Como os dados são contínuos seria necessário considerar uma distribuição contínua com mistura Bernoulli, veja uma proposta aqui: http://jupiter.est.ufmg.br/~posgrad/mestrado/dissertacao_mestrado_zaida%20_q... e um exemplo no script abaixo: ## hurdle model for zero-gamma (zero inflation and continuous outcome) require(INLA) ### rainfall data from Parana' state in Brazil data(PRprec) ### analize data from one location PRprec[301,1:3] y <- as.matrix(PRprec[,-(1:3)])[301,] summary(y) plot.ts(y) if (TRUE) dev.off() ### organize the dataset to build a model for ### smoothing the rainfall amount over the year ### accounting for a smoothed zero probability as well n <- length(y) dat <- list(resp=cbind( c((y>0)+0, rep(NA,n)), c(rep(NA,n), ifelse(y>0,y,NA))), t0=c(1:n, rep(NA,n)), t1=c(rep(NA,n), 1:n)) ### two choices to smooth the rainfall amount form1 <- resp ~ 0 + f(t0, model='rw2', cyclic=TRUE) + f(inla.group(t1, n=73), model='rw1', cyclic=TRUE) form2 <- resp ~ 0 + f(t0, model='rw2', cyclic=TRUE) + f(t1, model='rw2', cyclic=TRUE) inla.setOption(scale.model=TRUE) res1 <- inla(form1, family=c('binomial', 'gamma'), data=dat) res2 <- inla(form2, family=c('binomial', 'gamma'), data=dat) day <- seq(ISOdate(2011,1,1), ISOdate(2011,12,31), 24*3600) par(mfrow=c(2,1), mar=c(3,3,1,1), mgp=c(2,1,0)) plot(day, y>0) for (j in 4:6) { lines(day, 1/(1+exp(-res1$summary.ran$t0[,j])), lty=2) lines(day, 1/(1+exp(-res2$summary.ran$t0[,j])), lty=2, col=2) } plot(day, y, type='l') for (j in 4:6) { lines(day[seq(3,n,5)], exp(res1$summary.ran[[2]][,j]), lty=2) lines(day, exp(res2$summary.ran[[2]][,j]), lty=2, col=2) } Elias