does independence relation hold for autocovariance in a time series context I've been wondering the following for a while because it comes up on the dsp exchange site that I also belong to. Maybe it's obvious but not to me. Background: Clearly,  If $x$ and $y$ are independent random variables, then it is well known and pretty easy to show that $var(\alpha_1 x + \alpha_2 y ) = \alpha_1^2 \times  var(x) + \alpha_2^2 \times var(y) $. 
Actual Question: Does the same  relation  hold for autocovariances of the same variables in a time series context ? By this I mean that, if the same two variables are still independent, then does $cov(\alpha_1 x + \alpha_2 y )$ at lag $i$ = $ \alpha_1^2 \times $ ( autocov($x$) at lag i ) $ + \alpha_2^2  \times $ ( autocov($y$) at lag i)  ? Thanks. Any references or useful notes are also appreciated.
 A: Making the time dependence more explicit, I'll define:
$$Z_t := \alpha_1X_t + \alpha_2Y_t$$
We're looking for the autocovariance function of $Z_t$:
$$\text{Cov}(Z_t,Z_{t-i})=\text{Cov}(\alpha_1X_t + \alpha_2Y_t, \alpha_1X_{t-i} + \alpha_2Y_{t-i})$$
By the bilinearity property of the covariance, we can start by expanding the first term, and then the second:
\begin{align}
\text{Cov}(\alpha_1X_t + \alpha_2Y_t, \alpha_1X_{t-i} + \alpha_2Y_{t-i}) &= \alpha_1\text{Cov}(X_t, \alpha_1X_{t-i} + \alpha_2Y_{t-i}) \\
&+ \alpha_2\text{Cov}(Y_t, \alpha_1X_{t-i} + \alpha_2Y_{t-i}) \\
\\
&= \alpha_1^2 \text{Cov}(X_t,X_{t-i})+\alpha_1\alpha_2\text{Cov}(X_t,Y_{t-i})\\
&+ \alpha_2\alpha_1\text{Cov}(Y_t, X_{t-i}) + \alpha_2^2\text{Cov}(Y_t,Y_{t-i})
\end{align}
If the two series are independent at all lags, then the two cross terms $\text{Cov}(X_t,Y_{t-i})$ and $\text{Cov}(Y_t,X_{t-i})$ are both zero for any $i$ and $t$, so the result is, as you have claimed, that:
$$\text{Cov}(Z_t, Z_{t-i}) =\alpha_1^2 \text{Cov}(X_t,X_{t-i})+\alpha_2^2\text{Cov}(Y_t,Y_{t-i})$$
