Hi I'm facing lot of problems calculating the scaled deviance. generally, as per the text this is calculated as 2(Ls-Lm) Here Lm is the same as current model of likelihood so no need to change but how to calculate Ls. Q) An insurance portfolio consists of m group of individuals. in the ith group there ar en individuals aged xi. the no of claims from these ni individuals is a binomial random variable Yi with parameters ni and thetai(0<thetai<1). The random variables are independent 1) Derive MLE of theta1, theta2.... 2) If ln(thetai/(1-thetai))=a+bxi, calculate log likelihood and show that it is a*summation(yi)(i=1 to m)+b*(summation xiyi)((i=1 to m)- ni*ln(1+exp(a+bxi))+c wehre c doesn not depend on a or b 3) Derive the scaled deviance of model 2 Part 1 and 2 are easy but scaled deviace solution is saying it is 2*(part 1 ans-part 2). how come????
Scaled deviance is: -2[ log L(S) - log L(M) ] So, we are looking at the difference between the likelihood function of the saturated model and the model in question. Model 1 is the saturated model So, the scaled deviance of Model 2 is calculated, exactly as you have described, John
