Solve parameters in a three-state ill-death model

The ill-death model consists of three states, stable, progression, and death. It can be used to model the progression-free survival (PFS) and overall survival (OS) in clinical trial simulation. It models the correlation PFS and OS without assumptions on latent status and copula. Also, it does not assume PFS and OS satisfy the proportional hazard assumption simultaneously. The three-state ill-death model ensure the nice property that PFS <= OS with probability one. However, it requires three hazard parameters under the homogeneous Markov assumption. In practice, hazard parameters are hard to specify intuitively especially when no trial data is available at the planning stage.

This function reparametrizes the ill-death model in term of three parameters, i.e. median of PFS, median of OS, and correlation between PFS and OS. The output of this function, which consists of the three hazard parameters, can be used to generate PFS and OS with desired property. It can be used with the built-in data generator CorrelatedPfsAndOs3() when defining endpoints in TrialSimulator.

Usage

solveThreeStateModel(
  median_pfs,
  median_os,
  corr,
  h12 = seq(0.05, 0.2, length.out = 50)
)

Arguments

median_pfs

numeric. Median of PFS.

median_os

numeric. Median of OS.

corr

numeric vector. Pearson correlation coefficients between PFS and OS.

h12

numeric vector. A set of hazard from progression to death that may induce the target correlation corr given median_pfs and median_os. solveThreeStateModel will do a grid search to find the best hazard parameters that matches to the medians of PFS and OS, and their correlations.

Value

a data frame with columns:

corr

target Peason's correlation coefficients.

h01

hazard from stable to progression.

h02

hazard from stable to death.

h12

hazard from progression to death.

error

absolute error between target correlation and correlation derived from h01, h02, and h12.

Examples

dat <- CorrelatedPfsAndOs3(1e6, h01 = .1, h02 = .05, h12 = .12)

cor(dat$pfs, dat$os) ## 0.65
#> [1] 0.6467199

median(dat$pfs) ## 4.62
#> [1] 4.624702

median(dat$os) ## 9.61
#> [1] 9.601951

## find h01, h02, h12 that can match to median_pfs, median_os and corr
## should be close to h01 = 0.10, h02 = 0.05, h12 = 0.12 when corr = 0.65
ret <- solveThreeStateModel(median_pfs = 4.6, median_os = 9.6,
                            corr = seq(.5, .7, length.out=5))
ret
#>   corr        h01        h02        h12        error
#> 1 0.50 0.08077106 0.06991311 0.07755102 0.0019725523
#> 2 0.55 0.08619848 0.06448569 0.08979592 0.0008271392
#> 3 0.60 0.09334966 0.05733451 0.10510204 0.0027808278
#> 4 0.65 0.10091413 0.04977004 0.12040816 0.0055098914
#> 5 0.70 0.11388735 0.03679682 0.14489796 0.0002066345