If X is chi-square with df = m, then E(Xf(x)) = 2G(n/2+1)E(f(Y))/G(n/2), where Y is chi-square with df = m+2. G(.) denotes the gamma function. To find, E(ln(Y)), look Handbook of Mathematical Functions by Abromowitz and Stegun .
Kris ----- Original Message ----- From: Alan Miller <[EMAIL PROTECTED]> To: <[EMAIL PROTECTED]> Sent: Thursday, April 18, 2002 12:33 AM Subject: Re: Urgent!!! (Corrln. of X & ln(X)) > mike anderson wrote in message ... > >x is chi square with 2m d.o.f. what is the correlation between x and lnx? > >thanks > > > >-- > >Mike > > > > > Continuing from my previous response, I find that the correlation is: > 1/sqrt[m.zeta(2,m)] > That looks nice and simple, doesn't it? - until you start looking for > Riemann's zeta function, or what some authors call the generalized zeta > function (it has 2 arguments). > Now > zeta(2,m) = sum of squares of reciprocals of m, m+1, m+2, m+3, ... > This sum converges but painfully slowly. > However, it is known that zeta(2,1) = pi^2 / 6 = 1.644934067.. > Hence for small m, you can just subtract 1, 1/4, 1/9, 1/16, ... 1/(m-1)^2 > from this. > Finally this gives the numerical values which Colin Rose gave about 2 days > ago! > > Cheers > -- > Alan Miller (Honorary Research Fellow, CSIRO Mathematical > & Information Sciences) > http://www.ozemail.com.au/~milleraj > http://users.bigpond.net.au/amiller/ > > > > . > . > ================================================================= > Instructions for joining and leaving this list, remarks about the > problem of INAPPROPRIATE MESSAGES, and archives are available at: > . http://jse.stat.ncsu.edu/ . > ================================================================= > . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
