A 1-D random variable from a random distribution

I have a random variable $X$ that is drawn from the pdf $$f(x; \mu, \sigma, \sigma_{\mu}, \sigma_{\sigma}) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{|\hat{\sigma}|\sqrt{2\pi }} \frac{1}{ \sigma_\mu\sqrt{2\pi}} \frac{1}{\sigma_{\sigma}\sqrt{2\pi }} e^{\frac{-(x-\hat{\mu})^2}{2 \hat{\sigma}^2}}e^{\frac{-(\hat{\mu}-\mu)^2}{2\sigma_{\mu}^2}} e^{\frac{-(\hat{\sigma}-\sigma)^2}{2\sigma_{\sigma}^2}} d \hat{\sigma} d\hat{\mu},$$ (with a proper normalization constant).

In essence, I first draw $\hat{\mu}$ from $\mathcal{N}(\mu, \sigma_{\mu})$ and draw $\hat{\sigma}$ from $\mathcal{N}(\sigma, \sigma_{\sigma})$, and then I draw $X$ from the distribution $\mathcal{N}(\hat{\mu},|\hat{\sigma}|)$.

In general I have not seen much about the problem of random variables drawn from random distributions, and have a number of basic questions about this situation:

1. Is the pdf given above the right function to describe the situation under it?

2. Is this pdf a known distribution, or can it be manipulated into a known form?

3. What are the moments of this function?

4. Is there any way to find/estimate the values $\sigma_{\mu}$ and $\sigma_{\sigma}$?

• It is possible to do the $\hat\mu$ integral in the definition of $f$ explicitly. I wonder if the resulting $\hat\sigma$ integral could be done... It looks quite messy, but it might be possible. Have you tried doing the integrals? – Joonas Ilmavirta Oct 19 '14 at 18:54