How do you prove that co-variance is a linear operator?

Question

I mean how do you prove linearity property of co-variances with a proper domain and range?

Answer 1

In fact, it is a bilinear operator, meaning that its domain is the product of two vector spaces, and it is linear in each entry:

Let $L^2(\Omega;\mathbb{R})$ (just $L^2$ for short) be the vector space of real-valued random variables s.t. $E[\vert X \vert^2] < \infty$. Then we have $\text{Cov}: L^2 \times L^2 \rightarrow \mathbb{R}$ via $\text{Cov}(X,Y) = E[(X - E[X])(Y - E[Y]))$ is bilinear.

I will prove the linearity for the first entry - symmetry of the operator then implies linearity in the second.

Let $X_1, X_2 \in L^2$ and $\lambda \in \mathbb{R}$, then $$ \begin{align} \text{Cov}(\lambda X_1 + X_2,Y) &= E[(\lambda X_1 + X_2 - E[\lambda X_1 + X_2])(Y - E[Y])) \\ & = E[\lambda (X_1 - E[X_1])(Y - E[Y])) + (X_2 - E[X_2])(Y - E[Y])) = E[\lambda (X_1 - E[X_1])(Y - E[Y]))] + E[(X_2 - E[X_2])(Y - E[Y])) \\ &=\lambda \text{Cov}(X_1,Y) + \text{Cov}(X_2,Y) \end{align} $$

In essence, it comes down to the fact that the expected value is linear.