Intuitive understanding of variance of sum vs variance of difference $\newcommand{\Var}{\operatorname{Var}}\newcommand{Cov}{\operatorname{Cov}}$Mathematically, 
$\Var(X + Y) = \Var(X) + \Var(Y) + 2\Cov(X,Y)$ and
$\Var(X - Y) = \Var(X) + \Var(Y) - 2\Cov(X,Y)$
This implies that the variance of the sum of two positively correlated random variables is always larger than their difference. However, if the variables are uncorrelated, then the variance of their sum equals the variance of their difference. 
I am having trouble understanding intuitively how 


*

*The variances are equal if the variables are uncorrelated

*The variance of the sum is larger than the variance of the difference if they are correlated   

 A: Let $Z = -Y$ so that $X-Y = X+Z$. Then, $$\operatorname{var}(Z)= (-1)^2\operatorname{var}(Y) = \operatorname{var}(Y)$$ and 
$$\operatorname{cov}(X,Z) = \operatorname{cov}(X,-Y) = -\operatorname{cov}(X,Y)$$
so that 
\begin{align}
\operatorname{var}(X-Y) &= \operatorname{var}(X+Z)\\
&= \operatorname{var}(X)+\operatorname{var}(Z) + 2\operatorname{cov}(X,Z)\\
&= \operatorname{var}(X)+\operatorname{var}(Y) - 2\operatorname{cov}(X,Y)
\end{align}
In short, 
\begin{align}
\operatorname{var}(X+Y) &= \operatorname{var}(X) + \operatorname{var}(Y) + 2\operatorname{cov}(X,Y)\\
\operatorname{var}(X-Y) &= \operatorname{var}(X) + \operatorname{var}(Y) - 2\operatorname{cov}(X,Y)
\end{align}
really are the same formula, and which of the two variances is larger depends entirely on whether $\operatorname{cov}(X,Y)$ is positive or negative. In the case when $\operatorname{cov}(X,Y)$ equals $0$ (i.e. $X$ and $Y$ are uncorrelated random variables), $\operatorname{var}(X+Y)$ equals $\operatorname{var}(X) + \operatorname{var}(Y)$ as you have already noted.
As to intuition regarding why $\operatorname{var}(X+Y) > \operatorname{var}(X-Y)$ when $\operatorname{cov}(X,Y) > 0$, a geometric viewpoint might help.  $X$ and $Y$ can be regarded as vectors of lengths $\sigma_X = \sqrt{\operatorname{var}(X)}$ and $\sigma_X = \sqrt{\operatorname{var}(Y)}$ that are pointed roughly in the same direction when $\operatorname{cov}(X,Y) > 0$ and in roughly opposite direction when $\operatorname{cov}(X,Y) < 0$. Now, one would expect intuitively that the
(vector) sum of two vectors pointing in roughly the same direction is a longer vector than the (vector) difference of the two vectors, no?  A crudely drawn picture of the difference vector and the sum vector is in the diagram below.

Note that in the left-hand figure above, the vectors $X$ and $Y$ are shown as being at an acute angle $\theta$ and thus are "pointed in roughly the same direction".  In fact, the correlation coefficient $\rho$ of (random variables) $X$ and $Y$ is $\cos(\theta) > 0$. Now, in a triangle with vertices $A, B, C$, and opposite sides of lengths $a, b$ and $c$ respectively, we have the following result familiar from elementary geometry/trigonometry 
$$c^2 = a^2 + b^2 - 2ab\cos(\angle C)$$ which is equivalent to 
\begin{align}
\sigma_{X-Y}^2 &= \sigma_X^2 + \sigma_Y^2 - 2\sigma_X\sigma_Y\cos(\theta)\\
 \operatorname{var}(X-Y) &= \operatorname{var}(X) + \operatorname{var}(Y)
-2 \rho \sigma_X\sigma_Y\\
\operatorname{var}(X-Y) &= \operatorname{var}(X) + \operatorname{var}(Y)
-2 \operatorname{cov}(X,Y)
\end{align}
On the other hand, as shown in the right-hand figure above, for vector sums,
the included angle $\pi-\theta$ is obtuse, and so
$\cos(\pi-\theta) = -\cos(\theta) < 0$.  So, we get that
\begin{align}
\sigma_{X+Y}^2 &= \sigma_X^2 + \sigma_Y^2 + 2\sigma_X\sigma_Y\cos(\theta)\\
\operatorname{var}(X+Y) &= \operatorname{var}(X) + \operatorname{var}(Y)
+2 \operatorname{cov}(X,Y),
\end{align}
that is, $\operatorname{var}(X+Y) > \operatorname{var}(X-Y)$ for positively correlated random variables.
A: *

*Consider $X$ and $Y$ as uniform random variables between $0$-$1$. Sum of them varies in $[0,2]$ while difference of them varies in $[-1,1]$. Both have the same range length. The length of their range is not equal to variance of course, but it is an indicator of variability in some sense. 

*The variance of sum is not larger. You assume $cov(X,Y)>0$, which is not true in general.
