Linearity of Variance 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
 A: 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.
A: 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).
