# Mean of normal follows a T distribution

Suppose:

$$x \sim \mathcal{N}(x; \mu, \Sigma) \;\;\;$$ st. $$\;\;\; \mu \sim T_{v}(\mu; k, M)$$

Where $$T$$ is the $$t$$-distribution with v degrees of freedom, location $$k$$, and shape $$M$$.
Then, is there a clean posterior predictive for $$x$$ and/or a clean distribution for $$p([x, \mu])$$?

I have: \begin{aligned} p(x, \mu) &= p(\mu)p(x|\mu)\\ &= T_{v}(\mu; k, M)\mathcal{N}(x; \mu, \Sigma) \end{aligned} and \begin{aligned} p(x) &= \int p(\mu)p(x|\mu)\;d\mu\\ &= \int T_{v}(\mu; k, M)\mathcal{N}(x; \mu, \Sigma) \;d\mu \end{aligned}

However, I am not sure these can be simplified or have standard forms.

As, the normal is the conjugate prior to the mean of the normal, these have clean solutions when $$\mu \sim \mathcal{N}$$. However, I am not sure there will be closed form in the above case.

• I was hoping to find the analytic solution. Not approximate. It may be the case though that I must use numeric methods comparing the t(v)-distribution and Normal as you seem to be eluding to. Commented Jan 5 at 5:20
• I don't believe you understand the question. Again, I'm just asking if there's an analytic solution out there. I'm not about to pretend a T-distribution is a gaussian with n>30 Commented Jan 5 at 6:01
• Have you tried expressing the T distribution as a mixture of Normal distributions over their variance? Commented Jan 5 at 18:41
• @xI have tried expressing the T distribution as a normal with inverse-wishart distributed variance. This leads for my purposes to:\ $\int W^{-1}(M; T, v) \mathcal{N}(x; k, M + \Sigma) dM$\ However, I can't simplify this either... It is discouraging, but simply deriving the T-distribution is hard enough, so its not impossible I just haven't cracked it... Commented Jan 5 at 19:19