Is the normal distribution where the parameters themselves are normal random variables still normal?

Question

We denote by $\mathcal{N}(\mu, \sigma^2)$ the normal distribution where $\mu\in\mathbb{R}$ is the mean and $\sigma\in\mathbb{R}$ is the scale parameter.

Assume that the parameters $\mu,\sigma$ are themselves normal random variables, i.e., $\mu\sim \mathcal{N}(\mu_1, \sigma_1^2)$ and $\sigma\sim \mathcal{N}(\mu_2, \sigma_2^2)$. Is the distribution $\mathcal{N}(\mu,\sigma^2)$ still normal? I note that $\sigma\sim\mathcal{N}(\mu_2,\sigma_2^2)$ might be negative but this should not be a problem since $\mathcal{N}(\mu,\sigma^2)$ only depends on $|\sigma|$ but not on $\sigma$.

I plotted $\mathcal{N}(\mu,\sigma^2)$ for several values of $\mu_i, \sigma_i$, $i=1,2$. It has a strong resemblance to the normal distribution but it is probably not the normal distribution.

Edit: For clarification I also want to add that when $\sigma=0$ I assume that $\mathcal{N}(\mu,0)$ denotes the Dirac distribution on $\mu$. Hence, at least for $\sigma_1=\sigma_2=0$ the distribution $\mathcal{N}(\mu,\sigma^2)$ is indeed normal.

Answer 1

$X=\mu+\sigma Z$ where $\mu\sim N(0,\sigma_1^2),$ $\sigma\sim N(0,\sigma_2^2)$, $Z\sim N(0,1)$ and where $\mu$, $\sigma$ and $Z$ are independent and $\sigma_1$ and $\sigma_2$ are constant. Therefore $$E(e^{zX})=E(e^{z\mu})E(E(e^{z\sigma Z}|\sigma)) =e^{\sigma^2z^2/2}E(e^{\sigma^2 z^2/2})=e^{\sigma^2z^2/2}\frac{1}{\sqrt{1-\sigma_2^2 z^2}}.$$

Answer 2

Computing the Laplace transform is a way to answer to your question: is $X$ normal? Now I apologize for overlooking $\mu_1$ and $\mu_2.$ Now, let go, but adding a $Z_2\sim N(0,1)$ such that $\sigma =\mu_2+\sigma_2 Z_2.$

$$E(e^{sX})=e^{s\mu_1+\frac{s^2}{2}\sigma^2_1 }\times E(e^{\frac{s^2}{2}(\mu_2^2+2\mu_2 \sigma_2 Z_2+\sigma_2^2 Z_2^2)})\\=e^{s\mu_1+\frac{s^2}{2}(\sigma^2_1+\mu_2^2) }\times \frac{1}{\sqrt{2\pi}}\int_{R}e^{-\frac{z^2}{2}(1-s^2\sigma_2 )+\frac{s^2}{2}\mu_2\sigma_2 z}dz\\=e^{s\mu_1+\frac{s^2}{2}(\sigma^2_1+\mu_2^2) }\times\frac{1}{\sqrt{1-\sigma_2^2s^2}}\exp\left(\frac{1}{1-\sigma_2^2s^2}\times \frac{\mu_2^2\sigma_2^2}{8}s^4\right) .$$ Beurk, not normal isn't it?