September 2022 Paper A Q2

[Ray K]

In Q2iii it says in the ASET paper:

Differencing removed unit roots. So we need to determine when the process has a unit root and ensure that after differencing the remaining roots are larger than one in magnitude.

The process is I(1) if the characteristic polynomial of the process terms has one unit root and the other roots are larger than one in magnitude.

The characteristic polynomial of the process terms is:
1 - alpha * lambda - 0.5 lambda^2

My questions:
1) The process is I(1) if the characteristic polynomial of the process terms has one unit root
So my understanding of I(k) is that you must difference that number of times to become stationary. So why will it have one unit root? If it has a unit root, surely |lambda| <=1 as one solution, so it will no longer be stationary? I understand one unit root is removed, but one is still failing the test it seems.

2) Why is the characteristic polynomial of the process terms unchanged from part (i)?

 

[Andrew Martin]

Hello

1) I(k) (for k not equal to 0) processes are special types of non-stationary processes where we can obtain stationarity by differencing. In particular, as you say, we difference k times to get stationarity.

It turns out that this relates to there being k unit roots in the characteristic polynomial of the process terms for the original series. If a process has a unit root then it can't be stationary because we need each root to be (strictly) larger than one in magnitude. However, it turns out that each time we difference we get rid of a unit root. In order for the process to then be stationary after differencing the appropriate number of times, the remaining roots must be larger than one in magnitude. So, to begin with, for the original series, we must have some roots larger than one in magnitude and some unit roots, the latter being those that are removed when differencing.

If the original series has any roots less than one in magnitude (for the characteristic polynomial of the process terms) then it is not an I(k) process.

2) We are still considering the original process. We can see that the roots are 1 and -2. We can get rid of the 1 by differencing. The characteristic polynomial of the differenced series will have a single root of -2, which is larger than one in magnitude. So, the differenced series is stationary.

To see this, let's do the differencing. If we have:

Xt = 0.5Xt-1 + 0.5 Xt-2 + et + bet-1

Let Yt = Xt - Xt-1

Yt = 0.5Xt-1 + 0.5 Xt-2 + et + bet-1 - Xt-1
= -0.5Xt-1 + 0.5 Xt-2 + et + bet-1
= -0.5 * Yt-1 + et + bet-1

So, the characteristic polynomial of the series Yt is 1 + 0.5 * lambda. This has the single root of -2 as per the above.

We can also see this directly by considering Xt in terms of B:

(1 - 0.5B - 0.5B^2) Xt = et + bet-1

The sum of the coefficients on the LHS is 0 (1 - 0.5 - 0.5 = 0), which means we have a unit root. The characteristic polynomial is:

phi(lambda) = 1 - 0.5lambda - 0.5lambda^2
phi(1) = 1 - 0.5 - 0.5 = 0

Factorising out the root of 1, we get:

(1 - B)(1 + 0.5B) Xt = et + bet-1

Letting Yt = Xt - Xt-1 = (1-B)Xt, we get:

(1 - B)(1 + 0.5B) Xt = et + bet-1
=> (1 + 0.5B) (1 - B) Xt = et + bet-1
=> (1 + 0.5B) Yt = et + bet_1

or Yt = -0.5 * Yt-1 + et + bet-1

as per the above.

Hopefully this makes sense in terms of why differencing removes the unit root (as we take out a factor of (1-B), which when applied to the original series is differencing it).

Hope this helps!

Andy