Hi all,
If x1 & x2 are iid Poison (lambda1,lambda2) so, x1+x2 is Poisson
(lambda1+lambda2)
So you can generate 3 independant Poisson variables
X, Y, T with parameter lX, lY & lT
X+T and Y+T is possion and correlation is :
Correlation (X+T,Y+T)=lT/sqrt((lT+lX)*(lT+lY))
you will have to resolve
Final lambda1(known)=lX+lT
Final lambda2(known)=lY+lT
Correlation(known)=lT/sqrt((lT+lX)*(lT+lY))
Best regards
Naji
Le 15/02/05 22:55, ��[email protected]�� <[email protected]> a �crit�:
> In a message dated 2/15/2005 2:34:44 PM US Mountain Standard Time,
> [email protected] writes:
>> How can I generate a correlated Poisson sample in Stata? And the data
>> structure is like the following:
>>
>> Obs time1 time2 time3 time4
>> 1 2 4 5 2
>> 2 1 3 5 6
>> 3 2 3 1 5
>> 4 0 6 4 1
>> ...
>
>
> One of the random number generators created by Linde-Zwirble and myself
> about 10 years ago published in the STB included a Poisson with defined
> correlation. Lookup -rnd- for details.
>
> Joe Hilbe
>
> *
> * 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/
