non stationary covariance

Could someone explain why coeff b1 >1 makes the time series have non stationary covariance. I believe the term used is ‘explosive root’ (as opposed to unit root). also, i keep getting confused about when to use the SEE as opposed to Sf(std error of forecast) to calculate the Confidence interval of the dependent variable. please help. thanks dsylexic

I’m also struggling with this particular part of the curriculum right now.

For an easy example look at y(t) = c + b*y(t-1) + e(t) where y(t) = ozone level on day t in LA measured in ppm and e(t) = mean 0 normal error. Ozone doesn’t just go away overnight and then rebuild the next day - given the right weather it sticks around to rot your lungs day after day. So if I want to know ozone tomorrow, there is lots of info in the ozone today. First, covariance stationary means that if I want the covariance of y(100) and y(105), that covaraince is the same as the covariance between y(567432) and y(567432 + 5). It also means that the mean and variance of the y(t) does not depend on t. So intuitively suppose that b is some humoungous number like 500. That means that if we have some ozone today, we expect some massive amount of ozone tomorrow, and fifteen days from now we expect ozone to have turned LA into photochemical ooze. Clearly that’s not stationary. If b = 0, then our series is just y(t) = c + e(t) which means that each day is an independent normal which is about as stationary as things get. Somewhere in the middle is a boundary which turns out to be at b = 1 where the process is a random walk. That should seem pretty natural by just thinking of a simple random walk where you go +1 with a head and -1 with a tail. If you generate sample paths from this they diverge from 0 pretty quickly so the variance is not stable.

thanks Joey. the mean of the series is c/1-b and excepting for b =1, the series does have a finite,constant mean even if b >1. this is one of the conditions for cov. stationarity - i guess thats why i was confused and didnt recognize how the variance itself was not constant thus violating stationarity.

So just write it out: y(t) = c + b*y(t-1) + e(t) y(t) = c + b*(c + b*y(t-2) + e(t-1)) + e(t) y(t) = c + b*(c + b*(c + b*y(t-1) + e(t-2)) + e(t-1)) + e(t) =(c + b*c + b^2*c +…+b^t*c) + b^t*y(0) + (e(t) + b*e(t-1) + b^2*e(t-2) +…+b^t*e(0)) so there are three terms there. For |b|<1, the first goes to c/(1-b) , the second goes to 0, and the third goes to some normal mean 0 error term whose variance you can get. If b = 1, that first term just keeps getting bigger like c*t as does the variance (but I’m too lazy to figure out the formula for the variance). Remember covariance stationary is a much weaker condition than stationary and just means the first two moments don’t depend on t.

I’ve learned this term “covariance stationary” today studying for CFA. When I took Random processes we called it “weak stationary.” Using Joey’s example: y(t) = c + b*y(t-1)+e(t) Covariance stationary -> Var(y(t)) = Var(y(t-1)) = Var(y) Since c is a constant, e(t) is noise - independent of y(t-1) then Var(y) = b^2*Var(y)+Var(e) > b^2*Var(y) -> b^2 < 1 for y to be covariance stationary.