Variance of a random variable X+Y when X=Y

Question

I know that $Var(aX)=a^2Var(X)$ and that $Var(X+Y)=Var(X)+Var(Y)$.

When $X=Y$ why is the following not true?

$Var(X+Y)=Var(X+X)=Var(2X)$

$Var(X+X)=Var(X)+Var(X)$ and $Var(2X)=2^2Var(X)=4Var(X)$

$\therefore Var(X)+Var(X)=4Var(X)$

I understand that this could be that $X+X$ is not equal to $2X$ when it comes to variances but how do you know which to use?

Thanks

Answer 1

\begin{equation} var(X+Y) = E[((X+Y)-E(X+Y))((X+Y)-E(X+Y))'] \end{equation} Assuming $X,Y$ to be scalar random variables, we get, \begin{equation} \begin{split} var(X+Y) &= E[((X-E(X)) + (Y-E(Y)))^2]\\ &= E[(X-E(X))^2 + (Y-E(Y))^2 + 2\times (X-E(X)) (Y-E(Y))]\\ &=E[(X-E(X))^2] + E[(Y-E(Y))^2] + E[2\times (X-E(X)) (Y-E(Y))]\\ &=var(X)+var(Y)+2cov(X,Y) \end{split} \end{equation} so only when $cov(X,Y) = 0$, $var(X+Y) = var(X)+var(Y)$

Answer 2

The relation $Var(X+Y) = Var(X) + Var(Y)$ only holds for uncorrelated variables. Obviously, $X$ is correlated to $X$!

Answer 3

Just to complement the other answers, the general form of the formula you are trying to use is:

$$ \mathrm{Var}(aX+bY)=a^2\mathrm{Var}(X)+b^2\mathrm{Var}(Y)+2ab\, \mathrm{Cov}(X,Y), $$

which follows from the definitions of variance and covariance.