Thanks, this was the information I was looking for. The likelihood function is in terms of lambda1 and lambda2 only, but the quantity of interest to me is eta = (lambda1 - lambda2).
-dw [EMAIL PROTECTED] (Jason Owen) wrote in message news:<[EMAIL PROTECTED]>... > First question: by invariance of MLEs, yes. However, it > seems like your likelihood function is overparameterized > (i.e. the likelihood is really only a function of two distinct > parameters) and the MLEs you find will not be unique. > > Why can't you express the likelihood with only lambda1 and > lambda2 and find the MLEs numerically? > > > [EMAIL PROTECTED] (driver_writer) wrote in message news:<[EMAIL PROTECTED]>... > > If I had three parameters: eta, lambda1, lambda2 > > s.t. eta = (lambda1 - lambda2), > > then is it true that the MLE(eta) = MLE(lambda1) - MLE(lambda2)? > > > > Here the MLE (Maximum Likelihood Estimate) is derived from > > d(lambda1|y)/d(lambda1) = 0. > > Also, assume an average sample size (not too small, not too large). > > > > The reason I am asking this is because it is a little difficult to > > manipulate the likelihood expression which is terms of lambda1 and > > lambda2 to be an expression of eta. > > > > Any help would be greatly appreciated. > > > > -dw . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
