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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.

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    $\begingroup$ en.wikipedia.org/wiki/Hypoexponential_distribution $\endgroup$ – user17762 Mar 11 '12 at 23:57
  • $\begingroup$ @SivaramAmbikasaran: So there is no way to write the pdf of this? $\endgroup$ – alexm Mar 12 '12 at 0:09
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    $\begingroup$ @alexm: Sure, if the random variables are independent. Either use convolution directly, or find the cumulative distribution function first. The integration over the triangle that has $x+y \le w$ is straightforward. $\endgroup$ – André Nicolas Mar 12 '12 at 0:48
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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.

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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.

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