In CS2, chapter 13, at the bottom of page 31 there is an example that requires you to calculate the roots of the characteristic equation of the process $$ X_n = \frac{11}{6} X_{n-1} - X_{n-2} + \frac{1}{6} X_{n-3} + e_t $$ The characteristic equation for this process is $$ 1 - \frac{11}{6} z + z^2 - \frac{1}{6}z^3 $$ and these roots are $z=1$, $z=2$, and $z=3$ (this part is straightforward enough). However, the solution goes on to say that if we take the first difference of the process then the root 1 "clearly" disappears. I don't see why this would be the case. Can anyone please explain?
Hello I agree that the fact it disappears it not necessarily so obvious! Essentially it comes down to the fact that differencing the series is the same as applying the operation 1 - B, where B is the backwards shift operator. If we factorise the characteristic polynomial we have: \( 1 - 11/6 z + z^2 - 1/6 z ^3 = 1/6(1 - z)(2 - z)(3 - z) \) Writing the series in terms of the backwards shift operator and this factorisation: \( 1/6(1 - B)(2 - B)(3 - B) X_t = e_t \) Now, what we can do is to let \( Y_t = X_t - X_{t-1} = (1-B)X_t \) so we have: \( 1/6(2 - B)(3 - B)(1 - B) X_t = e_t \) \( 1/6(2 - B)(3 - B) Y_t = e_t \) So the first differenced series, \( Y_t \), has roots of 2 and 3 for its LHS characteristic polynomial. It no longer has a root of 1. Note that it is possible for the starting series to have multiple unit roots, in which case differencing once will remove one of the unit roots but the rest will still remain. Check We can check this as follows: \( Y_t = X_t - X_{t-1} = 11/6 X_{t-1} - X_{t-2} + 1/6 X_{t-3} + e_t - X_{t-1} \) \( Y_t = (11/6 - 1)( X_{t-1} - X_{t-2}) + (-1 + 11/6 - 1) * (X_{t-2} - X_{t-3})+ (1/6 -1 + 11/6 - 1) X_{t-3} + e_t \) Now \( (1/6 -1 + 11/6 - 1) = 0 \). This last \( X_t \) term (ie \( X_{t-p} \) ) will always be 0 for a series that has a unit root (see a general explanation below). Writing everything in terms of \( Y_t \): \( Y_t = 5/6 * Y_{t-1} -1/6 * Y_{t-2} + e_t \) Taking the \( Y_t \) terms to the LHS: \( Y_t - 5/6 * Y_{t-1} + 1/6 * Y_{t-2} = e_t \) So the LHS characteristic polynomial is: \( 1 - 5/6z + 1/6z^2 = 1/6(2 - z)(3 - z) \) ie the roots are 2 and 3 and we can write Y_t as we did above: \( 1/6(2 - B)(3 - B) Y_t = e_t \) General result (messy but if you are interested) In general, if we have the series: \( X_t = \alpha_1 * X_{t-1} + ... + \alpha_p * X_{t-p} + \beta_1 * e_{t-1} + ... + \beta_q * e_{t-q} + e_t \) Then we have a unit root for the characteristic polynomial of the process terms if \( 1- \alpha_1 - ... - \alpha_p = 0 \) . To see this, let's move the X terms to the LHS: \( X_t - \alpha_1 * X_{t-1} - ... - \alpha_p * X_{t-p} = \beta_1 * e_{t-1} + ... + \beta_q * e_{t-q} + e_t \) So, in terms of the backwards shift operator: \( (1 - \alpha_1 B - ... - \alpha_p B^p) X_t = \beta_1 * e_{t-1} + ... + \beta_q * e_{t-q} + e_t \). The LHS characteristic polynomial is then: \( 1 - \alpha_1 * \lambda - ... - \alpha_p * \lambda^p\) If there is a unit root, then this means that when \( \lambda = 1 \) this polynomial takes the value 0, ie: \( 1- \alpha_1 - ... - \alpha_p = 0 \) So, we can factorise this as follows: \( (1 - B)(1 + (1 - \alpha_1)B + (1 - \alpha_1 - \alpha_2)B^2 + ... + (1 - \alpha_1 - \alpha_2 - ... - \alpha_{p-1})B^{(p-1)}) X_t = \beta_1 * e_{t-1} + ... + \beta_q * e_{t-q} + e_t \) Where the final balancing term in the second bracket on the LHS, the \( (1 - \alpha_1 - \alpha_2 - ... - \alpha_{p-1} - \alpha_p)B^p \), disappears as \( 1- \alpha_1 - ... - \alpha_p = 0 \). Now let's consider the first difference for a series with a unit root, ie \( X_t - X_{t-1} \): \( X_t - X_{t-1} = \alpha_1 * X_{t-1} + ... + \alpha_p * X_{t-p} - X_{t-1} + \beta_1 * e_{t-1} + ... + \beta_q * e_{t-q} + e_t \) \( X_t - X_{t-1} = (\alpha_1 - 1) * (X_{t-1} - X_{t-2}) + (\alpha_2 + \alpha_1 - 1) * (X_{t-2} - X_{t-3}) + ... + \) \( (\alpha_{p-1} + \alpha_{p-2} + ... + \alpha_1 - 1) * (X_{t-(p-1)} - X_{t-p}) + (\alpha_{p} + \alpha_{p-1} + ... + \alpha_1 - 1) X_{t-p} + \beta_1 * e_{t-1} + ... + \beta_q * e_{t-q} + e_t \) Note that \( 1- \alpha_1 - ... - \alpha_p = 0 \) and so \( \alpha_p + ... + \alpha_1 - 1 = 0 \) and let \( Y_t = X_t - X_{t-1} \), then we get: \( Y_t = (\alpha_1 - 1) * Y_{t-1} + (\alpha_2 + \alpha_1 - 1) * Y_{t-2} + ... + (\alpha_{p-1} + \alpha_{p-2} + ... + \alpha_1 - 1) * Y_{t-(p-1)} + \beta_1 * e_{t-1} + ... + \beta_q * e_{t-q} + e_t \) Moving the process terms to the LHS: \( Y_t + (1 - \alpha_1) * Y_{t-1} + (1 - \alpha_2 - \alpha_1) * Y_{t-2} + ... + (1 - \alpha_{p-1} - \alpha_{p-2} - ... - \alpha_1) * Y_{t-(p-1)} = \beta_1 * e_{t-1} + ... + \beta_q * e_{t-q} + e_t \) This ties in with the expression for \( X_t \) we had above when we write in terms of B and we take \( (1-B)X_t \) as \( Y_t \). Hope this helps! Andy
