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I just wonder if pdf of Normal distribution with mean and variance which are normally distributed can be expressed in analytic formula, i.e.

$ \mathcal{N}(\mu, \sigma^2) $ where $ \mu $ ~ $ \mathcal{N}(m, d^2)$ and $\sigma$ ~ $Gamma(k, \theta)$

Let's say that I pick a random variable as time goes(t=1,2,.....,n). When I pick a random variable from $\mathcal{N}(\mu, \sigma^2)$ as time goes, each time the distribution is changed rather than fixed.

So, can $ \mathcal{N}(\mu, \sigma^2) $ where $ \mu $ ~ $ \mathcal{N}(m, d^2)$ and $\sigma$ ~ $Gamma(k, \theta)$ be expressed as analytic expression?

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  • $\begingroup$ Indeed $\sigma^2$ can't be a Normal, since $\sigma^2>0$ and support of any normal is $(-\infty,\infty)$ $\endgroup$ – sinbadh Jan 19 '16 at 2:51
  • $\begingroup$ Sorry. I was confused. I modified the error. Thank you. $\endgroup$ – qyong Jan 19 '16 at 3:13
  • $\begingroup$ Seems to me similar to a Bayesian problem obtaining normal posterior from normal prior and normal data. $\endgroup$ – BruceET Jan 19 '16 at 4:04

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