Proof. Both \(bk\) and \(tr\) count the units in the design, by blocks and by treatments. For the second, fix a treatment \(i\) and count the pairs (unit with a treatment \(l \ne i\), block) in which the block also contains \(i\). Treatment \(i\) is in \(r\) blocks, each with \(k - 1\) other units, giving \(r(k - 1)\); each of the \(t - 1\) other treatments meets \(i\) in \(\lambda \) blocks, giving \(\lambda (t - 1)\).

For the third (Fisher’s inequality), let \(N\) be the \(t \times b\) incidence matrix, \(n_{ij} = 1\) if treatment \(i\) is in block \(j\). Then \(NN^\top \) has diagonal entries \(r\) and off-diagonal entries \(\lambda \), so \(NN^\top = (r - \lambda )I + \lambda J\). Its eigenvalues are \(r - \lambda + \lambda t = rk\) (on the vector of ones, using the second identity) and \(r - \lambda \), with multiplicity \(t - 1\). Since \(k < t\), the second identity gives \(r > \lambda \), so \(NN^\top \) is non-singular and has rank \(t\). But its rank is at most the rank of \(N\), which is at most \(b\). □

Proof. (i) \(E(y_{i\cdot }) = r\mu + r\tau _i + \sum _jn_{ij}\beta _j\) and \(E(y_{\cdot j}) = k\mu + \sum _ln_{lj}\tau _l + k\beta _j\). Hence \[ \frac 1k\sum _jn_{ij}E(y_{\cdot j}) = r\mu + \frac 1k\sum _l\Bigl (\sum _jn_{ij}n_{lj}\Bigr )\tau _l + \sum _jn_{ij}\beta _j . \] The inner sum is the number of blocks containing both \(i\) and \(l\): \(r\) if \(l = i\) and \(\lambda \) otherwise. So the middle term is \(\frac 1k\bigl (r\tau _i + \lambda \sum _{l\ne i}\tau _l\bigr ) = \frac {(r - \lambda )\tau _i}{k}\), using \(\sum \tau _l = 0\), and \[ E(Q_i) = r\tau _i - \frac {(r - \lambda )\tau _i}{k} = \frac {r(k - 1) + \lambda }{k}\tau _i = \frac {\lambda (t - 1) + \lambda }{k}\tau _i = \frac {\lambda t}{k}\tau _i . \] The block effects have cancelled. Eliminating \(\mu \) and the \(\beta _j\) from the normal equations gives the reduced normal equations \(C\hat {\boldsymbol \tau } = \mathbf Q\) with \(C = rI - NN^\top /k = (\lambda t/k)(I - J/t)\), by the form of \(NN^\top \) found in Proposition 6.3.2; under \(\sum \hat \tau _i = 0\) their solution is \(\hat \tau _i = kQ_i/(\lambda t)\).

(ii) Using \(\Var (y_{i\cdot }) = r\sigma ^2\), \(\Cov (y_{i\cdot }, y_{\cdot j}) = n_{ij}\sigma ^2\), \(\Var (y_{\cdot j}) = k\sigma ^2\) and independence of different blocks, one finds \(\Cov (\mathbf Q) = \sigma ^2C\). Hence \[ \Var (\hat \tau _i - \hat \tau _l) = \Bigl (\frac {k}{\lambda t}\Bigr )^2\sigma ^2(C_{ii} + C_{ll} - 2C_{il}) = \Bigl (\frac {k}{\lambda t}\Bigr )^2\sigma ^2\cdot \frac {2\lambda t}{k} = \frac {2k\sigma ^2}{\lambda t} . \]

(iii) The reduction in the residual sum of squares from adding treatments to a model that already contains blocks is \(\hat {\boldsymbol \tau }^\top \mathbf Q = k\sum Q_i^2/(\lambda t)\), by the standard result that the extra regression sum of squares equals the estimate times the right-hand side of the reduced normal equations. □