I have the following question and I'd greatly appreciate any help!


Basically, I have an arbitrary probability distribution with pdf $f(x)$, we can assume it's continuous with support on $[0,\infty]$


Denote $g(x)$ as pdf of an exponential random variable with fixed, known rate $\lambda$, and $h(x) = f(x) \* g(x)$ where $\*$ is convolution.


My question is, does $h(x)$ uniquely determine $f(x)$? i.e. if I "deconvolve" $h(x)$ with an exponential random variable, do I recover $f(x)$ uniquely?


First time posting here. Hope my question makes sense and is clear enough.



Uniqueness of deconvolution after convolution?
The answer is yes. Let $\hat p$ denote the characteristic function of a pdf $p$, so that
$$\hat p(t)=\int\_{\mathbb R}e^{itx}p(x)\,dx$$
for real $t$. Then $\hat h=\hat f\,\hat g$ and
$$\hat g(t)=\frac1{1-it/\lambda}$$
for real $t$, so that
$$\hat f(t)=\hat h(t)/\hat g(t)=\hat h(t)(1-it/\lambda)$$
for real $t$. [Inverting](https://en.wikipedia.org/wiki/Characteristic_function_(probability_theory)#Inversion_formulae) now the characteristic function $\hat f$, we get $f$.


