2
$\begingroup$

This is a further question to my original question, where I did not get an helpful answer (at leas not helpful for me) :Methods of moments for t distribution

I want to fit a t distribution to my data (data can be found here: http://uploadeasy.net/upload/bo9j6.rar). The probability density function is given by: \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 ML with the following code (Fitting t distribtution to financial data):

# fit t distribution
library(MASS)

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
optim(c(0,0.1,2.5),loglik)

I get the following output:

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

and

$par
[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 do this with using the method of moments, I use the following part of a paper (http://ideas.repec.org/a/eme/jrfpps/v7y2006i3p292-300.html):

mean: \begin{align*} \mu=E(l) \end{align*} variance \begin{align*} \sigma^2 = V(l)= E((l-\mu)^2)=\frac{\beta \nu}{\nu-2} , \nu>2 \end{align*} excess kurtosis \begin{align*} \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 the 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?

$\endgroup$
  • 1
    $\begingroup$ Skewness is zero for the $t$-distribution, so this particular moment is not informative for any of its parameters. $\endgroup$ – StasK Mar 22 '13 at 18:34
  • 1
    $\begingroup$ 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 Jun 27 '17 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. $\endgroup$ – Stat Tistician Jul 16 '17 at 10:25
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.