Probability density of a sum of exponentially distibuted variables

Question

The $X_1, X_2, \dots\ X_n$ are identical independently distributed random variables. $X_i \sim \lambda e^{-\lambda x}$.
Find pdf of their sum.

Answer 1

It's distributed as $\text{Erlang}(n, \lambda)$; see https://en.wikipedia.org/wiki/Erlang_distribution for more details.

Answer 2

Let X be exponential distributed, i.e. $X \sim {\rm Exponential}(b)$ $$f_X(x) = be^{-bx} \qquad x,b > 0$$

Let $Y$ be Gamma distributed, i.e. $Y \sim {\rm Gamma}(a,b)$ $$f_Y(y) = \frac{b^a y^{a-1} e^{-by}}{\Gamma(a)} \qquad a,b,y>0$$

Note that for $a = 1$, the Gamma distribution $f_Y(y)$ coincides with exponential $f_X(x)$.

Let $Z = X+Y$. The PDF is $$\begin{align*} f_Z(z) &= \int_{y=0}^z f_Y(y) f_X(z-y) \, dy = \int_{y=0}^z \frac{b^{a+1} y^{a-1} e^{-by} e^{-b(z-y)}}{\Gamma(a)} \, dy \\ &= \frac{b^{a+1} e^{-bz}}{\Gamma(a)} \int_{y=0}^z y^{a-1} \, dy = \frac{b^{a+1} z^a e^{-bz}}{\Gamma(a+1)}. \end{align*}$$

Therefore, the sum $Z$ is gamma distributed with new parameter $a+1$ instead of $a$.

Finally, doing this by induction $n$ times, the sum of $n$ exponential variables with common rate parameter $\lambda$ is gamma with parameter $a = n$, and rate parameter $b = \lambda$.