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:


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 $p(Z|X,\theta)$?

[1] A Derivation of the EM Updates for Finding the MaximumLikelihood Parameter Estimates of the Student’s t Distribution, Carl Scheffler, 22 September 2008.

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    $\begingroup$ 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 $\endgroup$ Commented Oct 11, 2020 at 21:42
  • $\begingroup$ @kjetilbhalvorsen But for each $\eta_i$ there is a different distribution, so $x_i$ variables are not identically distributed. Isn't it correct? $\endgroup$
    – groove
    Commented Oct 11, 2020 at 22:55
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    $\begingroup$ 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 $(\nu/2,\nu/2).$ $\endgroup$
    – whuber
    Commented Oct 12, 2020 at 13:50
  • $\begingroup$ @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? $\endgroup$
    – groove
    Commented Oct 14, 2020 at 9:19
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    $\begingroup$ 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. $\endgroup$
    – Xi'an
    Commented Nov 20, 2020 at 15:55

1 Answer 1


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.


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