1. Conceptual and theoretical frameworks for decision modeling

Learning Objectives and Outline

Learning Objectives

-Construct and solve a decision problem by calculating an intervention’s expected value across competing strategies in a decision tree

-Determine the decision threshold across a range of scenarios

-Differentiate between joint and conditional probabilities and demonstrate their use in decision trees

Outline

Motivation
Examples of Decision Analysis
Structure Part 1: Decision Trees

Basic Concepts in Decision Analysis
Extending the Decision Modeling Toolkit

Motivation

The Past Two Decades …

Cured Hepatitis C
Significantly reduced incidence of HIV
Potential cure for relapsed/refractory leukemia & lymphoma
Perfected vaccines (e.g. HPV vaccine) to prevent diseases such as cervical & other cancers
Strides in preventing cardiovascular disease

Examples of Decision Analyses

Ex 1. HIV

You have been appointed as Director of a funding allocation committee responsible for prevention & treatment initiatives for HIV.

How will the committee decide on the proportion of funds for prevention efforts versus treatment?
Should any of the funds be used for research?
How do you respond to a member who argues that the funds are better spent on childhood vaccinations?

Ex 2. Birth Defects

A hypothetical birth defect is present in every 1 in 1,000 children born
Unless treated, this condition has a 50% fatality rate

Ex 2. Birth Defects

Should we test for this hypothetical birth defect?

A hypothetical birth defect is present in every 1 in 1,000 children born.
Unless treated, this condition has a 50% fatality rate.

Ex 2. Birth Defects

Should we test for this hypothetical birth defect?

Diagnostic test: Perfectly accurate
All newborns in whom the defect is identified can be successfully cured
BUT the test itself can be lethal:
- 4 in every 10,000 infants tested will die as a direct and observable result of the testing procedure

Ex 2. Birth Defects

Objective: Minimize total expected deaths

Ex 2. Birth Defects

Objective: Minimize total expected deaths

Consider a population of 100,000 newborns
Testing produces: (0.0004 x 100,000) = 40 expected deaths
No testing produces: (0.001 x 0.5 x 100,000) = 50 expected deaths

Ex 2. Birth Defects

Objective: Minimize total expected deaths

Consider a population of 100,000 newborns
Testing produces: (0.0004 x 100,000) = 40 expected deaths
No testing produces: (0.001 x 0.5 x 100,000) = 50 expected deaths

Looks like TESTING WINS!

Ex 2. Birth Defects

Objective: Minimize total expected deaths

Consider a population of 100,000 newborns
Testing produces: (0.0004 x 100,000) = 40 expected deaths
No testing produces: (0.001 x 0.5 x 100,000) = 50 expected deaths

Looks like TESTING WINS!

Anyone got a problem with this??

Different lives are lost

With testing, virtually all 40 deaths occur in infants born without the fatal condition.
With no testing, all 50 expected deaths occur from “natural causes” (i.e. unpreventable birth defect)

Different lives are lost

“Innocent deaths” inflicted on children who had “nothing to gain” from testing program
We may treat one child’s death as more tolerable than some other’s – even when we have no way, before the fact, of distinguishing one infant from the other.

Ex 3. Colon Cancer

3rd leading cause of cancer death.
Men, women, all races.
Early detection aids in prevention & improves health outcomes .
\(\sim80\%\) preventable.

Ex 3. Colon Cancer – Prevention

When to screen?
Which test & how often?
If a polyp is found/removed, when to repeat?

Ex 3. Colon Cancer – Prevention

Cost-effectiveness of alternative screening tests?
Alternative frequencies of screening?
How should screening vary by risk groups?

Ex 3. Colon Cancer - Estimating Probabilities

Predict colon cancer risk/incidence
- Family history, past polyps, other factors
- Probability of getting colon cancer
Predict impact from colon cancer
- Probability of cure
- Probability of death
- Probability of morbidity

Ex 3. Colon Cancer - Estimating Probabilities

Predict “benefit” from screening
- All previous probabilities
- Probability of early detection
- Probability of complication from testing
- Probability of false positive
- Probability of false negative
- Probability of positive tests = cancer
- Probability of negative test = no cancer

Estimating probabilities is fundamental to decision making

Cannot readily obtain needed probabilities
Varying time periods / lengths
Methods to estimate probabilities

Commonality of cases

Unavoidable tradeoffs
Different perspectives may lead to different conclusions
Multiple competing objectives
Complexity
Uncertainty

Decision Analysis

Aims to inform choice under uncertainty using an explicit, quantitative approach
Aims to identify, measure, & value the consequences of decisions + uncertainty when a decision needs to be made, most appropriately over time.

The Structure of a Decision Analysis

Should I go to the beach or stay home?

Possible States of the World:

At the beach with no rain.
At the beach with rain.
At home with no rain.
At home with rain.

Should I go to the beach or stay home?

Considerations:

Likelihood of rain.
My overall well being when
- At the beach with no rain.
- At the beach with rain.
- At home with no rain.
- At home with rain.

Should I go to the beach or stay home?

Considerations:

Likelihood of rain –> probabilities
My overall well being when
- At the beach with no rain.
- At the beach with rain.
- At home with no rain.
- At home with rain.

Should I go to the beach or stay home?

Considerations:

Likelihood of rain –> probabilities
My overall well being when
- At the beach with no rain. –> payoff
- At the beach with rain. –> payoff
- At home with no rain. –> payoff
- At home with rain. –> payoff

Decision Trees

A square decision node indicates a decision point between alternative options.
A circular chance node shows a point where two or more alternative events for a patient are possible.

Decision Trees

A square decision node indicates a decision point between alternative options.
A circular chance node shows a point where two or more alternative events for a patient are possible.

Decision Trees

Pathways are mutually exclusive sequences of events and are the routes through the tree.
Probabilities show the likelihood of particular event occurring at a chance node.

Should I go to the beach or stay home?

Decision Tree:

Probability of rain = 30%

Should I go to the beach or stay home?

Decision Tree:

Should I go to the beach or stay home?

Payoffs

Scenario	Payoff
At beach, no rain	1.0
At beach, rain	0.4
At home, no rain	0.8
At home, rain	0.6

Should I go to the beach or stay home?

Decision Tree:

What is the expected value of going to the beach?

\(\color{green}{0.82} = \underbrace{\color{red}{0.3} \cdot \color{blue}{0.4}}_{\text{Rain}} + \underbrace{\color{red}{0.7} \cdot \color{blue}{1.0}}_{\text{No Rain}}\)

Probabilities in red.
Payoffs in blue.
Expected value in green.

What is the expected value of staying home?

\(\color{green}{0.74} = \underbrace{\color{red}{0.3} \cdot \color{blue}{0.6}}_{\text{Rain}} + \underbrace{\color{red}{0.7} \cdot \color{blue}{0.8}}_{\text{No Rain}}\)

Probabilities in red.
Payoffs in blue.
Expected value in green.

Beach

EV(Beach)=0.82 > EV(Home)=0.74

Determining the Decision Threshold

At what probability \(p\) are the two choices equal?

Write the equation for each choice using a variable, p, for the probability in question
Set the equations equal to to one other and solve for p.

At what probability \(p\) are the two choices equal?

Beach: 0.82 = 0.3 x 0.4 + 0.7 x 1.0

Home: 0.74 = 0.3 x 0.6 + 0.7 x 0.8

At what probability \(p\) are the two choices equal?

\(\underbrace{0.3 * 0.4 + 0.7 * 1.0}_{\text{Beach}} = \underbrace{0.3 * 0.6 + 0.7 * 0.8}_{\text{Home}}\)

At what probability \(p\) are the two choices equal?

Replace probabilities with P and 1-P and solve for “P”

p * 0.4 + (1-p) * 1.0 = p * 0.6 + (1-p) * 0.8

\(\underbrace{0.3 * 0.4 + 0.7 * 1.0}_{\text{Beach}} = \underbrace{0.3 * 0.6 + 0.7 * 0.8}_{\text{Home}}\)

At what probability \(p\) are the two choices equal?

p * 0.4 + (1-p) * 1.0 = p * 0.6 + (1-p) * 0.8

At what probability \(p\) are the two choices equal?

p * 0.4 + (1-p) * 1.0 = p * 0.6 + (1-p) * 0.8

0.4p + 1-p = 0.6p + 0.8 - 0.8p

1-0.6p = 0.8 - 0.2p

1-0.8 = 0.6p - 0.2p

0.2 = 0.4 * p

0.5 = p

At what probability \(p\) are the two choices equal?

When the probability of rain is 50% at BOTH the beach and home, given how we weighted the outcomes, going to the beach would be the same as staying at home

In other words, you would be indifferent between the two – staying at home or going to the beach

At what probability (\(p_B\)) of rain for the beach are you indifferent between the two options?

At what probability (\(p_B\)) of rain for the beach are you indifferent between the two options?

Earlier, we solved for the expected payoff of remaining at home: 0.74
What would \(p_B\) need to be to yield an expected payoff at the beach of 0.74?

At what probability (\(p_B\)) of rain for the beach are you indifferent between the two options?

Set 0.74 (expected value of remaining at home) equal to the beach payoffs and solve for \(p_B\)

p_B * 0.4 + (1 - p_B) * 1.0 = 0.74

At what probability (\(p_B\)) of rain for the beach are you indifferent between the two options?

p_B * 0.4 + (1 - p_B) * 1.0 = 0.74

p_B * 0.4 + 1 - p~B = 0.74

At what probability (\(p_B\)) of rain for the beach are you indifferent between the two options?

p_B * 0.4 + 1 - p~B = 0.74

p_B * -0.6 = -0.26

At what probability (\(p_B\)) of rain for the beach are you indifferent between the two options?

p_B * -0.6 = -0.26

p_B = -0.26 / -0.6 = 0.43

At what probability (\(p_B\)) of rain for the beach are you indifferent between the two options?

When the probability of rain at the beach is 43% (probability of rain at home remains at 30%), we would be indifferent between staying at home & going to the beach.

If the probability of rain at the beach in > 43%, then we would stay home

Basic Concepts in Decision Analysis

Probabilities

The likelihood of an event taking place in the future.

Joint probability

P(A and B): The probability of two events occurring at the same time.

Conditional probability

P(A|B): The probability of an event A given that an event B is known to have occurred.

Probabilities

Moving from left to right, the first probabilities in the tree show the probability of an event.
Subsequent probabilities are conditional.
- The probability of an event given that an earlier event did or did not occur.
Multiplying probabilities along pathways estimates the pathway probability, which is a joint probability.