# covariance stationary/unit root question

Wireless Phone Minutes (WPM)t = bo + b1 WPMt-1 + Et ANOVA: Regression df= 1 sum of sq= 7,212.641 mean sq= 7,212.641 Error df= 26 sum of sq= 3,102.410 mean sq=119.324 Total df= 27 sum of sq= 10,315.051 Coefficients: Intercept Coefficient= -8.0237 Standard Error of the Coefficient=2.9023 WPM -1 Coefficient=1.0926 Standard Error of the Coefficient= 0.0673 The variance of the residuals from one time period within the time series is not dependent on the variance of the residuals in another. this was part of a few question vignette, Question ID#: 47804 if you want to see the charts in a much prettier format. here’s the question i just booted: Is the time series of WPM covariance stationary? A) Yes, because the computed t-statistic for a slope of 1 is not significant. B) No, because the computed t-statistic for a slope of 1 is not significant. C) Yes, because the computed t-statistic for a slope of 1 is significant. D) No, because the computed t-statistic for a slope of 1 is significant. I calc’d the T stat fine, got the significance part, but got the yes/no part wrong because I don’t think I really “get” what covariance stationary is or what a unit root is. can someone explain in simple terms a bit more about covariance stationary and unit roots. it’s 7:15am and i’m already studying… I plan to tackle quant today in a big way, prepare for a lot of probably basic questions coming to this forum all day long from me.

don’t look at this if you want to calc the answer without knowing the answer- i’ll just post the answer now so I don’t forget what qbank test this was on later- Your answer: A was incorrect. The correct answer was B) No, because the computed t-statistic for a slope of 1 is not significant. The t-statistic for the test of the slope equal to 1 is computed by subtracting 1.0 from the coefficient, 1.3759 [= (1.0926 − 1.0) / 0.0673], which is not significant at the 5% level. The time series has a unit root and is not covariance stationary. I got the T fine, that it was insignificant, but what does it mean then that it has a unit root and isn’t cov stationary? i’m sure someone here can explain it to me in basic terms better than the book. hook a sista up.

What’s up with this question? Aren’t you supposed to be doing Dickey-Fuller tests for unit roots? You can’t just do a standard issue t-test. Suppose that you owned immortal cats. The number of cats on any day = c(t). Then on average c(t) = 1 + 1.1*c(t-1) + error(t). That means if I ask how many cats you are likely to have in two years, you will tell me a bigger number than if I ask how many cats you will have in one year. That implies E(c(t)) increases with time. Furthermore Var(c(t)) increases with time which is not as easy to see but ought to be reasonable. I can predict with much more accuracy how mny cats you are going to have in a week than in a year. The villain is that 1.1. If the coefficient < 1 then the process has a constant mean and variance. At 1, it has constant mean but growing variance. Above 1, growing mean and variance. The unit root tests check whether or not it is < 1. Greater than or equal 1 means you’ve got work to do.

how do you do a dickey fuller test? quickly looking it up, says xt - xt-1 = b0 + (b1-1)xt-1 + e. says to t test if b1 - 1 is different than zero and if not, b1 = 1 and the series has a unit root. so pretty much i want to say that’s what we did, no? they call it a t test in the solution, seems like the dickey-fuller test is what we did, they maybe in question just didn’t call it what it was? or is the dickey fuller test something completely different? as for your example, thank you. i will remember that if i become a cat lady, i have some work to do. which is so true in life. that was a good illustration though. appreciate it.

oh my god every time I read a Quant question I absolutely don’t have the feeling that reading it 2 times has helped me at all.

Seems like a very screwy question to me. My understanding is to do a Duckey Fuller test: test if the transformed coefficient (b1-1) is different from zero using a modified t-test. If (b1-1) is not statistically different from zero, as the case in this problem apparently, then you cannot reject the null and the series has a unit root (is not covariance stationary). What doesn’t make sense is that they simply tested the b1 coefficient using a normal t-test, which sure doesn’t seem like a Dickey Fuller test to me.

for the normal T test, though, you’d do the estimated 1.0926 - the actual/std error of the coefficient. we didn’t have any actual here, so maybe it wasn’t just a t test…maybe i by accident performed a dickey-fuller test and didn’t even realize it b/c i did the 1.0926 -1/.0673 to get that T or modified T or whatever we’re calling it, etc… so is dickey-fuller just pretty much the same steps as doing a T test but instead of knowing the actual value of b1 you just take b1 -1 and go through the same sorts of steps? if insignificant, conclude that there’s a unit root and it’s not covariance stationary? my eyeball hint i want to say per joey would be b/c it’s over 1 so i probably have problems?

A Dickey-Fuller test is just about testing whether that coefficient is 1 or greater. The deal is that the t-test works fine when you are testing that coefficient = 0.5 or something but breaks down near 1 so you need a different test stat. Maybe you were doing Dickey Fuller, but I would always call it a unit root if my coefficient was > 1 without doing a test at all. If you don’t, you’re pretty much saying that you think everything about your model is messed but since ypu can’t prove it, you’re not going to fix it. Tell that to your investors.