\ \Rightarrow \mu =a_1\mu +a_2\mu +0\\ \Rightarrow \mu =0\end{array}$$

Now we apply the technique of Yule-Walker equations. Multiply the equation by $y_{t-s}fors=0,1,2,...$ and take expectation:
$E{y}_t{y}_t=a_{1}E{y}_{t-1}{y}_t+a_{2}E{y}_{t-2}{y}_t+E{\epsilon }_t{y}_t$
$E{y}_t{y}_{t-1}=a_{1}E{y}_{t-1}{y}_{t-1}+a_{2}E{y}_{t-2}{y}_{t-1}+E{\epsilon }_t{y}_{t-1}$
$E{y}_t{y}_{t-s}=a_{1}E{y}_{t-1}{y}_{t-s}+a_{2}E{y}_{t-2}{y}_{t-s}+E{\epsilon }_t{y}_{t-s}$
From the above equations, we have:
${\gamma }_0=a_1{\gamma }_1+a_2{\gamma }_2+{\sigma }^2$
${\gamma }_1=a_1{\gamma }_0+a_2{\gamma }_1+0$
${\gamma }_s=a_1{\gamma }_{s-1}+a_2{\gamma }_{s-2}+0$
So that,
$$\begin{gathered} {\rho }_1=a_1{\rho }_0+a_2{\rho }_1 \\ {\rho }_s=a_1{\rho }_{s-1}+a_2{\rho }_{s-2} \end{gathered}$$

By ${\rho }_0=1$, we derive that ${\rho }_1=a_1/(1-a_2)$ and then derive ${\rho }_s$ for all $s\ge 2$.
注意到，${\rho }_s=a_1{\rho }_{s-1}+a_2{\rho }_{s-2}$的特征方程与AR(2) model一致，均为$1-a_{1}L-a_2{L}^2=0$，因此方程${\rho }_s=a_1{\rho }_{s-1}+a_2{\rho }_{s-2}$也是平稳的，并且其收敛形式取决于$a_1,a_2$。