# Likelihood of Student's t distribution

The integral form of student's t distribution is given as follows [1]:

$$p(x|\mu,\lambda,\nu)=\int_0^\infty \textrm{Normal}(x|\mu,(\lambda\eta)^{-1})\textrm{Gamma}(\eta|\nu/2,\nu/2)d\eta~~~~~~~~~~~~$$ (1)

The complete data likelihood is written as follows:

$$P(X,Z|\theta)=\prod_{i=0}^N\textrm{Normal}(x_i|\mu,(\lambda\eta_i)^{-1})\textrm{Gamma}(\eta_i|\nu/2,\nu/2)~~~~~~~~~~$$(2)

I would like to ask why $$\eta$$ becomes $$\eta_i$$ when we're writing the likelihood. There are $$N$$ observations, single $$\mu$$, and $$N$$ instances of $$\eta$$. How can you see this in (1)? Is there a 1-to-1 relation between $$x_i$$ and $$\eta_i$$, or is it just a coincidence that there are $$N$$ instances of both? Lastly, can we see Gamma() here as a prior distribution over $$\eta$$ since when doing EM we find the posterior latent distribution $$p(Z|X,\theta)$$?

• supposedly there are iid observations from some common t distribution. If there were one only $\eta$, that would induce dependence, so to preserve independence independent $\eta_i$'s are necessary Oct 11, 2020 at 21:42
• @kjetilbhalvorsen But for each $\eta_i$ there is a different distribution, so $x_i$ variables are not identically distributed. Isn't it correct? Oct 11, 2020 at 22:55
• I don't see anything in your question that suggests the $\eta_i$ have different distributions. Indeed, (2) explicitly asserts they have a common Gamma distribution with parameters $(\nu/2,\nu/2).$
– whuber
Oct 12, 2020 at 13:50
• @whuber $\eta_i$ are identical, but for each $\eta_i$, the distribution of $x_i$ is different since the variance is scaled, is this correct? Oct 14, 2020 at 9:19
• The distribution of $x_1$ given $\eta_1$ is different from the distribution of $x_2$ given $\eta_2$ if $\eta_1\ne\eta_2$ but $x_1$ and $x_2$ are marginally iid. Nov 20, 2020 at 15:55

Since $$p(x|\mu,\lambda,\nu)=\int_0^\infty \varphi(x|\mu,(\lambda\eta)^{-1})\gamma(\eta|\nu/2,\nu/2)\,\text d\eta$$ the observed likelihood writes \begin{align}\prod_{i=1}^n p(x|\mu,\lambda,\nu)&=\prod_{i=1}^n\int_0^\infty \varphi(x_i|\mu,(\lambda\eta)^{-1})\gamma(\eta|\nu/2,\nu/2)\,\text d\eta\tag{1}\\ &=\prod_{i=1}^n\int_0^\infty \varphi(x_i|\mu,(\lambda\eta_i)^{-1})\gamma(\eta_i|\nu/2,\nu/2)\,\text d\eta_i\\ &=\int_{\mathbb R_+^n}\prod_{i=1}^n \varphi(x_i|\mu,(\lambda\eta_i)^{-1})\gamma(\eta_i|\nu/2,\nu/2)\,\text d\boldsymbol\eta \end{align} since $$\eta$$ is a dummy symbol in each of the $$n$$ integrals in (1), which can be written as $$\eta_i$$ for the $$i$$-th integral.
Thus the complete likelihood can be chosen as $$\prod_{i=1}^n \varphi(x_i|\mu,(\lambda\eta_i)^{-1})\gamma(\eta_i|\nu/2,\nu/2)$$
About the second question, it is counter-productive to see $$\eta$$ in a Bayesian light as a parameter with a Gamma prior since $$\eta$$ varies with the observation, which is issued from the marginal ($$\eta$$ integrated out) and not the conditional on $$\eta$$, as demonstrated above. An $$n$$ sample $$(x_1,\ldots,x_n)$$ comes with an $$n$$ latent sample $$(\eta_1,\ldots,\eta_n)$$. Contrary to a Bayesian sample, it is not possible to learn about $$\eta_i$$ from the sample.