# convolution of $n$ exponential distributions

Let $exp(k)$ be the exponential distribution, $k>0$. Then it has density

$$f(x)= \begin{cases} ke^{-kx} & \text{ if } 0\leq x < \infty\\ 0 &\text{otherwise} \end{cases}$$

I want to find the convolution of $n$ exponential distributions. For $n=2$ I have

$$\int_{\mathbb{R}} f(x-t)f(t) dt =\int_0^x (k e^{-k(x-t)}ke^{-kt}) dt=\int_0^x k^2e^{-kx} dt=k^2e^{-kx} \int_0^x dt= k^2xe^{-kx}.$$

For $n \geq 3$ I would like to take convolutions inductively, but I am not even sure what my inductive hypothesis would be. Some help?

• It is often easier to find the convolution by the moment generation function. Would such a move be acceptable or do you want to do it by the integral? – madprob Oct 30 '17 at 12:48
• @madprob I would prefer to do this via integrals, since I just learned the definition of a convolution. – Sarah Oct 30 '17 at 12:49
• Also, $f(x)=k^2 x \exp(-kx)$ is a known distribution. Do you know its name? – madprob Oct 30 '17 at 12:50
• @madprob No, but I can look it up. – Sarah Oct 30 '17 at 12:54
• @madprob Before seeing Jimmy's answer, I actually computed the convolution for $n=3$ and then $n=4$ to see the pattern. It makes sense that you get a Gamma distribution density, since, according to Wikepedia, the sum of exponential random variables follows a Gamma distribution. – Sarah Oct 30 '17 at 13:05

For $n=2$, you found that $$f_2(x)=k^2x^{2-1}e^{-kx}$$ The tricky part is that actually there is also a hidden $1/(2-1)!=1$. (you couldn't have known that, unless you calculated also the $n=3$ case). So, the inductive hypothesis for $n\ge 3$: $$f_n(x)=\frac{1}{(n-1)!}k^{n}x^{n-1}e^{-kx}$$ for $0\le x<+\infty$ and $f_n(x)=0$ otherwise. This is the Erlang distribution (or a particular instance of the Gamma distribution) with parameters: shape $n$ and rate $k$.