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THOUGHTS ON MODEL ASSESSMENT Porco, DAIDD, December 2012.

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1 THOUGHTS ON MODEL ASSESSMENT Porco, DAIDD, December 2012

2 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

3 Face Validity  Knowledge representation  Analogy  Does the model represent in some approximate sense facts about the system?

4 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

5 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

6 Prediction  Not everything is predictable  Predict individuals? Communities? Counterfactuals?  Does the model somehow correspond to reality?

7 California TB

8 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

9 Two approaches  Assessment of probabilistic forecasts  Brier Score  Information Score

10 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

11 Brier score

12  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

13 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]

14 Brier score  What is the smallest possible Brier score?  What is the largest possible Brier score?

15 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.EliminationScore

16 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%.

17 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?

18 Examples PredictionResultPredictionResult

19 Example  Here, we get a total Brier score of We have to add up (0.2-0)^2 + (0.2-1)^2+… (and not forget the factor of two).

20 Some terminology  Forecast levels: the different values possible for the forecast. Notation: f k 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.

21 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  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.

22 Reliability  So the total reliability score is going to be computed: This yields a total score of approximately for the reliability component REL. PredictionReliabilityRepeatsTotal

23 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.

24 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.

25 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 or so.

26 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

27 Script brier.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 }

28 Example ds <- 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] e-15

29 General form B: Brier score; K, number of forecast levels; n k the number of forecasts at level k Other quantities as defined earlier

30 More general forms  More general decompositions (Stephenson, five terms)  Continuous forms

31 Alternatives to Brier score  Decomposition into Reliability, Resolution, Uncertainty works for information measures (Weijs et al)

32 Information measures  Surprise  Expected surprise  Surprising a model

33 Surprisal  How much do you learn if two equally likely alternatives are disclosed to you?  Heads vs tails; 1 vs 2

34 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

35 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  …

36 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

37 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?  Log 2 (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

38 Example  Sequence of values of flips of a bent coin with P(heads)=1/3, P(tails)=2/3 ObservedProbability1/ProbabilitySurprisal 02/ / / / / / The average surprisal was or so.

39 Expected surprisal  Every time an event with probability p i happens, the surprisal is lg(1/p i ).  What is the expected surprisal?

40 Expected value

41 Expected square

42 Expected surprisal Shannon entropy

43  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)

44 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

45 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.

46 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 About 16% of patients had such an event.

47 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.

48 Example  For group 1, the chance of a bad outcome was 11%. The entropy given membership in this group is about  For group 2, the chance of a bad outcome was 21% or so. The entropy given membership in this group is about  The weighted average of these is the conditional entropy given the model. This gives us just

49 Example  So the entropy was 0.636, and now it’s 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 R 2 you get in logistic regression output.

50 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.


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