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Thus, consider the matrix cov( ˆ β$_{l}$ , ˆ β$_{l}$ | ω , ψ ) for some l ≤ L and any ( ω , ψ ). We compute values for L$_{lr}$ ( r = 1 , . . . , D ); the number of stages of the trial, up to analysis l , in which r treatments were remaining. Since we do not continue the trial unless at least one experimental treatment remains, L$_{l}$$_{1}$ = 0 always. It will be convenient however to still include this value. Moreover, it is clear that the L$_{lr}$ are uniquely determined given ( ω , ψ ). Now, cov( ˆ β$_{l}$ , ˆ β$_{l}$ | ω , ψ ) can always be decomposed to be a sum over the determined L$_{lr}$ and the pre-specified sequences S$_{r}$ (see Fitzmaurice et al. (2011) for details) 

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Here X$_{s}$$_{ri}$ is the uniquely defined r × (2 D − 1) design matrix for a single patient allocated to sequence s$_{ri}$ , and Σ$_{r}$ is the easily computed r × r covariance matrix of the responses for a single patient allocated r treatments in total. The factor n/ | S$_{r}$ | arises from the number of patients allocated to each sequence s$_{ri}$ by our choice of period balance. 

We now establish two key results about cov( ˆ β$_{l}$ , ˆ β$_{l}$ | L$_{l}$$_{1}$, . . . , L$_{lD}$ ). Following this, we provide a proof detailing how to strongly control the familywise error rate. 

Theorem 2.1. Let β = ( µ$_{0}$, π$_{2}$, . . . , π$_{D}$, τ$_{1}$, . . . , τ$_{D}$$_{−}$$_{1}$ ) $^{T}$. Consider an analysis to be performed after some number of stages l . Then 

1. We have 

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where 

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2. If q ≥ 2 is the largest integer such that L$_{lr}$ = 0 for r = 1 , . . . , q − 1 , then the covariance of the estimates of the fixed effects ˆ π$_{2}$$_{l}$, . . . , ˆ π$_{ql}$, ˆ τ$_{1}$$_{l}$, . . . , ˆ τ$_{q}$$_{−}$$_{1}$$_{l}$ is identical to that it would be for L$_{l}$$_{1}$ = · · · = L$_{lD}$$_{−}$$_{1}$ = 0 . Moreover, the covariance between the estimates of ˆ π$_{2}$$_{l}$, . . . , ˆ π$_{ql}$, ˆ τ$_{1}$$_{l}$, . . . , ˆ τ$_{q}$$_{−}$$_{1}$$_{l}$ and the estimates of ˆ π$_{q}$$_{+1}$$_{l}$, . . . , ˆ π$_{Dl}$, ˆ τ$_{ql}$, . . . , ˆ τ$_{D}$$_{−}$$_{1}$$_{l}$ is also identical to that it would be for L$_{l}$$_{1}$ = · · · = 

