Hi All, I just wanted to make sure I am understanding this correctly but generally when we are asked to find the scaled deviance, for the saturated model, we put in all the y's where the mu's/theta's are in the LogL. In this question, we found an estimate of theta, which was y/n, so instead of putting in y for the scaled deviance, we put in y/n. This was done because we actually found an estimate for the theta right? If we hadn't found an estimate for theta, would we have used y's in the saturated model?
Yes you are correct, under the saturated model every point has it's own theta since it fits the data perfectly (this will be the first term for scaled deviance). The second term is the model being considered, the MLE estimates go straight into the formula. I don't understand what you mean by your last comment, if you havn't estimated theta, you don't even have a model to consider.. You need an estimate of theta inorder to compare it to the "perfect" model to see how far off it is.
Let me try to explain further. This was a binomial question with n and theta as the parameters of the model. We were asked to find a MLE for theta which turned out to be y/n. We were also given the logit(theta) = alpha + beta (x). We also found MLE's for alpha and beta. Then we were asked for the deviance of the model. The original likelihood starts ylntheta, so instead of saying ylny, we put yln(y/n). So I was just wondering if we did this to because we found out earlier that the MLE for theta was y/n. Thanks Mr. Matthews
