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Communication

Estimating the Parameter of Direct Effects in Crossover Designs: The Case of 6 Periods and 2 Treatments

by
Miltiadis S. Chalikias
Department of Accounting and Finance, School of Business, Economics and Social Sciences, University of West Attica, 12244 Egaleo, Greece
Stats 2026, 9(1), 17; https://doi.org/10.3390/stats9010017
Submission received: 9 January 2026 / Revised: 2 February 2026 / Accepted: 10 February 2026 / Published: 12 February 2026
(This article belongs to the Section Statistical Methods)

Abstract

The present study investigates the derivation of optimal repeated measurement designs for two treatments, six periods, and n experimental units, focusing exclusively on the direct effects of the treatments. The optimal designs are determined for cases where n ≡ 0 or 1, 2, 3, 4 (mod 4). The adopted optimality criterion aims at minimizing the variance of the estimator of the direct effects, thereby ensuring maximum precision in parameter estimation and increased design efficiency. The results presented extend and complement earlier studies on optimal two-treatment repeated-measurement designs for a smaller number of periods, and are closely related to more recent work focusing on optimality with respect to direct effects. Overall, this work contributes to the theoretical framework of optimal design methodology by providing new insights into the structure and efficiency of repeated measurement designs, and lays the groundwork for future extensions incorporating treatment–period interactions.

1. Introduction

Crossover designs or Repeated measurement Designs are widely used in experimental research, particularly when variability between units is substantial. These designs involve observing the same experimental units over multiple periods, applying one treatment per period. By allowing each unit to serve as its own control, crossover designs improve precision and efficiency in estimating treatment effects [1].
In clinical trials, crossover designs enable subjects to receive multiple treatments across successive periods, facilitating direct within-subject comparisons and reducing variability due to individual differences [2]. Such designs are also applied in psychology, physiology, and industrial experiments, highlighting their broad utility [1].
A key challenge in crossover designs is accounting for carryover effects, where responses in later periods are influenced by earlier treatments. Ignoring these effects can bias estimates of direct treatment effects. Statistical models therefore distinguish between direct effects and carryover effects, and precise estimation requires balanced designs that control for subject, period, and treatment sequence effects [2,3,4].
Crossover designs for two-treatment experiments have been extensively studied in the statistical and clinical trial literature, owing to their efficiency in controlling between-subject variability and their ability to facilitate within-subject comparisons. Early methodological contributions emphasized both the potential advantages and the inherent limitations of two-treatment crossover trials, particularly in the presence of carryover effects and period imbalance [5]. Subsequent work has focused on the development of optimal design strategies, including variational approaches for minimizing the impact of carryover effects while preserving efficiency in the estimation of treatment contrasts [6]. More recent studies have provided systematic treatments of crossover designs tailored to two-treatment clinical trials, highlighting practical design considerations and assumptions required for valid inference [7,8]. Together, these contributions establish a solid methodological foundation for the study of optimal crossover designs and motivate further investigation of designs with an increased number of periods.
Circular crossover designs of two treatments have also attracted increasing attention in recent literature, as they allow for improved balance and symmetry in repeated measurement experiments. Recent contributions have investigated optimality properties and construction methods for two-treatment circular designs, particularly for small numbers of periods, and have established optimal and MSE-optimal solutions under various modeling assumptions [9,10]. Related work has further examined mean-squared error optimality in circular crossover settings, highlighting the practical advantages of circularity in controlling bias and variance [11].
Linear modeling frameworks for two-treatment designs with a smaller number of periods have been developed to formalize the estimation of direct and carryover effects. While designs with two, three [12], or four periods [13] are well understood, the complexity of achieving balance increases rapidly as the number of periods increases [14]. This paper extends this line of work to the case of six periods and opens a pathway toward addressing designs with an even larger number of periods.
The six-period case is particularly relevant for longitudinal clinical and pharmacological studies, where multiple periods are necessary to capture treatment dynamics and residual effects. Optimal design strategies for two-treatment, six-period experiments have not been investigated, especially regarding efficient estimation of the direct effect contrast τ = τ A τ B .
The present study addresses this gap by deriving conditions for optimal repeated measurement designs in the two-treatment, six-period setting. Using established modeling approaches, we identify classes of designs that allow efficient estimation of direct effects while accounting for subject and period variability [14]. This provides the first systematic characterization of optimal designs for direct effect estimation in this setting.
The remainder of the paper presents the statistical model and assumptions, describes the optimal designs and their efficiency properties, and concludes with a discussion and directions for future research.
The remainder of the paper is organized as follows. Section 2 presents the statistical model and assumptions. Section 3 describes the resulting optimal designs and their efficiency properties. Section 4 concludes with a discussion and directions for future research.

2. Methodology

2.1. The Model

Using two treatments and six periods, 64 sequences of treatments A and B exists are presented in the following Table 1 (where u i , i = 0 ,   1 , ,   63 is the number of experimental units that received the i -th sequence of treatments so u 0 + + u 63 = n , and all e.us are n). For example, the 19th sequence is BBAABA because B is in the positions 0, 1, 3, 4 and 19 = 2 0 + 2 1 + 2 4 .
The repeated measurement model for a two-treatment, multi-period design, according to [2,3], can be expressed as:
y i j k = μ + τ h + π j + δ i , j 1 + γ i + e i j k ,
where y i j k is the response recorded for unit k that follows sequence i in period j .
Index notation:
j = 1 ,   2 ,   3 ,   4 ,   5 ,   6 , indicates the period.
i = 0 , 1 , , 63 , represents the treatment sequence.
k = 1 ,   2 , , n , indexes the experimental unit.
Model Components:
  • τ h : denotes the direct effect of the treatment applied in period j (either A or B) τ A , τ B .
  • π j : captures the period effect, which accounts for systematic variation across different measurement periods.
  • δ A , δ B : represents the carryover effect from the treatment administered in the previous period.
  • γ i : is the sequence effect associated with the i-th sequence, reflecting the influence of treatment order
  • e i j k is the residual error term, assumed independent and normally distributed with mean zero, representing unexplained variability.
This formulation allows separate estimation of direct treatment effects, period effects, carryover effects, and sequence effects, while accommodating random variation between experimental units.
The above model (in overparameterized form) is written:
Y = τ A τ A + τ B τ B + δ A δ A + δ B δ B + π 1 π 1 + + π 6 π 6 + + γ 0 γ 0 + + γ 63 γ 63 + e
and Y , τ A , τ B , δ A , δ B , π 1 , π 6 , γ 0 , γ 63 , e are vectors of dimension 1 × 6 n .
The construction of the vectors is defined in [9] using 1 if the treatment is A, and zero if it is B. Hence τ A + τ B = 1 6 n , δ A + δ B + π 1 = 1 6 n , and π 1 + π 2 + π 3 + π 4 + π 5 + π 6 = 1 6 n . Also, it takes the value 1 when the i -th unit is employed and 0 otherwise, so that γ 0 + γ 1 + γ 2 + + γ 63 = 1 6 n .
For the uniqueness of the model we use the limits τ B = 0 , π 6 = 0 , γ 63 = 0 .
In a vector form:
Y = Χ b + e Y = X 1 X 2 b 1 b 2 ,
where Y is ( 6 n ) × 1 , b is s × 1 where s = 1 , Χ is ( 6 n ) × 1 , e is ( 6 n ) × 1 , b = b 1 b 2 , where b 1 the (vector of) direct effect τ A and b 2 the remaining parameters.
For the estimation of the difference of direct effect τ = τ A τ B after the limitation τ B = 0 of the direct effect τ A the variance is:
Var ( τ ^ A ) = σ 2 ( Χ 1 Τ Χ 1 Χ 1 Τ P X 1 ) 1 = σ 2 ( X 1 ( I 6 n P ) X 1 ) 1 = σ 2 Q 1 ,
where P = X 2 ( X 2 T X 2 ) 1 X 2 T is the 6 n × 6 n projection matrix of X 1 onto the space R ( X 2 ) , Q = Χ 1 Τ Χ Χ 1 Τ Ρ Χ 1 and X 1 = τ A , X 2 = μ , π 1 , π 2 , , π 5 , δ A , δ B , γ 0 , γ 1 , , γ 62 .
The ( 6 n ) × 1 vectors (columns) τ A , μ , π 1 , π 2 , , π 5 , δ A , δ B , γ 0 , γ 1 , , γ 62 correspond to the parameters τ A , μ , π 1 , π 2 , , π 5 , δ A , δ B , γ 0 , γ 1 , , γ 62 .
Proposition 1.
Linear space  R ( X 2 )  remains the same if we use the basis  π 1 , π 2 , π 3 , π 4 , π 5 , δ A δ B , γ 0 , γ 1 , , γ 63 .
Proof of Proposition 1.
We observe that 1 6 n = μ = π 1 + δ A + δ B , hence μ can be omitted. Moreover R ( X 2 ) does not change if the δ A and δ B replaced by δ A δ B , δ A + δ B . Finally, since γ 0 + γ 1 + + γ 63 = 1 6 n , δ A + δ B can be replaced with γ 63 . □
Proposition 2.
The quantity  Χ 1 Τ ( Ι 6 n P ) X 1 , where  P = X 2 ( X 2 T X 2 ) 1 X 2 T , does not change if, instead of  X 2 , we use the matrix with the vectors  π 1 , π 2 , π 3 , π 4 , π 5 , δ A δ B , γ 0 , γ 1 , , γ 63 .
Proof of Proposition 2.
Since R ( X 2 ) does not change and P X 1 is the orthogonal projection of X 1 onto R ( X 2 ) , then the orthogonal projection P X 1 remains unchanged. Moreover P is idempotent so P = P 2 . □
Matrices X 1 , X 2 are of dimensions ( 6 n ) × 1 and ( 6 n ) × ( 70 ) respectively:
X 1 = X 1 0 X 1 1 X 1 , 62 X 1 , 63 ,   X 2 = X 20 1 6 0 0 0 X 21 0 1 6 0 0 X 2 , 62 0 0 1 6 0 X 2 , 63 0 0 0 1 6 ,
where: X 1 , 0 = 1 1 1 1 1 1 ,   X 1 , 1 = 0 1 1 1 1 1 , ,   X 1 , 62 = 1 0 0 0 0 0 ,   X 1 , 63 = 0 0 0 0 0 0 ,
X 2 , 0 = π 1   π 2   π 3   π 4   π 5   δ 1 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 1 0 0 0 0 1 1 0 0 0 0 0 1 ,   X 2 , 1 = π 1   π 2   π 3   π 4   π 5   δ 1 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 1 0 0 0 0 1 1 0 0 0 0 0 1 , ,
X 2 , 62 = π 1   π 2   π 3   π 4   π 5   δ 1 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 1 0 0 0 0 1 1 0 0 0 0 0 1 , X 2 , 63 = π 1   π 2   π 3   π 4   π 5   δ 1 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 1 0 0 0 0 1 1 0 0 0 0 0 1 ,
with δ we notify δ A δ B .
The calculation of P is difficult, and it is better to calculate X 1 T ( I 6 n P ) X 1 , which is the square of the distance of vector X 1 P X 1 = ( I 6 n P ) X 1 , with different way of making calculations.

2.2. Calculation of Q

We need to calculate the minimum of the X 1 T ( I 6 n P ) X 1
Proposition 3.
If  w = ( x 1 , x 2 , x 3 , x 4 , x 5 , x 6 , z 0 , z 1 , , z 63 ) T ,  then  X 1 T I 6 n P X 1 = min w { ( X 1 X 2 w ) T ( X 1 X 2 w ) }
Proof of Proposition 3.
If T is a point of R ( X 2 ) with coordinates X 2 w , then ( X 1 X 2 w ) T ( X 1 X 2 w ) is the squared distance of X 1 from that point. Hence
Q = X 1 T ( I 6 n P ) X 1 = m i n w { ( X 1 X 2 w ) T X 1 X 2 w } .
The method of computation has been presented in [14]: We define the function
F ( x , z ) = ( X 1 X 2 w ) T X 1 X 2 w .
In terms of 70 variables ( x 1 , x 2 , x 3 , x 4 , x 5 , x 6 , z 0 , z 1 , , z 63 ) .
F ( x , z ) = ( X 1 X 2 w ) T ( X 1 X 2 w ) , then
F ( x , z ) = Χ 1 Τ Χ 1 2 Χ 1 Τ X 2 w + w T X 2 T X 2 w
where
X 1 T X 1 = u 0 X 10 T X 10 + u 1 X 11 T X 11 + + u 64 X 1 , 64 T X 1 , 64 X 1 T Χ 2 Τ w = u 0 X 10 T ( X 20 + z 0 1 6 ) + u 1 X 11 T ( X 21 + z 1 1 6 ) + + u 127 X 1 , 64 T ( X 2 , 64 + z 64 1 6 ) Χ 2 Τ Χ 2 Τ w = u 0 [ Χ 2 , 0 Τ X ~ 2 0 + 2 z 0 Χ 20 Τ 1 6 + z 0 2 1 6 T 1 6 ] + u 1 [ Χ 21 Τ X 21 + 2 z 1 X 21 1 7 + z 1 2 1 6 T 1 6 ] + + u 27 [ Χ 2 , 64 Τ X 2 , 64 + 2 z 64 X 2 , 64 1 6 + z 64 2 1 6 T 1 6 ]
The verticals are
F ( x , z ) , is a vector function of x and z and we minimize separately for x and z,
m i n { F ( x , z ) } = m i n x m i n z F x , z
So, we differ partially for z (for z 0 , z 1 , , z 63 )
F ( x , z ) z 0 = 2 u 0 [ X 10 T 1 6 + Χ 20 Τ 1 7 + 4 z 0 ] = 0 F ( x , z ) z 64 = 2 u 64 [ X 1 , 64 T 1 6 + Χ 2 , 64 Τ 1 6 + 4 z 64 ] = 0
So, we have
F ( x ) = u 0 X 1 0 T X 1 0 2 X 1 0 T X 20 + Χ 2 , 0 Τ X 2 0 ( X 1 0 T 1 6 Χ 2 , 0 Τ 1 6 ) 2 6 + + u 64 X 1 , 64 T X 1 , 64 2 X 1 , 64 T Χ 2 , 64 Τ + Χ 2 , 64 Τ X 2 , 64 ( X 1 , 64 T 1 7 Χ 2 , 64 Τ 1 6 ) 2 6
After substituting Χ 1 i and X 2 , i we obtain Q = F ( x ) = 1 24 [ R q T M 1 q ] , where:
R = 9 n ( u 3 + u 5 + u 6 + u 9 + u 10 + u 12 + u 15 + u 17 + u 18 + u 20 + u 23 + u 24 + u 27       + u 29 + u 30 + u 33 + u 34 + u 36 + u 39 + u 40 + u 43 + u 45 + u 46 + u 48      + u 51 + u 53 + u 57 + u 54 + u 58 + u 60 ) 4 ( u 1 + u 2 + u 4 + u 8 + u 16 + u 31 + u 32 + u 47 + u 55 + u 59 + u 61 + u 62 ) 9 ( u 0 + u 63 ) ,
q = q 1 q 2 q 3 q 4 q 5 q 6 ,           M = 5 n n n n n m 16 n 5 n n n n m 26 n n 5 n n n m 36 n n n 5 n n m 46 n n n n 5 n m 56 m 16 m 26 m 36 m 46 m 56 m 66 ,
with
q 1 = 3 u 7 u 56 3 u 11 u 52 3 u 13 u 50 + 3 u 14 u 49 3 u 19 u 44 3 u 21 u 42 + 3 u 22 u 41 3 u 25 u 38 + 3 u 26 u 37 + 3 u 28 u 35 ,
q 2 = 3 u 7 u 56 3 u 11 u 52 + 3 u 13 u 50 3 u 14 u 49 3 u 19 u 44 + 3 u 21 u 42 3 u 22 u 41 + 3 u 25 u 38 3 u 26 u 37 + 3 u 28 u 35 ,
q 3 = 3 u 7 u 56 + 3 u 11 u 52 3 u 13 u 50 3 u 14 u 49 + 3 u 19 u 44 3 u 21 u 42 3 u 22 u 41 + 3 u 25 u 38 + 3 u 26 u 37 3 u 28 u 35 ,
q 4 = 3 u 7 u 56 3 u 11 u 52 3 u 13 u 50 3 u 14 u 49 + 3 u 19 u 44 + 3 u 21 u 42 + 3 u 22 u 41 3 u 25 u 38 3 u 26 u 37 3 u 28 u 35 ,
q 5 = 3 u 7 u 56 + 3 u 11 u 52 + 3 u 13 u 50 + 3 u 14 u 49 3 u 19 u 44 3 u 21 u 42 3 u 22 u 41 3 u 25 u 38 3 u 26 u 37 3 u 28 u 35 ,
q 6 = 9 u 7 + u 56 3 u 11 + u 52 3 u 13 + u 50 + 3 u 14 + u 49 3 u 19 + u 44 15 u 21 + u 42 9 u 22 + u 41 3 u 25 + u 38 9 u 26 + u 37 + 3 u 28 + u 35 ,
m 16 = u 7 u 56 + u 11 u 52 + u 13 u 50 + u 14 u 49 + u 19 u 44 + u 21 u 42 + u 22 u 41 + u 25 u 38 + u 26 u 37 + u 28 u 35 ,
m 26 = 5 u 7 u 56 5 u 11 u 52 5 u 13 u 50 + 7 u 14 u 49 5 u 19 u 44 5 u 21 u 42 + 7 u 22 u 41 5 u 25 u 38 + 7 u 26 u 37 + 7 u 28 u 35 ,
m 36 = 5 u 7 u 56 5 u 11 u 52 + 7 u 13 u 50 5 u 14 u 49 5 u 19 u 44 + 7 u 21 u 42 5 u 22 u 41 + 7 u 25 u 38 5 u 26 u 37 + 7 u 28 u 35 ,
m 46 = 5 u 7 u 56 + 7 u 11 u 52 5 u 13 u 50 5 u 14 u 49 + 7 u 19 u 44 5 u 21 u 42 5 u 22 u 41 + 7 u 25 u 38 + 7 u 26 u 37 5 u 28 u 35 ,
m 56 = 7 u 7 u 56 5 u 11 u 52 5 u 13 u 50 5 u 14 u 49 + 7 u 19 u 44 + 7 u 21 u 42 + 7 u 22 u 41 5 u 25 u 38 5 u 26 u 37 5 u 28 u 35 ,
m 66 = 29 n .
Observation. 
If the column δ A δ B of X 2 replaced by the column δ B δ A , then q i , m i j , m i 2 ,   i , j { 1 ,   2 ,   3 ,   4 ,   5 } , m 6 2 do not change and the q 6 , m i 6 ,   i = 1 ,   2 , , 5 change sign. This is because X 2 i , i = 0 ,   1 , , 63 are of the form V + x 6 δ with V T = ( x 1 , x 2 , x 3 , x 4 , x 5 , 0 ) and δ = δ A δ B or δ = ( δ A δ B ) .

3. Results

Proposition 4.
The maximization of Q is achieved when the following values of  u i i = 0 , 1 , , 63  that minimize the variance of the direct effect  τ ^ A , that is, those that maximize the expression
Q = 1 24 [ R q T M 1 q ] ,   where
q T M 1 q = 1 6 n [ q 1 2 + q 2 2 + q 3 2 + q 4 2 + q 5 2 + ( q 1 + q 2 + q 3 + q 4 + q 5 ) 2 ] + [ q 6 1 6 n ( q 1 m 16 + + q 5 m 56 + ( q 1 + + q 5 ) ( m 16 + + m 56 ) ] 2 29 n 1 6 n [ m 16 2 + + m 56 2 + ( m 16 + + m 56 ) 2 ] ,
Proof of Proposition 4.
From the (3) Μ is of the form M = A B B T 29 n , where
A = n 5 1 1 1 1 5 1 1 1 5 1 1 5 1 5 A 1 = 1 6 n 2 1 1 1 1 1 2 1 1 1 1 1 2 1 1 1 1 1 2 1 1 1 1 1 2 M 1 = A 1 0 0 0 + 1 29 n B T A 1 B A 1 B 1 [ B T A 1 1 ] ,   B = m 16 m 26 m 36 m 46 m 56
After the replacement and the calculations, we have the result of (4). □
Proposition 5.
If  Q  is the maximum value of  Q = 1 24 R q T M 1 q   f o r   u i ,  i = 0 ,   1   , ,   63 , then the following cases arise
(a)
n 0  (mod4):  Q = 3 2 n  and the solutions satisfy
q 1 = q 2 = q 3 = q 4 = q 5 = q 6 = 0 ,
The solutions satisfy (6) are:
  • (i)
    u 7 = u 56 = n 8 ,  u i = u 63 i = x 0 , , 3 n 8 ,  u j = u 63 j = 3 n 8 x ,  i , j 11 ,   13 ,   19 ,   25 ,  i j ,  n 0  (mod8).
    (ii)
    u 7 = u 56 = x ,   u i = u 63 i = n 4 2 x ,   u j = u 63 i = n 4 + x , x 0 ,   1 , ,   n 8 ,  i { 14 ,   28 } ,  j { 11 ,   13 ,   19 ,   25 } ,  n 0  (mod4).
    (iii)
    u 7 = u 56 = n 4 , u i = u 63 i = x , u j = u 63 j = n 4 x ,   x 0 ,   1 , , n 4 ,  i , j 22 ,   26 ,  i j ,  n 0  (mod4).
(b)
n 2  (mod4):  Q = 1 6 9 n 6 2 29 n  and the solutions satisfy
q 1 = q 2 = q 3 = q 4 = q 5 = 0 , q 6 = ± 6 .
Some solutions are:
  • (i)
    u 7 = u 56 = n + 2 4 ,  u i = u 63 i = n 2 8 ,  i 11 ,   13 ,   19 ,   25 , for two values of  i , with  n 6  (mod8),  ( q 6 = 6 ) .
    (ii)
    u 7 = u 56 = n 2 4 ,  u i = u 63 i = n + 2 8 ,  i 11 ,   13 ,   19 ,   25  for two values of  i , with  n 2  (mod8),  ( q 6 = 6 ) .
    (iii)
    u 7 = u 56 = n + 2 4 ,  u i = u 63 i = n 2 4 ,  i 22 ,   26  for one value of  i , with  n 2  (mod8),  ( q ~ 6 = 6 ) .
    (iv)
    u 7 = u 56 = n 2 4 ,  u i = u 63 i = n + 2 8 ,  i { 22 ,   26 } ,  for two values of  i , with  n 2  (mod8),  ( q ~ 6 = 6 ) .
    (v)
    u 7 = u 56 1 = x ,  u 14 = u 49 + 1 = y ,  u 19 = u 44 + 1 = z ,  u 22 = u 41 1 = w ,  w { 0 , 1 , , n 10 4 } ,  y = n + 6 4 + w ,  x + z = n 6 4 w ,  z 1 ,  n 10 ,  n 2  (mod4),  ( q ~ 6 = 6 ) .
    (vi)
    u 7 = u 56 1 = x ,  u 14 = u 49 + 1 = y ,  u 19 = u 44 + 1 = z ,  u 22 = u 41 1 = w ,  w { 0 , 1 , , n 6 4 } ,  y = n + 2 4 + w ,  x + z = n 2 4 w ,  z 1 ,  n 6 ,  n 2  (mod4),  ( q ~ 6 = 6 ) .
(c)
n 1  (mod4):  Q = 1 6 9 n 9 n 9 ( n 5 ) 2 29 n ( n 2 1 )  and the solutions satisfy:
q 2 = q 4 = ± 3 , q 1 = q 3 = q 5 = 3 , q 6 = 3 .
  • (i)
    u 7 = u 56 = n 1 8 ,  u 11 = u 52 1 = x ,  u 13 = u 50 + 1 = y ,  u 19 = u 44 + 1 = z ,  x + y + z = 3 n + 5 8 ,  y 1 ,  z 1 ,  n 1  (mod8).
    (ii)
    u i = u 63 i = w ,  i { 11 ,   13 ,   25 } ,  u 19 = u 44 + 1 = x ,  u 26 = u 37 1 = y ,  u 28 = u 35 + 1 = z ,  x + y + z = n + 3 4 ,  y 1 ,  z 1 ,  n 1  (mod4).
(d)
n 3  (mod4):  Q = 1 6 9 n 9 n 9 ( n 3 ) 2 29 n ( n 1 ) ( n + 1 ) , and the solutions satisfy:
q 1 = q 3 = ± 3 , q 2 = q 4 = q 5 = 3 , q 6 = 3 q 1 = q 4 = ± 3 , q 2 = q 3 = q 5 = 3 , q 6 = 3 q 4 = q 5 = ± 3 , q 1 = q 2 = q 3 = 3 , q 6 = ± 3 .
Some solutions are:
  • (i)
    u 7 = u 56 1 = x ,  u 14 = u 49 + 1 = y ,  u 19 = u 44 + 1 = z ,  n 7 ,  x 0 ,   1 , , n 7 8 ,  y = n 3 4 2 x ,  z = n + 5 4 + x .
    (ii)
    u 7 = u 56 1 = x ,  u 14 = u 49 + 1 = y ,  u 19 = u 44 + 1 = z ,  u i = u 63 i = w ,  i { 11 , 13 , 25 } ,  y = n 3 4 2 x ,  z + w = n + 5 4 + x ,  z 1 ,  x 0 , 1 , , n 7 8 ,  n 7 .
    (iii)
    u 13 = u 50 1 = x ,  u 14 = u 49 + 1 = y ,  u 21 = u 42 + 1 = z ,  x = n + 5 4 3 z ,  y = n 3 4 + 2 z ,  z 1 , 2 , , n + 5 12 ,  n 7 .
    (iv)
    u 14 = u 49 + 1 = x ,  u 21 = u 42 + 1 = y ,  u 28 = u 35 1 = z ,  y = n + 5 12 ,  x + z = 5 n + 1 12 ,  x 1 ,  n 7  (mod12).
    (v)
    u 14 = u 49 + 1 = x ,  u 21 = u 42 + 1 = y ,  u 28 = u 35 1 = z ,  u i = u 63 i = w ,  i { 11 , 13 , 19 , 25 } ,  y 1 , 2 , , n + 5 12 ,  x + z = n 3 4 + 2 y ,  x 1 ,  w = n + 5 4 3 y ,  n 7 .
Proof of Proposition 5.
Observe that: q 1 + q 2 + q 3 + q 4 + q 5 = 3 m 16 ,
m 26 = 2 q 1 + m 16 , m 36 = 2 q 2 + m 16 , m 46 = 2 q 3 + m 16 , m 56 = 2 q 4 + m 16 .
and 1 5 m i 6 = 2 q 5 + 1 .
(a)
n 0 (mod4). The minimum of q T M 1 q is attained when q = 0   q 1 = q 2 = q 3 = q 4 = q 5 = q 6 = 0 since M is positive semidefinite. In this case Q = 3 n 2 . There are many solutions, some of which satisfy the relations given in the Proposition 5 (a).
(b)
n 2 (mod4). The quantities q i , i = 1 ,   2 , , 5 , defined in (3), are multiples of 3 and even numbers. This follows from the identity ( u 7 + u 56 ) + ( u 11 + u 52 ) + + ( u 26 + u 37 ) + ( u 28 + u 35 ) = n . This implies that among the ten terms ( u i + u 63 i ) in the sum, x terms are odd and y are even, with x + y = 10 , then x is even. Since the total sum is even, x must be even. Therefore, x of the terms ( u 7 u 56 ) , ( u 11 u 52 ) , + ( u 26 u 37 ) , ( u 28 u 35 ) are odd and the remaining y = 10 x are even. Consequently, q i , i = 1 ,   2 , , 5 are even because n is even; the same is for q 6 . Hence, q i 0 (mod6) = 0 , i = 1 ,   2 ,   3 ,   4 ,   5 ,   6 . The quantity q T M 1 q , as it is shown in (4), attains its minimum when q 1 = q 2 = q 3 = q 4 = q 5 = 0 . Otherwise, there exists at least one term q i 6 , i = 1 ,   2 , , 5 and then from (4) we obtain that q T M 1 q ( 6 ) 2 6 n = 6 n . In proportion, we have solutions Proposition 5 (b) satisfying
q T M 1 q = 6 2 29 n < 6 2 6 n
From (10) we have m i 6 = 0 , i = 1 ,   2 , , 5 and q 6 = 6 mod 1 2 . Therefore from (4) we have m i n { q T M 1 q } = min q 6 2 29 n = 6 2 29 n . In order to show q 6 6 (mod12) we use (3), so 0 = q 2 + q 3 + q 4 + q 5 = 6 [ ( u 14 u 49 ) + ( u 22 u 41 ) + ( u 26 u 37 ) + ( u 28 u 35 ) ] , then 0 = ( u 14 u 49 ) + ( u 22 u 41 ) + ( u 26 u 37 ) + ( u 28 u 35 ) = [ ( u 14 + u 49 ) + ( u 22 + u 41 ) + ( u 26 + u 37 ) + ( u 28 + u 35 ) ] 2 [ u 49 + u 41 + u 37 + u 35 ] . Then, [ ( u 14 + u 49 ) + ( u 22 + u 41 ) + ( u 26 + u 37 ) + ( u 28 + u 35 ) ] = 0 (mod2), and [ ( u 14 + u 49 ) ( u 22 + u 41 ) ( u 26 + u 37 ) + ( u 28 + u 35 ) ] = 0 (mod2). Finally, from (4) we have: q 6 + 3 n = 12 { ( u 7 + u 16 ) ( u 21 + u 42 ) ] + 6 [ ( u 14 + u 49 ) ( u 22 + u 41 ) ( u 26 + u 37 ) + ( u 28 + u 35 ) ] = 0 (mod12). So q 6 = 3 n + 0 (mod12)   = 6 (mod12)   | q 6 |   6 . The solutions satisfy the relations q i = 0 , i = 1 , 2 , 3 , 4 , 5 , q 6 = ± 6 as it is shown in the Proposition 5 (b).
(c)
n 1 (mod4). In this case, we will have that the quantities q i , i = 1 ,   2 , , 5 are multiples of 3 and odd, and the proof proceeds as in case (b), hence, q i 3 (mod6), i = 1 ,   2 ,   3 ,   4 ,   5 ,   6 and i = 1 5 q i = 3 (mod6). The optimal solution have the values q i = ± 3 , i = 1 5 q i = ± 3 , in other case there would be a term | q i | 9 , i = 1 ,   2 , , 5 or | i = 1 5 q i | 9 , and then q T M 1 q 3 2 5 + 9 2 6 n = 21 n . Therefore, in Proposition 5 (c) the solutions are q T M 1 q = 9 n + 9 ( n 5 ) 2 29 n ( n 2 1 ) < 21 n n > 0 . We show that at the optimal solution i = 1 5 q i = ± 3 , it is enough to examine the case where i = 1 5 q i = 3 because the spouse of the optimal solution is optimal solution also and has i = 1 5 q i = 3 . There are 10 cases that i = 1 5 q i = 3 in every case we calculate the terms m 16 , m 26 , , m 56 using Proportion 4, and we have, m i n { q T M 1 q } = 9 n + m i n { ( q 6 d n ) 2 29 ( n 1 n ) } , where d = 3 ,   9 ,   15 . The minimum value is when d = 15 and q 6 = 3 and gives: m i n { q T M 1 q } = 9 n + 9 ( n 5 ) 2 29 n ( n 2 1 ) which is satisfied when q 2 = q 4 = ± 3 , q 1 = q 3 = q 5 = 3 , q 6 = 3 . Some groups of optimal solutions are in the Proposition 5 (c).
(d)
n 3 (mod4). As in case n = 1 (mod4), we have q i = 3 (mod6), i = 1 , 2 , , 6 , i = 1 5 q i = ± 3 . The terms m i 6 , i = 1 , 2 , , 5 are calculated in (4). We will show that that the infimum, which is presented in case (c), cannot exist
Q = 1 6 9 n 9 6 6 n min q 6 9 n 2 29 n 1 n
then q 6 = ± 3 and
Q = 1 6 9 n 9 n 9 ( n 3 ) 2 29 n ( n 1 ) ( n + 1 )
The solutions which satisfy the optimal value of Q are presented in the Proportion 5 (d). The infimum m i n { q T M 1 q } = 9 n + 9 ( n 5 ) 2 29 n ( n 2 1 ) is achieved only when q 2 = q 4 = + 3 , q 1 = q 3 = q 5 = 3 , q 6 = 3 and their spouses. From (4) we have, q 6 + 3 n = 12 { ( u 7 + u 16 ) ( u 21 + u 42 ) ] + 6 [ ( u 14 + u 49 ) ( u 22 + u 41 ) ( u 26 + u 37 ) + ( u 28 + u 35 ) ] = 0 (mod12) and 0 = q 2 + q 3 + q 4 + q 5 6 [ ( u 14 u 49 ) + ( u 22 u 41 ) + ( u 26 u 37 ) + ( u 28 u 35 ) ] = 0 . Then as in (b), ( u 14 + u 49 ) ( u 22 + u 41 ) ( u 26 + u 37 ) + ( u 28 + u 35 ) is even then q 6 = 3 n + 0 (mod12) 3 (mod12). So, we cannot have q 6 = 3 which is necessary as infimum. □

4. Discussion

In the case of experiments involving six periods and considering direct effects, the optimization procedure is based on designs that satisfy the linear constraint
u i 0 , i = 7 ,   11 ,   13 ,   14 ,   19 ,   21 ,   22 ,   25 ,   26 ,   28 ,   35 ,   37 ,   38 ,   41 ,   42 ,   44 ,   49 ,   50 ,   52 ,   56 .
The terms appearing in the above sum correspond to treatment sequences that either consist of identical treatments across all six periods (3 treatments A and three treatments B), reflecting patterns analogous to those observed in optimal designs for direct effects with fewer periods. In particular, a directly comparable case is that of an even number of periods (4 periods), where we observe a correspondence between results [13]. Since the solutions depend on the value of n k (mod4), with k = 0 , 1 , 2 , 3 (see Proposition 4), the case of six periods differs from the previously examined cases with 2, 3, 4, and 5 periods, in which, only two different expressions for the variance arose, corresponding to even and odd values of n.
It is evident that the number of feasible solutions is substantially larger than in the cases of four or five periods [14], primarily due to the combinatorial expansion of possible sequences as the number of periods increases. These intermediate cases are particularly informative, as they may provide valuable insight toward the development of a general solution applicable to an arbitrary number of periods (p).
Repeated measurement designs focusing exclusively on direct effects constitute an important step in understanding more complex models in which interactions between periods or carryover effects are introduced.

Funding

The article is submitted after invitation from Stats.

Data Availability Statement

The data are contained within the article.

Conflicts of Interest

The author declares no conflict of interest.

References

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Table 1. The 64 sequences of treatments A, B.
Table 1. The 64 sequences of treatments A, B.
ABABABABABABABAB
AABBAABBAABBAABB
AAAABBBBAAAABBBB
AAAAAAAABBBBBBBB
AAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAA
u 0 u 1 u 2 u 3 u 4 u 5 u 6 u 7 u 8 u 9 u 10 u 11 u 12 u 13 u 14 u 15
ABABABABABABABAB
AABBAABBAABBAABB
AAAABBBBAAAABBBB
AAAAAAAABBBBBBBB
BBBBBBBBBBBBBBBB
AAAAAAAAAAAAAAAA
u 16 u 17 u 18 u 19 u 20 u 21 u 22 u 23 u 24 u 25 u 26 u 27 u 28 u 29 u 30 u 31
ABABABABABABABAB
AABBAABBAABBAABB
AAAABBBBAAAABBBB
AAAAAAAABBBBBBBB
AAAAAAAAAAAAAAAA
BBBBBBBBBBBBBBBB
u 32 u 33 u 34 u 35 u 36 u 37 u 38 u 39 u 40 u 41 u 42 u 43 u 44 u 45 u 46 u 47
ABABABABABABABAB
AABBAABBAABBAABB
AAAABBBBAAAABBBB
AAAAAAAABBBBBBBB
BBBBBBBBBBBBBBBB
BBBBBBBBBBBBBBBB
u 48 u 49 u 50 u 51 u 52 u 53 u 54 u 55 u 56 u 57 u 58 u 59 u 60 u 61 u 62 u 63
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Chalikias, M.S. Estimating the Parameter of Direct Effects in Crossover Designs: The Case of 6 Periods and 2 Treatments. Stats 2026, 9, 17. https://doi.org/10.3390/stats9010017

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Chalikias MS. Estimating the Parameter of Direct Effects in Crossover Designs: The Case of 6 Periods and 2 Treatments. Stats. 2026; 9(1):17. https://doi.org/10.3390/stats9010017

Chicago/Turabian Style

Chalikias, Miltiadis S. 2026. "Estimating the Parameter of Direct Effects in Crossover Designs: The Case of 6 Periods and 2 Treatments" Stats 9, no. 1: 17. https://doi.org/10.3390/stats9010017

APA Style

Chalikias, M. S. (2026). Estimating the Parameter of Direct Effects in Crossover Designs: The Case of 6 Periods and 2 Treatments. Stats, 9(1), 17. https://doi.org/10.3390/stats9010017

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