自己顶起来，希望高手解释下

Not quite sure, but my understanding is that

1.        it’s still AR(1) model, and as it says, the pattern seems to repeat every year, which is 4 quarters, so we have this lag factor (t-4), maybe it is more appropriate to note it as AR(1)4
2.        For AR(2) model, without the seasonality, y(t) will depend on both y(t-1) and y(t-2), which is not really the case here. For every year, the value only depends on the previous quarter, only that there is a pattern in it. If it is monthly data, we would have used (t-1) and (t-12) , but still it’s AR(1)12

3.    not sure about the observation….

首先感谢　forestrun　　的回复。
　　但是仍然不理解。 lnyt=b0+b1(lnyt-1)+b2(lnyt-4),中　lnyt-1 虽然不直接由lnyt-4决定，但是，从整个序列来讲lnyt-1仍然由lnyt-4决定。
　　若仅理解为此式对quarter4适用，或可认为可解释。
　　但从前后二张示例表的数据来看，每一期的lag数据均发生了变化，即此式适用用每个季度

　　哎，还是不甚明白。继续求教高人指点

你看看NOTES的勘误表吧：Book 1, pg. 233 - Correction
The correct number of observations in this problem is 36; this makes all the t-statistics slightly different.  Note that it is extremely unlikely you will have to calculate these t-statistics on the exam. ( Posted: 2010-03-23)
个人觉得那个36（也就是你一楼的38）不是因为是几阶的AR造成的，而是LAG的长度造成的，因为在 lnyt=b0+b1(lnyt-1)+b2(lnyt-4)的模型中需要t-4，所以可以用的样本数据就只剩下40-4个了。
而关于 lnyt=b0+b1(lnyt-1)+b2(lnyt-4)是AR(1)还是AR(2)，个人以为是要看年化的需要，是以年为单位，t-4正好是一年也就是说回归到自己所以只有t-1能算AR(1)，这样的话，如果模型是以月为单位，那么lnyt=b0+b1(lnyt-1)+b2(lnyt-4)就是AR(2)，而lnyt=b0+b1(lnyt-1)+b2(lnyt-12)就是AR(1)，而且40个样本数据只能用40-12=28个

个人的一点想法，不知大拿们怎么看。