mike anderson wrote in message ... >x is chi square with 2m d.o.f. what is the correlation between x and lnx? >thanks > >-- >Mike
This message has appeared previously on another newsgroup (sci.stat.math) and received responses from myself and from Colin Rose. My response was an approximation based upon simulation, but it turned out to be quite accurate except for very small m. My approx. was correlation = 1 - 1/(4m) I have consulted Gradshteyn & Ryzhik, `Tables of integrals, series, and products', 5th edition. Sections 4.352 and 4.358 contain some integrals which are useful here. For simplicity, let us write Y = ln(X), then: E(X) = 2m variance(X) = 4m E(Y) = psi(m) - ln(0.5) E(Y^2) = [psi(m) - ln(0.5)]^2 + zeta(2,m) Hence variance(Y) = zeta(2,m) E(X.Y) = 2m.[psi(m+1) - ln(0.2)] psi is Euler's psi function = 1st derivative of the log of the gamma function. zeta is Riemann's zeta function. This looks different from Colin's exact solution, but I guess that it evaluates to the same values. 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/ . =================================================================
