fe: Expected Number of Events Since First Subject Randomized

In phe9480/rgs: What the Package Does (One Line, Title Case)

Description

This function calculates the expected number of events under alternative hypothesis at calendar time t, which is calculated from first subject randomized. The function returns the expected number of events for each arm at time t, based on the provided enrollment distribution function and random lost-to-followup distribution if applicable. If the total sample size is not provided, then only the corresponding probability of event for each arm is provided.

Usage

fe(
  n = 450,
  DCO = 24,
  r = 1,
  h0 = function(t) {     log(2)/12 },
  S0 = function(t) {     exp(-log(2)/12 * t) },
  h1 = function(t) {     log(2)/12 * 0.7 },
  S1 = function(t) {     exp(-log(2)/12 * 0.7 * t) },
  Lambda = function(t) {     (t/18)^1.5 * as.numeric(t <= 18) + as.numeric(t > 18) },
  G0 = function(t) {     0 },
  G1 = function(t) {     0 }
)

Arguments

n	Total sample size for two arms.
DCO	Analysis time calculated from first subject randomization date.
r	Randomization ratio of experimental arm : control arm as r:1. When r = 1, it is equal allocation. Default r = 1.
h0	Hazard function of control arm. h0(t) = log(2)/m0 means T~exponential distribution with median m0.
S0	Survival function of control arm. In general, S0(t) = exp(- integral of h0(u) for u from 0 to t). but providing S0(t) can improves computational efficiency and usually the survival function is known in study design. The density function f0(t) = h0(t) * S0(t).
h1	Hazard function of experimental arm. h1(t) = log(2)/m1 means T~exponential distribution with median m0.
S1	Survival function of experimental arm. In general, S1(t) = exp(- integral of h1(u) for u from 0 to t). but providing S1(t) can improves computational efficiency and usually the survival function is known in study design. The density function f1(t) = h1(t) * S1(t).
Lambda	Distribution function of enrollment. For uniform enrollment, Lambda(t) = (t/A) where A is the enrollment period, i.e., Lambda(t) = t/A for 0<=t<=A, and Lambda(t) = 1 when t > A. For more general non-uniform enrollment with weight psi, Lambda(t) = (t/A)^psi*I(0<=t<=A) + I(t>A). Default Lambda is uniform distribution function.
G0	Distribution function of lost-to-follow-up censoring process for control arm.
G1	Distribution function of lost-to-follow-up censoring process for experimental arm.

Value

An object with a dataframe below.

• ne0: number of events for control group

• ne1: number of events for experimental group

• ne: total number of events for two groups

Examples

#Example (1) Trial scenario: 1:1 randomization, n = 450, enrollment follows non-uniform 
enrollment distribution with weight 1.5 and enrollment period is 18 months. 
Control arm ~ exponential distribution with median 12 months, and 
Experimental arm ~ exponential distribution (Proportional Hazards) with median 12 / 0.7 months.
Assuming no lost-to-followup. Find the expected number of events at calendar time 24 months, i.e.
6 months after last patient randomized.

HR = 0.65; delay = 6; lambda0 = log(2) / 12; 
h0 = function(t){lambda0}; S0 = function(t){exp(-lambda0 * t)}
Hazard function and survival function for experimental arm
lambda1 = lambda0 * HR
h1 = function(t){lambda1}; S1= function(t){exp(-lambda1 * t)}
Lambda = function(t){(t/18)^1.5*as.numeric(t <= 18) + as.numeric(t > 18)}

max.DCO = 30
nE = matrix(NA, nrow = max.DCO, ncol=3)
for (DCO in 1:max.DCO) {
  o = fe(DCO = DCO, r = 1, h0 = h0, S0 = S0, h1 = h1, S1 = S1, Lambda = Lambda, n = 450)
  nE[DCO, 1] = o$e0; nE[DCO, 2] = o$e1; nE[DCO, 3] = o$e
}
plot(1:max.DCO, nE[,3], type="n", xlab="Months", ylab = "Num of events")   
lines(1:max.DCO, nE[, 1], lty = 1, col=1)
lines(1:max.DCO, nE[, 2], lty = 2, col=2)
lines(1:max.DCO, nE[, 3], lty = 3, col=3)
legend(0, max(nE), c("Control", "Experimental", "Total"), col=1:3, lty=1:3, bty="n", cex=0.8)

#Example (2) Same trial set up as example (1) but assuming delayed effect for 
experimental arm. The delayed period is assumed 6 months, and after delay the
hazard ratio is assumed 0.65.

HR = 0.65; delay = 6; lambda0 = log(2) / 12; 
h0 = function(t){lambda0}; S0 = function(t){exp(-lambda0 * t)}
Hazard function and survival function for experimental arm
h1 = function(t){lambda0*as.numeric(t < delay)+HR*lambda0*as.numeric(t >= delay)}
c = exp(-delay*lambda0*(1-HR)); 
S1 = function(t){exp(-lambda0*t)*as.numeric(t<delay) + c*exp(-HR*lambda0*t)*as.numeric(t>=delay)}
Lambda = function(t){(t/18)^1.5*as.numeric(t <= 18) + as.numeric(t > 18)}

max.DCO = 30
nE = matrix(NA, nrow = max.DCO, ncol=3)
for (DCO in 1:max.DCO) {
  o = fe(DCO = DCO, r = 1, h0 = h0, S0 = S0, h1 = h1, S1 = S1, Lambda = Lambda, n = 450)
  nE[DCO, 1] = o$e0; nE[DCO, 2] = o$e1; nE[DCO, 3] = o$e
}
plot(1:max.DCO, nE[,3], type="n", xlab="Months", ylab = "Num of events")   
lines(1:max.DCO, nE[, 1], lty = 1, col=1)
lines(1:max.DCO, nE[, 2], lty = 2, col=2)
lines(1:max.DCO, nE[, 3], lty = 3, col=3)
legend(0, max(nE), c("Control", "Experimental", "Total"), col=1:3, lty=1:3, bty="n", cex=0.8)

phe9480/rgs documentation built on March 1, 2022, 12:26 a.m.