Linear trend models and mean reversion vs AR models

CFA Level II Quant
#1

I feel like I’m taking crazy pills 5 days before the exam, but somebody more sane please help.

Just by the way a linear trend model is formed, y = a + bt, isn’t it assuming no mean reversion? So if you have a mean reverting time series, your linear trend model will not be helpful and presumably have low explanatory power. Now if you have a time series that is NOT mean reverting, the linear trend model seems good but then an AR model doesn’t work since it isn’t covariance stationary (until you first difference).

Is this the correct way to think about it? Mean reverting time series = don’t use linear trend model?

#2

Linear trend model does not show mean reversion. Think of It as a linear line that either goes up or down.

Mean reversion is the tendency to move toward the mean and since there is no mean for a linear trend model, mean reversion is irrelevant

#3

Where are you getting this from? There absolutely can be a mean in a linear trend model (in fact, the entire function estimates the means of Y given the combination of independent variables [conditional mean], in this case at a certain point in time)…

#4

I meant since a linear trend is either decreasing or increasing one can not calculate a mean that dependent variable can revert to it. Of course you can always isolate some of the data and get the conditional mean for any time series but think of it this way: in a linear trend by any additional independent data point, dependent variable is moving away from the conditional mean you just mentioned.

#5

A time series isn’t mean reverting. An AR model is if the b1 is less than one.

For Xt = 2 + 0.5 (Xt-1)

20.00 12.00 12.00 8.00 8.00 6.00 6.00 5.00 5.00 4.50 4.50 4.25 4.25 4.13 4.13 4.06 4.06 4.03

2/0.5 = 4

#6

Right-- you did say there is no mean in a linear trend, which is untrue. A mean is different than a mean reverting level, which is definitely important to distinguish between. I only commented on it to make that clear.

#7

I have to disagree. It is like I tell you there is no “mean” for an exponential line and you argue that you can calculate conditional mean for a part of the exponential line!

#8

I’m not saying you could calculate it for a part of the line-- I’m saying the line consists of means. A true (population) regression function represent a line of true (population) means (all along the function). When we estimate the function, we are estimating the true line of means. This is embodied in the fact that we constantly use expectations (expected values, means).

If we fit Yi=b0 + b1*Ti + ei , we are also saying the relationship is this: E(Yi) = b0 + b1*Ti + E(ei) and E(ei) = 0 by our assumptions for the technique.

This idea is depicted in many texts as the PRF with distributions along the line, centered on the mean of the distribution. http://www.psychstat.missouristate.edu/introbook/sbk16.htm Scroll down to conditional distributions and you will see the picture I am referring to, even though this isn’t as nice as the image in some statistics texts.

A mean and a mean-reverting level are different ideas…

#9

I agree mean and mean-reverting level are different. I think I worded it poorly. When you are refering to conditional distributions you are talking about distribution of dependent variable based on a specific value of independent variable and in our linear trend it means the distribution of values for Y(t) based on a specific t. when I say “no mean” I am talking about dependent variable through out the whole trend. Again my bad for poor wording and I apologize for that but still I believe that you are talking about mean of distribution at a point in time otherwise it makes no sense to say a linear trend has mean

#10