Let's say $X$, $Y$ and $Z$ are three independent and identical random variables.
Let $V = 2X-Y$ and $W= 2X-2Z+5$.
I have to find covariance of $V$ and $W$.
I can find $E(V)$ and $E(W)$, but I am confused on how to find the expectation of $V*W$, which will be $E[(2X-Y)(2X-2Z+5)]$.
Can we apply basic algebraic rules here and multiply all variables by opening the brackets or is there any other way?
Rather than calculating $\operatorname{E}[VW]$, you can directly compute the covariance using the identity $$\operatorname{Cov}[aX + bY, cV + dW] = ac \operatorname{Cov}[X,V] + ad \operatorname{Cov}[X,W] + bc \operatorname{Cov}[Y,V] + bd \operatorname{Cov}[Y,W],$$ where $a, b, c, d$ are scalar constants and $X, Y, V, W$ are random variables. In other words, the covariance operator behaves like multiplication. The other property of covariance is that it is unchanged under location translations of random variables; i.e., $$\operatorname{Cov}[X + a, Y + b] = \operatorname{Cov}[X,Y],$$ for scalar constants $a, b$. This should be familiar if you already know that $$\operatorname{Var}[X + a] = \operatorname{Var}[X].$$
Then with these properties in mind, we have $$\begin{align} \operatorname{Cov}[2X - Y, 2X - 2Z + 5] &= \operatorname{Cov}[2X - Y, 2X - 2Z] \\ &= 4 \operatorname{Cov}[X, X] - 4 \operatorname{Cov}[X, Z] - 2 \operatorname{Cov}[Y, X] + 2 \operatorname{Cov}[Y, Z]. \end{align}$$ Finally, use the independence of $X, Y, Z$ to simplify, and recall $\operatorname{Cov}[X,X] = \operatorname{Var}[X]$.