Expressing variance of sum

Question

Let X and Y be two independent random variables with finite variances. If
, , are constants is there a way to simplify ar( + Y + ) in terms of ar() and ar(Y)

Answer 1

For a scalar random varible $z$ that has $\mathbb{E}(z) = \bar{z}$, where $\mathbb{E}$ means the expectation (mean), by definition of the variance,

$Var(z) = \mathbb{E}[(z - \bar{z})^2] = \mathbb{E}[z^2 - 2z\bar{z} + \bar{z}^2] = \mathbb{E}[z^2] - \bar{z}^2$

since the expectation is a linear operator.

Now, plug in $z = aX + bY + c$.

$$ \begin{aligned} Var(aX+bY+c) &= \mathbb{E}[(aX+bY+c)^2] - (a\overline{X} + b\overline{Y} + c)^2 \\ &= \mathbb{E}[a^2X^2 + b^2Y^2 + c^2 + 2abXY + 2aXc + 2bYc] \\ &\qquad - a^2\overline{X}^2-b^2\overline{Y}^2-c^2-2ab\overline{X}\overline{Y}-2a\overline{X}c-2b\overline{Y}c \\ &= a^2(\mathbb{E}[X^2]-\overline{X}^2) + b^2(\mathbb{E}[Y^2] - \overline{Y}^2)\\ &= a^2Var(X) + b^2Var(Y) \end{aligned} $$

Note that the second equality to the third equality is using the independence of $X$ and $Y$ to cancel terms, namely, $\mathbb{E}[XY] = \mathbb{E}[X]\cdot\mathbb{E}[Y]$.