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!
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
