"James K." <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] > This question is reposted since there was no answer, slightly > including one more point. > > On Fri, 28 Nov 2003 19:29:37 +0900, "James K." > <[EMAIL PROTECTED]> wrote: > > >Tell me the pdf of bigger one in two pigs, if we assume the pdf of each > >pig's weight is all Gaussian pdf with same mean (m) and variance (s^2). The > >weight value of bigger one is denoted by > > > > [EQ-1] w >= w1 >= w2, > > > >where w1 and w2 are the weights of two pigs, respectively. > > > >Base on the assumption, we have > > [EQ-2] E[w1] = E[w2] = m, > > [EQ-3] E[(w1-m)^2] = E[(w2-m)^2] = s^2, > > > >then, I guess, the mean and variance of w become > > [EQ-4] m(w) = m + s/Sqrt(Pi), > > [EQ-5] s(w)^2 = (-1+Pi)*s^2/Pi, > > > >while we are considering only the case s > 0, m > 0. > > My question is divided some particles, which are: > 1. Shall I be in correct way? > 2. If yes, are new mean and variance correct? > 3. After having these new mean and variance, is the pdf (new one) > still Gaussian? > > >BR, > > James K. ([EMAIL PROTECTED]) > - Any remarks, proposal and/or indicator would be respected. > - Private opinions: These are not the opinions from my affiliation. > [Home] http://myhome.naver.com/txdiversity
If f(x) is the p.d.f. of the normal distribution with mean m and variance s^2, then the pdf of the larger of two is 2.f(x).F(x) where F(x) is the probability that the other weight is less than x. This distribution is NOT Gaussian. Cheers -- Alan Miller http://users.bigpond.net.au/amiller Retired Statistician . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
