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