| [ QuizWit ] in KIDS 글 쓴 이(By): pinkrose (mirantz) 날 짜 (Date): 1998년 12월 3일 목요일 오전 08시 36분 07초 제 목(Title): [a3] poisson distribution. Oh yet another challenger from POSTECH? Are you the same idiot who posted the linear algebra problem? I will show why my previous posting[a2] is valid if a single phrase is added. I admit my mistake of not stating the actual probability space. ------------------------------------------------------------------------------- ----------- Suppose X is poisson(r1) ,Y is poisson(r2) and both of them independent. Assume further W.L.O.G. r1 - r2 > 0. then X - Y is not poisson in the probability space S= {0,1,2,3,4,5,...} which is the usuall probability measure space for poisson variable. However, if you define the characteristic function I_A (x) on a set A such that I_A(x) = 1 if x is in A 0 otherwise then (X-Y)*I_A is poisson distributed. ,where A={X>Y} is the subset of S^2. proof: Trivial exercise on conditional probability. THEREFORE, X - Y is poisson on the probabilty measure space A. HENCE, my previous posting is perfectly valid if the probability measure space is defined as A. This method of defining in a smaller space is what you might call a conditional probability in a basic probability course. Furthermore, if you generate outputs using computer , then each random sample X - Y is not poisson distributed because you can possibly have negative values, HOWEVER, if you get rid of all negative numbers, then the remaining random samples are poisson distributed. For those of avid fans who were confused rading my previous [a2] postings, please adviced that X - Y is poisson in the probability space A. Exercise: Find the distribution of |X-Y |. Hint: |X-Y| = (X-Y)*I_A + (Y-X)*I_A' where A' is a complement of A. ---------------------------------------------------- Wow... this is neater than the previous posting. Hmm.... I will put this problem in the exam. plese kids, don't even think about challenging me and try to learn something from my posting. okay? |