Let $Y$ be a random vector of $\mathbf{R}^p.$

Assume that $Y$ is $t_p(\mu,\Sigma,\nu)$ distributed. That is the density is given by $$f_p(y,\mu,\Sigma,\nu)={\displaystyle {\frac {\Gamma \left[(\nu +p)/2\right]}{\Gamma (\nu /2)\nu ^{d/2}\pi ^{p/2}\left|{\boldsymbol {\Sigma }}\right|^{1/2}}}\left[1+{\frac {1}{\nu }}({\mathbf {y} }-{\boldsymbol {\mu }})^{T}{\boldsymbol {\Sigma }}^{-1}({\mathbf {y} }-{\boldsymbol {\mu }})\right]^{-(\nu +p)/2}}$$

Suppose that $y_1,\ldots,y_n$ denote an observed random sample from $t_p(\mu,\Sigma,\nu)$ i.e. $$y=(y_1^T,\ldots,y_n^T)^T.$$

To perform the ML estimation, we compute the log likelihood $$\log(L(\Psi))=\sum_{i=1}^n\log f_p(y_i; \mu,\Sigma,\nu)$$ $$=-\frac{1}{2}np\log(\pi\nu)+n\bigl(\log \Gamma(\frac{\nu+p}{2})-\log \Gamma(\frac{\nu}{2})\bigr)-\frac{1}{2}n\log \vert \Sigma\vert+\frac{1}{2}n(\nu+p)\log \nu-\frac{1}{2}(\nu+p)\sum_{i=1}^n \log( \nu +\delta(y_i;\mu,\Sigma))$$

where $\delta(y_i;\mu,\Sigma)$ is the Mahalanobis square distance.

Question: Is there a closed form for the solution of the likelihood equation?

  • $\begingroup$ Not a 100% duplicate but closely linked: stats.stackexchange.com/questions/63647/… $\endgroup$ Feb 26, 2018 at 14:28
  • $\begingroup$ @FabianWerner Thanks, so it seems that closed form does not exist here. $\endgroup$
    – user188169
    Feb 26, 2018 at 14:31
  • $\begingroup$ @Xi'an is it possible to find a proof of that ? Thanks $\endgroup$
    – user188169
    Feb 26, 2018 at 17:58

1 Answer 1


When considering a Student's $t$ distribution with a fixed degree of freedom $\nu$, finding the MLE of $(\mu,\Sigma)$ is equivalent to finding the MLE for a Cauchy sample, which amounts to finding the minimum of $$|\Sigma|^{n/2}\prod_{i=1}^n \left[ 1+(x_i-\mu)^\text{T}\Sigma^{-1}(x_i-\mu)\right]$$This is a polynomial in $\mu$, which differential is also a polynomial, with no closed form for deriving its roots.


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