Is the following calculation of $\mathrm{Var}\left(\sum\limits_{i=0}^{k}(c_{x1}+c_{x2}X_i)(c_{y1}+c_{y2}Y_i)\right)$ correct?

Question

For $\mathrm{Var}(X_i)=1\land \mathrm{Var}(Y_i)=1$ where $X_i$ and $Y_i$ are i.i.d. $$\mathrm{Var}\left(\sum_{i=0}^{k}(c_{x1}+c_{x2}X_i)(c_{y1}+c_{y2}Y_i)\right)=k\left(c_{x1}^2c_{y2}^2+c_{x2}^2c_{y1}^2+c_{x2}^2c_{y2}^2\right) $$ where $c_{x1},c_{x2},c_{y1},c_{y2}$ are four constants.

Answer 1

Yes, your calculation is correct: \begin{align}\sum_{i=0}^k\left(c_{x1}+c_{x2}X_i\right)\left(c_{y1}+c_{y2}Y_i\right)&=kc_{x1}c_{y1}+c_{x2}c_{y1}\sum_{i=0}^kX_i+c_{x1}c_{y2}\sum_{i=0}^kY_i+c_{x2}c_{y2}\sum_{i=0}^kX_iY_i\end{align} Taking the variance $(\mathrm{Var})$ on both sides and assuming that $X_i,Y_i$ are independent we get \begin{align}\hspace{70pt}\dots&=\mathrm{Var}\left(kc_{x1}c_{y1}+c_{x2}c_{y1}\sum_{i=0}^kX_i+c_{x1}c_{y2}\sum_{i=0}^kY_i+c_{x2}c_{y2}\sum_{i=0}^kX_iY_i\right)\\[0.2cm]&=0+(c_{x_2}c_{y1})^2k\cdot(1)+(c_{x_1}c_{y2})^2k\cdot(1)+(c_{x_2}c_{y2})^2k\cdot(1)^2\\[0.4cm]&=k\left(c_{x2}^2c_{y1}^2+c_{x1}^2c_{y2}^2+c_{x2}^2c_{y2}^2\right)\end{align}