Can a (possibly infinite) mixture of Gaussians be Gaussian? Suppose we define a (possibly infinite) mixture of zero-mean Gaussians:
$$p(x) = \int_{\mathbb{R}^+} N(x; 0, \sigma^2)\ \pi(\sigma)\,\text d\sigma,$$
where $\pi$ defines the mixture components. Obviously, if $\pi$ is a point mass on some standard deviation $\sigma$, the resulting distribution is Gaussian with that variance.
Are there any other mixtures $\pi$ which will result in a Gaussian marginal?
Edit: it would actually still be helpful to characterize if/when this is possible even when the means are not required to be 0.
 A: Copying from the Wikipedia page on compound distributions

Gaussian scale mixtures:

*

*Compounding a normal distribution with variance distributed according to an inverse gamma distribution (or equivalently, with
precision distributed as a gamma distribution) yields a
non-standardized Student's t-distribution. This distribution has
the same symmetrical shape as a normal distribution with the same
central point, but has greater variance and heavy tails.


*Compounding a Gaussian distribution with variance distributed according to an exponential distribution (or with standard deviation
according to a Rayleigh distribution) yields a Laplace distribution.


*Compounding a Gaussian distribution with variance distributed according to an exponential distribution whose rate parameter is
itself distributed according to a gamma distribution yields a
Normal-exponential-gamma distribution. (This involves two compounding
stages. The variance itself then follows a Lomax distribution; see
below.)


*Compounding a Gaussian distribution with standard deviation distributed according to a (standard) inverse uniform distribution
yields a Slash distribution.

but I do not think there is a case outside the Dirac mass at $\sigma_0$ where the compound is also a Gaussian. This 2005 conference paper by Alecu et al. contains a proof of this result (among other things).
