convolution of $n$ exponential distributions

Question

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?

Answer 1

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 $k$.

Answer 2

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?

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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

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