convolutionestimationnormal distribution
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?
($F=Z/ Y => \mu_F=\mu_Z-\mu_Y$ and $\Sigma_F=\Sigma_Z-\Sigma_Y$ ?)
The question readily reduces to the case $\mu_X = \mu_Y = 0$ by looking at $X-\mu_X$ and $Y-\mu_Y$.
Clearly the conditional distributions are Normal. Thus, the mean, median, and mode of each are coincident. The modes will occur at the coordinates of a local maximum of the bivariate PDF of $X$ and $Y$ constrained to the curve $g(x,y) = x+y = c$. This implies the contour of the bivariate PDF at this location and the constraint curve have parallel tangents. (This is the theory of Lagrange multipliers.) Because the equation of any contour is of the form $f(x,y) = x^2/(2\sigma_X^2) + y^2/(2\sigma_Y^2) = \rho$ for some constant $\rho$ (that is, all contours are ellipses), their gradients must be parallel, whence there exists $\lambda$ such that
$$\left(\frac{x}{\sigma_X^2}, \frac{y}{\sigma_Y^2}\right) = \nabla f(x,y) = \lambda \nabla g(x,y) = \lambda(1,1).$$
It follows immediately that the modes of the conditional distributions (and therefore also the means) are determined by the ratio of the variances, not of the SDs.
This analysis works for correlated $X$ and $Y$ as well and it applies to any linear constraints, not just the sum.
In that case, you have to write (with hopefully clear notations) $$ \left(\begin{matrix}X\\Y \end{matrix}\right) \sim \mathcal{N}\left[ \left(\begin{matrix}\mu_X\\\mu_Y\end{matrix}\right), \Sigma_{X,Y} \right] $$ (edited: assuming joint normality of $(X,Y)$) Then $$ AX+BY=\left(\begin{matrix}A& B \end{matrix}\right) \left(\begin{matrix}X\\Y \end{matrix}\right) $$ and $$ AX+BY+C \sim \mathcal{N}\left[ \left(\begin{matrix}A& B \end{matrix}\right) \left(\begin{matrix}\mu_X\\\mu_Y\end{matrix}\right) + C, \left(\begin{matrix}A & B \end{matrix}\right)\Sigma_{X,Y} \left(\begin{matrix}A^T \\ B^T \end{matrix}\right)\right] $$ i.e. $$ AX+BY+C \sim \mathcal{N}\left[A\mu_X + B\mu_Y +C, A\Sigma_{XX}A^T+B\Sigma_{XY}^TA^T+A\Sigma_{XY}B^T+B\Sigma_{YY}B^T \right] $$
Best Answer
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.