# Moving average process notation: negative or positive $\theta$s?

I noticed that sometimes the moving average process is written in two different ways. I would like to know which one is correct.

$y_t = \epsilon_t + \theta \epsilon_{t-1}$

$y_t = \epsilon_t - \theta \epsilon_{t-1}$

• This is a matter of notational conventions. You can see immediately that the model is the same for $\tilde\theta=-\theta$. – Richard Hardy Jan 8 '16 at 16:35

Both are correct, where the first is, I would say, more common. They are equivalent because you can always transform $\theta$ to $-\theta$ and get the same process, with the same dynamics.
The variance is \begin{eqnarray} E(Y_t-\mu)^2&=&E(\epsilon_t+\theta\epsilon_{t-1})^2\notag\\ &=&E(\epsilon_t^2+2\epsilon_t\theta\epsilon_{t-1}+\theta^2\epsilon_{t-1}^2)\notag\\ &=&\sigma^2+0+\theta^2\sigma^2 \end{eqnarray} and the first autocovariance is \begin{eqnarray*} E(Y_t-\mu)(Y_{t-1}-\mu)&=&E(\epsilon_t+\theta\epsilon_{t-1})(\epsilon_{t-1}+\theta\epsilon_{t-2})\\ &=&E(\epsilon_t\epsilon_{t-1}+\theta\epsilon_{t-1}^2+\theta\epsilon_t\epsilon_{t-2}+\theta^2\epsilon_{t-1}\epsilon_{t-2})\\ &=&0+\theta\sigma^2+0+0, \end{eqnarray*} whether you define $\epsilon_t+\theta\epsilon_{t-1}$ or $\epsilon_t-(-\theta)\epsilon_{t-1}$.