Signature
EVPPI_phi = E_phi[max_d E_(psi|phi)[NB_d(phi,psi)]] - max_d E_(phi,psi)[NB_d(phi,psi)]
| Inputs | Definition | Unit |
|---|---|---|
NB_d | Net benefit of option d as a function of phi and psi, listed across draws and options | currency per person |
phi | Subset of model parameters, such as a group of treatment effects, whose values would be learnt | units of the parameters concerned |
psi | All other model parameters, which remain uncertain after phi is learnt | units of the parameters concerned |
EVPPI_phi | Expected value of perfect information about the parameters phi alone | currency per person |
|---|
Function
Value of information function
Maps the joint distribution of net benefit across the options in a decision to the expected gain from resolving some or all of the uncertainty before choosing, compared with choosing the option with the highest expected net benefit on current information.
Implementations
Excel
EVPPI for a discrete parameter
With draws in rows, the value of phi in the named range PhiGroup and the net benefits of options A and B in NB_A and NB_B, the formula in I2, filled down beside the distinct phi values in column H, returns the best conditional mean. For phi values with equal numbers of draws, EVPPI is then =AVERAGE(I2:I3)-MAX(AVERAGE(NB_A),AVERAGE(NB_B)).
=MAX(AVERAGEIFS(NB_A,PhiGroup,H2),AVERAGEIFS(NB_B,PhiGroup,H2))
Assumptions
Conditional expectation estimated adequately
The inner expectation is estimated by nested simulation or by regression of net benefit on phi, as in the generalised additive model and Gaussian process methods. With few inner draws the maximisation biases a nested estimate upwards.
Ordering with EVPI
EVPPI lies between zero and the EVPI for all parameters, apart from small violations caused by numerical error. Parameters are grouped by the research question that could inform them.
Worked examples
Four equally likely draws with two parameters
Parameter phi takes values 1 or 2 and psi takes values 1 or 2, independently and with equal probability. Options A and B have net benefits of £1,000 and £600 at phi 1 and psi 1, £600 and £800 at phi 1 and psi 2, £200 and £800 at phi 2 and psi 1, and £200 and £600 at phi 2 and psi 2. Knowing phi, A is chosen at phi 1 with a mean of £800 and B at phi 2 with a mean of £700, an average of £750. On current information B is chosen with £700, so EVPPI for phi is £50, half the EVPI of £100 for both parameters.
phi = [1,1,2,2]; psi = [1,2,1,2]; NB_d = [[1000,600],[600,800],[200,800],[200,600]]; EVPPI_phi = 50
Common errors
Maximising before averaging over psi
Taking the best option in each draw before averaging over psi values perfect information about all parameters, so the result is EVPI rather than EVPPI for phi. In the worked example this gives £100 instead of £50.
Sources
Partial EVPI from a probabilistic sample
Strong M, Oakley JE, Brennan A. Estimating multiparameter partial expected value of perfect information from a probabilistic sensitivity analysis sample: a nonparametric regression approach. Medical Decision Making. 2014;34(3):311-326.
ISPOR methods for EVPPI and sample information
Rothery C, Strong M, Koffijberg HE, Basu A, Ghabri S, Knies S, Murray JF, Sanders Schmidler GD, Steuten L, Fenwick E. Value of information analytical methods. Report 2 of the ISPOR Value of Information Analysis Emerging Good Practices Task Force. Value in Health. 2020;23(3):277-286.
Canonical Identity
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