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Gary Weissman posted an excellent illustration of how we may calculate the maximum Brier score, which we need for the scaled Brier score. Three approaches were found to be equivalent, while the intuition for the maximum score may be best for the first formulation, as discussed at |Twitter.

scaled_brier_score_1 <- function(obs, pred) { 1 - (brier_score(obs, pred) / brier_score(obs, mean(obs))) } scaled_brier_score_2 <- function(obs, pred) { 1 - (brier_score(obs, pred) / (mean(obs) * (1 - mean(obs)))) } scaled_brier_score_3 <- function(obs, pred) { 1 - (brier_score(obs, pred) / (mean(obs) * (1 - mean(obs))^2 + (1 - mean(obs)) * mean(obs)^2)) }

# Jan 07, Ewout Steyerberg # function to perform Hosmer-Lemeshow test for external validation # instead of chi square and corresponding p-value, this function provides the number of subjects per group, # and the mean values of p and y per group. ########################## # p : predicted probability # Y : outcome variable # g : number of groups to calculate H-L (10 is default) # # NB: the library Hmisc need to be attached in order to be able to run hl.ext2 ########################## hl.ext2<-function(p,y,g=10, df=g-1) { matres <-matrix(NA,nrow=g,ncol=5) sor <-order(p) p <-p[sor] y <-y[sor] groep <-cut2(p,g=g) #g more or less equal sized groups len <-tapply(y,groep,length) #n per group sump <-tapply(p,groep,sum) #expected per group sumy <-tapply(y,groep,sum) #observed per group meanp <-tapply(p,groep,mean) #mean probability per group meany <-tapply(y,groep,mean) #mean observed per group matres <-cbind(len,meanp,meany, sump, sumy) #matrix for results contr<-((sumy-sump)^2)/(len*meanp*(1-meanp)) #contribution per group to chi square chisqr<-sum(contr) #chi square total pval<-1-pchisq(chisqr,df) #p-value corresponding to chi square with df degrees of freedom cat("\nChi-square",chisqr," p-value", pval,"\n") dimnames(matres) <-list(c(1:g),Cs(n,avg(p),avg(y), Nexp, Nobs)) result <- list(table(groep), matres,chisqr,pval) }