2
$\begingroup$

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.

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ While Iosif's solution works for general convolution kernels, in this particular case there is a simpler solution: $\lambda^{-1} e^{\lambda x} h(x)$ turns out to be the integral of $e^{\lambda x} f(x)$, and hence $f(x) = \lambda^{-1} e^{-\lambda x} (e^{\lambda x} h(x))'$. $\endgroup$ Commented Feb 9, 2021 at 8:28
  • $\begingroup$ Interesting! Thank you Mateusz! $\endgroup$ Commented Feb 9, 2021 at 17:53

1 Answer 1

Reset to default
2
$\begingroup$

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 now the characteristic function $\hat f$, we get $f$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks a lot Iosif! I was thinking the similar via mgf but realized mgf does not necessarily exist. This is greatly helpful. Thanks again! $\endgroup$ Commented Feb 9, 2021 at 1:09

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.