Covariance of sums of time series random variables [duplicate]

Question

I find the following in my Cryer and Chan Time Series Analysis with application in R textbook:

Let $c_{1}, c_{2},\ldots, c_{m}$ and $d_{1}, d_{2},\ldots, d_{n}$ be constants and $t_{1}, t_{2},\ldots, t_{m}$ and $s_{1}, s_{2},\ldots, s_{n}$ be time points. Further let $Y$ be a random variable indexed with the previously mentioned time points.

Then:

$$ Cov\:\left(\sum_{i = 1}^{m}c_{i}Y_{t_{i}} \:,\:\sum_{j = 1}^{n}d_{j}Y_{s_{j}}\right) = \sum_{i = 1}^{m} \sum_{j = 1}^{n}c_{i} d_{j}\:Cov(Y_{t_{i}} \: Y_{s_{j}}) $$

How is this be proven? The book says that it follows from the linear properties of expectation, but I'm not sure how this was used here.

Further, the book states:

$$ Var\:\left(\sum_{i=1}^{m}c_{i}Y_{t_{i}}\right) = \sum_{i = 1}^{m}c_{i}^{2}\:Var(Y_{t_{i}}) + 2 \sum_{i = 2}^{m} \sum_{j = 1}^{i - 1}c_{i}c_{j}\:Cov(Y_{t_{i}} , Y_{t_{j}}) $$

Which is a special case of the first result. Why is this a special case of the first result?

Answer 1

By definition, $Cov(Y_{t_i},Y_{s_j}) = E(Y_{t_i}Y_{s_j}) - E(Y_{t_i})E(Y_{s_j})$. Applying this definition to the first equation gives \begin{eqnarray*} Cov \left(\sum_{i=1}^m c_i Y_{t_i},\sum_{j=1}^n d_j Y_{s_j}\right) &=& E \left(\sum_{i=1}^m c_i Y_{t_i} \sum_{j=1}^n d_j Y_{s_j}\right) - E\left(\sum_{i=1}^m c_i Y_{t_i}\right) E\left( \sum_{j=1}^n d_j Y_{s_j}\right) \\ &=& \sum_{i=1}^m \sum_{j=1}^n c_i d_j E\left(Y_{t_i}Y_{s_j}\right) - \sum_{i=1}^m \sum_{j=1}^n c_i d_j E\left(Y_{t_i}\right)E\left(Y_{s_j}\right) \\ &=& \sum_{i=1}^m\sum_{j=1}^n c_i d_j Cov(Y_{t_i},Y_{s_j}). \end{eqnarray*}

The second equation you give is a special case of the above since \begin{eqnarray*} Var \left(\sum_{i=1}^m c_i Y_{t_i}\right) &=& Cov\left(\sum_{i=1}^m c_i Y_{t_i},\sum_{i=1}^m c_i Y_{t_i}\right). \end{eqnarray*} Use the previous formula to simplify the above and separate the double summation into two cases $(i=j)$ and $(i \ne j)$.