Optimal crossover designs in a model with self and mixed carryover effects with correlated errors

Abstract

We determine optimal crossover designs for the estimation of direct treatment effects in a model with mixed and self carryover effects. The model also assumes that the errors within each experimental unit are correlated following a stationary first-order autoregressive process. The paper considers situations where the number of periods for each experimental unit is at least four and the number of treatments is greater or equal to the number of periods.

Appendix: Proofs

Lemma 1

Consider the matrix $W_\lambda$, defined in (4). Assume $p \ge 4$ and $\lambda \in [\lambda ^*(p),1)$ with $\lambda ^*(p)$ as in Proposition 1. We then get for the entries $w_{ij}$ of $W_\lambda$ that $w_{ii} > 0$ for $1 \le i \le p$ and $w_{ij} \le 0$ for $1 \le i \ne j \le p$.

Proof

It was shown by Kunert (1985) for $\lambda \in [\lambda ^*(p),1)$ that $w_{ij}\le 0$ for $i \ne j$ and $w_{ii} \ge 0$.

To see that the diagonal elements are in fact positive, define $(1-\lambda )^k =: L_k$. We observe $\mathbf{1}_\mathbf{p}^\mathbf{T} {\varvec{\varLambda }^{-1}} = [L_1,L_2, \dots ,L_2,L_1]$ and $\mathbf{1}_\mathbf{p}^\mathbf{T} {\varvec{\varLambda }^{-1}} \mathbf{1}_\mathbf{p} = L_1(p-\lambda (p-2)) = z_p,$ say. For $2 \le i \le p-1$ the substitution $p=v+4$ yields

$$\begin{aligned} w_{ii} = 1+\lambda ^2-\frac{L_4}{z_p}= \frac{(1-\lambda )(1+\lambda ^2)v-\lambda ^3+\lambda ^2+\lambda +3}{(1-\lambda )v-2\lambda +4} \end{aligned}$$

which is positive for $v\ge 0$ and therefore also for $p \ge 4$.

The entries $w_{11}$ and $w_{pp}$ are positive because, using the substitution $p=v+4$ again, we have

$$\begin{aligned} w_{11} = w_{pp} = 1- \frac{L_2}{z_p}= \frac{(1-\lambda )v-\lambda +3}{(1-\lambda )v-2\lambda +4} \end{aligned}$$

which is positive. $\square$

Lemma 2

For $t \ge p \ge 4$ and $\lambda \in [\lambda ^*(p),1)$ consider an arbitrary sequence class $\ell$. Then $c_{11}(1) \ge c_{11}(\ell )$ and $c_{11}(1)+2c_{12}(1) \ge c_{11}(\ell )+2c_{12}(\ell ).$

Proof

Assume $u$ is a unit receiving a sequence $s = [s_1,...,s_p]$ from class $\ell$. Then

$$\begin{aligned} c_{11}(\ell )={{\mathrm{tr}}}(\mathbf{B}_\mathbf{t}\mathbf{T}_\mathbf{du}^\mathbf{T}\mathbf{W}_{\lambda } \mathbf{T}_\mathbf{du}) \end{aligned}$$

and

$$\begin{aligned} c_{12}(\ell )={{\mathrm{tr}}}(\mathbf{B}_\mathbf{t}\mathbf{T}_\mathbf{du}^\mathbf{T}\mathbf{W}_{\lambda } \mathbf{M}_\mathbf{du}). \end{aligned}$$

Since $\mathbf{W}_{\lambda }$ has column-sums zero and since $\mathbf{T}_\mathbf{du}\mathbf{1}_\mathbf{t} = \mathbf{1}_\mathbf{p}$, it follows that

$$\begin{aligned} c_{11}(\ell )={{\mathrm{tr}}}(\mathbf{T}_\mathbf{du}^\mathbf{T}\mathbf{W}_{\lambda } \mathbf{T}_\mathbf{du}) \quad {\text {and}} \quad c_{12}(\ell )={{\mathrm{tr}}}(\mathbf{T}_\mathbf{du}^\mathbf{T}\mathbf{W}_{\lambda } \mathbf{M}_\mathbf{du}). \end{aligned}$$

For $1 \le i \le t$ denote the $i$-th column of $\mathbf{T}_\mathbf{du}$ by $\mathbf{t}_\mathbf{du}^\mathbf{(i)}$ and the $i$-th column of $\mathbf{M}_\mathbf{du}$ by $\mathbf{m}_\mathbf{du}^\mathbf{(i)}$. Observe that

• the $j$-th entry of $\mathbf{t}_\mathbf{du}^\mathbf{(i)}$ is 1 if $s_j=i$ and 0 otherwise,

• the $j$-th entry of $\mathbf{m}_\mathbf{du}^{(i)}$ is 1 if $j\ge 2, s_{j-1}=i$ and $s_{j-1} \ne s_j$. In all other cases it is 0.

For any given $i$ it follows that

$$\begin{aligned} \mathbf{t}_\mathbf{du}^\mathbf{(i)T}\mathbf{W}_{\lambda } \mathbf{t}_\mathbf{du}^\mathbf{(i)} = \sum _{j=1}^p I(s_j=i)\sum _{r=1}^p w_{jr} I(s_r=i), \end{aligned}$$

where $I(statement)$ is 1 if $statement$ is true and 0 otherwise. Hence,

$$\begin{aligned} c_{11}(\ell )&= \sum _{i=1}^t \mathbf{t}_\mathbf{du}^\mathbf{(i)T}\mathbf{W}_{\lambda } \mathbf{t}_\mathbf{du}^\mathbf{(i)} \nonumber \\&= \sum _{i=1}^t\left( \sum _{j=1}^p\sum _{r=1}^p w_{jr} I(s_j=s_r=i)\right) =\sum _{j=1}^p\sum _{r=1}^p w_{jr} I(s_j=s_r) \nonumber \\&= \sum _{j=1}^p w_{jj} + 2\sum _{j=1}^{p-1}\sum _{r=j+1}^p w_{jr} I(s_j=s_r) \\&\le \sum _{j=1}^p w_{jj}, \nonumber \end{aligned}$$

(12)

because all $w_{jr}\le 0$ if $j\ne r$. It is easy to see that $c_{11}(1)=\sum _{j=1}^p w_{jj}$ and, therefore, we have proved that

$$\begin{aligned} c_{11}(\ell ) \le c_{11}(1). \end{aligned}$$

We can also use Eq. (12) to derive the slightly sharper bound

$$\begin{aligned} c_{11}(\ell ) \le \sum _{j=1}^p w_{jj} +2 \sum _{j=1}^{p-1} w_{j,j+1} I(s_j=s_{j+1}). \end{aligned}$$

(13)

On the other hand,

$$\begin{aligned} \mathbf{t}_\mathbf{du}^\mathbf{(i)T}\mathbf{W}_{\lambda } \mathbf{m}_\mathbf{du}^\mathbf{(i)}=\sum _{j=1}^p I(s_j=i)\sum _{r=2}^p w_{jr} I(s_r \ne s_{r-1}=i). \end{aligned}$$

Therefore,

$$\begin{aligned} c_{12}(\ell )&= \sum _{i=1}^t\left( \sum _{j=1}^p\sum _{r=2}^p w_{jr} I(s_j=i,s_r \ne s_{r-1}=i)\right) \\&= \sum _{j=1}^p\sum _{r=2}^p w_{jr} I(s_j=s_{r-1} \ne s_r)\\&= \sum _{j=1}^p\sum _{r \ne j} w_{jr} I(s_j=s_{r-1} \ne s_r), \end{aligned}$$

because $s_j = s_{r-1} \ne s_r$ can never hold for $r = j$. On the other hand, if $r = j+1$, then $s_j = s_{r-1} \ne s_r$ becomes $s_j \ne s_{j+1}$. Making use of the fact that all $w_{jr} \le 0$ for all $r \ne j$, we conclude that

$$\begin{aligned} c_{12}(\ell ) \le \sum _{j=1}^{p-1} w_{j,j+1} I(s_j \ne s_{j+1}). \end{aligned}$$

Combining this with Eq. (13), we conclude that

$$\begin{aligned} c_{11}(\ell )+2c_{12}(\ell ) \le \sum _{j=1}^p w_{jj} \!+\!2 \sum _{j=1}^{p-1} w_{j,j+1} I(s_j=s_{j+1}) \!+\! 2 \sum _{j=1}^{p-1} w_{j,j+1} I(s_j \ne s_{j+1}). \end{aligned}$$

Since it is easy to see that

$$\begin{aligned} c_{12}(1) = \sum _{j=1}^{p-1} w_{j,j+1}, \end{aligned}$$

(14)

we have proved that $c_{11}(\ell )+2c_{12}(\ell ) \le c_{11}(1)+2c_{12}(1)$. $\square$

Lemma 3

Under the conditions of Lemma 2 we have $c_{22}(1) \ge c_{22}(\ell )$.

Proof

If $\mathbf{M}_\mathbf{du}$ is as in the proof of Lemma 2, then

$$\begin{aligned} c_{22}(\ell )={{\mathrm{tr}}}(\mathbf{B}_\mathbf{t} \mathbf{M}_\mathbf{du}^\mathbf{T} \mathbf{W}_{\lambda } \mathbf{M}_\mathbf{du})={{\mathrm{tr}}}(\mathbf{M}_\mathbf{du}^\mathbf{T} \mathbf{W}_{\lambda } \mathbf{M}_\mathbf{du}) - \frac{1}{t}\mathbf{1}_\mathbf{t}^\mathbf{T} \mathbf{M}_\mathbf{du}^\mathbf{T} \mathbf{W}_{\lambda } \mathbf{M}_\mathbf{du} \mathbf{1}_\mathbf{t}. \end{aligned}$$

Observe that $\mathbf{M}_\mathbf{du} \mathbf{1}_\mathbf{t}$ is a $p$-dimensional vector with entries 1 or 0. The first element of $\mathbf{M}_\mathbf{du} \mathbf{1}_\mathbf{t}$ is always 0, the $j$-th element, for $j \ge 2,$ is 0, if $s_{j-1}=s_j,$ and 1, otherwise. Hence,

$$\begin{aligned} \mathbf{1}_\mathbf{t}^\mathbf{T} \mathbf{M}_\mathbf{du}^\mathbf{T} \mathbf{W}_{\lambda } \mathbf{M}_\mathbf{du} \mathbf{1}_\mathbf{t}\!&= \!\sum _{j=2}^p\sum _{r=2}^p w_{jr} I(s_{j-1}\ne s_j)I(s_{r-1}\ne s_r)\\ \!&= \!\sum _{j\!=\!2}^p w_{jj} I(s_{j\!-\!1}\ne s_j) \!+\! 2\sum _{j=2}^{p\!-\!1}\sum _{r=j\!+\!1}^p w_{jr} I(s_{j\!-\!1}\ne s_j)I(s_{r\!-\!1}\ne s_r). \end{aligned}$$

Defining $\mathbf{m}_\mathbf{du}^\mathbf{(i)}$ as in the proof of Lemma 2, we get

$$\begin{aligned} {{\mathrm{tr}}}(\mathbf{M}_\mathbf{du}^\mathbf{T} \mathbf{W}_{\lambda } \mathbf{M}_\mathbf{du})&= \sum _{i=1}^t \mathbf{m}_\mathbf{du}^\mathbf{(i)T}\mathbf{W}_{\lambda } \mathbf{m}_\mathbf{du}^\mathbf{(i)}\\&= \sum _{i=1}^t \left( \sum _{j=2}^p\sum _{r=2}^p w_{jr} I(s_j \ne s_{j-1}=i=s_{r-1}\ne s_r)\right) \\&= \sum _{j=2}^p\sum _{r=2}^p w_{jr} I(s_j\ne s_{j-1}= s_{r-1} \ne s_r)\\&= \sum _{j=2}^p w_{jj} I(s_j\ne s_{j-1}) \!+\! 2\sum _{j=2}^{p-1}\sum _{r=j+1}^p w_{jr} I(s_j\ne s_{j-1}\!=\! s_{r-1} \ne s_r)\\&\le \sum _{j=2}^p w_{jj} I(s_j\ne s_{j-1}). \end{aligned}$$

Combining the two parts, we get

$$\begin{aligned} c_{22}(\ell )&\le \sum _{j=2}^p w_{jj} I(s_j\ne s_{j-1}) - \frac{1}{t}\left( \sum _{j=2}^p w_{jj} I(s_j\ne s_{j-1}) \right. \\&\left. + \,2\sum _{j=2}^{p-1}\sum _{r=j+1}^p w_{jr} I(s_{j-1}\ne s_j)I(s_{r-1}\ne s_r) \right) \\&= \frac{t-1}{t} \sum _{j=2}^p w_{jj} I(s_j\ne s_{j-1}) - \frac{2}{t}\sum _{j=2}^{p-1}\sum _{r=j+1}^p w_{jr}I(s_{j-1}\ne s_j)I(s_{r-1}\ne s_r)\\&\le \frac{t-1}{t} \sum _{j=2}^p w_{jj} - \frac{2}{t}\sum _{j=2}^{p-1}\sum _{r=j+1}^p w_{jr}, \end{aligned}$$

where the last inequality made use of the fact that all $w_{jj}>0$ and all $w_{jr} \le 0$, for $j \ne r$. It is easy to see that this bound for $c_{22}(\ell )$ equals $c_{22}(1)$. $\square$

Lemma 4

Under the conditions of Lemma 2 consider an $x \in [0,1]$. We then have that

$$\begin{aligned} c_{11}(1) + 2xc_{12}(1) \ge c_{11}(\ell ) + 2xc_{12}(\ell ). \end{aligned}$$

Proof

Define $f_{(\ell )}(x) := c_{11}(\ell ) + 2xc_{12}(\ell )$. Due to the linearity of the function $f_{(\ell )}(x),$ it is sufficient to show that $f_{(1)}(0) \ge f_{(\ell )}(0)$ and $f_{(1)}(1) \ge f_{(\ell )}(1), 1 \le \ell \le K$. The inequalities are valid because of Lemma 2. $\square$

Lemma 5

Under the conditions of Lemma 2, we have that $c_{22}(1) > 0$ and the function $h_{1}{(x,0)}$ has an unique minimum at

$$\begin{aligned} x^* = -\frac{c_{12}(1)}{c_{22}(1)} \in [0,1]. \end{aligned}$$

Proof

It follows from the proof of Lemma 3 that

$$\begin{aligned} c_{22}(1) = \frac{t-1}{t} \sum _{j=1}^p w_{jj} - \frac{2}{t} \sum _{j=2}^{p-1} \sum _{r=j+1}^p w_{jr}. \end{aligned}$$

Making use of Lemma 1, we see that $c_{22}(1) > 0.$ This implies that $h_1(x,0)=c_{11}(1)+2xc_{12}(1)+x^2c_{22}(1)$ has a unique minimum at the point $x^*$. It only remains to show that $0 \le x^* \le 1$.

For each $j$, it holds that $\sum _{r=1}^p w_{jr} = 0.$ This implies that

$$\begin{aligned} \frac{t-1}{t} \sum _{j=2}^p w_{jj}&= -\frac{t-1}{t} \sum _{j=2}^p \left( \sum _{r=1}^{j-1}w_{jr}+\sum _{r=j+1}^{p}w_{jr}\right) \\&= -\frac{t-1}{t} \sum _{j=2}^p \sum _{r=1}^{j-1}w_{jr} -\frac{t-1}{t} \sum _{j=2}^{p-1} \sum _{r=j+1}^{p}w_{jr}. \end{aligned}$$

We therefore can rewrite $c_{22}(1)$ as

$$\begin{aligned} c_{22}(1) = -\frac{t-1}{t}\sum _{j=2}^p \sum _{r=1}^{j-1}w_{jr} - \frac{t+1}{t}\sum _{j=2}^{p-1} \sum _{r=j+1}^{p}w_{jr}. \end{aligned}$$

Since all $w_{jr}$ in this sum are non-positive, we get

$$\begin{aligned} c_{22}(1)&\ge -\frac{t-1}{t}w_{21} - \frac{t+1}{t}\sum _{j=2}^{p-1} w_{j,j+1}\\&\ge -\frac{t-1}{t}w_{21} - \sum _{j=2}^{p-1} w_{j,j+1} - \frac{1}{t}w_{p-1,p}. \end{aligned}$$

Observing that $w_{p-1,p}=w_{21}=w_{12}$, we conclude that

$$\begin{aligned} c_{22}(1) \ge - \sum _{j=1}^{p-1} w_{j,j+1}. \end{aligned}$$

Equation (14) then shows that $0 \le -c_{12} (1) \le c_{22} (1)$. This completes the proof. $\square$

Lemma 6

Under the conditions of Lemma 2 we get for any $x \in [0,1]$ that

$$\begin{aligned} h_{1}{(x,0)} \ge h_{\ell }{(x,0)}. \end{aligned}$$

Proof

Making use of Lemmas 3 and 4 we observe that

$$\begin{aligned} h_{1}{(x,0)}&= c_{11}(1) + 2xc_{12}(1) + x^2c_{22}(1) \\&\ge c_{11}(\ell ) + 2xc_{12}(\ell ) + x^2c_{22}(\ell ) \\&= h_{\ell }{(x,0)}. \end{aligned}$$

$\square$