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差分方程的解析解

例题.求$y_t=a_0+a_1y_{t-1}+\epsilon_t$的解析解
当初始条件已知,即$y_0$已知时,迭代算法如下: * 向前迭代算法
$$y_1=a_0+a_1y_0+\epsilon_1$$ $$y_2=a_0+a_1y_1+\epsilon_2$$ $$y_3=a_0+a_1y_2+\epsilon_3$$ $$...$$ $$y_t=a_0+a_1y_{t-1}+\epsilon_t$$
以上列举了该差分方程的所有项,我们要做的就是通过这些递推关系来逐步迭代,当迭代完所有项之后,最终可以得到目标解析解.
$$y_1=a_0+a_1y_0+\epsilon_1$$ $$y_2=a_0+a_1y_1+\epsilon_2=a_0+a_1(a_0+a_1y_0+\epsilon_1)+\epsilon_2=$$ $$a_0+a_0a_1+{a_1}^2y_0+a_1\epsilon_1+\epsilon_2=[a_0(1+a_1)]+[a_1^2y_0]+[a_1\epsilon_1+\epsilon_2]$$ $$y_3=a_0+a_1y_2+\epsilon_3=a_0+a_1(a_0+a_0a_1+{a_1}^2y_0+a_1\epsilon_1+\epsilon_2)+\epsilon_3=$$ $$[a_0+a_0a_1+a_0{a_1}^2]+[a_1^3y_0]+[a_1^2\epsilon_1+a_1\epsilon_2+\epsilon_3]=[a_0(1+a_1+a_1^2)]+[a_1^3y_0]+[a_1^2\epsilon_1+a_1\epsilon_2+\epsilon_3]$$ $$...$$ 由数学归纳法,递推下去,则有:
$$y_t=[a_0(a_1+a_1^2+...+a_1^{t-1})]+[a_1^ty_0]+[a_1^{t-1}\epsilon_1+a_1^{t-2}\epsilon_2+...+a_1^0\epsilon_t]=$$ $$a_0\sum_{i=0}^{t-1}a_1^i+a_1^ty_0+\sum_{i=0}^{t-1}a_1^i\epsilon_{t-i}$$ 上式便是目标解析解,因为$y_0$已知,即不含有任何未知项.
之所以称之为“向前迭代法”,是因为迭代的索引是从$y_1$开始,一步一步往前面跑去,逐步迭代得到到目标式$y_n$的.
你肯定会想到,既然有向前迭代,那么一定有向后迭代了,没错,向后迭代就是从最后一项$y_t$逐渐向$y_0$进行迭代.
向后迭代
同样是求解$y_t=a_0+a_1y_{t-1}+\epsilon_t$的解析解: $$y_t=a_0+a_1y_{t-1}+\epsilon_t$$ $$y_{t-1}=a_0+a_1y_{t-2}+\epsilon_{t-1}$$ $$y_{t-2}=a_0+a_1y_{t-3}+\epsilon_{t-2}$$ $$...$$ $$y_2=a_0+a_1y_1+\epsilon_2$$ $$y_1=a_0+a_1y_0+\epsilon_1$$ $$y_0=a_0+a_1y_{-1}+\epsilon_0$$ 开始迭代:(前面写了两次迭代过程,总共迭代$t-1$次)
$$y_t=a_0+a_1y_{t-1}+\epsilon_t=$$ $$a_0+a_1(a_0+a_1y_{t-2}+\epsilon_{t-1})+\epsilon_t=$$ $$a_0+a_1a_0+a_1^2y_{t-2}+a_1\epsilon_{t-1}+\epsilon_t=$$ $$[a_0(1+a_1)]+[a_1^2y_{t-2}]+[a_1\epsilon_{t-1}+\epsilon_t]$$ $$a_0+a_1a_0+a_1^2(a_0+a_1y_{t-3}+\epsilon_{t-2})+a_1\epsilon_{t-1}+\epsilon_t=$$ $$[a_0+a_1a_0+a_1^2a_0]+[a_1^3y_{t-3}]+[a_1^2\epsilon_{t-2}+a_1\epsilon_{t-1}+\epsilon_t]=$$ $$[a_0(1+a_1+a_1^2)]+[a_1^3y_{t-3}]+[a_1^2\epsilon_{t-2}+a_1\epsilon_{t-1}+\epsilon_t]=$$ $$...$$ $$=[a_0(1+a_1+a_1^2+...+a_1^{t-1})]+[a_1^ty_0]+[a_1^{t-1}\epsilon_{1}+a_1^{t-2}\epsilon_{2}+...+a_1^0\epsilon_t]=$$ $$a_0\sum_{i=0}^{t-1}a_1^i+a_1^ty_0+\sum_{i=0}^{t-1}a_1^i\epsilon_{t-i}$$
而当初始条件未知,即$y_0$未知时,我们需要对上面得到的差分方程的解继续化:
$$y_t=a_0\sum_{i=0}^{t-1}a_1^i+a_1^ty_0+\sum_{i=0}^{t-1}a_1^i\epsilon_{t-i}=$$ $$a_0\sum_{i=0}^{t-1}a_1^i+a_1^t(a_0+a_1y_{-1}+\epsilon_0)+\sum_{i=0}^{t-1}a_1^i\epsilon_{t-i}=$$ $$a_0\sum_{i=0}^{t-1}a_1^i+(a_0a_1^t+a_1^{t+1}y_{-1}+a_1^t\epsilon_0)+\sum_{i=0}^{t-1}a_1^i\epsilon_{t-i}=$$ $$a_0\sum_{i=0}^{t}a_1^i+a_1^{t+1}y_{-1}+\sum_{i=0}^{t}a_1^i\epsilon_{t-i}=$$ $$...$$ $$=a_0\sum_{i=0}^{(t-1)+m}a_1^i+a_1^{t+m}y_{-m}+\sum_{i=0}^{(t-1)+m}a_1^i\epsilon_{t-i}$$
由于初始条件未知,我们现在来讨论解是否存在,即解的敛散性:
当$|a_1|