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In our last note we mentioned the possibility of “fully calibrated models.” This note is an example of a probability model that is calibrated in the traditional sense, but not fully calibrated in a finer grained sense.

First let’s attach our packages and generate our example data in R.

library(wrapr)
d <- build_frame(
"x1"  , "x2", "y"   |
1   , 1   , TRUE  |
1   , 0   , FALSE |
1   , 0   , TRUE  |
1   , 1   , FALSE |
0   , 0   , TRUE  |
0   , 1   , TRUE  |
0   , 1   , FALSE |
0   , 0   , FALSE |
0   , 0   , TRUE  )
# cat(wrapr::draw_frame(d))

knitr::kable(d)
x1 x2 y
1 1 TRUE
1 0 FALSE
1 0 TRUE
1 1 FALSE
0 0 TRUE
0 1 TRUE
0 1 FALSE
0 0 FALSE
0 0 TRUE

Now, we fit our logistic regression model.

model <- glm(y ~ x1 + x2,
data = d,
family = binomial())
summary(model)
##
## Call:
## glm(formula = y ~ x1 + x2, family = binomial(), data = d)
##
## Deviance Residuals:
##     Min       1Q   Median       3Q      Max
## -1.4213  -1.2572   0.9517   1.0996   1.2572
##
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)
## (Intercept)   0.5572     1.0784   0.517    0.605
## x1           -0.3715     1.3644  -0.272    0.785
## x2           -0.3715     1.3644  -0.272    0.785
##
## (Dispersion parameter for binomial family taken to be 1)
##
##     Null deviance: 12.365  on 8  degrees of freedom
## Residual deviance: 12.201  on 6  degrees of freedom
## AIC: 18.201
##
## Number of Fisher Scoring iterations: 4

We land our model predictions as a new column.

d$prediction <- predict(model, newdata = d, type = 'response') knitr::kable(d) x1 x2 y prediction 1 1 TRUE 0.4537010 1 0 FALSE 0.5462990 1 0 TRUE 0.5462990 1 1 FALSE 0.4537010 0 0 TRUE 0.6358007 0 1 TRUE 0.5462990 0 1 FALSE 0.5462990 0 0 FALSE 0.6358007 0 0 TRUE 0.6358007 We can see this model is calibrated or unbiased in the sense E[prediction] = E[outcome]. colMeans(d[, qc(y, prediction)]) %.>% knitr::kable(.) x y 0.5555556 prediction 0.5555556 And it is even calibrated in the sense we expect for logistic regression, E[prediction * x] = E[outcome * x] (where x is any explanatory variable). for(v in qc(x1, x2)) { print(paste0( v, ' diff: ', mean(d[[v]] * d$prediction) - mean(d[[v]] * d$y))) } ## [1] "x1 diff: 2.77555756156289e-17" ## [1] "x2 diff: 5.55111512312578e-17" However, we can see this model is not “fully calibrated” in an additional sense requiring that E[outcome | prediction] = prediction for all observed values of prediction. cal <- aggregate(y ~ prediction, data = d, FUN = mean) knitr::kable(cal) prediction y 0.4537010 0.5000000 0.5462990 0.5000000 0.6358007 0.6666667 We can re-calibrate such a model (in practice you would want to do this on out of sample data using isotonic regression, or using cross-frame methods to avoid nested model bias). cal_map <- cal$prediction := cal$y d$calibrated <- cal_map[as.character(d$prediction)] knitr::kable(d) x1 x2 y prediction calibrated 1 1 TRUE 0.4537010 0.5000000 1 0 FALSE 0.5462990 0.5000000 1 0 TRUE 0.5462990 0.5000000 1 1 FALSE 0.4537010 0.5000000 0 0 TRUE 0.6358007 0.6666667 0 1 TRUE 0.5462990 0.5000000 0 1 FALSE 0.5462990 0.5000000 0 0 FALSE 0.6358007 0.6666667 0 0 TRUE 0.6358007 0.6666667 This new calibrated prediction is also calibrated in the standard sense. colMeans(d[, qc(y, prediction, calibrated)]) %.>% knitr::kable(.) x y 0.5555556 prediction 0.5555556 calibrated 0.5555556 And, at least in this case, still obeys the explanatory roll-up conditions. for(v in qc(x1, x2)) { print(paste0( v, ' diff: ', mean(d[[v]] * d$calibrated) - mean(d[[v]] * d$y))) } ## [1] "x1 diff: 0" ## [1] "x2 diff: 0" The new calibrated predictions are even of lower deviance than the original predictions. deviance <- function(prediction, truth) { sum(-2 * (truth * log(prediction) + (1 - truth) * log(1 - prediction))) } deviance(prediction = d$prediction, truth = d$y) ## [1] 12.20102 deviance(prediction = d$calibrated, truth = d\$y)
## [1] 12.13685

The reason the original logistic model could not make the calibrated predictions is: the calibrated predictions are not a linear function of the explanatory variables in link space.