Lebesgue density of a sum of i.i.d. exponentially distributed random variables

Question

Let $X$ be an exponentially distributed random variable with parameter $\lambda$, i.e., its Lebesgue density is given by

$$ f(x) = \begin{cases} \lambda e^{- \lambda x} & \text{if} \quad x \geq 0 \\ 0 & \text{if} \quad x < 0 \end{cases} $$ The characteristic function of $X$ can be then computed through $$ \varphi(z) = E[e^{izX}] = \int_{0}^{\infty} e^{izx} \lambda e^{ - \lambda x} d x = \lambda \int_{0}^{\infty} \frac{ e^{ (iz - \lambda)x} }{iz - \lambda } d ((iz - \lambda)x) = \lambda \frac{ e^{ (iz - \lambda) x} }{i z - \lambda} = \frac{\lambda}{\lambda - iz } $$ If we further take $n$ i.i.d. copies of $X$, then the characterisitic function of their sum is given by $$ \frac{ \lambda^n}{ {( \lambda - i z )}^n} \tag{1} $$ Now, consider a somewhat reverse problem, namely, given the characteristic function in $(1)$, can one come up with a Lebesgue density? I would be interested in an explicit computation. Can this be helpful?

Answer 1

We can compute the density of $S_n=\sum_{i=1}^n X_i$ directly by convolution. Assume that for some $n\geqslant 1$ that $$f_{S_n}(x) = \frac{\lambda(\lambda x)^{n-1}}{(n-1)!} e^{-\lambda x}. $$ Then \begin{align} f_{S_{n+1}}(x) &= f_{S_{n}}\star f_{X_{n+1}}(x)\\ &= \int_0^x \frac{\lambda(\lambda y)^{n-1}}{(n-1)!} e^{-\lambda y} \lambda e^{-\lambda(x-y)}\ \mathsf dy\\ &= \frac{\lambda e^{-\lambda x}}{(n-1)!}\int_0^x\lambda(\lambda y)^{n-1}\ \mathsf dy\\ &= \frac{\lambda e^{-\lambda x}}{(n-1)!}\left[\frac{(\lambda y)^n}n \right]_0^x\\ &= \frac{\lambda (\lambda x)^n}{n!}e^{-\lambda x}. \end{align} By induction, we see that $f_{S_n} = \frac{\lambda(\lambda x)^{n-1}}{(n-1)!} e^{-\lambda x}$ for all positive integers $n$.

Answer 2

I doubt this is possible in general.

Here is a useful result that may work sometimes.

Let $X: \Omega \to \mathbb{R}^k$ be a random vector with characteristic function $\phi$ such that $$\int_{\mathbb{R}^k} |\phi(t)| dt < \infty$$

Then $\mathbb{P}_X$ is absolutely continuous w.r.t. Lebesgue measure with density function

$$x \mapsto \frac{1}{(2 \pi)^k} \int_{\mathbb{R}^k} \exp(-i \langle t,x \rangle) \phi(t)dt $$