Ankush Bhargava wrote:
Thank you everyone for responding.
David,
3.75 in my example was equivalent to the mean of the values, which i thought
was too much a coincidence...
What do you think the significance of "(0.03343)". What is this value?
Kjetil,
Are you saying that mean(x) is same as the MLE for the poisson lambda?
Yes, the mle for lambda is mean(x) [easy to derive from the
likelihood function] and the estimated sd for the mle is
sqrt(mean(x)/length(x)) which is 0.03343 in your case.
Cheers,
-Peter Ehlers
Thanks again!
Ankush
________________________________
From: Peter Ehlers <ehl...@ucalgary.ca>
To: Kjetil Halvorsen <kjetilbrinchmannhalvor...@gmail.com>
Cc: r-h...@stat.math.ethz.ch; ankush...@yahoo.com
Sent: Tue, October 27, 2009 10:15:31 AM
Subject: Re: [R] MLE for lambda of Poisson distribution using fitdistr
Kjetil Halvorsen wrote:
What is wrong with using
mean(x)
to get the MLE of the poisson lambda?
and
mean(x)/length(x)
to get its estimated variance.
-Peter Ehlers
Kjetil
On Tue, Oct 27, 2009 at 9:17 AM, David Winsemius <dwinsem...@comcast.net> wrote:
On Oct 26, 2009, at 11:25 PM, ankush...@yahoo.com wrote:
Hi,
I am using the fitdistr of MASS to get the MLE for the lambda of a Poisson
distribution.
When i run the fitdistr command, i get an output that looks like -
lambda
3.750000
(0.03343)
Couple of questions -
1. is the MLE 0.03343 for the lambda of the given distribution then?
2. How would I calculate the variance of the MLE for the lambda?
It would be more typical statistical usage to have output of the form:
estimate ( se(estimate) ) .... so I was expecting 3.75 to be the estimate.
Looking at the help page and running str(.) on the fitdistr object of the
first example confirms my expectations. Why did you think the help page was
suggesting otherwise?
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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______________________________________________
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