individual when there are r treatments remaining. Then using ∑ s δ$_{sq}$ = ∑ s δ$_{ps}$ = 1. Lemma 4.2. Take the vector of fixed effects β to be Then we have the following result where 0$_{m,n}$ is a matrix of zeroes of dimension m × n , and Proof. Denote the columns of X$_{s}$$_{ri}$ by Thus Π$_{jr}$ is the column corresponding to the period effect π$_{j}$ , T$_{ds}$$_{ri}$ to the treatment effect τ$_{d}$ , and 1$_{r}$ to the intercept µ$_{0}$ . Using this representation, and Lemma 4.1, we have 20 Σ$_{rps}$Σ − 1 rsq = 1 σ 2 $_{e}$( σ 2 e + rσ 2 $_{b}$) r ∑ s =1 $^{{}$σ 2 b + σ 2 $_{e}$δ$_{ps}$ $^{} {}$( σ 2 e + rσ 2 $_{b}$) δ$_{sq}$ − σ 2 b $^{}}$, = 1 σ 2 $_{e}$( σ 2 e + rσ 2 $_{b}$) r ∑ s =1 $^{{}$σ 2 $_{b}$( σ 2 e + rσ 2 $_{b}$) δ$_{sq}$ − σ 4 b + σ 2 $_{e}$( σ 2 e + rσ 2 $_{b}$) δ$_{pq}$ − σ 2 $_{e}$σ 2 $_{b}$δ$_{ps}$ $^{}}$, = 1 σ 2 $_{e}$( σ 2 e + rσ 2 $_{b}$) $^{{}$σ 2 $_{b}$( σ 2 e + rσ 2 $_{b}$) − rσ 4 b + σ 2 $_{e}$( σ 2 e + rσ 2 $_{b}$) δ$_{pq}$ − σ 2 $_{e}$σ 2 b $^{}}$, = δ$_{pq}$, β = ( µ$_{0}$, π$_{2}$, . . . , π$_{D}$, τ$_{2}$, . . . , τ$_{D}$$_{−}$$_{1}$ ) $^{T}$. | S r | ∑ i =1 n | S r | X T s ri Σ − 1 r X s ri = n$^{}$        A B T 0 1 ,D − r B T 0 1 ,D − r B C 0 r − 1 ,D − r E 0 r − 1 ,D − r 0 D − r, 1 0 D − r,r − 1 0 D − r,D − r 0 D − r,r − 1 0 D − r,D − r B E 0 r − 1 ,D − r C 0 r − 1 ,D − r 0 D − r, 1 0 D − r,r − 1 0 D − r,D − r 0 D − r,r − 1 0 D − r,D − r$^{}$        , (4.3) A = rσ 2 e σ 2 $_{e}$( σ 2 e + rσ 2 $_{b}$) , B$_{pq}$ = rσ 2 e σ 2 $_{e}$( σ 2 e + rσ 2 $_{b}$) ( p = 1 , . . . , r − 1; q = 1) , C$_{pq}$ = 1 σ 2 $_{e}$( σ 2 e + rσ 2 $_{b}$) { ( σ 2 e + rσ 2 $_{b}$) δ$_{pq}$ − σ 2 $_{b}$} ( p = 1 , . . . , r − 1; q = 1 , . . . , r − 1) , E$_{pq}$ = 1 r σ 2 e σ 2 $_{e}$( σ 2 e + rσ 2 $_{b}$) ( p = 1 , . . . , r − 1; q = 1 , . . . , r − 1) . X$_{s}$$_{ri}$ = ( 1$_{r}$ Π$_{2}$$_{r}$ . . . Π$_{Dr}$ T$_{1}$$_{s}$$_{ri}$ . . . T$_{D}$$_{−}$$_{1}$$_{s}$$_{ri}$$^{)}$ .