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
Aims to inform choice under uncertainty using an explicit, quantitative approach
Aims to identify, measure, & value the consequences of decisions (risks/benefits) & uncertainty when a decision needs to be made, most appropriately over time
Probability of rain = 30%
Scenario | Payoff |
---|---|
At beach, no rain | 1.0 |
At beach, rain | 0.4 |
At home, no rain | 0.8 |
At home, rain | 0.6 |
At beach, no rain > At home, no rain > At home, rain > At beach, rain
\color{green}{0.82} = \underbrace{\color{red}{0.3} * \color{blue}{0.4}}_{\text{Rain}} + \underbrace{\color{red}{0.7} * \color{blue}{1.0}}_{\text{No Rain}}
\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}}
EV(Beach)=0.82 > EV(Home)=0.74
The expected value within the context of decision trees are the “payoffs” weighted by their preceding probabilities
What we get is: the result that is expected ON AVERAGE for any one decision alternative (e.g. length of life, quality of life, lifetime costs)
Maximizing expected value is a reasonable criterion for choice given uncertain prospects; though it does not necessarily promise the best results for any one individual
Suppose we want to know at what probability of rain p we are indifferent between going to the beach vs. staying at home…
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.
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
\underbrace{0.3 * 0.4 + 0.7 * 1.0}_{\text{Beach}} = \underbrace{0.3 * 0.6 + 0.7 * 0.8}_{\text{Home}}
Replace probability of rain 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}}
p * 0.4 + (1-p) * 1.0 = p * 0.6 + (1-p) * 0.8
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
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
Earlier, we solved for the expected payoff of remaining at home: 0.74 (which was a lower expected value than going to the beach when the chance of rain at both was 30%)
What would p_B need to be to yield an expected payoff at the beach of 0.74?
Set 0.74 (expected value of remaining at home) equal to the beach payoffs and solve for p_B
pB * 0.4 + (1 - pB) * 1.0 = 0.74
pB * 0.4 + (1 - pB) * 1.0 = 0.74
pB * 0.4 + 1 - pB = 0.74
pB * 0.4 + 1 - pB = 0.74
pB * -0.6 = -0.26
pB * -0.6 = -0.26
pB = -0.26 / -0.6 = 0.43
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
Mutually exclusive events
Mutually exclusive events
Mutually exclusive events
P(A or B) = P(A) + P(B)
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.
Example: What is the conditional probability of death within a year of birth, given the infant has a mother who smokes?
(Probability of an event occurring (B) given that another event occurred (A))
P(A|B) = P(A and B) / P(B)
What is the conditional probability of death within a year of birth, given the infant has a mother who smokes?
*A= death in first year; B=mother who smokes
P(A|B) = 16,712 / (1,197,142 + 16,712)
= 14 per 1,000 births
Or, if we wanted to use the conditional probability equation
P(A|B) = P(A and B)/P(B)
*A= death in first year; B=mother who smokes
P(A and B) = 16,712 / 4,111,059 = 0.0041
P(B) = 1,213,854/4,111,059 = 0.295
P(A|B) = 0.0041/0.295 = 0.014
Probability of A|B is different from that of B|A
If A = death in first year; B=normal birth weight infant, then the conditional probability of P(A|B) = the probability of an infant death, given that the child has a normal birth weight
What is the conditional probability of an infant death, given that the child has a normal birth weight
If A = death in first year; B=normal birth weight infant
P(A|B) = 14,442 / (14,442+ 3,804,294)
= 3.8 deaths per 1,000 births
Or, if we wanted to use equation
P(A|B) = P(A and B)/P(B)
P(A and B) = 14,442
P(B) = 3,818,736
P(A|B) = 14,442/3,818,736 = 0.0038
On the other hand, the conditional probability of P(B|A) is the probability that an infant had normal birth weight, given that the infant died within 1 year from birth
*B=normal birth weight infant; A = death in first year
Solve P(B|A) – the probability that an infant had normal birthweight, given that the infant died within 1 year from birth
P(B|A) = 14,442 / (14,442+ 21,054) = 0.41
Maximizing expected value is a reasonable criterion for choice given uncertain prospects; though it does not necessarily promise the best results for any one individual.
Mini-max regret
Maxi-max
Expected utility
Maximizing expected value is a reasonable criterion for choice given uncertain prospects; though it does not necessarily promise the best results for any one individual.
Mini-max regret
Maxi-max gain
Expected utility
Maximizing expected value is a reasonable criterion for choice given uncertain prospects; though it does not necessarily promise the best results for any one individual.
Maximizing expected value is a reasonable criterion for choice given uncertain prospects; though it does not necessarily promise the best results for any one individual.