# Why is the variance of a moving average process calculated this way?

I have a moving average process that looks like:

$$Y_{t}=\frac{e_{t}+e_{t-1}}{2}$$

And I can see that the variance has been calculated as follows:

$${Var}\left ( Y_{t} \right )={Var}\left \{ \frac{e_{t}+e_{t-1}}{2} \right \}=\frac{{Var}\left ( e_{t} \right )+{Var}\left ( e_{t-1} \right )}{4}$$

Which I cannot seem to understand. The rules of variance say that the variance of the sum of two random variables... Well it doesn't look anything like that, it has a Covariance term in it.

If someone could help explain how this step is possible, I would be very grateful.

EDIT: After some further research I think that this has to do with the fact that the two variables are independent?

Suppose $X$ and $Y$ are random variables and $a$ and $b$ are constants. Then:
$\newcommand{\Var}{\rm Var} \newcommand{\Cov}{\rm Cov} \Var(aX+bY) = a^2\Var(X)+b^2\Var(Y) + 2ab \Cov(X,Y)$.
In your case, $a = b = \frac{1}{2}$, $X = e_{t}$, and $Y=e_{t-1}$. Since $e_{t}$, and $e_{t-1}$ are independent, the covariance is just 0. Hence:
$\Var\left(\frac{e_{t}+e_{t-1}}{2}\right) = \Var\left(\frac{1}{2}e_{t} + \frac{1}{2}e_{t-1} \right)=(\frac{1}{2})^2\Var(e_{t})+(\frac{1}{2})^2\Var(e_{t-1}) = \frac{\Var(e_{t})+\Var(e_{t-1})}{4}$.