Determining variance from sum of two random correlated variables I understand that the variance of the sum of two independent normally distributed random variables is the sum of the variances, but how does this change when the two random variables are correlated?  

Consider a function of two variables,
$ z = f(x, y)
$. Then the variation of z, $\delta z$, is 
$$\tag{1} \delta z = \frac{df}{dx} \ \delta x $$
where 
$$ \frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{ dy}{dx}. $$ 
Squaring equation (1) we get 
$$ (\delta z)^2 = \Big[ \left( \frac{\partial f}{\partial x} \right)^2 + 2 \frac{\partial f}{\partial x} \frac{\partial f}{\partial y} \frac{dy}{dx} + \left( \frac{\partial f}{\partial y}\right)^2 \left( \frac{dy}{dx} \right)^2 \Big] (\delta x)^2. $$ Multiplying this out we get $$ (\delta z)^2 =  \left( \frac{\partial f}{\partial x} \right)^2 (\delta x)^2+ 2 \frac{\partial f}{\partial x} \frac{\partial f}{\partial y} \delta x \delta y + \left( \frac{\partial f}{\partial y}\right)^2 (\delta y)^2, $$ where we have used that $\delta y = \frac{dy}{dx} \delta x$. Now we can identify the quadratic variation terms with the variances and covariance of random variables: $$ \text{Var}(z) = \left( \frac{\partial f}{\partial x} \right)^2 \text{Var}(x) +  2 \frac{\partial f}{\partial x} \frac{\partial f}{\partial y} \text{Cov}(x,y) +   \left( \frac{\partial f}{\partial y}\right)^2  \text{Var}(y). $$ When the function $f$ is just a sum of $x$ and $y$ then the partial derivative terms are all equal to one, giving $$\text{Var}(z) =  \text{Var}(x) +  2\ \text{Cov}(x,y) +  \text{Var}(y). $$ 
