CT6 2010 Q3(i) - Time series

[omurice]

Hi,

From the Examiner's report, why does gamma0 not include a variance term? And how is the conclusion of rk = 0 unless k is divisible by 4 arrived at from the equations?

Thank you in advance!

 

[Andrew Martin]

Hello

There appears to be a typo in the Examiner's report here. Gamma_0 should be alpha * gamma_4 + sig^2.

The general pattern is that gamma_k = alpha * gamma_k-4 for k >=1. For the first few, assuming stationarity this gives:

gamma_1 = alpha * gamma_(-3) = alpha * gamma_3
gamma_2 = alpha * gamma_(-2) = alpha * gamma_2
gamma_3 = alpha * gamma_(-1) = alpha * gamma_1
gamma_4 = alpha * gamma_0
gamma_5 = alpha * gamma_1
gamma_6 = alpha * gamma_2
gamma_7 = alpha * gamma_3
etc

The first and the third equation tell us that gamma_3 = alpha^2 * gamma_3. As |alpha| < 1 (assuming stationarity) the only way this works is if gamma_3 = 0. Equation 1 tells us that gamma_1 = 0.

Equation 2 is g_2 = a * g_2. Again, if |alpha| < 1, we must have g_2 = 0.

So, we know that g_1 = g_2 = g_3 = 0. Given that gamma_k = alpha * gamma_k-4 for k >=1, this tells us that g_(1 + 4 * r) = g_(2 + 4 * r)= g_(3 + 4 * r) = 0 for r >= 0 where r is an integer. In other words, gamma_k = 0 unless k is a multiple of 4.

We know that g_4 = a * g_0, which must be non-zero assuming a positive variance (ie positive g_0). Then we know that g_8 = a * g_4 and g_12 = a * g_8 etc.

Hope this helps!

Andy

 

[omurice]

Thank you so much for the explanation Andy!