# 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

• $Var(aX)=a^2Var(X)$ is clearly an indication of non-linearity.
– user65203
Oct 3 '16 at 18:44

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)^2+\mathsf E(Y-\mathsf EY)^2+2\mathsf E(X-\mathsf EX)\mathsf E(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.

• Uhm, yes, you can. If $\mathsf {Var}(X+Y)=\mathsf {Var}(X)+\mathsf {Var}(Y)$, then $\mathsf {Cov}(X,Y)=0$ and the variables are uncorrelated (since that is what it means). $~$ (What you cannot conclude is: independence.) Oct 4 '16 at 2:27
• @GrahamKemp Honestly, I was referring to the general case $\text{Var}(\sum X_i)$ in the last part.
– msm
Oct 4 '16 at 2:32
• Oh, yes, when you have more than two variables, then the sums of covariances being zero cannot be used to infer the individual covariances are. Oct 4 '16 at 2:43

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).