# MA(1) model - prove that correlation of order two equals to zero

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: $Cov(Y_{t},Y_{t-1})=Cov(-\theta u_{t-1}, u_{t-1})$

• Can you share your working for the autocorrelation of order two? Jan 15, 2017 at 12:40
• I put the $Cov(Y_{t},Y_{t-2})$ term. I think i found it. Jan 15, 2017 at 12:51
• I think you correctly answered your own question. Jan 15, 2017 at 15:49

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