Assuming $X$ and $Y$ are two Gaussians with parameters of $\mu_X,\Sigma_X$ and $\mu_Y,\Sigma_Y$ then for their convolution we know that (reference) :

$Z=X*Y$ is also a Gaussian with parameters of $\mu_Z=\mu_X+\mu_Y$ and $\Sigma_Z=\Sigma_X+\Sigma_Y$

what can we say about deconvolution of $Z$ and $Y$? Is the deconvolution of two Gaussians, also a Gaussian? Are the parameters of $Z$ can be similarly computed? ($F=Z/ Y \implies \mu_F=\mu_Z-\mu_Y$ and $\Sigma_F=\Sigma_Z-\Sigma_Y$ ?)


1 Answer 1


I think so.

The Fourier transform of a Gaussian is a Gaussian, so we can go to the f domain.

Furthermore, dividing a Gaussian by a Gaussian (deconvolution in the f domain) also yields a Gaussian, so we now have a Gaussian f domain quotient.

The inverse Fourier of a Gaussian is also a Gaussian.

Voila (I won't say QED as this is far from formal!!!)

Regarding the parameters I would say yes as well.

  • $\begingroup$ makes sense. Are you also agree with the parameters? subtraction of mean and covariances? Are you aware of any proof for the parameters? I could consider $F*Y=X$ then F should have parameters as I described. But not sure if it's enough as a proof. $\endgroup$
    – PickleRick
    Jan 21, 2016 at 16:00
  • 1
    $\begingroup$ No, haven't seen a rigorous proof of this. However, what if we assumed the opposite and tried to work backwards. If the weren't as you suggest we wouldn't get back to the same point if we reconvolved. A little reductio ad absurdum $\endgroup$
    – MikeP
    Jan 21, 2016 at 16:02
  • $\begingroup$ I think so. If this is true, then it means if the parameters of two Gaussians subtract, their PDF is deconvolved. Am I right? $\endgroup$
    – PickleRick
    Jan 21, 2016 at 16:04
  • $\begingroup$ yup, i would say so $\endgroup$
    – MikeP
    Jan 21, 2016 at 16:05
  • $\begingroup$ so we got an answer! $\endgroup$
    – PickleRick
    Jan 21, 2016 at 17:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.