Hi, below γ represents autocovariance Q 12.20: Derive: γ0 = ∝1 γ1 + ∝2 γ2 + ∝3 γ3 + σ ^2 Solution: γ0 = cov ( Xt , Xt ) = cov( ∝1 Xt-1 + ∝2 Xt-2 + ∝3 Xt-3 + et , Xt ) = ∝1γ1 + ∝2γ2 + ∝3γ3 + cov ( et , ∝1 Xt-1 + ∝2 Xt-2 + ∝3 Xt-3 + et ) = ∝1γ1 + ∝2γ2 + ∝3γ3 + σ ^2 I was wondering why µ from Xt = µ + ∝1(Xt-1 - µ) + ... ∝p(Xt-p - µ) + et is not used ; I think because cov ( µ, Xt) = 0, so eventually it is of no use to do so. Am I correct ? What about cov ( et , ∝1 Xt-1) etc, is it = 0 ? Why ? Is it b'coz Xt is dependent on past et only so cov ( et , Xt-1 ) = 0 ??? Thanks,
The covariance between a constant and a variable is indeed 0. The defn of ets is such they are independent. By defn of Xt-1 [= µ + ∝1(Xt-2 - µ) + ... ∝p(Xt-1-p - µ) + et-1] will only have an et-1 in it. Therefore cov ( et , Xt-1 ) = cov[ et ,µ + ∝1(Xt-2 - µ) + ... ∝p(Xt-1-p - µ) + et-1] = 0
