Presentation on theme: "Decision theory and Bayesian statistics. More repetition Tron Anders Moger 22.11.2006."— Presentation transcript:
Decision theory and Bayesian statistics. More repetition Tron Anders Moger 22.11.2006
Overview Statistical desicion theory Bayesian theory and research in health economics Review of previous slides
Statistical decision theory Statistics in this course often focus on estimating parameters and testing hypotheses. The real issue is often how to choose between actions, so that the outcome is likely to be as good as possible, in situations with uncertainty In such situations, the interpretation of probability as describing uncertain knowledge (i.e., Bayesian probability) is central.
Decision theory: Setup The unknown future is classified into H possible states of nature: s 1, s 2, …, s H. We can choose one of K actions: a 1, a 2, …, a K. For each combination of action i and state j, we get a ”payoff” (or opposite: ”loss”) M ij. To get the (simple) theory to work, all ”payoffs” must be measured on the same (monetary) scale. We would like to choose an action so to maximize the payoff. Each state s i has an associated probability p i.
Desicion theory: Concepts If action a 1 never can give a worse payoff, but may give a better payoff, than action a 2, then a 1 dominates a 2. a 2 is then inadmissible The maximin criterion for choosing actions The minimax regret criterion for choosing actions The expected monetary value criterion for choosing actions
Example No birdflu outbreak Small birdflu outbreak Birdflu pandemic A: No extra precautions 0-500-100000 B: Some extra precautions -100-10000 C: Vaccination of whole pop. -1000 states actions
Maximin and minimax Maximin: Maximize the minimum payoff: 1.For each row, compute the minimum 2.Maximize over the actions Minimax regret: Minimize the maximum regret possible 1.Compute the regrets in each column, by finding differences to max numbers 2.Maximize over the rows 3.Find action that minimizes these maxima.
Example No birdflu outbreak Small birdflu outbreak Birdflu pandemic A: No extra precautions 040099000 B: Some extra precautions 1090000 C: Vaccination of whole pop. 10009000 states actions Find that action C is preferred under the maximin criterion Regret table: Action C is also preferred under the minimax criterion
Expected monetary value criterion Need probabilities for each state Assume P(no outbreak)=P 1 =95%, P(small outbreak)=P 2 =4.5%, P(pandemic)=P 3 =0.5% EMV(A)=P 1 *M 11 +P 2 *M 12 +P 3 *M 13 = 0*0.95-500*0.045-100000*0.005= -522.5 EMV(B)=-55.45 EMV(C)=-1000 Should choose action B
Decision trees Contains node (square junction) for each choice of action Contains node (circular junction) for each selection of states Generally contains several layers of choices and outcomes Can be used to illustrate decision theoretic computations Computations go from bottom to top (or left to right in the book) of tree
Example: Action A Action C *Action B EMV=-522.5 EMV=-55.45 EMV=-1000 Small outbreak (0.045) No outbreak (0.95) Pandemic (0.005) 0 -100000 -500 -100 -1000 -10000 -1000 Pandemic (0.005) Small outbreak (0.045) No outbreak (0.95)
Updating probabilities by aquired information To improve the predictions about the true states of the future, new information may be aquired, and used to update the probabilities, using Bayes theorem. If the resulting posterior probabilities give a different optimal action than the prior probabilities, then the value of that particular information equals the change in the expected monetary value But what is the expected value of new information, before we get it?
Example: Prior probabilities: P(no outbreak)=95%, P(small outbreak)=4.5%, P(pandemic)=0.5%. Assume the probabilities are based on whether the virus has a low or high mutation rate. A scientific study can update the probabilities of the virus mutation rate. As a result, the probabilities for no birdflu, some birdflu, or a pandemic, are updated to posterior probabilities: We might get, for example:
The new information might affect what action we would take But not in this example: –If we find out that birdflu virus has high mutation rate, we would still choose action B! –EMV(A)=-5075, EMV(B)=-515.8, EMV(C)=- 1000 –If we find out that birdflu virus has low mutation rate, we would still choose action B! –EMV(A)=-104.5, EMV(B)=-11.9, EMV(C)=- 1000
Expected value of perfect information If we know the true (or future) state of nature, it is easy to choose optimal action, it will give a certain payoff For each state, find the difference between this payoff and the payoff under the action found using the expected value criterion The expectation of this difference, under the prior probabilities, is the expected value of perfect information
Example: Found that action B was best using the prior probabilities However, if there is no outbreak, action A is one unit better than B Similarily, if there is a pandemic, action C is 9000 units better than B The expected value of perfect information is then EVPI=0.95*1+0.045*0+0.005*9000=45.95
Expected value of sample information What is the expected value of obtaining updated probabilities using a sample? –Find the probability for each possible sample –For each possible sample, find the posterior probabilities for the states, the optimal action, and the difference in payoff compared to original optimal action –Find the expectation of this difference, using the probabilities of obtaining the different samples.
Utility When all outcomes are measured in monetary value, computations like those above are easy to implement and use Central problem: Translating all ”values” to the same scale In health economics: How do we translate different health outcomes, and different costs, to same scale? General concept: Utility Utility may be non-linear function of money value
Risk and (health) insurance When utility is rising slower than monetary value, we talk about risk aversion When utility is rising faster than monetary value, we talk about risk preference If you buy any insurance policy, you should expect to lose money in the long run But the negative utility of, say, an accident, more than outweigh the small negative utility of a policy payment.
Desicion theory and Bayesian theory in health economics research As health economics is often about making optimal desicions under uncertainty, decision theory is increasingly used. The central problem is to translate both costs and health results to the same scale: –All health results are translated into ”quality adjusted life years” –The ”price” for one ”quality adjusted life year” is a parameter called ”willingness to pay”.
Curves for probability of cost effectiveness given willingness to pay One widely used way of presenting a cost-effectiveness analysis is through the Cost- Effectiveness Acceptability Curve (CEAC) Introduced by van Hout et al (1994). For each value of the threshold willingness to pay λ, the CEAC plots the probability that one treatment is more cost-effective than another.
Repetition: What is relevant for the exam Probability theory Expected values and variance Distributions Tests, regression, one-way ANOVA and at least an understanding of two-way ANOVA are all relevant (obviously) Interpretation of a time-series regression model might also show up Do not forget how to interpret SPSS output (including graphs and figures)!! Also, do not forget the chi-square test!!
Conditional probability If the event B already has occurred, the conditional probability of A given B is: Can be interpreted as follows: The knowledge that B has occurred, limit the sample space to B. The relative probabilities are the same, but they are scaled up so that they sum to 1.
Probability postulates 3 Multiplication rule: For general outcomes A and B: P(A B)=P(A|B)P(B)=P(B|A)P(A) Indepedence: A and B are statistically independent if P(A B)=P(A)P(B) –Implies that
The law of total probability - twins A= Twins have the same gender B= Twins are monozygotic = Twins are heterozygotic What is P(A)? The law of total probability P(A)=P(A|B)P(B)+P(A| )P( ) For twins: P(B)=1/3 P( )=2/3 P(A)=1 · 1/3+1/2 · 2/3=2/3
Bayes theorem Frequently used to estimate the probability that a patient is ill on the basis of a diagnostic Uncorrect diagnoses are common for rare diseases
Example: Cervical cancer B=Cervical cancer A=Positive test P(B)=0.0001P(A|B)=0.9 P(A| )=0.001 Only 8% of women with positive tests are ill
Probability postulates 4 Assume that the events A 1, A 2,..., A n are independent. Then P(A 1 A 2 .... A n )=P(A 1 )·P(A 2 ) ·.... · P(A n ) This rule is very handy when all P(A i ) are equal The complement rule: P(A)+P( )=1
Example: Doping tests Let’s say a doping test has 0.2% probability of being positive when the athlete is not using steroids The athlete is tested 50 times What is the probability that at least one test is positive, even though the athlete is clean? Define A=at least one test is positive Complement rule Rule of independence 50 terms
Expected values and variance Remember the formulas E(aX+b) = aE(X)+b and How do you calculate expectation and variance for a categorical variable? For a continuous variable? How do you construct a standard normal variable from a general normal variable? Finding probabilities for a general normal variable?
Distributions Distributions we’ve talked about in detail Binomial Poisson Normal Approximations to normal distributions? Other distributions are there just to allow us to make test statistics, but you need to know how to use them
Remember this slide? (This was difficult) The probabilities for –A: Rain tomorrow –B: Wind tomorrow are given in the following table: 0.10.20.050.01 0.050.10.150.04 0.050.1 0.05 No rain Light rain Heavy rain No wind Some windStrong windStorm
And this one? Marginal probability of no rain: 0.1+0.2+0.05+0.01=0.36 Similarily, marg. prob. of light and heavy rain: 0.34 and 0.3. Hence marginal dist. of rain is a PDF! Conditional probability of no rain given storm: 0.01/(0.01+0.04+0.05)=0.1 Similarily, cond. prob. of light and heavy rain given storm: 0.4 and 0.5. Hence conditional dist. of rain given storm is a PDF ! Are rain and wind independent? Marg. prob. of no wind: 0.1+0.05+0.05=0.2 P(no rain,no wind)=0.36*0.2=0.072≠0.1
Think wheat fields! Wheat field was a bivariate distribution of wheat and fertilizer Only: Continuous outcome instead of categorical Calculations on previous incomprehensible slide is exactly the same as we did for the wheat field! Mean wheat crop for wheat 1 regardless of fertilizer->Marginal mean!! Mean crop for wheat 1 given that you use fertilizer ->Conditional mean!! (corresponds to mean for a single cell in our field)
Chi-square test: Expected cell values: Abortion/op.nurses: 13*36/70=6.7 Abortion/other nurses: 13*34/70=6.3 No abortion/op.nurses: 57*36/70=29.3 No abortion/other nurses: 57*34/70=27.7 Can be easily extendend to more groups of nurses As long as you have only two possible outcomes, this is equal to comparing proportions in more than two groups (think one-way ANOVA) AbortionsNo abortionsTotal Op.nurses102636 Other nurses33134 Total135770
We get: This has a chi-square distribution with (2-1)*(2-1)=1 d.f. Want to test H 0 : No association between abortions and type of nurse at 5%-level Find from table 7, p. 869, that the 95%-percentile is 3.84 This gives you a two-sided test! Reject H 0 : No association Same result as the test for different proportions in Lecture 4!
In SPSS: Check Expected under Cells, Chi-square under statistics, and Display clustered bar charts!
Next time: Find some topics you don’t understand, and we can talk about them
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