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?
 A: 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 $k$.
A: We have that for a R.V. $X$ exponentially distributed that it´s density function is given by:
$$
\\ f_X(x)=\begin{cases}
\lambda e^{-\lambda x} & x\geq 0\\ 
0 & x<0
\end{cases}
$$
Define $S_n=\sum_{k=1}^n X_k$.
Then, as You have already shown, the sum of two exponential R.V´s has density
$$
\begin{aligned}
f_{S_2}&=f_{X_1+X_2}(t)\\
&=\int_0^t \lambda e^{-\lambda(t-s)}\lambda e^{-\lambda s} \,ds\\
&=\lambda^2  e^{-\lambda t}\int_0^t   \,ds\\
&=\lambda^2  t e^{-\lambda t} \qquad \blacksquare
\end{aligned}
$$
For the sum of three R.V´s
$$
\begin{aligned}
f_{X_1+X_2+X_3}(t)&=f_{S_2+X_3}(t)\\
&=\int_{-\infty}^\infty f_{S_2}(t-s)f_{X_3}(s)\,ds \\
&=\int_{-\infty}^\infty f_{X_3}(t-s)f_{S_2}(s)\,ds \\
&=\int_0^t \lambda e^{-\lambda(t-s)}\lambda^2 s\, e^{-\lambda s} \,ds\\
&=\lambda^3  e^{-\lambda t}\int_0^t  s \,ds\\
&=\frac{\lambda^3  t^2 e^{-\lambda t}}{2!}  \qquad \blacksquare
\end{aligned}
$$
Where we used the result for $f_{S_2}=\lambda^2  t e^{-\lambda t}$ and that $F_1*F_2=F_2*F_1$.
For the sum of four R.V´s we have:
$$
\begin{aligned}
f_{X_1+X_2+X_3+X_4}(t)&=f_{S_3+X_4}(t)\\
&=\int_{-\infty}^\infty f_{S_3}(t-s)f_{X_4}(s)\,ds \\
&=\int_{-\infty}^\infty f_{X_4}(t-s)f_{S_3}(s)\,ds \\
&=\int_0^t \lambda e^{-\lambda(t-s)} \frac{\lambda^3 s^2\, e^{-\lambda s} }{2!}\,ds\\
&=\frac{\lambda^4  e^{-\lambda t}}{2!}\int_0^t  s^2 \,ds\\
&=\frac{\lambda^4  t^3 e^{-\lambda t}}{3!}  \qquad \blacksquare
\end{aligned}
$$
Can you see a pattern emerging?

Claim:
$$f_{S_n}(t)=f_{X_1+\cdots+X_n}(t)=\frac{\lambda^n t^{n-1}\,e^{-\lambda t}}{(n-1)!}$$
Proof:
Assume that
$$f_{S_{n-1}}(t)=f_{X_1+\cdots+X_{n-1}}(t)=\frac{\lambda^{n-1} t^{n-2}\,e^{-\lambda t}}{(n-2)!}$$
Than
$$
\begin{aligned}
f_{S_{n}}(t)&=f_{X_1+\cdots+X_n}(t)\\
&=f_{S_{n-1}+X_n}(t)\\
&=\int_{-\infty}^\infty f_{S_{n-1}}(t-s)f_{X_n}(s)\,ds \\
&=\int_{-\infty}^\infty f_{X_n}(t-s)f_{S_{n-1}}(s)\,ds \\
&=\int_0^t \lambda e^{-\lambda(t-s)} \frac{\lambda^{n-1} s^{n-2}\, e^{-\lambda s} }{(n-2)!}\,ds\\
&=\frac{\lambda^n  e^{-\lambda t}}{(n-2)!}\int_0^t  s^{n-2} \,ds\\
&=\frac{\lambda^n  t^{n-1} e^{-\lambda t}}{(n-1)!} \\
&=\frac{\lambda^n  t^{n-1} e^{-\lambda t}}{\Gamma(n)}  \qquad \blacksquare
\end{aligned}
$$
