Can a (possibly infinite) mixture of Gaussians be Gaussian?

Question

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.

Answer 1

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).