Mean and variance under balanced biological and technical replication

Summing the model in [F1] over $i,j$ gives $\bar{Y}=\mu +\bar{\tau }+\bar{B}+\bar{E}$. Each summand is integrable, so [F6] and the zero means in [F1] give $\mathbb{E}[\bar{Y}]=\mu +\bar{\tau }$ when $\mathit{\tau }$ is fixed, and $\mathbb{E}[\bar{Y}]=\mu +\mathbb{E}[\bar{\tau }]$ when it is random.

For distinct biological indices $i\mathrm{\ne }k$, distinct technical indices $(i,j)\mathrm{\ne }(k,\mathrm{\ell })$, and any biological-technical pair $(i,(k,\mathrm{\ell }))$, the variables in each pair are independent by [F1] and [F5]. They are integrable because they are square-integrable. Applying [F4] to identity functions gives expectation of the product equal to the product of their zero means; [F2] therefore makes every such off-diagonal covariance zero. By [F3], bilinearity over the finite index families then gives $\mathrm{Cov}(\bar{B},\bar{E})=\frac{1}{n^{2}m}{\sum }_{i=1}^n{\sum }_{k=1}^n{\sum }_{\mathrm{\ell }=1}^m\mathrm{Cov}(B_i,E_{k\mathrm{\ell }})=0$.

Bilinearity [F3] expands $\mathrm{Var}(\bar{B})$ into $n^2$ covariance terms with coefficient $n^{-2}$. By 1.2 the off-diagonal terms vanish, and the $n$ diagonal terms equal ${\sigma }_b^2$ by [F1] and [F2]. Hence $\mathrm{Var}(\bar{B})=n^{-2}n{\sigma }_b^2={\sigma }_b^2\mathrm{/}n$.

Bilinearity [F3] expands $\mathrm{Var}(\bar{E})$ into $(nm{)}^2$ covariance terms with coefficient $(nm{)}^{-2}$. By 1.2 the off-diagonal terms vanish, and the $nm$ diagonal terms equal ${\sigma }_e^2$ by [F1] and [F2]. Hence $\mathrm{Var}(\bar{E})=(nm{)}^{-2}(nm){\sigma }_e^2={\sigma }_e^2\mathrm{/}(nm)$.

With fixed $\mathit{\tau }$, $\mu +\bar{\tau }$ is constant and has zero covariance with every square-integrable random variable by [F2]. Expand the variance of $\bar{B}+\bar{E}$ using [F3]; its cross-covariance is zero by 1.2. Substituting 2.1 and 2.2 proves the fixed-schedule formula.

With random $\mathit{\tau }$, expand the variance of $\bar{\tau }+\bar{B}+\bar{E}$ by [F3]. The $\bar{B}$-$\bar{E}$ covariance is zero by 1.2, and 2.1–2.2 give the two component variances, yielding exactly the displayed general formula with both treatment cross-covariances retained. If $\bar{\tau }$ is independent of $\bar{B}$ and $\bar{E}$, [F4] makes those cross-covariances zero. Expanding $\mathrm{Var}(n^{-1}{\sum }_i{\tau }_i)$ by [F3] gives the stated covariance sum; when the ${\tau }_i$ are mutually independent with common variance ${\sigma }_{\tau }^2$, only its $n$ diagonal terms remain, giving ${\sigma }_{\tau }^2\mathrm{/}n$.