TrialSimulator (version 1.7.0)

solveMixtureExponentialDistribution: Solve Parameters in a Mixture Exponential Distribution

Description

This is a helper function to explore parameters for endpoint generator, likely in an enrichment design.

Assume that the overall population in an arm is a mixture of two exponential distributions with medians median1 (\(m_1\)) and median2 (\(m_2\)). Given the proportion of the first component (\(p_1\)) and the overall median \(m\), we have

$$p_1 (1 - e^{-\log(2)m/m_1}) + (1 - p_1) (1 - e^{-\log(2)m/m_2}) = 1/2$$

This function computes \(m_2\) or \(m\) given \(p_1\) and \(m_1\). These parameters can be used in custom random number generator to define exponential distributed endpoints.

Note that the math formula above may not be displayed correctly on a html page. You can read it with better format by running ?solveMixtureExponentialDistribution.

Usage

solveMixtureExponentialDistribution(
  weight1,
  median1,
  median2 = NULL,
  overall_median = NULL
)

Value

a named vector of median2 or overall_median.

Arguments

weight1

numeric. The proportion of the first component.

median1

numeric. Median of the first component.

median2

numeric. Median of the second component. If NULL, then overall_median must be specified, and this function will calculate and return median2.

overall_median

numeric. Median of the overall population. If NULL, then median2 must be specified, and this function will calculate and return overall_median.

Examples

library(dplyr)

median2 <-
  solveMixtureExponentialDistribution(
    weight1 = .3,
    median1 = 10,
    overall_median = 8)

median2

n <- 1e6
ifelse(
  runif(n) < .3,
  rexp(n, rate=log(2)/10),
  rexp(n, rate=log(2)/median2)) %>%
  median() ## should be close to 8

overall_median <-
  solveMixtureExponentialDistribution(
    weight1 = .4,
    median1 = 12,
    median2 = 4)

overall_median

ifelse(
  runif(n) < .4,
  rexp(n, rate=log(2)/12),
  rexp(n, rate=log(2)/4)) %>%
  median() ## should be close to overall_median