Variance of Sums

Question

Why does $Var(cX) = c^2Var(X)$ even for independent $X$?

Can't we do

$Var(2X) = Var(X + X) = Var(X) + Var(X) = 2Var(X)$ since $X$ is independent?

Answer 1

The square comes directly from the definition of variance, the expectation of the square of deviation from the mean. The first equality in your string of equalities is true. However, if $X$ is nondegenerate, then $X$ is never independent of itself! So the second equality is false.

Answer 2

Note that in general $$ Var(X+Y)=Var(X)+Var(Y)+2Cov(X,Y). $$ Thus $$ Var(2X)=Var(X)+Var(X)+2Cov(X,X)=4Var(X) $$ since $Cov(X,X)=EX^2-(EX)^2=Var(X)$. In particular $$ Var(X+Y)=Var(X)+Var(Y) $$ iff $Cov(X,Y)=0$. Thus your equality $$ Var(2X) = Var(X + X)=Var(X)+Var(X) $$ is false in general unless $Cov(X,X)=Var(X)=0$ iff $X\stackrel{\text{w.p.1}}{=}EX$. Bottom line $X$ and $X$ may have nonzero covariance and hence are not independent.