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Here is Fitting the t-Distribution by Maximum Likelihood t-method in book Statistics and Data Analysis for Financial Engineering with R examples page 113 and 168.

But I cannot understand

For the univariate case (first picture):

  1. How does the fitting work? Is it the method called expectation–maximization (EM) algorithm?

  2. And I understood as first we obtain $\mu,\sigma$ by fitting, then use MLE to obtain the parameter $\nu,$ is it correct?

For the multivariate case (second picture):

  1. Does that mean we first compute $\nu,$ then use MLE to compute $\mu$ and $\Lambda?$

  2. If so, then how could we compute $\nu$ with unknown $\mu$ and $\Lambda?$ And the logic totally reverses compared with the univariate case.

  3. I think the parameter estimation should be consistent between univariate and multivariate cases and I must have a big misunderstanding.

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user6703592
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1 Answers1

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This is just a partial answer in that it addresses the univariate questions.

The method in that textbook tells you to maximize the log of the likelihood so it is using maximum likelihood (using odd notation in my opinion as it seems to assume unjustifiably that $\mu$ and $\sigma$ are known and that it displays those two parameters that have nothing to do with a "standardized" t-distribution - maybe that's a typo in the book).

One can also use the method of moments where one equates the sample moments to the associated expected moments. For this distribution one could use $E(X)=\mu$, $Var(X)=\nu \sigma^2/(\nu-2)$ (for $\nu>2$), and $E[(X-\mu)^4]=\frac{3 \nu ^2 \sigma ^4}{(\nu -4) (\nu -2)}$ (for $\nu>2$). If $\bar{x}$ is the sample mean, $s^2$ is the sample variance, and $m_4$ is the sample fourth central moment, then $\hat{\mu}=\bar{x}$, $\hat{\sigma}^2=\frac{m_4 s^2}{2 m_4-3 s^4}$, and $\hat{\nu}=\frac{2 \left(2 m_4-3 s^4\right)}{m_4-3 s^4}$. (The distribution is symmetric about $\mu$ so the third central moment doesn't help us estimate the parameters. In other words, the expected third central moment is zero and therefore not dependent on any of the parameters.)

These values can be used as starting values for an iterative maximum likelihood procedure (although I think that $\bar{x}$ will be the maximum likelihood estimator for $\mu$).

JimB
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    The MLE for $\mu$ won't be $\bar{X}$. – Glen_b Aug 07 '19 at 22:35
  • @Glen_b Uh, oh. Good thing I said "I think" (even though I apparently wasn't thinking). – JimB Aug 07 '19 at 22:45
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    No problem. The heavier tail on the t makes sample means inefficient. – Glen_b Aug 07 '19 at 22:47
  • @Glen_b An extremely feeble comeback: $\bar{x}$ appears to be the maximum likelihood estimator for $\mu$ when the sample size is 1 or 2 (but no other sample size) - or at least the derivative of the log likelihood is zero when $\hat{\mu}=\bar{x}$ for those two sample sizes. – JimB Aug 07 '19 at 23:03
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    I really don't think so, but I am happy to be convinced otherwise. Can you say what led you to this conclusion (for either one or the other of those two)? – Glen_b Aug 07 '19 at 23:06
  • @Glen_b Mathematica: logL = LogLikelihood[ StudentTDistribution[\[Mu], \[Sigma], \[Nu]], {x[1]}]; FullSimplify[Solve[D[logL, \[Mu]] == 0, \[Mu]]] and logL = LogLikelihood[ StudentTDistribution[\[Mu], \[Sigma], \[Nu]], {x[1], x[2]}]; FullSimplify[Solve[D[logL, \[Mu]] == 0, \[Mu]]]. – JimB Aug 07 '19 at 23:08
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    Yes, the partial derivative of the log-likelihood would be zero (missed the last part of your comment before; just saw the comment about it being a max). For the second case when $\nu=1$ (or even just sufficiently small) and the two x-values are not close together, $\bar{x}$ is a local minimum in the $\mu$-direction. – Glen_b Aug 07 '19 at 23:19
  • Let us continue this discussion in chat. – JimB Aug 07 '19 at 23:45