This is a further question to my original question, where I did not get a helpful answer. I want to fit a t-distribution to my data, whose probability density function is: \begin{align*} f(l|\nu ,\mu ,\beta) = \frac{\Gamma (\frac{\nu+1}{2})}{\Gamma (\frac{\nu}{2}) \sqrt{\pi \nu} \beta} \left(1+\frac{1}{\nu}\left(\frac{l - \mu}{\beta}\right)^2 \right)^{-\frac{1+\nu}{2}} \end{align*}

First of all, I use maximum-likelihood estimation of the parameters (ML) with the following code (Fitting t distribtution to financial data):

# fit t distribution

fitdistr(alvsloss, "t")

# or

# log-likelihood function
loglik <-function(par){
if(par[2]>0 & par[3]>0) return(-sum(log(dt((alvsloss-par[1])/par[2],df=par[3])/par[2])))
else return(Inf)

# optimisation step

I get the following output:

   m               s               df     
  -0.0004919768    0.0130128873    2.6340459185 
 ( 0.0003182568) ( 0.0003453702) ( 0.1620424078)


[1] -0.0004451138  0.0129659465  2.6182237477

which is more or less the same, I guess differences are due to the precision of the numerical procedures.

Now I want to use the method of moments, based on this paper, where mean, variance and kurtosis are as follows:

\begin{align} \mu=&E(l)\\ \sigma^2 =& V(l)= E((l-\mu)^2)=\frac{\beta \nu}{\nu-2} , \nu>2\\ \kappa=&\frac{6}{\nu-4} , \nu > 4 \end{align}

my first questions:

  1. Why are they using the excess kurtosis and not the third moment, skewness?
  2. What values do I have to insert, is the following correct?:

mean: \begin{align*} \mu=E(l)=\bar{l} \end{align*} variance \begin{align*} \sigma^2 = V(l)= E((l-\mu)^2)=\frac{\beta \nu}{\nu-2} = \frac{1}{n}\sum_{i=1}^n (l_i-\bar{l})^2, \nu>2 \end{align*} excess kurtosis \begin{align*} \kappa=\frac{6}{\nu-4} = \frac{1}{n} \sum_{i=0}^n \left(\frac{l_i-\bar{l}}{s}\right)^4-3, \nu > 4 \end{align*}

this gives: \begin{align} \hat{\mu}_{MM}=&\bar{l}\\ \hat{\nu}_{MM} =& \frac{6}{\left(\frac{1}{n} \sum_{i=1}^n \left(\frac{l_i-\bar{l}}{s}\right)^4-3\right)} + 4\\ \hat{\beta}_{MM} =& \left(\frac{1}{n}\sum_{i=1}^n (l_i-\bar{l})^2\right) * \frac{(\hat{\nu}-2)}{\hat{\nu}} \end{align}

so I am using the sample mean, sample variance and sample excess kurtosis. Is this correct?

And my main question: The output of ML tells me (the column df), that $\nu$<4 ($\nu$ is the number of degrees of freedom, df), but in MM I need $\nu$ to be greater than 4 or? So what does this mean? Is MM not usable? Or does it not matter?

  • 3
    $\begingroup$ Skewness is zero for the $t$-distribution, so this particular moment is not informative for any of its parameters. $\endgroup$
    – StasK
    Commented Mar 22, 2013 at 18:34
  • 2
    $\begingroup$ The repec.org link is broken. This is why it's important to give full references! There's some hope of finding it with a full reference, but with broken links there's often little one can do $\endgroup$
    – Glen_b
    Commented Jun 27, 2017 at 1:41
  • $\begingroup$ @Glen_b You are right the link is broken. Thanks for the hint. Yes, better to give full reference, I will do so in the future. $\endgroup$ Commented Jul 16, 2017 at 10:25
  • $\begingroup$ the link to the article still hasn't been fixed. are $\sigma^2_{MM}$, $\nu_{MM}$, $\kappa_{MM}$, etc written in the question the correct derivations for the t-distribution parameters following the method of moments? $\endgroup$
    – develarist
    Commented Dec 7, 2020 at 17:31

1 Answer 1


In your previous question it was already noted that

First of all, for the MM to work, you will need to have higher order moments to ensure that the sums necessary for the MM converge.

In this case the MLE indicates that the $\nu<3$. Then, it does not makes sense at all to use the method of moments. The method of moments is very restrictive and, in this case, the MLE approach is giving you a good fit.

If you want to consider an alternative method you could use Bayesian inference.


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