Theoretical variance of a sum of random variables

Question

$W_1$ and $W_2$ would be $0$ since by using the fact that $E(X_1) - E(X_2)$ this gives us $0$?

Answer 1

The expectation operator is linear, hence $\mathbb{E}\sum X_i = \sum \mathbb{E}X_i$. For the variance use the covariance between each pair, i.e., $$ \operatorname{var}(\sum X_i) = \sum \operatorname{var}(X_i) + 2 \sum_{i>j}\operatorname{cov}(X_i, X_j). $$

In matrix:

Let $X = (X_1, X_2, X_3) ^ T $ and $ c_1 = (1, -1, 0) ^ T $ and $c_2 = (1, 1, -2) ^ T$. Then, $W_1 = c_1 ^ T X$ and $W_2 = c_2 ^ T X$, as such $$ \mathbb{E}[W_1] = c_1 ^ T \mathbb{E} X = c_1 ^ T \underline{\mu} = \mu - \mu=0. $$ where $\underline{\mu} = (\mu_1, \mu_2, \mu_3) ^ T $. And for the variance $$ \mathbb{V}(W_1) = c_1 ^ T\Sigma_{X}c_1 \, , $$ where $\operatorname{var}(X) = \Sigma_X \in \mathcal{M}_{2 \times 2}(\mathbb{R})$. Same is true for $W_2$.