Of course, if p1=p2 the answer is Binomial(n1+n2,p1).
Otherwise, there is no "easy" answer (i.e., no standard distribution).
Jon Cryer
At 05:45 AM 2/5/01 GMT, you wrote:
>Kumara Sastry wrote in message <[EMAIL PROTECTED]>...
>>Suppose, X ~ Binomial(n1,p1), Y~Binomial(n2,p2) , and X and Y are
>>independent. Also, Z = X+Y. Can anyone please comment on what the pdf
>>of Z is?
>>
>>Thanks
>>Kumara
>>
>
>Pr(Z=z) = Sum from l to u of p1(r).p2(z-r)
>
>where p1 is the first binomial probability, and p2 is the second.
>The upper & lower limits of summation, l & u, are not necessarily
>0 and z but:
>l = max(0, z-n2) and u = min(z, n1)
>
>I hope I have got that right.
>I doubt if the sum simplifies much.
>
>
>--
>Alan Miller, Retired Scientist (Statistician)
>CSIRO Mathematical & Information Sciences
>Alan.Miller -at- vic.cmis.csiro.au
>http://www.ozemail.com.au/~milleraj
>http://users.bigpond.net.au/amiller/
>
>
>
>
>
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