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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

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

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  • $\begingroup$ 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.) $\endgroup$ Oct 4 '16 at 2:27
  • $\begingroup$ @GrahamKemp Honestly, I was referring to the general case $\text{Var}(\sum X_i)$ in the last part. $\endgroup$
    – msm
    Oct 4 '16 at 2:32
  • $\begingroup$ 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. $\endgroup$ Oct 4 '16 at 2:43
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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).

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