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Presentation on theme: "THOUGHTS ON MODEL ASSESSMENT Porco, DAIDD, December 2012."— Presentation transcript:


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 { "@context": "", "@type": "ImageObject", "contentUrl": "", "name": "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 0.810.08 0.100.02 0.51 0.410.72 0.801.28

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 0.200.81 0.210.80 0.200.80 0.200.80 0.200.81 0.210.81

19 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).

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 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.

22 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. PredictionReliabilityRepeatsTotal 0.20.035660.2133 0.80.222261.08

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 5.8333 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] -1.776357e-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 02/31.50.585 02/31.50.585 11/331.585 02/31.50.585 11/331.585 11/331.585 The average surprisal was 1.085 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 0.971.

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 0.637. 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 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.

49 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 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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