You can use the incomplete gamma function. If Y~Poisson(lambda), then
prob(Y<=k)=Q(k+1, lambda), where Q is the upper incomplete gamma
function.
So in Stata code the cumulative probability prob(Y<=k) is
1-gammap(k+1, lambda)
Jamie Griffin
>>> [email protected] 08/30/06 12:14 am >>>
Dear all,
The MLE for poisson function is L= exp(-lambda)Lambda^Y/Y!
I found the following code for poisson on a website (by David Todd) as
follows:
program define poisreg2
args lnf theta
quietly replace `lnf' = -exp(`theta') +
$ML_y1*(`theta')-lnfact($ML_y1)
end
Now, I want to truncate the distribution. The new MLE function is
L2 = L/prob(y<=ymax).
i.e. divide the likelihood function by prob(y<=ymax).
i.e subtract the log function by CDF_POISSON(lambda,ymax).
For that I need a poisson cumulative distribution function which I am
not able to find out.
What is the command for finding a cumulative distribution function? It
should use two parameters and should be something like, f=
poisson(lambda,ymax)
Thanks,
Prabhu
*
* For searches and help try:
* http://www.stata.com/support/faqs/res/findit.html
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
*
* For searches and help try:
* http://www.stata.com/support/faqs/res/findit.html
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
