FOCE objective function

2 messages 2 people Latest: Nov 13, 2001

FOCE objective function

From: Matt Hutmacher Date: November 12, 2001 technical
From: "HUTMACHER, MATTHEW [Non-Pharmacia/1820]" <matthew.hutmacher@pharmacia.com> Subject: [NMusers] FOCE objective function Date: Mon, 12 Nov 2001 15:02:44 -0600 Hello all, I am sure I am missing something simple here, but could someone explain to me where the last term in the objective function written on page 4 (Chapter II) of Manuel 7 comes from? It is the term, "-(0.5*G+O^(-1)*E)`(O^-1+0.5*H)^(-1)(0.5*G+O^-1*E)" where G=gradient, O=omega, and H=hessian. I was just working through the derivation for my on edification and I could only get the first four pieces. Using the notation from Chapter I and suppressing the subscript i, then l(psi,eta) is the conditional and arbitrary likelihood and h(eta;O) is the multivariate normal distribution with mean 0 and covariance matrix O. Let M=l(psi,eta)h(eta;O) and m()=log{M()}=log{l(psi,eta)}+ log{h(eta:O)}. By the general Laplacian approximation method, the marginal likelihood is represented by the integral (let INT stand for taking the integral), INT{exp(m)*d(eta)}=k*|-m"(etahat)|^(-0.5)*exp(m(etahat)), where etahat maximizes m() and k is a constant. When I take -2log{INT{exp(m)*d(eta)}}, I get --> -2log{l(psi,eta)}+log|O|+log|O^(-1)+0.5*H|+etahat`O^(-1)etahat, where H=l"(psi,eta). Could someone please explain to me my error? Thanks for your time. Matt

RE: FOCE objective function

From: Lars Erichsen Date: November 13, 2001 technical
From: "LSEN (Lars Erichsen)" <lsen@novonordisk.com> Subject: RE: [NMusers] FOCE objective function Date: Tue, 13 Nov 2001 12:47:41 +0100 In the usual Laplace approximation of the integral, the first derivative of the exponent is 0 in etahat, because etahat maximises the exponent. This is not assumed in the derivation in the NONMEM manual Part VII on page 4 which is why the last term is not zero. The integral is L=Int[exp(-0.5*(Phi(eta)+eta'*O*eta))] where O=inverse of Omega If etahat minimises Phi(eta)+eta'*O*eta we have that (Gamma=derivative of Phi): Gamma(etahat)+2*O*etahat=0, so that 0.5*Gamma(etahat)+O*etahat=0. I hope this is of use. Lars Erichsen Pharmacokinetics, Pharmacometrics Novo Nordisk Denmark.