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Let $H=X(X^{T}X{)}^{-1}{X}^T$ be the orthogonal projection onto the design column space. Linear transformations of the Gaussian errors giveThe residual distribution is a singular multivariate normal supported on the orthogonal complement of that column space. Their cross-covariance is ${\sigma }^2(X^{T}X{)}^{-1}{X}^T(I-H)=0$. Joint normality even makes the two vectors independent.

Order the observations as $11,12,21,22$. The two normal linear models have design matricesRemoving each row mean leaves the information for the common slope equal to the sum of within-row squared fertilizer deviations. These sums are $1$ and $4$, respectively. ThereforeThe second design estimates the slope four times as precisely in variance terms.

The observed data in the second design are fitted exactly by $\hat{\theta }=(100,300,100{)}^T$. With $a=(2,2,0{)}^T$, the no-fertilizer total forecast isThe future four independent errors contribute variance $4{\sigma }^2$. Parameter-estimation uncertainty contributes ${\sigma }^2{a}^T(X_{II}^T{X}_{II}{)}^{-1}a=13{\sigma }^2$, so the prediction-error variance is $17{\sigma }^2$, not merely the variance of the estimated mean. Normally, with $s^2=\Vert \widehat{ϵ}{\Vert }^2/(4-3)$, the prediction interval in a normal linear model isHere the printed data have zero residual and hence $s=0$. The formal plug-in interval collapses to $[800,800]$, exhibiting zero-residual degeneracy of regression prediction. Under the stated model with unknown positive variance, this exact fit is a probability-zero sample and the Studentized statistic is undefined at it; the collapsed expression is not evidence that future yield is certain. No positive-width numerical 95% interval can be obtained from these exact data by the usual residual-variance method without extra variance information or an explicit observation-rounding model.

The Rao-Blackwell theorem says that if $S$ is a sufficient statistic and $\delta$ a finite-variance estimator, then ${\delta }^{\ast }=\mathbb{E}(\delta \mid S)$ is an estimator with the same expectation and no larger variance. Sufficiency ensures the conditional distribution, and hence this formula as a function of the data, does not involve the unknown parameter. The tower property of conditional expectation gives $\mathbb{E}{\delta }^{\ast }=\mathbb{E}\delta$, while the law of total variance givesThis proves the theorem and shows that equality holds exactly when $\delta$ is already determined by $S$, almost surely. In particular, unbiasedness is preserved.

Assume $n\ge 3$. Independence gives $\mathbb{E}\hat{\theta }=p_0{p}_1{p}_2=\theta$, so the proposed indicator is unbiased and has variance $\theta (1-\theta )$. The type counts $S=(n_0,n_1,n_2)$ are a sufficient statistic: conditional on them, every ordering of the labels is equally likely. Consequently,For example, the conditional probability of the ordered first three labels is $(n_0/n)(n_1/(n-1))(n_2/(n-2))$. This is Rao-Blackwellization of a multinomial probability product.

If all three $p_i$ are positive, there is positive probability that all three counts are positive. On such an event the conditional indicator has probability strictly between zero and one, so its conditional variance is positive. Thus $\mathrm{Var}({\theta }^{\ast })<\theta (1-\theta )$ for interior parameter values. If any $p_i=0$, then $\theta =0$ and both estimators have zero variance; the printed strict inequality is impossible at those boundary parameters. The valid unrestricted statement is the non-strict inequality.

Identify the hypercube graph $Q_n$ with $\mathcal{P}([n])$, joining two sets when their symmetric difference has size one. Write $N[A]$ for the closed graph neighbourhood of $A$, and $N_t[A]$ for its radius-$t$ neighbourhood in Hamming distance. In the simplicial order on the discrete cube, smaller sets come first, with lexicographic order within each rank: the smallest differing coordinate belongs to the earlier set. Let $I_m$ be the first $m$ vertices in this order. Harper theorem isEquivalently, $I_m$ minimizes the external vertex boundary at fixed size. We will also obtain $|N_t[A]|\ge |N_t[I_{|A|}]|$ for every integer $t\ge 0$.

Here is a complete proof by simplicial section compression. First note two properties of the simplicial order on the discrete cube. Its restriction to either face obtained by fixing one coordinate is the corresponding order on the smaller hypercube graph. Also, the closed graph neighbourhood of an initial segment is an initial segment. To verify the latter, a nonempty initial segment contains every level below its last level $r$, and a lexicographic prefix $\mathcal{L}$ at level $r$. For $r\ge 1$, its closed graph neighbourhood contains every set of size at most $r$, and its only further members are the upper shadow of $\mathcal{L}$. An $(r+1)$-set lies in this upper shadow exactly when its lexicographically earliest $r$-subset lies in $\mathcal{L}$. That earliest subset is obtained by deleting the largest element. These earliest subsets vary monotonically in lexicographic order, so the upper shadow is another lexicographic prefix. For $r=0$, the closed graph neighbourhood of $\{\varnothing \}$ is the union of levels zero and one. The empty initial segment causes no exception.

Induct on $n$, with $n=0,1$ immediate. Fix a coordinate $i$ and write the sections of $A$ as $A_0,A_1\subseteq Q_{n-1}$, deleting $i$ from the latter section. Replace them by the initial segments $I_a,I_b$ with $a=|A_0|$, $b=|A_1|$. The sections of the old closed graph neighbourhood areBy induction their sizes are at least $\max \{|N[I_a]|,b\}$ and $\max \{|N[I_b]|,a\}$, respectively. These are exactly the sizes of the new sections, because both sets in each new union are initial segments and hence nested. Thus simplicial section compression preserves size and never enlarges the closed graph neighbourhood.

Repeatedly compress any section that is not already an initial segment. Every change strictly reduces the sum of the global simplicial positions of the included vertices. This nonnegative integer decreases only finitely often, so the process ends at a family $B$ whose sections in every coordinate are initial segments. If $B$ itself is an initial segment, the induction is complete. Otherwise choose the earliest absent vertex $X$ and the latest present vertex $Y$, with $X\prec Y$. They cannot agree in any coordinate: agreement would put them in one compressed section, where an earlier absent vertex cannot precede a later present one. Hence $Y=X^c$. Nor can a vertex $Z$ lie between them: if $Z$ is present, pairing it with $X$ would force $Z=X^c=Y$; if absent, pairing it with $Y$ would force $Z=Y^c=X$. Thus $X,Y$ are consecutive, and $B$ is obtained from $I_{|B|}$ by replacing its last vertex $X$ with the next vertex $Y$.

There are only two possibilities for these terminal families for simplicial section compression. If $|X|<|Y|$, consecutiveness makes $X$ the last $k$-set and $Y$ the first $(k+1)$-set. Complementarity forces $n=2k+1$, $X=\{k+2,\dots ,2k+1\}$ and $Y=\{1,\dots ,k+1\}$. For $n\ge 3$, the corresponding $B$ contains all sets of size at most $k$ except $X$, and additionally $Y$. Every $(k+1)$-set has at least two $k$-subsets, so at least one belongs to $B$. The missing $X$ has a $(k-1)$-subset in $B$. Consequently $N[B]$ contains every set of size at most $k+1$, which is exactly $N[I_{|B|}]$.

If $|X|=|Y|$, then $n=2k$. Since $X,Y$ are complementary, the smaller one contains coordinate $1$. Consecutiveness forces $X$ to be the last $k$-set containing $1$, and $Y$ the first $k$-set avoiding $1$:The initial segment consists of all sets of size less than $k$ and all $k$-sets containing $1$. Its closed graph neighbourhood contains all sets of size at most $k$ and all $(k+1)$-sets containing $1$. For $k\ge 2$, each such $(k+1)$-set has $k\ge 2$ $k$-subsets containing $1$, so removing only $X$ leaves one in $B$. All sets of size at most $k$ are still in $N[B]$, because all levels below $k$ belong to $B$. Hence again $N[I_{|B|}]\subseteq N[B]$. For $k=1$, the two families are related by exchanging coordinates and both have the whole $Q_2$ as their closed graph neighbourhood. These comparisons finish the induction and prove Harper theorem.

For the radius-$t$ version, repeatedly use the one-step result. At each step the comparison family remains an initial segment, and its closed graph neighbourhood is monotone in its size. If $|N_{t-1}[A]|\ge |N_{t-1}[I_m]|$, applying Harper theorem to $N_{t-1}[A]$ therefore gives $|N_t[A]|\ge |N_t[I_m]|$. Induction on $t$ proves the asserted extension.

A Lévy family of graphs has vanishing concentration of measure functions after distances have been scaled by the graph diameter. More explicitly, for connected finite graphs $G_n$ of diameter $D_n$, use normalized graph distance $d_n=d_{G_n}/D_n$ and uniform probability measure. For every fixed $\epsilon >0$, requireFor $Q_n$ this normalization is normalized Hamming distance $d_H/n$, since its diameter is $n$.

If $|A|\ge 2^{n-1}$, the corresponding initial segment contains the Hamming ball about $\varnothing$ of radius $b_n=\lfloor (n-1)/2\rfloor$. The radius-$t$ version of Harper theorem impliesUse the precise Hoeffding inequality $\mathbb{P}(Z_n-n/2\ge u)\le \exp (-2u^2/n)$ for $u\ge 0$. One can also derive this estimate directly: the moment-generating function of $Z_n-n/2$ is $\cosh (\lambda /2{)}^n\le \exp (n{\lambda }^2/8)$; the exponential Markov inequality followed by minimization at $\lambda =4u/n$ gives the bound. With $t=\lfloor \epsilon n\rfloor$, we have $b_n+t\ge n/2+\epsilon n-2$. Thus, whenever $\epsilon n>2$,This proves that the hypercube graphs form a Lévy family of graphs. The stronger fixed-positive-mass formulation follows as well. For any $a>0$, choose $c$ with $1/(4c^2)<a$. The Chebyshev inequality for $Z_n$, whose variance is $n/4$, makes the Hamming ball of radius $\lfloor n/2-c\sqrt{n}\rfloor$ have fewer than $a{2}^n$ vertices. An initial segment of size at least $a{2}^n$ contains that ball. Enlarging by $\lfloor \epsilon n\rfloor$ and using the same Hoeffding inequality gives a complement proportion tending to zero uniformly over all such $A$. The normalization matters: with unscaled unit Hamming distance, cubes do not satisfy the metric version at every fixed radius; for a radius below one, a half-cube has no enlargement.

Equality in Harper theorem does not determine a down-set up to isomorphism. In $Q_4$, takeThis is a down-set of size nine. Its closed graph neighbourhood contains every set of size at most two, because all singletons belong to $A$. Every triple contains an edge of the four-cycle among the listed pairs, so every triple also belongs to $N[A]$. The four-element set does not. Hence $|N[A]|=1+4+6+4=15$.

The size-nine simplicial order on the discrete cube initial segment is all sets of size at most one together with the pairs $12,13,14,23$. Every triple contains one of these pairs, and the four-element set is again absent from its closed graph neighbourhood. Its neighbourhood therefore also has size $15$, proving extremality of $A$. Nevertheless the induced graphs are not isomorphic: in $A$, the empty set has degree four, the four singletons have degree three, and the four pairs have degree two. The initial segment has two vertices of degree four, namely $\varnothing$ and $\{1\}$. Since a graph automorphism preserves induced degree of a vertex, no automorphism of the cube can carry one family to the other. This gives nonunique down-set extremizers for Harper theorem even under the stronger test of induced graph isomorphism.

The Skorokhod embedding of a centered random walk states that a random walk with independent identically distributed centered steps of finite variance ${\sigma }^2$ can be realized on an appropriate probability space aswhere $B$ is a standard Brownian motion and the $T_n$ are finite stopping times. More precisely, the stopped positions have the same joint law as the given random walk, and the pairsmay be chosen independent and identically distributed. Their spatial component has the step law, and $\mathbb{E}(T_n-T_{n-1})={\sigma }^2$. Repeating the one-step Skorokhod embedding theorem with the Strong Markov property gives this formulation. In the present normalization, the mean time increment is one, and the strong law of large numbers gives $T_n/n\to 1$ almost surely.

The Donsker invariance principle states that the linearly interpolated diffusively rescaled random walkconverges weakly as a random element of $C[0,1]$, equipped with the uniform norm, to standard Brownian motion restricted to $[0,1]$. At $t=1$ the fractional term is zero. The only step assumptions needed here are zero mean, unit variance, and independent identical distributions; a higher moment or bounded support is not required. This is a functional central limit theorem, concerning the entire interpolated path rather than only its endpoint.

For quota share reinsurance the insurer retains the same fraction of each claim, soThe annual retained loss is consequently $T_I=\alpha T$. Each transformed risk severity has probability density function $f_i(y/\alpha )/\alpha$ for $y>0$. The mixed transformed severity, with the same weights ${\lambda }_i/({\lambda }_1+{\lambda }_2)$, still gives a compound Poisson distribution. Scaling the expected value and variance givesThis agrees with the general retained-claim formulas because $\mathbb{E}g(X_i)=\alpha {\mu }_i$ and $\mathbb{E}[g(X_i{)}^2]={\alpha }^2({\sigma }_i^2+{\mu }_i^2)$.

For excess of loss reinsurance the insurer pays each claim up to its retention level:The cap applies separately to every claim. In particular the retained annual loss is ${\sum }_j\min (X_j,M)$, rather than a single cap on the annual total.

Let $F_i$ be the cumulative distribution function for the claim size on risk $i$, and put ${\overline{F}}_i=1-F_i$. The retained severity on that risk has the original probability density function on $0<y<M$ and an atom of a measure at $M$ of mass ${\overline{F}}_i(M)$. Thus $T_I$ has a compound Poisson distribution with rate ${\lambda }_1+{\lambda }_2$ and the mixture of these capped severity laws. The mixture's mass at $M$ is ${\sum }_i{\lambda }_i{\overline{F}}_i(M)/({\lambda }_1+{\lambda }_2)$.

For the capped claim moments, use the tail integral formula for moments. Since $\mathbb{P}(\min (X_i,M)>x)={\overline{F}}_i(x)$ for $0\le x<M$ and is zero for $x\ge M$,Substitution into the compound Poisson distribution moment formulas givesEquivalently, the integrals are ${\int }_0^{M}x{f}_i(x)dx+M{\overline{F}}_i(M)$ and ${\int }_0^M{x}^2{f}_i(x)dx+M^2{\overline{F}}_i(M)$. The annual variance uses the retained raw second moments; subtracting their squared means would omit the variation in the Poisson distribution count.

For an aggregate claims model, let $X$ denote one claim size, with expected value $\mu$ and variance ${\sigma }^2$. The total $S$ is zero when the claim count is zero. Given $N=n$, independence givesThe law of total expectation and law of total variance, together with $\mathbb{E}N=\mathrm{Var}(N)=\lambda$ for a Poisson distribution, implyThe raw second moment appears because the random count itself contributes $\lambda {\mu }^2$ to the variance.

Conditional on $N$, the moment-generating function of the sum is $M_X(t{)}^N$. Averaging with the Poisson distribution therefore gives the compound Poisson distribution transformThis is valid wherever $M_X(t)$ is finite. Finite first and second moments alone do not ensure positive exponential moments; for positive claims the corresponding Laplace transform always exists.

For the independent portfolios put $\Lambda ={\lambda }_1+{\lambda }_2>0$ and $w_i={\lambda }_i/\Lambda$. Their Poisson distribution counts add to a Poisson distribution with parameter $\Lambda$. By Poisson-multinomial conditioning, conditional on the total count $n$ the first risk count has a binomial distribution with parameters $n,w_1$, and the second is the remaining count. Thus one can generate the same total loss by drawing $n$ independent risk labels with these weights, then drawing each claim from its label's law. The merged severity has mixture distributionIt follows that $T$ has a compound Poisson distribution with count parameter $\Lambda$ and this severity law. This is the fixed-year version of Poisson superposition of insurance portfolios. Its expected value and variance areAlternatively, multiplying the two independent aggregate Laplace transforms yields $\exp \{\Lambda (w_1{M}_{X_1}(t)+w_2{M}_{X_2}(t)-1)\}$ wherever finite, confirming the same compound Poisson distribution.

For the retained compound Poisson aggregate under per-claim reinsurance, replace each claim $X$ by its retained payment $g(X)$. Assume the retention is measurable, with $0\le g(x)\le x$ as usual. The count parameter remains $\Lambda$, and the severity law is the pushforward measure of the mixture severity under $g$. Thus the retained compound Poisson aggregate $T_I$ has a compound Poisson distribution with that transformed severity andIts moment-generating function is $\exp \{{\sum }_i{\lambda }_i(M_{g(X_i)}(t)-1)\}$ on its finite domain. For a general retention $g$, zero retained payments can occur and are allowed as compound-Poisson marks; they may equivalently be removed by Poisson thinning. The two specified contracts retain strictly positive payments for strictly positive claims.

An invertible time-series representation recovers the driving white noise from current and past observations. In the inverse series the support condition is thereforeAgain the series must converge. Stable invertibility uses ${\sum }_{r\ge 0}|b_r|<\infty$, which ensures mean-square convergence when $X_t$ has finite variance. Merely writing a bilateral inverse is not invertibility in this one-sided sense: it may require future observations. For a general correlated input $X$, square summability of $b_r$ alone is not the same sufficient condition as it is for a white noise input.

A causal time-series representation uses only the present and past driving white noise. Thus the coefficient condition isThe series must have its stated convergence meaning. For centered white noise of positive finite variance, ${\sum }_{r\ge 0}|a_r{|}^2<\infty$ is sufficient and necessary for mean-square convergence. In the usual stable-filter convention one imposes the stronger ${\sum }_{r\ge 0}|a_r|<\infty$. A bilateral stationary linear process need not be causal: terms with $r<0$ involve future driving values.

For an autoregressive moving-average model, the driving-to-output transfer function is $A(z)=\theta (z)/\varphi (z)$, and the inverse transfer function is $A(z{)}^{-1}=\varphi (z)/\theta (z)$. The causality and invertibility root criteria for an ARMA model require these respective rational functions to have power series about zero converging on a disk larger than the unit disk. If the two polynomials have no common factor, the conditions becomeIndeed, outside-disk roots leave a radius of convergence greater than one, so the coefficients decay geometrically and are absolutely summable. Conversely an uncancelled pole inside or on the unit disk prevents the required stable power series. If factors are common, apply the criterion after cancellation, to the noise-driven solution rather than additional homogeneous components. The backshift operator translates these power series into the desired one-sided filters.

The original representation has $\varphi (z)=1-2z$ and $\theta (z)=1+2z$, with roots $1/2$ and $-1/2$. There is no cancellation. Hence it is neither causal nor invertible relative to its specified driving noise. Stationarity is nevertheless possible through a two-sided solution: expanding the autoregressive inverse in negative powers givesThis is an anticausal time series representation with square-summable coefficients.

To establish the alternative representation on the same process, defineThe inverse of $1+\frac{1}{2}B$ is a stable one-sided filter. For $|z|=1$, the identities $|1+2z{|}^2=4|1+z/2{|}^2$ and $|1-2z{|}^2=4|1-z/2{|}^2$ giveThe time-series spectral density of $X$ is ${\sigma }^2|1+2e^{-i\omega }{|}^2/(2\pi |1-2e^{-i\omega }{|}^2)$; therefore the defined $\eta$ has constant time-series spectral density ${\sigma }^2/(2\pi )$. Its mean is zero, its variance is ${\sigma }^2$, and all its nonzero-lag autocovariances vanish. It is thus weak white noise. Its definition directly givesThis root reflection of an ARMA representation has roots $2,-2$, so is causal and invertible. A constant spectrum proves whiteness, not independence of non-Gaussian coordinates; no Gaussian assumption is needed for the required white noise representation.

For best linear prediction from an infinite past, let ${\mathcal{H}}_0$ be the closed linear span of $X_t$ with $t\le 0$. The causal representation expresses every such $X_t$ in present and past $\eta$ values, while invertibility puts every ${\eta }_t$ with $t\le 0$ in ${\mathcal{H}}_0$. In particular ${\eta }_1$ is orthogonal to ${\mathcal{H}}_0$. The representation at time one then gives the orthogonal projectionConsequently the best linear predictor and its error variance areOrthogonality proves optimality among linear predictors in the closed past span, without asserting that the predictor must be the conditional mean for a non-Gaussian process.

For a real weakly stationary process, extend its autocovariance by ${\gamma }_{-k}={\gamma }_k$. The exact existence of a time-series spectral density condition is that its spectral measure of a stationary time series be absolutely continuous with respect to Lebesgue measure. Absolute summability ${\gamma }_0+2{\sum }_{k\ge 1}|{\gamma }_k|<\infty$ is a useful sufficient condition, not a necessary one.

We use the conventional angular-frequency density on $[-\pi ,\pi ]$, restricted to $[0,\pi ]$ by symmetry. Under the absolute-summability condition, the Fourier series and its inverse relation areThus $2{\int }_0^{\pi }f={\gamma }_0$. If a one-sided density is instead normalized to integrate to the full variance, use $g=2f$ and omit the factor two in the inverse formula. This is the positive-frequency spectral normalization convention difference. With merely an integrable spectral density, the inverse relation remains valid; one must not assume pointwise convergence of the unweighted Fourier series. Its Fejér sums recover the density in $L^1$:

If $X$ and $Y$ are independent stationary processes, the cross covariances vanish, soLinearity of the inverse Fourier series relation, or addition of the spectral measures of a stationary time series, gives $f_Z=f_X+f_Y$. Independence can in fact be weakened to zero cross-covariances at every lag.

For the ARMA$(1,1)$ representation, away from an uncancelled unit root the transfer function givesThe usual causal case has $|\varphi |<1$; the same expression holds for the two-sided stationary solution when $|\varphi |>1$. We work in the nondegenerate case $v,w>0$ and $\varphi \ne \pm 1$; cancellations and zero-noise cases are obtained by the appropriate reduced representation or limits. Since independent white noise of variance $w$ contributes $w/(2\pi )$, putThenThe white-noise addition to an ARMA(1,1) process problem is therefore the factorizationDefineBoth are positive under the stated nondegeneracy conditions. An invertible choice isThese satisfy $|\alpha |<1$, $\lambda (1+\alpha {)}^2=P$, and $\lambda (1-\alpha {)}^2=Q$. HenceIn particular, it is the autoregressive coefficient $\varphi$ that enters the terms from the added observation noise.

To obtain a representation of the actual $Z$, defineThe stable inverse exists because $|\alpha |<1$, and spectral filtering gives $f_{\xi }=\lambda /(2\pi )$. Thus $\xi$ is weak white noise, andIts variance can equivalently be writtenWhen $\alpha \ne 0$, also $\lambda =(v\theta -w\varphi )/\alpha$; when $\alpha =0$, necessarily $C=0$ and $\lambda =A$, so the latter quotient should not be used. The model may reduce in order through cancellation. Boundary limits with $P=0$ or $Q=0$ give $|\alpha |=1$ and need not be stably invertible; the displayed stable reconstruction applies to the nondegenerate case above.