Can the covariance of a linear combination be written as:
$\sum_{i=1}^n\sum_{j=1}^n a_ia_jCov[X_i,Y_j]$
My problem is I'm trying to figure out an exact property for finding:
$Cov[\frac{1}{2}\epsilon_t + \frac{1}{2}\epsilon_{t-1}, \frac{1}{2}\epsilon_{t-1} + \frac{1}{2}\epsilon_{t-2}]$ where $\epsilon_t$ is i.i.d. $(0, \sigma^2)$ and independent white noise.
So I am thinking I can just do the following:
$(0.5 \cdot0.5)Cov[\epsilon_t, \epsilon_{t-1}] + (0.5 \cdot0.5)Cov[\epsilon_t, \epsilon_{t-2}] + (0.5 \cdot0.5)Cov[\epsilon_{t-1}, \epsilon_{t-1}] + (0.5 \cdot0.5)Cov[\epsilon_{t-1}, \epsilon_{t-2}]$
Which then comes out to:
$(0.5 \cdot 0.5)Cov[\epsilon_{t-1}, \epsilon_{t-1}] = \frac{1}{4}\sigma^2$ because the covariance between lags should just be $0$ so three of the four terms fall out.
Is all of this correct or am I missing something?
Yes, that's correct. Covariance has the property you mentioned.$Cov(aX+bY,cZ+dW)=abCov(X,Y)+acCov(X,Z)+bcCov(Y,Z)+bdCov(Y,W)$.
For your case, $$\begin{split}.25cov(e_t, e_{t-1})+.25cov(e_t, e_{t-2})+.25cov(e_{t-1}, e_{t-1})+ .25cov(e_{t-1}, e_{t-2})&=0+0+.25var(e_{t-1})+0\\ &=.25\sigma^2\end{split}$$