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Im trying to prove that autocorrelation of order two of MA(1) process is equal to zero. We use somewhat different approach than what I have found on the internet

Given the MA(1) model: $$ Y_{t} = a + u_{t} - \theta u_{t-1} $$ where $a$ is constant, $u_{t}$ is white noise process and $\theta$ is weight on past innovation.

The correlation of order one has following equation:

$$ Cov(Y_{t},Y_{t-1}) = Cov(a + u_{t} - \theta u_{t-1}, a + u_{t-1} - \theta u_{t-2}) $$ The constant $a$ can be taken out and if we multiply the equation, we get: $$ Cov(Y_{t},Y_{t-1}) = Cov(u_{t},u_{t-1}) + Cov(u_{t},-\theta u_{t-2}) + Cov(-\theta u_{t-1},u_{t-1}) + Cov(-\theta u_{t-1},-\theta u_{t-2}) $$ Covariances with white noise process are zero, therefore we get:

$$ Cov(Y_{t},Y_{t-1}) = -\theta Var(u_{t}) = -\theta \sigma^2 $$

This implies that the autocorrelation of order one will be: $$ \rho_{1}=Corr(Y_{t}, Y_{t-1})=\frac{Cov(Y_{t},Y_{t-1})}{Var(Y_{t})} = \frac{-\theta \sigma^2}{\sigma^2(1+\theta^2)}=\frac{-\theta}{1+\theta^2} $$

If then i move to autocorrelation of order two, I can replicate the same process with $Cov(Y_{t},Y_{t-2})$ and $Corr(Y_{t},Y_{t-2})$, but I get the same results. I know it should be be equal to zero, but I dont know why.

EDIT

Covariance of order two:

$$ Cov(Y_{t},Y_{t-2}) = Cov(u_{t},u_{t-2}) + Cov(u_{t},-\theta u_{t-3}) + Cov(-\theta u_{t-1},u_{t-2}) + Cov(-\theta u_{t-1},-\theta u_{t-3}) $$ Now i think this equation will produce zero because all the $u_{t}$ are in different time. Which is not the case for $Cov(Y_{t},Y_{t-1})$, where there are two terms in the same time: $Cov(Y_{t},Y_{t-1})=Cov(-\theta u_{t-1}, u_{t-1})$

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  • $\begingroup$ Can you share your working for the autocorrelation of order two? $\endgroup$ – Comp_Warrior Jan 15 '17 at 12:40
  • $\begingroup$ I put the $Cov(Y_{t},Y_{t-2})$ term. I think i found it. $\endgroup$ – HonzaB Jan 15 '17 at 12:51
  • $\begingroup$ I think you correctly answered your own question. $\endgroup$ – Michael R. Chernick Jan 15 '17 at 15:49
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Given the covariance of order two, all the terms are uncorrelated ($u_{s}$ is a white process, hence there is no correlation), therefore the covariance, respective correlation, is zero $$ Cov(Y_{t},Y_{t-2}) = Cov(u_{t},u_{t-2}) + Cov(u_{t},-\theta u_{t-3}) + Cov(-\theta u_{t-1},u_{t-2}) + Cov(-\theta u_{t-1},-\theta u_{t-3}) = 0 $$ Which is not the case of covariance of order one, where there is one term with same time: $$ Cov(Y_{t},Y_{t-1})=Cov(-\theta u_{t-1}, u_{t-1}) = - \theta \sigma^2 $$

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