Solve parameters in a mixture exponential distribution — solveMixtureExponentialDistribution • TrialSimulator

Assume that the overall population 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 $m_1$.

Usage

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

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.

Value

a named vector of median2 or overall_median.

Examples


library(dplyr)
#> 
#> Attaching package: ‘dplyr’
#> The following objects are masked from ‘package:stats’:
#> 
#>     filter, lag
#> The following objects are masked from ‘package:base’:
#> 
#>     intersect, setdiff, setequal, union

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

median2
#>  median2 
#> 7.305935 

n <- 1e6
ifelse(
  runif(n) < .3,
  rexp(n, rate=log(2)/10),
  rexp(n, rate=log(2)/median2)) %>%
  median() ## should be close to 8
#> [1] 8.010029

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

overall_median
#> overall_median 
#>        5.90597 

ifelse(
  runif(n) < .4,
  rexp(n, rate=log(2)/12),
  rexp(n, rate=log(2)/4)) %>%
  median() ## should be close to overall_median
#> [1] 5.901173