"Isavuconazole versus voriconazole for primary treatment of invasive mould disease caused by Aspergillus and other filamentous fungi (SECURE): a phase 3, randomised-controlled, non-inferiority trial" (Maertens et al., 2016)

https://www.ncbi.nlm.nih.gov/pubmed/26684607

• Novel antifungal (isavuconazole): want to demonstrate similar efficacy but improved safety (fewer adverse events) compared to voriconazole

• "1:1 allocation"; "Roughly 255 patients"; "80% power"; "upper limit of 95% CI"; "treatment difference of 10% or less (prespecified non-inferiority margin)"; "20% mortality rate was assumed for both drugs"

#Per-arm sample size approximately matches
ceiling(2*(qnorm(0.975) + qnorm(0.8))^2/(0.10^2)*(0.2*0.8));

## [1] 252

"Tenofovir alafenamide versus tenofovir disoproxil fumarate, coformulated with elvitegravir, cobicistat, and emtricitabine, for initial treatment of HIV-1 infection: two randomised, double-blind, phase 3, non-inferiority trials" (Sax et al., 2015)

https://www.ncbi.nlm.nih.gov/pubmed/25890673

• Study of anti-retroviral treatment in untreated HIV-infected patients. Standard formulation of tenofovir causes substantial renal (kidney) toxicity. Two separate studies

• "two-sided 95% CI"; "840 patients in each study"; "at least 95% power"; "overall response rate of 85% for viral suppression at week 48"; "with a prespecified noninferiority margin of 12%"

#Expect close to 420
ceiling(2*(qnorm(0.975) + qnorm(0.95))^2/(0.12^2)*(0.15*0.85));

## [1] 231

#Closer to their answer if we assume maximum binomial variance?
ceiling(2*(qnorm(0.975) + qnorm(0.95))^2/(0.12^2)*(0.50*0.50));

## [1] 452

• Let's do some sanity checks:

na = nb = 420;#Stated sample size
nsim = 2e4;
true_pa = 0.85; 
true_pb = 0.85;
noninf_boundary = 0.12;
sim_arma = rbinom(nsim,na,true_pa);
sim_armb = rbinom(nsim,nb,true_pb);
mean_pa = sim_arma / na;
mean_pb = sim_armb / nb;
#Power
mean(((mean_pa - mean_pb) - qnorm(0.975) * sqrt(mean_pa * (1 - mean_pa) / na + mean_pb * (1 - mean_pb) / nb)) > (-noninf_boundary));

## [1] 0.998

#Avg width of CI
mean(2 * qnorm(0.975) * sqrt(mean_pa * (1 - mean_pa) / na + mean_pb * (1 - mean_pb) / nb));

## [1] 0.0965

na = nb = 230;#Our calculation
nsim = 2e4;
true_pa = 0.85; 
true_pb = 0.85;
noninf_boundary = 0.12;
sim_arma = rbinom(nsim,na,true_pa);
sim_armb = rbinom(nsim,nb,true_pb);
mean_pa = sim_arma / na;
mean_pb = sim_armb / nb;
#Power
mean(((mean_pa - mean_pb) - qnorm(0.975) * sqrt(mean_pa * (1 - mean_pa) / na + mean_pb * (1 - mean_pb) / nb)) > (-noninf_boundary));

## [1] 0.952

#Avg width of CI
mean(2 * qnorm(0.975) * sqrt(mean_pa * (1 - mean_pa) / na + mean_pb * (1 - mean_pb) / nb));

## [1] 0.13

• Magnitude of these differences slightly disconcerting

1. Calculate Wald-type test statistic

2. Find its distribution under alternative hypothesis, often a location-shifted Normal

3. Solve for \(n\)

• We've derived (many) Normal-based sample size formulas

• Often, outcomes will be yes/no or time-to-event

• Simplest strategy for sample size calculations:

• Many (most) phase 3 outcomes are times, e.g.
  • time to disease progression or death
  • time to recurrence
  • time to death

• Not all patients will experience the outcome of interest and so are not equally informative

• Question is not "how many patients do I need?" but "how much information do I need?" or "how many events do I need?"

• Using central limit theorem, sample size formulas we've already studied provide basis for suitable answers

• Usually denoted by \(\lambda(t)\), hazard is rate of occurence; instantaneous probability

• Mathematically, \[ \begin{aligned} \lambda(t) &= \lim_{\epsilon\rightarrow 0} \dfrac{ \Pr(t<X<t+\epsilon|X>t)}{\epsilon}\\ &= \dfrac{1}{\Pr(X>t)} \lim_{\epsilon\rightarrow 0} \dfrac{\Pr(X>t) - \Pr(X>t+\epsilon)}{\epsilon}\\ &= \dfrac{1}{\Pr(X>t)}\left(-\dfrac{d}{dt} \Pr(X>t)\right)\\ &= \dfrac{f(t)}{\Pr(X>t)} \end{aligned} \]

• \(\lambda(t)>0\). Larger values indicate events, i.e. death, disease progression, recurrence, exacerbation of symptoms, occurring at higher rate

• Formally, \(\Pr(X>t) = \exp\{-\int_0^t \lambda(u) du\}\)

• Formally, \(\Pr(X>t) = \exp\{-\int_0^t \lambda(u) du\}\)

• Do events occur at higher rate in arm B than arm A?

• \(\lambda_B(t)/\lambda_A(t)>1\Leftrightarrow\log\lambda_B(t)-\log\lambda_A(t)>0\)

• Assume complete follow-up, time-independent hazards, i.e. \(\lambda_A(t)\equiv\lambda_A\)

• Collect \(V_1,\ldots,V_{n_A} \sim \text{Exp}(\lambda_A)\) from \(n_A\) iid subjects (experimental arm)

• Collect \(W_1,\ldots,W_{n_B} \sim \text{Exp}(\lambda_B)\) from \(n_B\) iid subjects (standard of care/control/placebo)

• \(H_0\): \(\log\lambda_B-\log\lambda_A = 0\)
• \(H_1\): \(\log\lambda_B-\log\lambda_A = \delta\)

• MLE for \(\lambda_A\) is \(\hat\lambda_A = 1/ \bar V\). For large \(n\), \[ \begin{aligned} \hat\lambda_A & \overset{\cdot}{\sim} N\left(\lambda_A ,\dfrac{1}{n_AI(\lambda_A)}\right)\\ & \overset{d}{=} N\left(\lambda_A ,\dfrac{\lambda_A^2}{n_A}\right)\\ \end{aligned} \]

\(I(\lambda_A)\) is expected Fisher information from one subject for estimating \(\lambda_A\).

\((\hat\lambda_A - \lambda_A)/(\lambda_A/\sqrt{n_A})\overset{\cdot}{\sim} N(0,1)\)

set.seed(1);
lambda = 0.025;
base_draws = matrix(rexp(100 * 1e4), nrow = 1e4);
nsamp = 5;
ggplot() + 
  geom_histogram(aes(x = ( 1/rowMeans(base_draws[,1:nsamp,drop=F]/lambda) - lambda)/ (lambda / sqrt(nsamp)),
                     y = ..density..),
                 bins=30) + 
  stat_function(aes(x = c(-2,2)), fun = dnorm, n = 301, args = list(mean = 0, sd = 1)) + 
  theme(text = element_text(size = 14)) + 
  labs(x = "n = 5", y = "");

\[ \begin{aligned} -\log \hat\lambda_A = \log \bar V &\overset{\cdot}{\sim}N\left(-\log \lambda_A,E(V_1)^2\dfrac{\lambda_A^2}{n_A}\right)\\ &\overset{d}{=} N(-\log\lambda_A,1/n_A) \end{aligned} \] Similarly, \(\log \bar W \overset{\cdot}{\sim} N(-\log\lambda_B,1/n_B)\)

• \(T = \dfrac{\log \bar V - \log \bar W}{\sqrt{1/n_A + 1/n_B}}\)

• Let \(n_B = rn_A\)

• Under \(H_0\), \(T\sim N(0,1)\);

• Under \(H_1\), \(T\sim N\left(\delta\sqrt{n_A}/\sqrt{1 + 1/r},1\right)\)

• \(n_A = \dfrac{(z_{1-\alpha} + z_{1-\beta} )^2}{\delta^2}(1 + 1/r)\)

• \(n_B = r n_A\)

Sample size formulae assumes that event time is measured for all subjects. Usually cannot count on this:

1. Trial may end before all subjects experience event ("administrative censoring")

2. Subjects will not enroll instantaneously ("staggered enrollment")

3. Subjects may be lost to follow-up before experiencing event ("lost to follow-up")

• Less follow-up per subject requires enrolling more subjects.

• Trial ends at some time \(t_0\)
• Some subjects will not have event by time \(t_0\)
• Data are \(R_i = \min\{V_i, t_0\}\) and \(S_i =\min\{W_i,t_0\}\)

• Likelihood contribution from individual \(i\) in arm A is \(\lambda_A \exp\{-\lambda_A R_i \}\) if \(V_i < t_0\) (exponential density) and \(\Pr(V_i > t_0)= \exp\{-\lambda_A t_0\}\) if \(V_i \geq t_0\)

• Use this to show that MLE for hazard is \(\hat\lambda_A = \dfrac{\sum_i 1_{[V_i<t_0]}}{\sum_i R_i}\) (Observed events divided by total follow-up)

• (Expected) Fisher information is \((1-\exp\{-\lambda_A t_0\})/\lambda_A^2\), so that \[ \hat\lambda_A \overset{\cdot}{\sim} N\left(\lambda_A, \dfrac{1}{n_A}\dfrac{\lambda_A^2}{1-\exp\{-\lambda_A t_0\}} \right) \]

Applying Delta method,

\[ \begin{aligned} -\log\hat\lambda_A &\overset{\cdot}{\sim} N\left(-\log\lambda_A,\left(\dfrac{1}{\lambda_A}\right)^2\dfrac{1}{n_A}\dfrac{\lambda_A^2}{1-\exp\{-\lambda_A t_0\}} \right)\\ &\overset{d}{=} N\left(-\log\lambda_A,\dfrac{1}{n_A}\dfrac{1}{1-\exp\{-\lambda_A t_0\}} \right) \end{aligned} \]

Compare to no-censoring case: asymptotic variance of \(\log\hat\lambda_A\) multiplicatively increases by \(\left(1-\exp\{-\lambda_A t_0\}\right)^{-1}\), i.e. inverse of probability of arm A subject experiencing event.

As \(t_0\) increases, amount of censoring […]?

• \(T = \dfrac{-\log \hat\lambda_A + \log \hat\lambda_B}{\sqrt{\dfrac{1}{n_A(1- \exp\{-\lambda_A t_0\})} + \dfrac{1}{n_B(1- \exp\{-\lambda_B t_0\})}}}\)

• Under \(H_0\), \(T\sim N(0,1)\)

• \(1- \exp\{-\lambda_A t_0\} = \Pr(V_i < t_0)\)

\[ \begin{aligned} n_A &= \dfrac{(z_{1-\alpha} + z_{1-\beta} )^2}{\delta^2}\times\\ &\quad([1- \exp\{-\lambda_A t_0\}]^{-1} + [1- \exp\{-\lambda_B t_0\}]^{-1}/r) \end{aligned} \]

For simplicity, may be reasonable to assume

\[ ([1- \exp\{-\lambda_A t_0\}]^{-1} + [1- \exp\{-\lambda_B t_0\}]^{-1}/r)\approx (1+1/r)/\pi, \]

where

\[ \pi = \dfrac{1}{1+r}\left(1 - \exp\{-\lambda_A t_0\}\right) + \dfrac{r}{1+r}\left(1-\exp\{-\lambda_B t_0\}\right) \]

so that

\[ \begin{aligned} n_A &= \dfrac{(z_{1-\alpha} + z_{1-\beta} )^2}{\pi\delta^2}(1 + 1/r) \end{aligned} \]

• \(\pi\) is crude approximation for average probability of observing event in single subject under \(H_1\)

• If \(\pi\) is small, more subjects are required

• Denote \(\pi^{-1}\) by \(\text{IF}\), which always exceeds 1 and stands for "inflation factor".

• Test of monoclonal antibody belimumab in autoimmune disease lupus
• Primary efficacy endpoint was improvement in the Systemic Lupus Erythematosus Responder Index (SRI) at week 52
• 1:1:1 randomization to one of two dose levels or placebo
• Powered for comparison of one dose level against placebo arm

• Text your calculated sample size via PollEverywhere

ceiling(2*((qnorm(0.975) + qnorm(0.90))^2/(0.14^2)) * 0.5^2);

## [1] 269

Reference: "Efficacy and safety of belimumab in patients with active systemic lupus erythematosus: a randomised, placebo-controlled, phase 3 trial" (Navarra et al., 2011)

Assumption 1: Censoring mechanism is independent of event mechanism

Assumption 2: Hazard is constant (will discuss this soon)

1. Apply appropriate Normal-based sample size formula (tells you how many events needed)

2. Identify estimate of \(\text{IF}=\pi^{-1}\)

3. Multiplicatively increase (1) by (2) (tells you how many subjects to enroll to get needed events`)

• Let \(C_i\) indicate censoring time, \(R_i = \min\left\{V_i, C_i\right\}\) be observed follow-up, and \(\varepsilon_i = 1_{[V_i<C_i]}\) be event indicator
• Assume \(C_i\perp V_i\) and \(V_i\) is exponential with rate \(\lambda_A\)
• Observed information is \(\tfrac{\sum_i \varepsilon_i}{\hat\lambda_A^2}\), where \(\hat\lambda_A = \sum_i \varepsilon_i / \sum_i R_i\). The average information contribution then becomes

\[ \begin{aligned} \dfrac{1}{n_A}\dfrac{\sum_i \varepsilon_i}{\hat\lambda_A^2}\rightarrow_P \dfrac{\Pr(V_i < C_i)}{\lambda_A^2} \equiv \dfrac{\Pr(\text{observing event})}{\lambda_A^2} \end{aligned} \]

• Can also be written as

\[ \begin{aligned} \dfrac{1}{n_A}\dfrac{\sum_i \varepsilon_i}{\hat\lambda_A^2} = \dfrac{1}{n_A}\dfrac{\sum_i R_i}{\hat\lambda_A} \rightarrow_P \dfrac{E[R_i]}{\lambda_A} \end{aligned} \]

• Note that \(\lambda_A = 1/E[V_i]\), i.e. inverse of expected follow-up under no censoring or simply expected event time. Putting these together, we have \[ \begin{aligned} \Pr(\text{observing event})=\dfrac{E[R_i]}{E[V_i]} \end{aligned} \]

• Patients enroll throughout the trial (rather than time zero), \({E_{iA}}\sim \text{Unif}(0,t_0)\), \({E_{iB}}\sim \text{Unif}(0,t_0)\)

• Data are \(R_i = \min\{V_i, t_0-{E_{iA}}\} \overset{d}{=}\min\{V_i, {E_{iA}}\}\) and \(S_i =\min\{W_i,t_0-{E_{iB}}\}\overset{d}{=}\min\{W_i,{E_{iB}}\}\)

Expected follow-up time (length of "stick") in arm A is \[ \begin{aligned} E[R_i] &= \dfrac{t_0\lambda_A - (1 - \exp\{-\lambda_A t_0\})}{t_0\lambda_A^2} \end{aligned} \] and probability of observing event is \[ \begin{aligned} \Pr(\varepsilon_i = 1) = \Pr(V_i < t_0 - E_{iA}) = \dfrac{t_0\lambda_A - (1 - \exp\{-\lambda_A t_0\})}{t_0\lambda_A} \end{aligned} \]

Similar calculation for Arm B, then calculate \(\pi\) as before:

\[ \begin{aligned} \pi = \dfrac{1}{1+r}\Pr(V_i < t_0 - E_{iA}) + \dfrac{r}{1+r}\Pr(W_i < t_0 - E_{iB}) \end{aligned} \]

• All subjects followed for additional period of \(t_1-t_0\), to ensure minimum possible follow-up
  
• Data are \(R_i = \min\{V_i, t_1-{E_{iA}}\}\) and \(S_i =\min\{W_i,t_1-{E_{iB}}\}\)

Expected follow-up time in arm A \[ \begin{aligned} E[R_i] &= \dfrac{t_0\lambda_A - \exp\{-\lambda_A(t_1-t_0)\}(1 - \exp\{-\lambda_A t_0\})}{t_0\lambda_A^2} \end{aligned} \] and probability of observing event is \[ \begin{aligned} \Pr(V_i< t_1-E_{iA}) = \dfrac{t_0\lambda_A - \exp\{-\lambda_A(t_1-t_0)\}(1 - \exp\{-\lambda_A t_0\})}{t_0\lambda_A} \end{aligned} \]

• "Dabrafenib in BRAF-mutated metastatic melanoma: a multicentre, open-label, phase 3 randomised controlled trial" (Hauschild et al., 2012)

https://www.ncbi.nlm.nih.gov/pubmed/22735384

• BRAF-inhibitor (arm A) versus chemotherapy (arm B).

• Wanted 3:1 randomization ratio in favor of BRAF-inhibitor arm: \(n_B = (1/3)n_A\), or \(r=1/3\)

• Median time to progression or death is 2 months on chemotherapy. Expect increase to 6 months on dabrafenib. For exponential distribution, \(\text{median} = \log(2)/\lambda\). So aim to establish

\[ \delta = \log \lambda_B - \log \lambda_A = \log(\log(2)/2) - \log(\log(2)/6) = \log(3) \]

• Targeting power \(1-\beta = 0.997\); one-sided type I error rate \(\alpha = 0.02\).

r=1/3;
lambdaA = log(2)/6;
lambdaB = log(2)/2;
#na
(na = ceiling((1+1/r)*(qnorm(0.98) + qnorm(0.997))^2/log(lambdaB/lambdaA)^2));

## [1] 77

#nb
(nb = ceiling(r * na));

## [1] 26

• Our calculation of 103 is close to authors' 102

• Planned to enroll and randomize 200 patients (implies \(\pi = 102/200=0.51\) and \(\text{IF}=1.96\))

• Enrollment period from Dec 23, 2010 to Sept 1, 2011 (\(t_0=8.25\) months)

• Administratively censored on Dec 19, 2011 (\(t_1-t_0=3.5\) months)

t0 = 8.25;t1 = t0 + 3.5;
piA = (t0*lambdaA - exp(-lambdaA*(t1-t0))*
                 (1-exp(-t0*lambdaA)))/(t0*lambdaA);
piB = (t0*lambdaB - exp(-lambdaB*(t1-t0))*
                 (1-exp(-t0*lambdaB)))/(t0*lambdaB);
pi = piA/(1+r) + piB*r/(1+r);
1/pi;#IF

## [1] 1.53

• We'll revisit this soon

• "Afatinib versus cisplatin plus gemcitabine for first-line treatment of Asian patients with advanced non-small-cell lung cancer harbouring EGFR mutations (LUX-Lung 6): an open-label, randomised phase 3 trial" (Wu et al., 2014)

https://www.ncbi.nlm.nih.gov/pubmed/24439929

• Treatment for non-small cell lung cancer.

• Testing for \(\delta = \log(1.57)\) (median survival of 7 months in arm A versus 11 months in arm B)

• Targeting power \(1-\beta = 0.90\); type I error rate \(\alpha = 0.025\).

r=1/2;
lambdaA = log(2)/11;
lambdaB = log(2)/7;
#na 
(na = ceiling((1+1/r)*(qnorm(0.975) + qnorm(0.90))^2/log(lambdaB/lambdaA)^2));

## [1] 155

#nb
(nb = ceiling(r * na));

## [1] 78

• Thus, require 233 events (authors calculate 217)

• Planned to enroll 330 patients (implies \(\text{IF}=330/217=1.52\))

• Enrollment period from April 27, 2010 to Nov 16, 2011 (\(t_0=18.5\) months)

• Administratively censored on Oct 29, 2012 (\(t_1-t_0=11.5\) months)

t0 = 18.5;t1 = t0 + 11.5;
piA = (t0*lambdaA - exp(-lambdaA*(t1-t0))*
                 (1-exp(-t0*lambdaA)))/(t0*lambdaA);
piB = (t0*lambdaB - exp(-lambdaB*(t1-t0))*
                 (1-exp(-t0*lambdaB)))/(t0*lambdaB);
pi = piA/(1+r) + piB*r/(1+r);
1/pi;#IF

## [1] 1.32

Why were our calculated IFs smaller than papers' IFs? One possible reason: subjects will drop out of studies with long follow-up (for reasons unrelated to outcome), e.g. \(F_{iA}\sim \text{Exp}(\eta)\)

\[ \begin{aligned} R_i &= \min\{V_i, {F_{iA}}, t_1-{E_{iA}}\}\\ &\overset{d}{=}\min\{\min\{V_i,{F_{iA}}\}, t_1-{E_{iA}}\}\\ &\overset{d}{=}\min\{V^*_i, t_1-{E_{iA}}\} \end{aligned} \]

What is distribution of \(V^*_i\)?

\[ \begin{aligned} \Pr(V^*_i > u) &= \Pr(V_i>u, F_{iA} >u)\\ &= \exp\{-\lambda_A u\} \exp\{-\eta u\}\\ &= \exp\{-(\lambda_A+\eta) u\}\\ \Rightarrow V^*_i&\sim\text{Exp}(\lambda_A+\eta) \end{aligned} \]

Expected follow-up time in arm A is \[ \begin{aligned} E[R_i] &= \dfrac{t_0(\lambda_A+\eta) - \exp\{-(\lambda_A+\eta)(t_1-t_0)\}(1 - \exp\{-(\lambda_A+\eta) t_0\})}{t_0(\lambda_A+\eta)^2} \end{aligned} \]

and probability of observing event is

\[ \begin{aligned} &\Pr(V_i< \min\{F_{iA}, t_1- E_{iA}\}) =\\ &=\dfrac{\lambda_A\left[t_0(\lambda_A+\eta) - \exp\{-(\lambda_A+\eta)(t_1-t_0)\}(1 - \exp\{-(\lambda_A+\eta) t_0\})\right]}{t_0(\lambda_A+\eta)^2} \end{aligned} \]

• So far, have assumed \(\lambda_A(t)\equiv \lambda_A\) (likewise for \(\lambda_B\)).

• Can relax this to assume that hazards change over time but remain proportional:

  • \(H_0\): \(\log\lambda_B(t)-\log\lambda_A(t) = 0\)
  • \(H_1\): \(\log\lambda_B(t)-\log\lambda_A(t) = \delta\)

• Very common test for equality of survival curves (or hazards) is non-parametric "log-rank test"

• Assuming \(H_0\), i.e. equality of curves, and conditional on data, each event should occur with probability proportional to the number at risk (uncensored and free of event) at that time.

• Let \(j=1,\ldots,\pi(n_A+n_B)\) index the number of events observed out of \(n_A+n_B\) total patients

• If \(n_{Aj}\) arm A subjects and \(n_{Bj}\) arm B subjects at risk at time \(t_j\) (\(n_j = n_{Aj} + n_{Bj}\)), then, knowing that subject will experience event at \(t_j\), expect \(p_{Aj}=n_{Aj}/n_j\) events in arm A and \(p_{Bj}=n_{Bj}/n_j\) events in arm B.

• We observe event in arm B (suppose), so discrepancy between observed and expected at time \(j\) is \(O_{Bj} - p_{Bj} = 1 - n_{Bj}/n_j= n_{Aj}/n_j\)

• Variance of this discrepancy is \(V_j = p_{Bj}(1-p_{Bj})\)

• Under \(H_0\), each \(O_{Bj} - p_{Bj}\) has mean zero and no correlation with \(O_{Bj'} - p_{Bj'}\), for \(j\neq j'\)

\[ \begin{aligned} T = \dfrac{\sum_j O_{Bj} - p_{Bj}}{\sqrt{\sum_j p_{Bj}(1-p_{Bj}}} \end{aligned} \]

• Under \(H_0\), \(T \sim N\left(0,1\right)\)

Under \(H_1\),

\[ \begin{aligned} E[O_{Bj} - p_{Bj}] &= \dfrac{\lambda_B(t_j) n_{Bj}}{\lambda_A(t) n_{Aj} + \lambda_B(t_j) n_{Bj}} - \dfrac{ n_{Bj}}{n_{Aj} + n_{Bj}}\\ &= \dfrac{e^\delta n_{Bj}}{ n_{Aj} + e^\delta n_{Bj}}- \dfrac{ n_{Bj}}{n_{Aj} + n_{Bj}}\\ &= \text{...algebra...}\\ &= \left(\dfrac{e^\delta - 1}{n_{Aj} + n_{Bj} e^\delta}\right)\left(\dfrac{n_{Aj}n_{Bj}}{n_{Aj}+n_{Bj}}\right)\\ &\approx \left(\dfrac{\delta}{n_{Aj} + n_{Bj}}\right)\left(\dfrac{n_{Aj}n_{Bj}}{n_{Aj}+n_{Bj}}\right)\quad\text{(next slide)}\\ &=\delta p_{Bj}p_{Aj}=\delta p_{Bj}(1-p_{Bj}) \end{aligned} \]

\((n_{Aj} + n_{Bj})\left(\dfrac{e^\delta - 1}{n_{Aj} + n_{Bj} e^\delta}\right)\) versus \(\delta\) for \(n_{Aj}=n_{Bj}=50\)

Asymptotically, under \(H_1\),

\[ \begin{aligned} E[T] &\approx \delta \dfrac{\sum_j p_{Bj}(1-p_{Bj}) }{\sqrt{\sum_j p_{Bj}(1-p_{Bj}) }} \\ &= \delta \sqrt{\sum_j p_{Bj}(1-p_{Bj})}\\ &\approx \delta \sqrt{\pi (n_A+n_B)\dfrac{n_An_B}{(n_A+n_B)^2}}\quad\left(\text{using } p_{Bj}\approx n_B/(n_A+n_B)\right)\\ &= \delta \sqrt{\dfrac{\pi n_A}{1+1/r}}\\ \end{aligned} \]

• \(T \overset{\cdot}{\sim} N\left(\delta\sqrt{\pi n_A}/\sqrt{1 + 1/r},1\right)\)

• Sample size formula is identical (!):

• \(n_A = \dfrac{(z_{1-\alpha} + z_{1-\beta} )^2}{\pi \delta^2}(1 + 1/r)\)

• \(n_B = r n_A\)

(Schoenfeld, 1981), (Schoenfeld, 1983)

1. Our approach emphasizes difference between number of events required and number of subjects to enroll in order to observe events.

   • Adds layer of approximation, uncertainty.
   • Actual values of hazards (rather than only relative relationship) matter if patients are censored

2. Can be difficult to re-create others' power calculations. Important to provide details on assumptions used to calculate power/sample size

3. Simulation studies to confirm, explore power characteristics very useful

4. Implict assumption is that censoring mechanism is (i) unrelated to event mechanism and (ii) non-differential across arms

See Chapter 4 (Cook and DeMets, 2007)