- 90 Views
- Uploaded on
- Presentation posted in: General

Decision theory and Bayesian statistics. More repetition

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Decision theory and Bayesian statistics. More repetition

Tron Anders Moger

22.11.2006

- Statistical desicion theory
- Bayesian theory and research in health economics
- Review of previous slides

- 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.

- The unknown future is classified into H possible states of nature: s1, s2, …, sH.
- We can choose one of K actions: a1, a2, …, aK.
- For each combination of action i and state j, we get a ”payoff” (or opposite: ”loss”) Mij.
- 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 si has an associated probability pi.

- If action a1 never can give a worse payoff, but may give a better payoff, than action a2, then a1 dominates a2.
- a2 is then inadmissible
- The maximin criterion for choosing actions
- The minimax regret criterion for choosing actions
- The expected monetary value criterion for choosing actions

states

actions

- Maximin: Maximize the minimum payoff:
- For each row, compute the minimum
- Maximize over the actions
- Minimax regret: Minimize the maximum regret possible
- Compute the regrets in each column, by finding differences to max numbers
- Maximize over the rows
- Find action that minimizes these maxima.

Find that action C is preferred under the maximin criterion

Regret table:

states

actions

Action C is also preferred under the minimax criterion

- Need probabilities for each state
- Assume P(no outbreak)=P1=95%, P(small outbreak)=P2=4.5%, P(pandemic)=P3=0.5%
- EMV(A)=P1*M11+P2*M12+P3*M13=
0*0.95-500*0.045-100000*0.005= -522.5

- EMV(B)=-55.45
- EMV(C)=-1000
- Should choose action B

- 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

No outbreak (0.95)

0

Action A

Small outbreak (0.045)

-500

Pandemic (0.005)

EMV=-522.5

-100000

No outbreak (0.95)

EMV=-55.45

-1

*Action B

Small outbreak (0.045)

-100

Pandemic (0.005)

-10000

No outbreak (0.95)

EMV=-1000

-1000

Small outbreak (0.045)

-1000

Action C

Pandemic (0.005)

-1000

- 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?

- 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:

- 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

- 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

- 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

- 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.

- 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

- 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.

- 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”.

- 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.

- 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!!

- 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.

- Multiplication rule: For general outcomes A and B:
P(AB)=P(A|B)P(B)=P(B|A)P(A)

- Indepedence: A and B are statistically independent if P(AB)=P(A)P(B)
- Implies that

- 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

- Frequently used to estimate the probability that a patient is ill on the basis of a diagnostic
- Uncorrect diagnoses are common for rare diseases

- 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

- Assume that the events
A1, A2 ,..., An are independent. Then P(A1A2....An)=P(A1)·P(A2) ·.... ·P(An)

This rule is very handy when all P(Ai) are equal

- The complement rule: P(A)+P( )=1

- 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

- 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 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

- The probabilities for
- A: Rain tomorrow
- B: Wind tomorrow
are given in the following table:

Some wind

Strong wind

Storm

No wind

No rain

Light rain

Heavy rain

- 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

- 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)

- 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)

- This has a chi-square distribution with (2-1)*(2-1)=1 d.f.
- Want to test H0: 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 H0: No association
- Same result as the test for different proportions in Lecture 4!

Check Expected under Cells, Chi-square under statistics, and

Display clustered bar charts!

- Find some topics you don’t understand, and we can talk about them