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Write $n=bk$, let $I_n$ be the identity matrix, let $J_m$ be the $m$-by-$m$ all-ones matrix, and put $B=\mathrm{diag}(J_k,\dots ,J_k)$, with one diagonal block per experimental block. The variance-covariance matrix isIts orthogonal decomposition is obtained by first subtracting each block's sample mean, then subtracting the grand sample mean from the block means. More explicitly, writing $x_{jt}$ for coordinate $t$ in block $j$, the three invariant subspaces areThey are pairwise orthogonal, with dimensions $b(k-1)$, $b-1$ and $1$, and their direct sum is ${\mathbb{R}}^n$. On $W$, both $B$ and $J_n$ vanish. On $U$, $B=kI$ and $J_n=0$. On $G$, $B=kI$ and $J_n=nI$. Thus the eigenvalues and corresponding invariant subspaces areIf some eigenvalues coincide, their eigenspace is the direct sum of the listed subspaces with that value. Zero-dimensional rows are omitted when $b=1$ or $k=1$. For an admissible covariance matrix the eigenvalues on nonzero subspaces must be nonnegative; these conditions are also sufficient for positive semidefiniteness. The expectation parameters do not affect this calculation.

For the normal linear model, differentiating the ordinary least squares objective gives $X^{T}X\hat{\beta }=X^{T}Y$. Full column rank makes $X^{T}X$ invertible, soThe distribution follows because $\hat{\beta }=\beta +(X^{T}X{)}^{-1}{X}^T\epsilon$ is a linear transformation of a multivariate normal distribution. Its expectation is $\beta$, and direct multiplication gives the displayed covariance matrix.

Let $H=X(X^{T}X{)}^{-1}{X}^T$, the orthogonal projection onto the design's column space. The residual sum of squares is $RSS=\Vert Y-X\hat{\beta }{\Vert }^2=Y^T(I-H)Y$. Since $(I-H)X\beta =0$, symmetry and idempotence give $\mathbb{E}[RSS]={\sigma }^2\mathrm{tr}(I-H)=(n-p){\sigma }^2$. Thusis an unbiased estimator.

For the two independent regressions choose $\beta =(a,b,c,d{)}^T$. The first $m$ rows of the design matrix are $(1,u_i,0,0)$ and the last $m$ rows are $(0,0,1,w_i)$. The centering conditions and distinct predictors giveIt follows thatThe two estimated slopes are independent normal random variables, as they depend on independent error blocks. ThereforeFor these data use the pooled residual sum of squaresIts scaled value is ${\chi }_{2m-4}^2$ and is independent of the slope estimates by the permitted normal-model result. Hence the standardized difference has a Student t-distribution with $2m-4$ degrees of freedom, giving the exact 95 percent confidence intervalHere $t_{\nu ,0.975}$ is the 0.975 quantile of that distribution. This is a confidence interval for the difference of independent regression slopes. As required by the original $p<n$ framework, this specialization needs $m>2$. Distinct predictors alone allow $m=2$, but then the fit is saturated, $RSS=0$ and there are no residual degrees of freedom to estimate the unknown variance or form this interval.

For independent identically distributed observations $Y_1,\dots ,Y_n$, the likelihood function and log-likelihood areA maximum-likelihood estimator is a measurable choice ${\hat{\theta }}_n\in {\mathrm{argmax}}_{\theta \in \Theta }{L}_n(\theta )$, when a maximizer exists. It need not be unique. The one-observation score function is $s_{\theta }(y)={\nabla }_{\theta }\log f(y,\theta )$, and the Fisher information matrix isUnder the regularity assumptions, differentiation beneath the integral gives ${\mathbb{E}}_{\theta }{s}_{\theta }={\nabla }_{\theta }\int f=0$. Differentiating again gives the second information identity. Independent observations have total Fisher information $nI(\theta )$.

Assume fixed parameter dimension, an interior true parameter ${\theta }_0$, a positive-definite matrix $I({\theta }_0)$, the standard differentiability and integrability conditions, and statistical consistency of ${\hat{\theta }}_n$. Then the asymptotic normality of a maximum likelihood estimator isHere $I$ is the information per observation. Thus the leading covariance matrix of the estimator itself is $I({\theta }_0{)}^{-1}/n$.

To prove this, statistical consistency places the estimator in an interior ball about ${\theta }_0$ with probability tending to one, where its score function vanishes. Write $d_n={\hat{\theta }}_n-{\theta }_0$ and use the integral first-order Taylor formula for a vector map:This integral matrix is needed in a vector problem; one does not have to assert a common scalar mean-value point for every score component. PutThe regular local uniform law of large numbers, continuity of the expected Hessian matrix, and statistical consistency give $J_n\stackrel{P}{\to }I({\theta }_0)$. The multivariate central limit theorem givesThe limiting information matrix is nonsingular, so $J_n^{-1}\stackrel{P}{\to }I({\theta }_0{)}^{-1}$. Solving the Taylor identity and applying the Slutsky theorem proves the displayed normal limit, since $I^{-1}I{I}^{-1}=I^{-1}$. Boundary parameters or singular information are outside this regular theorem.

Set $z=z_{1-\alpha }$. Under the null hypothesis rejection requires both $W_1\ge f$ and $W_2>z$. Since $W_2$ has the standard normal distribution,The bivariate normal distribution in the preceding calculation has a nonsingular covariance matrix and strictly positive statistical probability density everywhere. For every finite futility boundary $f$ and $0<\alpha <1$, the open rectangle $\{W_1<f,W_2>z\}$ has positive probability. ThereforeFor an explicit expression, conditional on $W_1=w$ the full-sample statistic is $N(w/\sqrt{2},1/2)$ under the null hypothesis, yieldingThe Type I error is reduced because some otherwise rejecting paths stop for futility. Equality is approached as $f\to -\infty$, but does not hold at a finite futility boundary.

Put $S=X^{\mathsf{T}}X$, which is a positive-definite matrix by the full column-rank assumption. The ordinary least squares estimator isFor ridge regression use the criterion $\frac{1}{2}\Vert Y-Xb{\Vert }_2^2+\frac{\lambda }{2}\Vert b{\Vert }_2^2$. Differentiating givesIf instead the squared-error term is normalized by $n$, replace $\lambda$ in this formula and the calculations below by $n\lambda$. Only the scale of the tuning parameter changes.

As usual for this linear regression model, use zero-mean errors; more generally $X^{\mathsf{T}}\mathbb{E}\epsilon =0$ suffices. Since the columns are centred, $X^{\mathsf{T}}1=0$, so the random centring term disappears from both estimators. ThusLet $A=(S+\lambda I{)}^{-1}$. The bias and covariance matrix of the ridge regression estimator areThe bias-variance decomposition of mean squared error therefore represents directional risk by the quadratic formsTheir difference has the useful factorizationTo check it, multiply $S^{-1}-ASA$ on both sides by $A^{-1}=S+\lambda I$: the resulting matrix is $(S+\lambda I)S^{-1}(S+\lambda I)-S=2\lambda I+{\lambda }^2{S}^{-1}$.

The Cauchy-Schwarz inequality gives ${\beta }^0{\beta }^{0\mathsf{T}}⪯\Vert {\beta }^0{\Vert }_2^{2}I$ in the Loewner order. Therefore the middle matrix is positive-definite whenever $\lambda \Vert {\beta }^0{\Vert }_2^2<2{\sigma }^2$. For example,works, independently of the eigenvalues of $S$. Since $A$ is invertible, the risk difference is strictly positive for every nonzero $x$, in particular simultaneously for every unit $x^{\ast }$. This proves uniform directional risk improvement by ridge regression. For the normalized-loss convention, take the displayed choice divided by $n$.

For the final assertion, fix any $\lambda >0$ and choose a unit eigenvector $v$ of $S$ with eigenvalue $d>0$. Put $x^{\ast }=v$ and ${\beta }^0=Bv$. Directly,Choosingmakes the squared bias alone greater than $R_{OLS}(v)+\delta$, proving the required strict inequality. Thus unbounded directional risk of fixed ridge shrinkage is compatible with the existence of a beneficial signal-dependent small penalty: no fixed positive penalty uniformly improves risk over all possible signals.

The mean assumption matters if the printed covariance condition is read in isolation. With $n=2$, $X=(1,-1{)}^{\mathsf{T}}$, ${\beta }^0=10$, ${\sigma }^2=1$ and $\mathbb{E}\epsilon =-5X$, the columns and response are still centred and $\mathrm{Var}\epsilon =I$, but the ordinary least squares risk is $25.5$. Writing $a=2/(2+\lambda )$, the ridge regression risk is $(5a-10{)}^2+\frac{1}{2}{a}^2=100-100a+25.5a^2>25.5$ for every $0<a<1$. Hence the standard zero-mean regression-error convention, or its stated orthogonality version, is necessary for the first risk comparison; covariance alone is insufficient.

The ordinary least squares objective has gradient $2X^{T}X\beta -2X^{T}Y$. The normal equation therefore givesThe Gram matrix $X^{T}X$ is a positive-definite matrix because $X$ has full column rank. More explicitly, for any $b$,since $X^T(Y-X\hat{\beta })=0$. This proves that the displayed solution is the unique global minimum.

The fitted values are $\hat{Y}=X\hat{\beta }=HY$, where the hat matrix isThe inverse of the symmetric matrix $X^{T}X$ is symmetric, so $H^T=H$. Multiplication gives $H^2=X(X^{T}X{)}^{-1}(X^{T}X)(X^{T}X{)}^{-1}{X}^T=H$. Thus $H$ is the orthogonal projection onto the column space of the design matrix.

The regression residual vector is $e=Y-\hat{Y}=GY$ with $G=I_n-H$. Consequently $G^T=G$, $G^2=I_n-2H+H^2=G$, and the residual sum of squares isBy fitted-residual orthogonality, $GH=0$. Using $\mathrm{Cov}(Y)={\sigma }^2{I}_n$, the cross-covariance matrix isBoth vectors are jointly normal, so they are also independent by independence of uncorrelated jointly normal variables. The zero covariance calculation itself only needs the common error variance and lack of error correlations.

For the first wind fit, let $v_i$ be wind velocity and $Y_i$ electrical output. The simple linear regression is $Y_i=\alpha +\beta v_i+{\epsilon }_i$, with independent ${\epsilon }_i\sim N(0,{\sigma }^2)$. It has two fitted mean parameters and $25-2=23$ residual degrees of freedom. The missing analysis of variance entries are therefore velocity degrees of freedom $1$, velocity mean square $8.9296$, and residual degrees of freedom $23$. The residual mean square is $1.2816/23\simeq 0.05572$, and $8.9296/(1.2816/23)\simeq 160.25$, consistent with the printed value after rounding.

Because the model includes a regression intercept, the coefficient of determination is the explained sum of squares divided by the total centered sum of squares:A large coefficient of determination does not rule out a wrong mean function. In the PDF's first residual-versus-fitted plot, residuals are negative at both ends and positive in the middle. This curved pattern agrees with the visibly flattening output-versus-velocity relationship and motivates a polynomial regression with a quadratic term.

The second wind model is $Y_i=\alpha +{\beta }_1{v}_i+{\beta }_2{v}_i^2+{\epsilon }_i$, again with independent common-variance normal errors. Its fitted mean is $-1.155898+0.722936v_i-0.038121v_i^2$. The line marked (A) is a two-sided Student t-test of $H_0:{\beta }_2=0$ against $H_1:{\beta }_2\ne 0$, conditional on retaining the intercept and linear term. Under the null hypothesis,Its two-sided p-value is $6.59\times 10^{-8}$. This is far below $0.05$, so reject a purely linear mean in favour of the quadratic fit. Equivalently, the extra-term nested-model F-test has $F=t^2\simeq 63.15$ and null law $F_{1,22}$.

The third fit is a reciprocal-predictor regression, $Y_i=a+b/v_i+{\epsilon }_i$, with estimated mean $2.9789-6.9345/v_i$. It has two mean parameters rather than the quadratic model's three. Its residual standard error is smaller, $0.09417$ versus $0.1227$, and its coefficient of determination is larger, $0.9800$ versus $0.9676$. The corresponding residual sums of squares are approximately $23(0.09417{)}^2=0.20396$ and $22(0.1227{)}^2=0.33122$. The reciprocal fit is preferable on both these fit measures and parsimony. The two models are not nested, so an ordinary extra-term F-test between them is inappropriate. On the common normal-error likelihood, the difference ${\mathrm{AIC}}_3-{\mathrm{AIC}}_2=25\log (0.20396/0.33122)-2\simeq -14.12$ also favours the third model.

Neither lower residual sum of squares nor a higher coefficient of determination establishes adequate assumptions. The quadratic residual-versus-fitted plot removes the original pronounced curvature; the reciprocal plot also has no comparably obvious mean trend. The low-output residuals appear somewhat more spread out, so check scale-location plots and residuals against velocity for heteroscedasticity. Q-Q plots assess normality; residuals against observation order assess serial dependence; regression leverage and Cook's distance identify influential observations. Further cross-validation, replicate observations at comparable velocities, and prediction errors would help choose between their extrapolation behaviours. Both fitted shapes should be judged principally over the observed positive-velocity range.

The statistician first models the age- and gender-dependent mean by linear regression. Its fitted value is $27.01675-0.10483\mathrm{age}+1.47031\mathrm{gender}$. At fixed gender, a year of age is associated with a decrease of $0.10483$ BMI units; the group coded gender $1$ has a fitted mean $1.47031$ units above the group coded $0$ at the same age. This question does not state which gender receives which code. The intercept refers to age zero in the reference group, outside a typical adult-patient range, so its substantive interpretation is limited. Both slope Student t-tests have small p-values, and the overall F-test supports an age/gender-dependent mean. The coefficient of determination is only $0.18$, leaving considerable individual variation to investigate.

The residual diagnostics check whether mean adjustment is plausible and whether a single normal error law is adequate. The residual-versus-fitted plot has no strong curved mean trend, although its smooth curve is not perfectly flat. The scale-location plot suggests some decline in spread as fitted BMI increases, so common residual variance is a working approximation rather than an established fact. The normal Q-Q plot has systematic departures, notably shorter tails than the normal reference, suggesting the residual law is not exactly normal. The regression leverage plot does not display an obviously extreme leverage point, but the marked observations and Cook's distance should still be checked for influence. Independence between patients cannot be diagnosed from these four plots alone.

Next the statistician extracts the regression residuals and fits an intercept-only normal model as a baseline residual distribution. With an intercept in the original ordinary least squares model, the residuals sum to zero by the normal equation; the near-zero fitted residual mean and its $p=1$ are therefore automatic, not new evidence of good fit. The $2.698$ standard error in this refit uses $499$ degrees of freedom, whereas the original $2.704$ uses $497$ after estimating three regression coefficients. The reported log-likelihood for the single-normal residual model is $-1205.25$.

The histogram suggests a shape worth exploring beyond one normal density, with broad shoulders and some asymmetry, without showing unambiguous separated clusters. The statistician fits two- and three-component finite Gaussian mixtures with a common variance by the expectation-maximization algorithm, starting from separated means and positive weights. Successive likelihoods increase and the displayed final iterations stabilize, as expected from EM likelihood monotonicity. This is evidence of numerical convergence from those starts, not proof of a global maximum.

The two-component fit assigns weights $(0.3317,0.6683)$, residual means $(-2.7844,1.3819)$ and common variance $3.4175$. The three-component fit assigns weights $(0.2051,0.3926,0.4023)$, residual means $(-3.7306,-0.4417,2.3329)$ and common variance $2.1448$. The code correctly passes the square roots of those variances to dnorm and overlays the resulting weighted densities. Both curves broadly follow the histogram and are very similar. In particular, a two-component mixture need not have two distinct visible modes.

These fits explore residual mixture clustering: groups differ in BMI relative to the same age/gender-adjusted mean, rather than simply in raw BMI. Membership can be summarized by the fitted mixture responsibilities, retaining uncertainty instead of asserting a certain label for each patient. The common-variance assumption is economical, but should be checked; the common regression slope assumption and constancy of mixture proportions across age and gender are also substantive. A residual mixture alone does not establish genuine biological subpopulations; a flexible continuous distribution, missing predictors, nonlinear mean effects or heteroscedasticity could explain a similar marginal shape.

There is also a two-stage residual mixture fitting limitation. Fitted regression residuals are not exactly independent identically distributed observations: even under a normal regression, their covariance matrix is ${\sigma }^2(I-H)$, and estimating the mean adjustment introduces uncertainty. With $500$ subjects and three initial regression coefficients, treating them as independent is a useful approximation for exploring shape, but ordinary mixture likelihoods omit the first-stage uncertainty. A joint mixture regression with shared slopes, or a suitable bootstrap of the whole analysis, would give a firmer basis for parameter uncertainty and cluster selection. Examine component membership versus age/gender and repeat the algorithm from multiple starts before making substantive clustering claims.

Let initial wealth be $w$ and let ${\theta }_i$ be the number of units held of risky asset $i$; the bank investment is $w-{\theta }^T{S}_0$. Put $b=\mu -(1+r)S_0$. Terminal wealth is normal, with mean $(1+r)w+{\theta }^{T}b$ and variance ${\theta }^{T}V\theta$. The expectation of the exponential utility isFor $\gamma >0$ and positive definite covariance matrix $V$, maximizing this is equivalent to maximizing the certainty equivalent ${\theta }^{T}b-\gamma {\theta }^{T}V\theta /2$. Thus the optimal investment isAll investors have proportional risky holdings. A market portfolio can therefore be represented by ${\theta }_M=V^{-1}b$, with any positive scalar multiple giving the same return. Suppose its initial value $P={\theta }_M^T{S}_0$ is positive and $b\ne 0$. Set $B=b^T{V}^{-1}b$. For returns $R_i=S_1^i/S_0^i-1$ and $R_M={\theta }_M^T{S}_1/P-1$,The beta of an asset is its covariance with the market return divided by the market variance. ConsequentlyThis is the capital asset pricing model. Since ${\beta }_i={\rho }_i{\sigma }_i/{\sigma }_M$, dividing the excess-return identity by ${\sigma }_i$ gives the Sharpe ratio identityThese ratio statements require nonzero variances and nonzero initial portfolio value. If $V$ is singular and $b$ is in its range, replace $V^{-1}$ by its pseudoinverse; any kernel holding can be added to an optimizer. If $b$ has a component in the kernel, it gives a deterministic excess gain that can be scaled without bound, so there is no finite optimal portfolio. If $b=0$, there is no distinguished risky market portfolio and its beta ratios are undefined.

Write $R_j=I-P_{-j}$. Minimizing the residual sum of squares first over the coefficients of $X_{-j}$ reduces the problem to minimizing $\Vert R_{j}Y-b{R}_j{X}_j{\Vert }^2$. Full column rank ensures $R_j{X}_j\ne 0$. Thus the partial regression formula and its variance areThis uses $R_j^T=R_j=R_j^2$ and the isotropic error covariance matrix ${\sigma }^{2}I$.

Since the span of $X_k$ is contained in the column space of $X_{-j}$, orthogonal projection onto the latter removes at least as much squared norm. Consequentlyand inversion proves the requested lower bound. Full rank makes the denominator positive.

Adding $c{1}_n$ to column $j\ge 2$ leaves the column space unchanged. Every old fitted vector is reproduced by keeping all slopes unchanged and replacing the regression intercept by ${\hat{\beta }}_1-c{\hat{\beta }}_j$. Uniqueness of the ordinary least squares estimators then proves that slope $j$ is unchanged. In particular all columns can be centered by such changes. Applying the previous projection argument to the centered design proves the bound with $X_j-{\bar{X}}_j{1}_n$ and $X_k-{\bar{X}}_k{1}_n$; the normalized inner product is now the usual sample correlation.

Each reported coefficient test is the two-sided Student's t-test of $H_0:{\beta }_j=0$ against $H_1:{\beta }_j\ne 0$, conditional on the other predictors. Under the null, ${\hat{\beta }}_j/(\hat{\sigma }\sqrt{(X^{T}X{)}_{jj}^{-1}})$ has a Student's t-distribution with $100-3=97$ degrees of freedom. At five percent the intercept is significant, but neither individual test slope is. The final F-test compares the full model against an intercept-only model: $H_0:{\beta }_{\mathrm{Maths}}={\beta }_{\mathrm{Stats}}=0$. It has $(2,97)$ degrees of freedom and rejects at five percent, since its $p$-value is $0.003166$. Thus the predictors jointly explain variation, without strong evidence separating their individual conditional effects.

The two test predictors have sample correlation $0.9371630$, although each has only moderate correlation with the final exam response. The variance inflation factor is $1/(1-0.9371630^2)\simeq 8.22$. Their nearly shared direction produces substantial multicollinearity, inflating the individual slope standard errors. The joint F-test can detect this common predictive direction even though either coefficient, given the other, has a large $p$-value.

Let $\xi =x-⟨x⟩$ and $C_{ij}=⟨{\xi }_i{\xi }_j⟩$. In the linear noise approximation, $A_i=a_{ik}{\xi }_k$ and $B_{ij}=b_{ij}$. At the stationary mean, the preceding moment identity gives ${\dot{C}}_{ij}=a_{ik}{C}_{kj}+a_{jk}{C}_{ik}+b_{ij}$. Hence the stationary covariance matrix solves the Continuous Lyapunov equationwhich is precisely the requested component formula because $C$ is symmetric.

Write $V=\{1,\dots ,p\}$. The global Markov property for a directed acyclic graph says that, for every three pairwise disjoint vertex sets $A,B,C$,Thus graphical D-separation implies probabilistic conditional independence. Causal minimality means that $P$ has this global Markov property for a directed acyclic graph, but has it for no proper subgraph of $\mathcal{G}$ obtained by deleting edges while keeping all vertices. Faithfulness of a directed acyclic graph means that the displayed implication is an equivalence: the conditional independences of $P$ are exactly the D-separations of $\mathcal{G}$. In particular it includes the global Markov property for a directed acyclic graph.

A causally minimal distribution can nevertheless have an extra independence. Here is a causally minimal nonfaithful Gaussian model. Take three independent random variables $E_1,E_2,E_3$, each with the standard normal distribution, and setUse the Directed acyclic graph with edges $1\to 2$, $1\to 3$ and $2\to 3$. The equations give a directed acyclic graph factorization, so the global Markov property for a directed acyclic graph holds. Alternatively this complete acyclic graph imposes no nontrivial D-separations. But $Z_3=E_2+E_3$ is independent of $Z_1=E_1$, even though the direct edge $1\to 3$ prevents their D-separation. Hence faithfulness of a directed acyclic graph fails.

To check causal minimality explicitly, the covariance matrix isThe linear transformation from $(E_1,E_2,E_3)$ to $(Z_1,Z_2,Z_3)$ is invertible, so this is a nonsingular multivariate normal distribution. Deleting $1\to 2$ would require $Z_1\perp Z_2$, whereas $\mathrm{Cov}(Z_1,Z_2)=1$. Deleting $1\to 3$ would require $Z_1\perp Z_3\mid Z_2$, whereas the conditional covariance isDeleting $2\to 3$ would require $Z_2\perp Z_3\mid Z_1$, whereas its conditional covariance is $1-1\cdot 0=1$. By Gaussian conditional independence, each nonzero conditional covariance rules out the required independence. Every proper subgraph is contained in a graph obtained by a single edge deletion, and further deletion can only add Markov restrictions. Thus no proper subgraph is Markov for $P$, proving causal minimality.

For the Markov blanket, use its standard convention that the focal vertex is excluded:We prove Markov blanket D-separation by examining an arbitrary path from $k$ to $v\notin B\cup \{k\}$. If its first edge points into $k$, its next vertex is a parent of $k$, hence a conditioned noncollider in $B$, and the path is blocked. Otherwise the path starts $k\to h$ at a child $h\in B$. If the next edge points out of $h$, then $h$ is again a conditioned noncollider. If it points into $h$, the path begins $k\to h\leftarrow u$. Although conditioning on the collider $h$ opens that local segment, $u$ is another parent of $h$, hence belongs to $B$. The outgoing arrow $u\to h$ makes $u$ a noncollider on the path, so conditioning on $u$ blocks it. Since $v$ is outside $B$, none of these blocking vertices can be the endpoint.

Every path is therefore blocked by $B$. Applying the global Markov property for a directed acyclic graph givesThis proves the requested screening property without any positivity assumption on $P$.

Finally assume faithfulness of a directed acyclic graph and let $A$ have the stated screening property. Then $k$ and every vertex outside $A\cup \{k\}$ must be D-separated by $A$. If a parent or child $h$ of $k$ were missing from $A$, the one-edge path between them would be active, a contradiction. Thus every parent and child is in $A$. If another parent $u$ of a child $h$ were missing from $A$, the path $k\to h\leftarrow u$ would be active: its only interior vertex is the collider $h$, already in $A$. This is again a contradiction. ConsequentlyUnder faithfulness the graphical blanket is contained in every screening set. This is the minimal Markov blanket under faithfulness; the preceding argument also shows that the blanket itself is a screening set.

The Gaussian likelihood has log-likelihood, apart from no omitted parameter-dependent terms,For fixed ${\sigma }^2$, maximizing it is ordinary least squares. Differentiating with respect to $\beta$ gives the normal equation $X^{T}X\hat{\beta }=X^{T}Y$. Full column rank makes the quadratic strictly convex, soAt this value, differentiating with respect to $v={\sigma }^2$ gives $-n/(2v)+\mathrm{RSS}/(2v^2)=0$, soFor positive RSS this is the global likelihood maximum in $v>0$. RSS is positive almost surely under the model since $p<n$; an exactly zero observed RSS instead gives a boundary supremum as $v\downarrow 0$.

As a linear image of a multivariate normal vector, the coefficient estimate satisfiesIndeed, $\hat{\beta }-\beta =(X^{T}X{)}^{-1}{X}^T\epsilon$, and direct matrix multiplication gives its covariance matrix.

The fitted values are $\hat{Y}=X\hat{\beta }$, so the hat matrix isIt is symmetric, andIt fixes every vector in the column space of $X$, and annihilates its orthogonal complement. Thus $H$ is the orthogonal projection onto that column space, has matrix rank and trace $p$, and $I-H$ is the complementary orthogonal projection. ConsequentlySince $(I-H)X=0$, the covariance matrix between the regression residual vector and the coefficient estimate isThese quantities have a joint normal distribution, so they are also independent, a stronger conclusion than zero covariance.

Now $\hat{e}=(I-H)\epsilon$ has mean zero. Using the expectation of a quadratic form,Thus the maximum-likelihood estimator hasThe unbiased alternative isThe orthogonal projection of a Gaussian vector also yields $\mathrm{RSS}/{\sigma }^2\sim {\chi }_{n-p}^2$, explaining the residual degrees of freedom.

Assume the known covariance matrix $\Sigma$ is positive definite. The Gaussian likelihood, up to constants, isIts gradient is $X^T{\Sigma }^{-1}(Y-X\beta )$. The negative Hessian matrix $X^T{\Sigma }^{-1}X$ is positive definite because $X$ has full column rank. Thus the unique maximum-likelihood estimator is the generalized least squares estimatorIt is unbiased, and multiplication of its linear coefficient matrix by $\Sigma$ givesIn fact ${\hat{\beta }}_{\mathrm{GLS}}\sim N_p(\beta ,(X^T{\Sigma }^{-1}X{)}^{-1})$. Equivalently, a whitening transformation premultiplies $Y$ and $X$ by ${\Sigma }^{-1/2}$, after which ordinary least squares applies. There is no extra ${\sigma }^2$ factor when $\Sigma$ is the complete known covariance; such a factor appears if only a shape matrix $\Omega$ is known and $\Sigma ={\sigma }^2\Omega$.

Fix $u$. Put $m=\phi (u)$ and form the centered real vectorsThey are bounded independent and identically distributed random variables, so the multivariate central limit theorem applies to $n^{-1/2}{\sum }_j{Y}_j=(\mathrm{Re}{\mathcal{C}}_n(u),\mathrm{Im}{\mathcal{C}}_n(u){)}^T$.

To identify its limiting covariance matrix, put $Z=e^{iu{X}_1}-m$. Its complex covariance and pseudo-covariance are respectivelyFor $Z=A+iB$, $|Z{|}^2=A^2+B^2$ and $Z^2=A^2-B^2+2iAB$. ThusThe two conditions on $\Gamma (u)$ identify exactly this real covariance matrix: the second is ${\mathrm{Cov}}_{\mathbb{C}}(\Gamma (u),\overline{\Gamma (u)})=\mathbb{E}\Gamma (u{)}^2$, with a conjugate in its second argument. A centered multivariate normal distribution is determined by its covariance matrix, including when that matrix is singular. ThereforeThis proves convergence in distribution at each fixed $u$; it does not assert convergence of the entire empirical characteristic process.

Because the scalar signal and the noise vector are independent random variables with normal distributions, their sum is a Gaussian random vector. Its expected value is zero, and the covariance matrix isThus this is a rank-one covariance spike. Each summand $X_i{X}_i^{\top }$ has expected value ${\Sigma }_S$, so $\mathbb{E}\hat{\Sigma }={\Sigma }_S$. Here the empirical matrix is an uncentered empirical second-moment matrix; because the population expected value is known to be zero, it is an unbiased estimator of the covariance matrix. The TeX's apparent factorial denominator is a transcription error.

Use an ideal carrier-sense multiple access scheme on the interference graph. An active station finishes its transmission after an independent unit-rate exponential distribution waiting time. An inactive station has an independent attempt clock with an exponential distribution of rate $e^{{\theta }_r}$; an attempt succeeds only when all its neighbors are inactive. Attempts that are blocked do not change the state. Thus, on the independent sets represented by $S$, the continuous-time Markov chain has transition ratesThe empty independent set can be reached by successive deactivations, and any other independent set can be reached from it by successive activations. Hence this finite chain is an irreducible Markov chain. The weight $a_{\theta }(n)=e^{\theta \cdot n}$ satisfiesBy detailed balance for a continuous-time Markov chain, its unique stationary law isThis is a finite exponential family. Its natural parameter of an exponential family is $\theta$ and its mean parameter of an exponential family is the vector of active fractions.