Let $S, T$ be two independent random variables both with the exponential distribution and the same parameter $\lambda > 0$.
I would like to find the density function of $S+T$.
$S$ and $T$ both have the density function $f(t) = \lambda \cdot e^{-\lambda t}$ where $t>0$. How do I find the density function of $S+T$? I don't know how to begin, please help me.
Best Answer
Suppose that $\left(X_{i}\right)_{1\leq i\leq n}$ is independant and identically distributed according to an exponential law with a parameter $\lambda>0$ . For all $x\in\mathbb{R}$ , we have$$\mathbb{P}\left[X_{i}\leq x\right]=1-e^{-\lambda x}.$$ Now, it is straightforward to use the characteristic functions : if $S_{n}=\sum_{i=1}^{n}X_{i}$ , then we have for all $t\in\mathbb{R}$ $$\chi_{_{S_{n}}}\left(t\right)=\prod_{i=1}^{n}\chi_{_{X_{i}}}\left(t\right)=\left(\chi_{_{X_{1}}}\left(t\right)\right)^{n}=\left(\left(1-i\lambda t\right)^{-1}\right)^{n}=\left(1-i\lambda t\right)^{-n}$$ which is the characteristic function of a random variable that follow a gamma law of parameters $\left(n,\lambda\right)$ (I used independance and identical distribution here). Thus, the density of the law of $S_{n}$ is the density of the gamma law (see for example http://en.wikipedia.org/wiki/Gamma_distribution).
You can also show by induction that the density of the sum of INDEPENDANT random variables is the convolution of the densities.