Exponential variables Suppose we have two exponential random variables $X_1$ and $X_2$ with parameters $\lambda_1$ and $\lambda_2$. Would the sum of them have any recognized distribution? If they have the same parameter $\lambda$, then the sum is a gamma random variable.
 A: Let $W=X+Y$.  The fact that $X$ and $Y$ are known exponentials is not enough to determine the distribution of $W$.  We will assume that $X$ and $Y$ are independent.
We find the cumulative distribution function $F_W(w)$ of $W$ more or less from basic principles.  This is $P(W \le w)$. It is clearly $0$ when $w<0$.  To make typing easier, let the parameters of $X$ and $Y$ be $\alpha$ and $\beta$.  By independence, the joint density function of $X$ and $Y$ is $\alpha e^{-\alpha x}\beta e^{-\beta y}$ (for $x\ge 0$, $y\ge 0$).
The probability that $W \le w$ is the integral of the joint density over the triangle bounded by the axes and the line $x+y=w$. So 
$$P(W \le w)=\int_{x=0}^w \alpha e^{-\alpha x} \left(\int_{y=0}^{w-x} \beta e^{-\beta y}\,dy\right)\, dx.$$
The inner integral is $1-e^{-\beta(w-x)}$, that is, $1-e^{-\beta w}e^{\beta x}$. So now we need to find
$$\int_0^w \left( \alpha e^{-\alpha x}  -\alpha e^{-\beta w}e^{\beta x-\alpha x}\right)\,dx.$$
Again, we are just integrating an exponential.  After some simplification we find that if $\alpha\ne \beta$, then
$$P(W \le w)=1 -\frac{\beta e^{-\alpha w}-\alpha e^{-\beta w}}{\beta -\alpha}$$
(for $w \ge 0$).  Differentiate to get the density function $f_W(w)$. This is $0$ if $w<0$, and 
$$\frac{\alpha\beta}{\beta -\alpha}\left( e^{-\alpha w}- e^{-\beta w}\right)$$
when $w \ge 0$.
A special case: Note that the formula only applies if $\beta \ne \alpha$. When $\beta=\alpha$, the second integral is simply
$$\int_0^w \left(\alpha e^{-\alpha x} -\alpha e^{-\alpha w}\right)\,dx.$$
So we get that in the case of equality,
$$F_W(w)=1-e^{-\alpha w} -\alpha w e^{-\alpha w}$$
(for $w \ge 0$).  Again, differentiate to get the density. 
The two formulas are not quite as different as they look.  If we find the limit of the density function as $\beta$ approaches $\alpha$, we will get the density function of the special case.
A: Let $Z = X_1 + X_2$. Let $\mathcal{L}_Z(t) = \mathbb{E}(\mathrm{e}^{-t Z})$ be the Laplace transform of the distribution density of $Z$. Notice that the Laplace transform is related to the moment generating function $\mathcal{L}_Z(t) = \mathcal{M}_Z(-t)$.
The moment generating function of $Z$ is the product of those of summands, assuming that $X_1$ and $X_2$ are independent, i.e. $\mathcal{M}_Z(t) = \mathcal{M}_{X_1}(t) \mathcal{M}_{X_2}(t)$:
$$
   \mathcal{M}_Z(t) = \frac{\lambda_1}{\lambda_1 - t} \cdot  \frac{\lambda_2}{\lambda_2 - t}
$$
Assuming $\lambda_1 \not= \lambda_2$ we can perform the partial fraction decomposition:
$$
   \mathcal{M}_Z(t) = \frac{\lambda_2}{\lambda_2 -\lambda_1} \cdot \frac{\lambda_1}{\lambda_1 - t} - \frac{\lambda_1}{\lambda_2 -\lambda_1} \cdot \frac{\lambda_2}{\lambda_2 - t}
$$
Applying the inverse Laplace transform we get deduce the probability generating function:
$$
   f_Z(x) = \frac{\lambda_2 f_{X_1}(x) - \lambda_1 f_{X_2}(x)}{\lambda_2 -\lambda_1} = 
           \lambda_1 \lambda_2 \frac{\mathrm{e}^{-\lambda_1 x} - \mathrm{e}^{-\lambda_2 x} }{\lambda_2 - \lambda_1} \cdot [ x \geqslant 0 ]
$$
This form shows that $f_Z(x) > 0$ for $x \geqslant 0$.
Variable $Z$ is said to be hypoexponential as Siviram has already commented.
