Sum of correlated exponential random variables
The pdf $f_Z(z)$ of the sum $Z=X+Y$ of any two jointly continuous random variables $X$ and $Y$ with joint pdf $f_{X,Y}(x,y)$ is as follows: 
$$\text{For all } z, -\infty &lt; z &lt; \infty, ~~ f_Z(z) = \int_{-\infty}^\infty f_{X,Y}(x,z-x) \, \mathrm dx.\tag{1}$$

For the special case when $X$ and $Y$ are nonnegative random variables (including as a special case, exponential random variables) and so take on nonnegative values only, $f_{X,Y}(x,y)$ has value $0$ if at least one of $x$ and $y$ is smaller than $0$. Hence, in this case, the integrand $f_{X,Y}(x,z-x)$ in $(1)$ has value $0$ if $x &lt; 0$ or if $z &lt; x$. Consequently, if $z$ is a negative number, then the integrand in $(1)$ is always $0$ regardless of the value of $x$ and therefore so is the integral. All of which is just a long-winded way of saying that $f_Z(z)$ has value $0$ when $z&lt;0$, that is, $Z$ takes on nonegative values only, which any idiot could have deduced from the fact that $Z=X+Y$ and both $X$ and $Y$ are nonnegative. But the approach is useful even for $z&gt;0$ since now we have that the integrand in $(1)$ is zero when $x&lt;0$ or when $x &gt;z$ and so for nonnegative $X$ and $Y$, we can simplify $(1)$ to
$$f_Z(z) = \begin{cases}\displaystyle\int_0^z f_{X,Y}(x,z-x) \, \mathrm dx, &amp; z \geq 0,\\\quad\\
0, &amp; z &lt; 0\end{cases} \tag{2}$$


  No further simplification of $(2)$ is possible in general.


For the special case when $X$ and $Y$ are independent random variables, $f_{X,Y}(x,y)$ factors into $f_X(x)f_Y(y)$ and so $(1)$ becomes the familiar convolution integral and $(2)$ the somewhat-less-familiar convolution integral for causal signals. But no such simplification is possible for nonindependent random variables $X$ and $Y$; we need the joint pdf to calculate $f_Z(z)$ and just knowing that $X$ and $Y$ are correlated random variables (whether exponential or Gaussian or whatever) is not enough.