Linearity of Variance

Question

Is $Var(X + Y) = Var(X) + Var(Y)$ generally, for two random variables $X, Y$? Is $Var(aX) = a^2Var(X)$ generally?

Important: In which cases, $Var(X + Y) \leq Var(X) + Var(Y)?$

$Var$ = Variance

Answer 1

Of course it is not linear. But if you just want to know when $\text{Var}(X+Y)=\text{Var}(X)+\text{Var}(Y)$ holds, you can see the fllowing $$\begin{align}\text{Var}(X+Y)&=\mathsf E(X+Y-\mathsf E(X+Y))^2\\ &=\mathsf E(X+Y-\mathsf EX-\mathsf EY)^2\\ &=\mathsf E((X-\mathsf EX)+(Y -\mathsf EY))^2\\ &=\mathsf E(((X-\mathsf EX)+(Y -\mathsf EY))((X-\mathsf EX)+(Y -\mathsf EY)))\\ &=\mathsf E((X-\mathsf EX)^2+ \mathsf (X-\mathsf EX)\mathsf (Y-\mathsf EY)+ \mathsf (Y-\mathsf EY)\mathsf (X-\mathsf EX)+ \mathsf (Y-\mathsf EY)^2)\\ &=\mathsf E(X-\mathsf EX)^2+ \mathsf E(Y-\mathsf EY)^2 +\mathsf 2E((X-\mathsf EX)\mathsf (Y-\mathsf EY))\\ &=\text{Var}(X)+\text{Var}(Y)+2\text{Cov}(X,Y) \end{align}$$

Therefore, $\text{Var}(X+Y)=\text{Var}(X)+\text{Var}(Y)$ only when $\text{Cov}(X,Y)=0$, that is when $X$ and $Y$ are uncorrelated.

Answer 2

Hint: $\mathbb{V}(X) = \mathbb{E}[(X-\mathbb{E}[X])^2]$ (1)

By invoking the linearity of $X \mapsto \mathbb{E}[X]$ (which results from linearity of the integral), one has $\mathbb{V}(X) = \mathbb{E}[X^2] - \mathbb{E}[X]^2$. Then $\mathbb{V}(aX) = a^2\mathbb{E}[X^2] - a^2\mathbb{E}[X]^2 = a^2\mathbb{V}(X)$. Now see for yourself if $\mathbb{V}(X + Y) = \mathbb{V}(X) + \mathbb{V}(Y)$ using (1) (spoiler: it is not).