R help - Feb 2006 - Fitdistr and MLE for parameter lambda of Poisson distribution

Hello!

I would like to get MLE for parameter lambda of Poisson distribution. I 
can use fitdistr() for this. After looking a bit into the code of this 
function I can see that value for lambda and its standard error is 
estimated via

estimate <- mean(x)
sds <- sqrt(estimate/n)

Is this MLE? With my poor math/stat knowledge I thought that MLE for 
Poisson parameter is (in mixture of LaTeX code)

l(\lambda|x) \propto \sum^n_{i=1}(-\lambda + x_iln(\lambda)).

Is this really equal to (\sum^n_{i=1} x_i) / n

-- 
Lep pozdrav / With regards,
     Gregor Gorjanc

----------------------------------------------------------------------
University of Ljubljana     PhD student
Biotechnical Faculty
Zootechnical Department     URI: http://www.bfro.uni-lj.si/MR/ggorjan
Groblje 3                   mail: gregor.gorjanc <at> bfro.uni-lj.si 

SI-1230 Domzale             tel: +386 (0)1 72 17 861
Slovenia, Europe            fax: +386 (0)1 72 17 888 

----------------------------------------------------------------------
"One must learn by doing the thing; for though you think you know it,
  you have no certainty until you try." Sophocles ~ 450 B.C.

Gregor Gorjanc <gregor.gorjanc at bfro.uni-lj.si> writes:
> Hello!
> 
> I would like to get MLE for parameter lambda of Poisson distribution. I 
> can use fitdistr() for this. After looking a bit into the code of this 
> function I can see that value for lambda and its standard error is 
> estimated via
> 
> estimate <- mean(x)
> sds <- sqrt(estimate/n)
> 
> Is this MLE? With my poor math/stat knowledge I thought that MLE for 
> Poisson parameter is (in mixture of LaTeX code)
> 
> l(\lambda|x) \propto \sum^n_{i=1}(-\lambda + x_iln(\lambda)).
> 
> Is this really equal to (\sum^n_{i=1} x_i) / n
Yes....

Maximizing l(lambda) is the same as maximizing

sum(x)/n ln lambda - lambda

Now either take the derivative and set equal to zero, or
 
rewrite further as equivalent to

ln (lambda/(sum(x)/n)) -  (lambda/(sum(x)/n))

and notice that ln(x) - x has a global maximum at x=1 (since ln is
strictly concave and the tangent at x=1 is the line y = x - 1)
 

(I think this is in the first 20 pages I ever read on theoretical
statistics ...)
> -- 
> Lep pozdrav / With regards,
>      Gregor Gorjanc
> 
> ----------------------------------------------------------------------
> University of Ljubljana     PhD student
> Biotechnical Faculty
> Zootechnical Department     URI: http://www.bfro.uni-lj.si/MR/ggorjan
> Groblje 3                   mail: gregor.gorjanc <at> bfro.uni-lj.si 
> 
> SI-1230 Domzale             tel: +386 (0)1 72 17 861
> Slovenia, Europe            fax: +386 (0)1 72 17 888 
> 
> ----------------------------------------------------------------------
> "One must learn by doing the thing; for though you think you know it,
>   you have no certainty until you try." Sophocles ~ 450 B.C.
> 
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
> 
-- 
   O__  ---- Peter Dalgaard             ??ster Farimagsgade 5, Entr.B
  c/ /'_ --- Dept. of Biostatistics     PO Box 2099, 1014 Cph. K
 (*) \(*) -- University of Copenhagen   Denmark          Ph:  (+45) 35327918
~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk)                  FAX: (+45) 35327907

Peter Dalgaard wrote:
> Gregor Gorjanc <gregor.gorjanc at bfro.uni-lj.si> writes:
> 
> 
>>Hello!
>>
>>I would like to get MLE for parameter lambda of Poisson distribution. I 
>>can use fitdistr() for this. After looking a bit into the code of this 
>>function I can see that value for lambda and its standard error is 
>>estimated via
>>
>>estimate <- mean(x)
>>sds <- sqrt(estimate/n)
>>
>>Is this MLE? With my poor math/stat knowledge I thought that MLE for 
>>Poisson parameter is (in mixture of LaTeX code)
>>
>>l(\lambda|x) \propto \sum^n_{i=1}(-\lambda + x_iln(\lambda)).
>>
>>Is this really equal to (\sum^n_{i=1} x_i) / n
> 
> 
> Yes....
> 
> Maximizing l(lambda) is the same as maximizing
> 
> sum(x)/n ln lambda - lambda
> 
> Now either take the derivative and set equal to zero, or
>  
> rewrite further as equivalent to
> 
> ln (lambda/(sum(x)/n)) -  (lambda/(sum(x)/n))
> 
> and notice that ln(x) - x has a global maximum at x=1 (since ln is
> strictly concave and the tangent at x=1 is the line y = x - 1)
>  
> 
> (I think this is in the first 20 pages I ever read on theoretical
> statistics ...)
Thank you very much for this. It shows, how much I still need to learn.

-- 
Lep pozdrav / With regards,
    Gregor Gorjanc

----------------------------------------------------------------------
University of Ljubljana     PhD student
Biotechnical Faculty
Zootechnical Department     URI: http://www.bfro.uni-lj.si/MR/ggorjan
Groblje 3                   mail: gregor.gorjanc <at> bfro.uni-lj.si

SI-1230 Domzale             tel: +386 (0)1 72 17 861
Slovenia, Europe            fax: +386 (0)1 72 17 888

----------------------------------------------------------------------
"One must learn by doing the thing; for though you think you know it,
 you have no certainty until you try." Sophocles ~ 450 B.C.

Bernardo Rangel tura wrote:
> At 09:35 AM 2/10/2006, Gregor Gorjanc wrote:
> 
>> Hello!
>>
>> I would like to get MLE for parameter lambda of Poisson distribution. I
>> can use fitdistr() for this. After looking a bit into the code of this
>> function I can see that value for lambda and its standard error is
>> estimated via
>>
>> estimate <- mean(x)
>> sds <- sqrt(estimate/n)
>>
>> Is this MLE? With my poor math/stat knowledge I thought that MLE for
>> Poisson parameter is (in mixture of LaTeX code)
>>
>> l(\lambda|x) \propto \sum^n_{i=1}(-\lambda + x_iln(\lambda)).
>>
>> Is this really equal to (\sum^n_{i=1} x_i) / n
>>
>> -- 
>> Lep pozdrav / With regards,
>>      Gregor Gorjanc
> 
> 
> Gregor,
> 
> If I understood your LaTeX You is rigth.
> 
> If you don??t know have a command wich make this for you:  fitdistr()
> 
> Look:
> 
> 
>> d<- rpois(50,5)
>> d
>  [1]  6  4  6  4  5  5  4 11  7  5  7  3  5 10  4  9  4  2  4  5  4  4 
> 9  3 10
> [26]  4  3  9  6  7  5  4  2  7  3  6  7  8  6  6  3  3  3  2  5  4  3 
> 8  5  7
>> library(MASS)
>> fitdistr(d,"Poisson")
>     lambda
>   5.3200000
>  (0.3261901)
Thanks for this, but I have already said in the first mail, that
fitdistr can help me with this. I was just "surprised" or knowledge
undernourished, that there is closed form solution for this. Look into
the source of fitdistr.

-- 
Lep pozdrav / With regards,
    Gregor Gorjanc

----------------------------------------------------------------------
University of Ljubljana     PhD student
Biotechnical Faculty
Zootechnical Department     URI: http://www.bfro.uni-lj.si/MR/ggorjan
Groblje 3                   mail: gregor.gorjanc <at> bfro.uni-lj.si

SI-1230 Domzale             tel: +386 (0)1 72 17 861
Slovenia, Europe            fax: +386 (0)1 72 17 888

----------------------------------------------------------------------
"One must learn by doing the thing; for though you think you know it,
 you have no certainty until you try." Sophocles ~ 450 B.C.

At 09:35 AM 2/10/2006, Gregor Gorjanc wrote:
>Hello!
>
>I would like to get MLE for parameter lambda of Poisson distribution. I
>can use fitdistr() for this. After looking a bit into the code of this
>function I can see that value for lambda and its standard error is
>estimated via
>
>estimate <- mean(x)
>sds <- sqrt(estimate/n)
>
>Is this MLE? With my poor math/stat knowledge I thought that MLE for
>Poisson parameter is (in mixture of LaTeX code)
>
>l(\lambda|x) \propto \sum^n_{i=1}(-\lambda + x_iln(\lambda)).
>
>Is this really equal to (\sum^n_{i=1} x_i) / n
>
>--
>Lep pozdrav / With regards,
>      Gregor Gorjanc
Gregor,

If I understood your LaTeX You is rigth.

If you don??t know have a command wich make this for you:  fitdistr()

Look:


 > d<- rpois(50,5)
 > d
  [1]  6  4  6  4  5  5  4 11  7  5  7  3  5 
10  4  9  4  2  4  5  4  4  9  3 10
[26]  4  3  9  6  7  5  4  2  7  3  6  7  8  6  6  3  3  3  2  5  4  3  8  5  7
 > library(MASS)
 > fitdistr(d,"Poisson")
     lambda
   5.3200000
  (0.3261901)






[]s
Tura

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?

Thanks much!
Ankush

--
This message was sent on behalf of ankush2kn at yahoo.com at openSubscriber.com
http://www.opensubscriber.com/message/r-help at stat.math.ethz.ch/3316818.html

On Oct 26, 2009, at 11:25 PM, ankush2kn at 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

What is wrong with using
mean(x)
to get the MLE of the poisson lambda?

Kjetil

On Tue, Oct 27, 2009 at 9:17 AM, David Winsemius <dwinsemius at
comcast.net> wrote:
>
> On Oct 26, 2009, at 11:25 PM, ankush2kn at 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
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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 <dwinsemius at
comcast.net> wrote:
>> On Oct 26, 2009, at 11:25 PM, ankush2kn at 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
>>
>> ______________________________________________
>> R-help at r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
> 
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.