Transcript expected value of perfect information

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: 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.
Desicion theory: Concepts
• 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
Example
states
actions
No birdflu
outbreak
Small birdflu
outbreak
Birdflu
pandemic
A: No extra
precautions
0
-500
-100000
B: Some extra
precautions
-1
-100
-10000
C: Vaccination
of whole pop.
-1000
-1000
-1000
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
Find that action C is preferred under the maximin criterion
Regret table:
states
actions
No birdflu
outbreak
Small birdflu
outbreak
Birdflu
pandemic
A: No extra
precautions
0
400
99000
B: Some extra
precautions
1
0
90000
C: Vaccination
of whole pop.
1000
900
0
Action C is also preferred under the minimax criterion
Expected monetary value 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
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:
No outbreak (0.95)
0
Action A
Small outbreak (0.045)
-500
EMV=-522.5
EMV=-55.45
*Action B
Pandemic (0.005)
-100000
No outbreak (0.95)
Small outbreak (0.045)
-1
-100
Pandemic (0.005)
-10000
EMV=-1000
No outbreak (0.95)
-1000
Small outbreak (0.045)
-1000
Action C
Pandemic (0.005)
-1000
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:
P (none | high _ mutation)  80%
P ( some | high _ mutation)  15%
P (none | low _ mutation)  99%
P ( some | low _ mutation)  0.9%
P ( pand . | high _ mutation)  5%
P ( pand . | low _ mutation)  0.1%
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 CostEffectiveness 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:
P( A  B )
P( A | B ) 
P( B )
• 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
P( A  B) P( A) P( B)
P( A | B ) 

 P( A)
P( B )
P( B )
The law of total probability - twins
•
•
•
•
A= Twins have the same gender
B= Twins are monozygotic
B= Twins are heterozygotic
What is P(A)?
• The law of total probability
P(A)=P(A|B)P(B)+P(A|B)P(B )
For twins: P(B)=1/3 P(B )=2/3
P(A)=1 · 1/3+1/2 · 2/3=2/3
Bayes theorem
P( B ) P( A | B )
P( B | A) 
P( B ) P( A | B )  P( B ) P( A | B )
• 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)=0.0001 P(A|B)=0.9
P(A|B)=0.001
P( A | B ) P( B )
P( B | A) 
P( A | B ) P( B )  P( A | B ) P( B )
0.9 * 0.0001

 0.08
0.9 * 0.0001  0.001 * 0.9999
• Only 8% of women with positive tests are ill
Probability postulates 4
• Assume that the events
A1, A2 ,..., An are independent. Then
P(A1A2....An)=P(A1)·P(A2) ·.... · P(An)
This rule is very handy when all P(Ai) are
equal
• The complement rule: P(A)+P(A)=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
P( A)  1  P( A )  1  (1  0.002) * .... * (1  0.002)
 1  (1  0.002)50  0.095  9.5%
Expected values and variance
• Remember the formulas E(aX+b) =
aE(X)+b and Var (aX  b)  a 2Var ( X )
• 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:
No wind
Some wind Strong wind
Storm
No rain
0.1
0.2
0.05
0.01
Light rain
0.05
0.1
0.15
0.04
Heavy rain
0.05
0.1
0.1
0.05
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:
Op.nurses
Abortions
10
No abortions
26
Total
36
Other nurses
3
31
34
Total
13
57
70
• 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)
We get:
(10  6.7) 2 (3  6.3) 2 (26  29.3) 2 (31  27.7) 2



 4.2
6.7
6.3
29.3
27.7
• 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!
In SPSS:
Abortions * Group Crosstabulation
Abortions
No abortion
Abortion
Total
Count
Expected Count
Count
Expected Count
Count
Expected Count
Group
Op.nurs es Othe nurs es
26
31
29,3
27,7
10
3
6,7
6,3
36
34
36,0
34,0
Total
57
57,0
13
13,0
70
70,0
Bar Chart
Group
Op.nurses
Othe nurses
30
Pears on Chi-Square
Continuity Correctiona
Likelihood Ratio
Fisher's Exact Test
Linear-by-Linear
Ass ociation
N of Valid Cas es
Value
4,154 b
2,995
4,359
4,095
df
1
1
1
1
Asymp. Sig.
(2-s ided)
,042
,084
,037
Exact Sig.
(2-s ided)
Exact Sig.
(1-s ided)
,064
,040
Count
Chi-Square Tests
20
10
,043
70
a. Computed only for a 2x2 table
b. 0 cells (,0%) have expected count les s than 5. The minimum expected count is
6,31.
0
Abortion
No abortion
Abortions
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