# Page 31 of Chapter 13 [Time Series 1] - question on roots of an equation

Discussion in 'CS2' started by Stephan Iliffe, Mar 10, 2019.

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1. ### Stephan IliffeMember

Hi all

On page 31 of chapter 13 of CS2, there is the following in the first solution:

“The polynomial can be factorized as (1+3z)(1-z), so it’s roots are 1/3 and 1.”

I understand that we are interested in the absolute value of the roots so in this case there is one less than 1 and the process is not stationary.

Surely it is still incorrect to say the root is 1/3 and not -1/3.

I do not see any corrections for this and was just wondering if my logic is wrong.

Regards
Stephan Iliffe

2. ### Stephan IliffeMember

Also, in the next question, the solution says it is easy to see that [with roots 1, 2 and 3] the differencing process will eliminate the root 1. So the process will be stationary.

Is it really obvious that the root 1 will be eliminated or do you actually have to do the differencing to see this?

3. ### Sid KumarActive Member

For the purpose of this exam focus on absolute value of roots, they should be greater than one for stationarity test.

With the next question, since only one root is failing the test, you can assume that the process needs to be differenced only once. You can try and work out the roots for the differenced characteristic poly, I suspect they will have roots >1

4. ### Stephan IliffeMember

Thanks Sid

Does this mean that the root failing the test is always the root that will be eliminated when differencing?

Why is this?

I can see that it happens in this example but why would we assume in advance this to be true?

Regards

5. ### Sid KumarActive Member

I think that is the basic idea behind differencing. I havent come across a question that suggests otherwise. Generally it will ask you to check for stationarity and a simple way to eliminate 1 as a root is to put it in the characteristic equation to see if it equals zero (I have seen this to be the case in a few questions).

6. ### Stephan IliffeMember

Thanks Sid

I found this on Wikipedia:
If the other roots of the characteristic equation lie inside the unit circle—that is, have a modulus (absolute value) less than one—then the first difference of the process will be stationary; otherwise, the process will need to be differenced multiple times to become stationary.[1] Due to this characteristic, unit root processes are also called difference stationary.[2][3]

https://en.m.wikipedia.org/wiki/Unit_root

So I suppose it is in the specific case.

7. ### Stephan IliffeMember

I suppose it makes sense since a root of one means there is a lag of 1.

The roots less than one do not contradict the stationarity of the process.

So differencing once will remove the lag of one and the process will be stationary.

8. ### Sid KumarActive Member

Right, good one. Cheers