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A regression factor is a categorical predictor represented by level indicators and contrasts. Treating year as a regression factor permits arbitrary differences between the three yearly means, reflecting conditions such as weather, rather than imposing a linear time trend.

Let $Y_{ijk}$ be the yield for seed variety $i\in \{1,2\}$, fertiliser level $j\in \{L,H\}$, and year $k\in \{2004,2005,2006\}$. A treatment-coded two-factor normal linear model with additive year effects iswith independent $N(0,{\sigma }^2)$ errors. The unrestricted coefficients are ${\beta }_0,a,b,c_{2005},c_{2006},d$: six parameters for twelve observations. Equivalently use baseline-zero variety, fertiliser, and interaction arrays. There are no year-treatment interactions in this fit.

Dividing each sum of squares by its statistical degrees of freedom gives the completed analysis of variance:In particular, the missing interaction statistic is $16.0083/(2.8414/6)\simeq 33.803$. Its null hypothesis is $H_0:d=0$, against $H_1:d\ne 0$: the change in mean yield from high fertiliser is the same for both varieties. Under the null hypothesis and the independent homoscedastic normal linear model, this nested-model F-test has distribution $F_{1,6}$. The printed $p$-value 0.001137 strongly rejects no interaction.

Do not drop fertiliser just because its main-effect ANOVA row has $p=0.7011$. In this balanced factorial design that row measures a fertiliser effect averaged over varieties; large opposite effects can cancel. The strongly supported interaction requires retaining its associated main effects by regression model hierarchy. Seed and year also have evidence of effects in their ANOVA rows, so this table alone gives no compelling simplification of the fitted terms.

The interaction contrast is $-4.6200$, with standard error 0.7946. The fitted high-minus-low fertiliser effect is $+2.4700$ for variety 1 but $2.4700-4.6200=-2.1500$ for variety 2. The fitted variety-2-minus-variety-1 contrast is $+3.6500$ under low fertiliser and $3.6500-4.6200=-0.9700$ under high fertiliser. Thus variety 2 performs best under low fertiliser, while high fertiliser benefits variety 1 and reduces variety 2's fitted yield. The marginal fertiliser effect is only $(2.4700-2.1500)/2=0.1600$.

The four fitted means in the baseline year are $5.5050,7.9750,9.1550,7.0050$ for $(1,L),(1,H),(2,L),(2,H)$ respectively. Add $1.5775$ in 2005 and $0.0875$ in 2006 to every fitted mean. The 2005 coefficient has $p=0.017646$ under a $t_6$ test of a zero contrast with 2004; the analogous 2006 contrast has $p=0.863216$. The latter is lack of evidence for a difference, not proof that the yearly means coincide. Standard errors for the combined contrasts require the coefficient covariance matrix. The experiment is small: residual standard error is 0.6882 on only six statistical degrees of freedom. If the same fields were reused, possible within-field dependence and field allocation would need checking before treating the fitted error independence as established.

The moment-based approach specifies marginal means, variances, and within-student covariances, without supplying a joint probability distribution for each count profile. It is a quasi-likelihood or generalized estimating equation strategy. The parameter interpreted as a structural-zero fraction is not automatically an actual probability of a structural-zero component merely because it appears in the moment formulas. A valid positive-definite working covariance and parameter identifiability must also be checked.

The printed covariance specification is not admissible for all the stated parameter values. For example, take $\pi =1/2$, $\tau =1$, and all three component means equal to 10. It gives diagonal variance 80 and off-diagonal covariance 100, so $\mathrm{Var}(Y_{i1}-Y_{i2})=80+80-2(100)=-40$. A covariance matrix cannot have a negative variance in any direction. Thus the moment approach requires additional admissibility restrictions or a valid working covariance; the parameter ranges alone do not define a valid model. This does not alter the hierarchical model, whose covariance derived below is positive semidefinite.

The alternative gives a full hierarchical mixture distribution, specifically a shared zero-inflated Gamma-Poisson count model: a common student random effect is zero with probability $\pi$, and otherwise has a Gamma distribution with mean 1 and variance $\tau$. Conditional on this effect, the three counts are independent Poisson random variables. Integrating it out induces both excess zeros and positive within-student dependence. Zero inflation is shared for the whole student, rather than independently reselected in each term. This full distribution supports likelihood-based inference but is more sensitive to the distributional assumptions.

The marginal mean in the hierarchical model has the same form as the proposed moment mean. Its marginal covariance does not, in general, equal the printed working covariance; the law of total covariance calculation below makes the distinction explicit. At $\tau =0$ the Gamma component is interpreted as its point-mass-at-one limit. At $\pi =1$ all counts are zero, and the log-mean parameterization and regression effects are not identifiable.

The autocovariance of an MA(1) process comes directly from shared noise terms:Work with the zero-mean white noise convention and ${\sigma }^2>0$. Put ${\mathcal{H}}_{t-1}=\mathrm{span}(X_1,\dots ,X_{t-1})$. Suppose the preceding finite-past innovations $U_1,\dots ,U_{t-1}$ are mutually orthogonal. Their triangular relation to the observations means that they span the same finite past. For $s\le t-2$, $U_s$ is a linear combination of $X_1,\dots ,X_s$, so $\mathrm{Cov}(X_t,U_s)=0$. Moreover,Projection onto this orthogonal basis therefore has only its last coefficient nonzero:The residual is orthogonal to the entire preceding span, proving the induction. Expanding its variance givesThe initialization is $U_1=X_1$, ${\phi }_1={\gamma }_0$, ${\lambda }_1=0$ and $U_0=0$. Each finite covariance matrix is a positive-definite matrix: the last observation in any nonzero finite linear combination contains a noise term absent from earlier observations. Thus all the projection denominators are positive. These are the finite-sample innovations of an MA(1) process, requiring no Gaussian assumption.

Substituting the covariances into the recursion givesThe denominator has a positive limit when ${\lambda }_{t-1}\to \lambda \in [-1,1]$. Taking limits gives $(\lambda -\theta )(\theta \lambda -1)=0$. The permitted interval selectsFor $\theta =0$, every coefficient is zero; at $\theta =\pm 1$, the roots coincide. This is the limiting MA(1) innovations coefficient. The limiting innovation variance is correspondingly ${\sigma }^2\max (1,{\theta }^2)$.

For the observed numerical case, ${\gamma }_0=2$ and ${\gamma }_1=-1$. The exact innovation calculations areTherefore the predictor of $X_3$ based on the two observations and its error variance areFor $X_4$, the available information is still only ${\mathcal{H}}_2$. Since its covariances with both $X_1$ and $X_2$ are zero, its orthogonal projection onto that span is zero:This is two-step prediction for an MA(1) process. Applying the next one-step recursion would instead use an observed $X_3$, which is not available here.

For any $r_1,\dots ,r_n\in \mathbb{R}$, the linearity from part (a) givesalmost surely, and the right side has a normal distribution. This is the defining linear-combination criterion for a multivariate normal distribution; singular covariance matrices are allowed.

Passing to the limit in the inner products of the partial sums givesThe passage to the limit is justified by Cauchy-Schwarz inequality and $L^2$ convergence. Equivalently, the vector's characteristic function isThus both joint normality and the complete covariance matrix follow from the Hilbert-space inner product.

The ordinary least squares estimator is $\hat{\beta }=(X^{T}X{)}^{-1}{X}^{T}Y$. Consequently the fitted values and regression residuals areAn affine transformation of a multivariate normal distribution is again multivariate normal, possibly with a singular covariance matrix. For a random vector with covariance matrix $\Sigma$, its transformed covariance matrix is $A\Sigma A^T$. Since $PX=X$, $P^2=P=P^T$ and $(I-P{)}^2=I-P$, these results giveBoth multivariate normal distributions are supported on their respective projected subspaces. In particular, ${\hat{Y}}_i$ has variance ${\sigma }^2{P}_{ii}$ and $e_i$ has variance ${\sigma }^2(1-P_{ii})$; the regression residuals need not be mutually independent.

Because both vectors are linear transformations of the same multivariate normal response, $(\hat{Y},e)$ is jointly multivariate normal. Its cross-covariance matrix isZero cross-covariance implies independence for jointly multivariate normal vectors, including singular ones. Therefore the fitted values and the entire vector of regression residuals are independent. The fitted-residual orthogonality identity gives the zero covariance; the normal distribution assumption is what upgrades it to independence.