>>>>> "l" == lagreene <lagreene...@gmail.com>
>>>>> on Fri, 15 May 2009 04:22:59 -0700 (PDT) writes:
l> Thanks Jorge,
l> but I still don't understand where they come from. when I use:
l> fitdistr(mydata, "t", df = 9) and get values for m and s, and the
variance
l> of my data should be the df/s?
definitely *not*; How did you get to this completely wrong formula?
l> I jsut want to be able to confirm how m and s are calculated
by maximum likelihood.
And, of course, only for the normal (aka Gaussian) are the ML
estimates of mu the artithmetic mean and of sigma (n-1)/n * sd(x)
{i.e. even *there* the ML estimate of s is *not* the SD}
As you can read on ?dt,
the variance of a (0,1)-t-distribution is df / (df - 2)
and hence only defined for df > 2.
Consequently, the variance of a (mu,sigma)-t-distribution is
sigma^2 * df / (df - 2)
l> mydt <- function(x, m, s, df) dt((x-m)/s, df)/s
l> fitdistr(x2, mydt, list(m = 0, s = 1), df = 9, lower = c(-Inf, 0))
{this is copy-pasted from example(dt);
the examples have nice comments there....}
l> Jorge Ivan Velez wrote:
>>
>> Dear lagreene,
>> See the second example in
>>
>> require(MASS)
>> ?fitdistr
>>
>> HTH,
>>
>> Jorge
>>
>>
>> On Thu, May 14, 2009 at 7:15 PM, lagreene <lagreene...@gmail.com> wrote:
>>
>>>
>>> Hi,
>>> I was wondering if anyone could tell me how m and s are calculated for a
>>> t
>>> distribution?
>>>
>>> I thought m was the sample mean and s the standard deviation- but
>>> obviously
>>> I'm wrong as this doesn'y give the same answer.
>>>
>>> Thank you
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