求助！！！关于ARCH(1)

对于ARCH(1): Xt=ct*Zt
                        ct^2=a+b*X(t-1) ^2
若b<1则方差c固定为a/(1-b)

那么，是否可以将b<1时ARCH(1)表示为 Xt=Zt * [a/(1-b)]^0.5，为何不可？谢谢

没人能指点下吗？？？

This is a time series problem, X(t)=Zt * [a+b*X(t-1) ^2]^0.5. I guess the reason why you are wrong is that the X(t-1) is still a random variable. No matter it has constant variance or not. Hope this could help.

yeah, I find the answer, just because ct is a process!
strictly,  xt is also a process instead of a rv
But thank u all the same!

And give u the money!

只要区别条件方差和方差就可以了，事实上，条件异方差并不意味着（无条件）异方差

楼上可否详细点，谢谢