A randomized two stage allocation for continuous response clinical trials

Abstract

A randomized two treatment allocation design, conducted in two stages, is proposed for a class of continuous response trials. Patients are assigned to each treatment in equal numbers in the first stage and p value of a test of equality of treatment effects based on these data is used to determine the assignment probability of second stage patients. Relevant properties of the proposed allocation design are investigated and compared with suitable competitors.

Appendix

Result 4.1

Suppose the response to treatment k has a normal distribution with mean $\mu _{k}$ and variance $\sigma _{k}^2$. Then, under any ${{\varvec{\xi }}}=(\mu _A,\mu _B,\sigma _A,\sigma _B)$ with $\mu _A\ne \mu _B,$ as $m\rightarrow \infty$

$$\begin{aligned} p_{m}\rightarrow 1 \text{ or } 0 \end{aligned}$$

in probability, according as $\mu _{A}>$ or $<\mu _{B}$.

Proof

Consider two statistics $T_{0m}$ and $T_{1m}$ where the distribution of $T_{0m}$ under any ${{\varvec{\xi }}}$ is the same as that of $T_{m}$ under $\mu _A=\mu _B.$ Further the distributions of $T_{1m}$ and $T_{m}$ are identical under any ${{\varvec{\xi }}}$. Then $p_m$ can be expressed as $p_m=P_{\varvec{\xi }}(T_{0m}\ge T_{1m})$. Since the first stage of the allocation design results in independent and identically distributed observations, it follows from the strong law of large numbers (Chapter 6, Billingsley 1995) that as $m\rightarrow \infty$

$$\begin{aligned} \bar{X}_{km}\rightarrow \mu _{k} \text{ and } s^{2}_{km}\rightarrow \sigma _{k}^{2} \end{aligned}$$

almost surely. Therefore, for any ${{\varvec{\xi }}}$,

$$\begin{aligned} \frac{T_{0m}}{\sqrt{m}}-\frac{T_{1m}}{\sqrt{m}}\rightarrow \frac{\mu _B-\mu _A}{\sqrt{\sigma _{A}^{2}+\sigma _{B}^{2}}} \end{aligned}$$

almost surely and consequently, as $m\rightarrow \infty$

$$\begin{aligned} p_m=P_{\varvec{\xi }}\left( \frac{T_{0m}}{\sqrt{m}}- \frac{T_{1m}}{\sqrt{m}}\ge 0\right) \rightarrow 1 \text{ or } 0 \end{aligned}$$

in probability according as $\mu _{A}>$ or $<\mu _{B}.$ Hence the result follows. $\square$

Note

If the response to treatment k has an exponential distribution with mean $\mu _{k}$, then for any $(\mu _{A},\mu _{B})$

$$\begin{aligned} \frac{T_{0m}}{T_{1m}}\rightarrow \frac{\mu _B}{\mu _A} \end{aligned}$$

almost surely, as $m\rightarrow \infty .$ The rest of the proof is exactly the same as above with ${\varvec{\xi }}=(\mu _{A},\mu _{B})$.

Result 4.2

If (2.3) holds, then under any ${{\varvec{\xi }}}$ with $\mu _A\ne \mu _B,$ as $n\rightarrow \infty$

$$\begin{aligned} \frac{N_{An}}{n}\rightarrow 1-\theta \text{ or } \theta \end{aligned}$$

in probability, according as $\mu _{A}>$ or $<\mu _{B}$.

Proof

Assume $\mu _{A}>\mu _B$. Then it follows from Result 1 that as $m\rightarrow \infty$

$$\begin{aligned} p_{m}\rightarrow 0 \end{aligned}$$

in probability. Then from (2.1) and (2.2) it follows easily that

$$\begin{aligned} E_{\varvec{\xi }}\left( \frac{N_{An}}{n}\right) \rightarrow 1-\theta \end{aligned}$$

and

$$\begin{aligned} Var_{\varvec{\xi }}\left( \frac{N_{An}}{n}\right) \rightarrow 0. \end{aligned}$$

as $n\rightarrow \infty$. Consequently, as $n\rightarrow \infty$

$$\begin{aligned} \frac{N_{An}}{n}\rightarrow 1-\theta \end{aligned}$$

in probability. A similar argument leads to the proof with $\mu _A<\mu _B$. This completes the proof. $\square$

Note

The above result holds good for the exponentially distributed responses if we redefine ${\varvec{\xi }}=(\mu _{A},\mu _{B})$.