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Use the normal linear model $Y=X\beta +\epsilon$, with fixed full-column-rank design matrix $X$, independent errors ${\epsilon }_i\sim N(0,{\sigma }^2)$, and ${\sigma }^2>0$. Its log-likelihood isFor fixed ${\sigma }^2$, differentiating with respect to $\beta$ gives $X^T(Y-X\hat{\beta })=0$. Since $X^{T}X$ is positive-definite, the unique least-squares estimate isLet $\mathrm{RSS}=\Vert Y-X\hat{\beta }{\Vert }^2$. The variance score equation is $-n/(2{\sigma }^2)+\mathrm{RSS}/(2{\sigma }^4)=0$, givingThis is the maximum-likelihood estimator, whereas the residual standard error reported by linear-regression software is $s=\sqrt{\mathrm{RSS}/(n-p)}$. The latter uses the unbiased residual estimate of Gaussian noise variance. These formulas assume $n>p$ and positive RSS; exact interpolation gives a variance-boundary degeneracy rather than an interior maximum.

In the species-only model, the log odds for adult species A is the intercept, so ${\hat{\theta }}_A=0.91191$ with standard error 0.07824. This also follows directly by pooling its vials: there are 570 other-species eggs among 799 eggs eaten, giving ${\hat{p}}_A=570/799$ andFor a regular maximum-likelihood estimator, asymptotic normality of a maximum likelihood estimator gives ${\hat{\theta }}_A\stackrel{\mathrm{approx}}{\sim }N({\theta }_A,I_A^{-1})$, with information $I_A=N_A{p}_A(1-p_A)$. Substitution of ${\hat{p}}_A$ gives $\mathrm{se}({\hat{\theta }}_A)=\sqrt{1/570+1/229}\simeq 0.07824$. An approximate 95% Wald confidence interval isThis interval treats the selected species-only model as fixed; it does not account separately for uncertainty from choosing that model.

For the Poisson regression, the log-likelihood isSince ${\mu }_i=e^{x_i^T\beta }$, the score function and its derivative areConsequently a finite maximum-likelihood estimator satisfiesThese nonlinear equations are normally solved by iteratively reweighted least squares. If $X$ has full column rank, the Hessian is negative definite and any finite solution is the unique maximizer. Full rank does not by itself rule out a boundary solution at infinite coefficients for every conceivable data set. The Poisson generalized linear model has dispersion one, so there is no separate scale parameter in this likelihood.

A sufficient statistic is one for which the conditional distribution of the full sample, given its value, is independent of the parameter. For independent observations the likelihood is ${\prod }_{i=1}^6{p}_i^{n_i}$, depending on the data only through the counts. The factorization criterion for sufficiency therefore makes $(n_1,\dots ,n_6)$ sufficient. Directly, all ordered samples with given counts have the same probability, so their conditional distribution is uniform over the $n!/\prod n_i!$ possible arrangements, independently of the $p_i$.

The unrestricted maximum-likelihood estimator is ${\hat{p}}_i=n_i/n$. Thus the generalized likelihood-ratio test useswith zero-count terms interpreted by continuity. Reject for small $\Lambda$, equivalently large $D$.

Write $n_i=n/6+{\delta }_i$, so $\sum {\delta }_i=0$. A Taylor series expansion of $(n/6+{\delta }_i)\log (1+6{\delta }_i/n)$ gives ${\delta }_i+3{\delta }_i^2/n+O(|{\delta }_i{|}^3/n^2)$. Summing cancels the linear terms, and yieldsUnder the null hypothesis, the multinomial central limit theorem places the standardized count deviations in the five-dimensional subspace with coordinate sum zero, with identity covariance on that subspace. Their squared norm consequently tends to the chi-squared distribution with five degrees of freedom. This gives the Pearson chi-squared goodness-of-fit test.

For $T=8.12$, the asymptotic p-value is about $0.150$. The 5% critical value of ${\chi }_5^2$ is approximately $11.07$. Do not reject at the usual 5% level, or at 10%. A significance level is not specified in the paper, so an unconditional yes-or-no decision is not determined: using this approximation, rejection would require a chosen level at least about $15\%$.

With the same squared-error normalization as in Question 2, the ridge estimator solvesThe unpenalized intercept in ridge regression is $\hat{a}=\overline{Y}$ because the columns of $X$ are centered. Differentiating the remaining quadratic objective gives $(X^{T}X+n\lambda I_p){\hat{\beta }}_{\lambda }^R=X^T{Y}_c$. Since $\lambda >0$, the left matrix is positive definite even when $X$ is rank-deficient. Thus the closed-form ridge regression estimator isThe second equality again uses centering. If the penalty convention is $\lambda \Vert b{\Vert }_2^2/2$ with an unnormalized squared-error term $\Vert Y_c-Xb{\Vert }_2^2/2$, replace $n\lambda$ by $\lambda$ throughout; the limiting assertion is unchanged.

For the full-column-rank case, take a singular value decomposition $X=UD{V}^T$ with positive singular values $d_1,\dots ,d_p$. The ordinary least squares estimator, which is the slope maximum-likelihood estimator in this normal linear model, and the ridge regression estimator are respectivelyConsequently principal-component shrinkage by ridge regression multiplies the least-squares coordinate in direction $v_j$ by $d_j^2/(d_j^2+n\lambda )$. Ridge shrinks most strongly in directions with the smallest singular values. These are combinations of predictors that the data distinguish least well.

Near multicollinearity creates small singular values, and the factor $1/d_j$ in ordinary least squares greatly amplifies noise in those directions. Its coefficient variance there is ${\sigma }^2/d_j^2$, while the ridge variance iswhich is smaller. The price is bias: the expectation of the ridge coordinate is $d_j^2/(d_j^2+n\lambda )$ times the corresponding true coefficient. This bias-variance tradeoff can substantially reduce estimation and prediction error when the poorly identified directions do not contain an excessively large signal. It is not a guarantee of improvement for every coefficient vector. Adding $n\lambda$ to every eigenvalue also improves the condition number of the normal matrix and makes numerical inversion more stable.

To handle every rank and every relation between $n$ and $p$, put $G=X^{T}X$ and use its spectral theorem for real symmetric matrices. Choose an orthonormal eigenbasis $v_1,\dots ,v_p$ with eigenvalues ${\gamma }_j\ge 0$, and put $z=X^{T}Y$. ThenIf ${\gamma }_j=0$, then $\Vert X{v}_j{\Vert }_2^2=v_j^{T}G{v}_j=0$, so $X{v}_j=0$ and $v_j^{T}z=(X{v}_j{)}^{T}Y=0$. Thus these terms are exactly zero for every positive $\lambda$; there is no divergent component in the null space. Each remaining term converges to $(v_j^{T}z/{\gamma }_j)v_j$. By the definition of the Moore-Penrose pseudoinverse,and thereforeThis proves the vanishing-penalty ridge limit without an invertibility assumption. The limit is the minimum-norm least-squares solution: every other least-squares solution differs by a vector in $\ker X$, orthogonal to the displayed solution, and therefore has at least as large a Euclidean norm.

At stage 1, the maximum-likelihood estimators of the two normal arm means are their sample means. Their difference estimates the treatment effect:Each arm has $n_1$ patients and variance ${\sigma }^2/n_1$, so independence across arms givesThe quantity $I_1$ is the Fisher information for the mean difference when the common variance is known. The PDF's preliminary expression uses $n_i$ for the variance denominator, although $i$ indexes arms and $n_0$ is not defined. The stage-specific sample size is $n_j$; the calculation uses this intended reading, consistent with the specified $n_1,n_2$ per-arm recruitment and the later weighted estimator. Fresh stage-2 patients are independent of stage-1 patients in the usual trial model; the cumulative stage means themselves are not independent across stages.

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.