THOUGHTS ON MODEL ASSESSMENTPorco, DAIDD, December 2012

Assessment Some models are so wrong as to be

useless. Not really possible to argue that a model

is “true” Treacherous concept of “validation”: has

a model that has been “validated” been shown to be “valid”?

Philosophy of science

Face Validity Knowledge representation Analogy Does the model represent in some

approximate sense facts about the system?

Verification Does your simulation actually simulate

what you think it simulates? How do you know your program is not

generating nonsense? Unit testing Boundary cases “Test harnesses” Formal software testing techniques

Thought “…the gold standard of model

performance is predictive power on data that wasn’t used to fit your model…”

--Conway and White, Machine Learning for Hackers

Prediction Not everything is predictable Predict individuals? Communities?

Counterfactuals? Does the model somehow correspond to

reality?

California TB

Probabilistic forecasting Weather Forecast expressed as a probability Correctly quantify uncertainty Uncertainty is needed to make the best

use of a forecast If you don’t know, don’t say you do Not bet-hedging, but honesty

Two approaches Assessment of probabilistic forecasts

Brier Score Information Score

Assessing simple forecasts Binary event: 0 for no, 1 for yes Forecast: p is the probability it occurred If p=1, you forecast it with certainty If p=0, you forecast that it was certain

not to have occurred If 0<p<1, you were somewhere in the

middle

Brier score

Brier score Squared error of a probabilistic forecast BS = (p-1)2 + ((1-p)-0)2

BS = 2(1-p)2

Brier score has a negative orientation (like in golf, smaller is better)

Some authors do not use the factor of 2

Brier score Example. Suppose I compute a forecast

that trachoma will be eliminated in village A after two years as measured by pooled PCR for DNA in 50 randomly selected children. Suppose I say the elimination probability is 0.8.

If trachoma is in fact eliminated, the Brier score for this is (0.8-1)^2 + (0.2-0)^2 = [you work it out]

Brier score What is the smallest possible Brier

score? What is the largest possible Brier score?

Brier score Suppose we now forecast elimination in

five villages. We can compute an overall Brier score by just adding up separate scores for each village.

Elimination Prob.

Elimination Score

0.8 1 0.080.1 0 0.020.5 1 0.50.4 1 0.720.8 0 1.28

Murphy decomposition Classical form applies to binary

predictions Predictions are probabilities Finite set of possible probabilities Example: chance of a meteorological

event happening might be predicted as 0%, 10%, 20%, 30%, 40%, 50%, 60%, 70%, 80%, 90%, and 100%.

Reliability (sensu Murphy) Similar to Brier score – less is better Looking at all identical forecasts, how

similar is what really happened to what was forecast?

Examples

Prediction Result Prediction Result0.2 0 0.8 10.2 1 0.8 00.2 0 0.8 00.2 0 0.8 00.2 0 0.8 10.2 1 0.8 1

Example Here, we get a total Brier score of 6.96.

We have to add up (0.2-0)^2 + (0.2-1)^2+… (and not forget the factor of two).

Some terminology Forecast levels: the different values

possible for the forecast. Notation: fk for the levels of k

Observed means at the forecast levels: averaging the number of times the event happened stratifying on forecast level. Notation:

We have two forecast levels in the example, 0.2 and 0.8. For forecast level 0.2, we have an observed mean of 2/6. For the forecast level 0.8, we have an observed mean of 3/6.

Examples We had 2/6 that were YES when we gave

a 20% chance. Let’s compare the prediction to the successes just for the 20% predictions. For each one we have the same thing: (0.2 – 2/6)2+(0.8-4/6)2 which is about 0.0356. We had 3/6 that were YES when we gave

an 80% chance. We do a similar computation and we get (0.8-3/6)2 + (0.2-3/6)2=0.222 or so.

Reliability So the total reliability score is going to

be computed:

This yields a total score of approximately 1.2933 for the reliability component REL.

Prediction Reliability Repeats Total0.2 0.0356 6 0.21330.8 0.2222 6 1.08

Resolution When we said different things would

happen, how different really were they? We’re going to contrast the observed means at different forecast levels.

Specifically, we want the variance of the distribution of the observed means at different forecast levels.

Example Using the simple example: the overall

mean is 5/12. Then compute 2*(2/6 – 5/12)2 for every observation in the first group, and 2*(3/6-5/12)2 for every observation in the second group.

The total resolution component RES is 6*2*(2/6 – 5/12)2 +2*(3/6-5/12)2 = 1/6.

Uncertainty In the classical Murphy formula, this is

computed by calculating the mean observation times one minus the mean—and adding this up for each observation.

For our example, the uncertainty is 5.8333 or so.

Murphy decomposition Murphy (1973) showed that the Brier score can

be decomposed as follows: BS=REL-RES+UNC N.B. the negative sign in front of resolution High uncertainty contributes to high Brier score

(all else being equal) High discrepancy of the observed means at each

level from the forecasts raises the Brier score But having those observed means at each level

separate from each other lowers the Brier score

Scriptbrier.component <- function(forecast,observation) { 2*(forecast-observation)^2}reliability <- function(fk,ok) { 2*(fk-ok)^2}resolution <- function(obar,ok) { 2*(ok-obar)^2}

gen.obark <- function(predictions,outcome,key) { tmp <- data.frame(key=key,outcome=outcome,pred=predictions) f1 <- merge(tmp, ddply( tmp, .(pred), function(x){mean(x$outcome)} ), by="pred") f1 <- f1[order(f1$key),] f1$V1}

Exampleds <- c(0,1,0,0,0,1,1,0,0,0,1,1)fs <- c(rep(0.2,6),rep(0.8,6))obark <- gen.obark(fs,ds,1:12)rel <- sum(reliability(fs,obark))res <- sum(resolution(mean(ds),obark))unc <- 12*2*(mean(ds)*(1-mean(ds)))brs <- sum(brier.component(fs,ds))(rel-res+unc)-brs[1] -1.776357e-15

General form

B: Brier score; K, number of forecast levels; nk the number of forecasts at level kOther quantities as defined earlier

More general forms More general decompositions

(Stephenson, five terms) Continuous forms

Alternatives to Brier score Decomposition into Reliability,

Resolution, Uncertainty works for information measures (Weijs et al)

Information measures Surprise Expected surprise Surprising a model

Surprisal How much do you learn if two equally

likely alternatives are disclosed to you? Heads vs tails; 1 vs 2

Surprisal How much do you learn if 3 are

disclosed? A, B, or C How much if I disclose one of 26 equally

likely outcomes? You should have learned more

Surprisal Standard example: Now combine two independent things. If

we had 1 vs 2, and A, B, or C: A/1 C/2 B/1 B/2 …

Surprisal Six equally likely things s(2 x 3) = s(2) + s(3) Uniquely useful way to do this: define

surprisal as log(1/p), log of the reciprocal probability

Stevens information tutorial online

Surprisal Different bases can be used for the logarithm Log base 2 is traditional How much information is transmitted if you learn

which of two equally likely outcomes happened? Log2(1/(1/2)) = 1 If we use base two, then the value is 1, referred

to as 1 bit. Disclosure of one of two equally likely outcomes

reveals one bit of information. Notation: log base 2 often written lg

Example Sequence of values of flips of a bent coin

with P(heads)=1/3, P(tails)=2/3Observed Probabilit

y1/Probability

Surprisal

0 2/3 1.5 0.5850 2/3 1.5 0.5851 1/3 3 1.5850 2/3 1.5 0.5851 1/3 3 1.5851 1/3 3 1.585

The average surprisal was 1.085 or so.

Expected surprisal Every time an event with probability pi

happens, the surprisal is lg(1/pi). What is the expected surprisal?

Expected value

Expected square

Expected surprisal

Shannon entropy

Shannon entropy Shannon entropy can be used to quantify

the amount of information provided by a model for a categorical outcome, in much the same way that the squared multiple correlation coefficient can quantify the amount of variance explained by a continuous model

Use of entropy and related measures to assess probabilistic predictions is sometimes recommended (Weijs)

Estimated entropy If you have a Bernoulli trial, what is the

entropy? The success probability is p. H=-(1-p)lg(1-p) – p lg(p) If we estimate p from data and plug this

in, we get an estimator of the entropy. Example: if we observe 8 successes in

20 trials, the observed frequency is 0.4, and the estimated entropy would be about 0.971.

What does a model do? If we now have a statistical model that

provides a better prediction, we can compute the conditional entropy.

Without the model, our best estimate of the chance of something is just the observed relative frequency.

But with the model, maybe we have a better estimate.

Example Logistic regression: use data from the

Mycotic Ulcer Treatment Trial (Prajna et al 2012, Lietman group).

Using just the chance of having a transplant or perforation, the estimated entropy is 0.637. About 16% of patients had such an event.

Example Now, let’s use a covariate, such as what

treatment a person got. Now, we have a better (we hope)

estimate of the chance of a bad outcome for each person. These are 11% for treatment 1 and 21% for treatment 2.

Example For group 1, the chance of a bad

outcome was 11%. The entropy given membership in this group is about 0.503.

For group 2, the chance of a bad outcome was 21% or so. The entropy given membership in this group is about 0.744.

The weighted average of these is the conditional entropy given the model. This gives us just 0.623.

Example So the entropy was 0.636, and now it’s

0.623. The treatment model does not predict very much.

If we look at the proportional reduction, we get 2.2%; this model explained 2.2% of the uncertainty.

This is exactly the McFadden R2 you get in logistic regression output.

Summary Probabilistic forecasts can be assessed

using the Brier score. The Brier score can be decomposed into

reliability, resolution, and uncertainty components.

Information theoretic measures can also be used in assessment.

Information theoretic measures are useful in other biostatistical applications as well.