Codex @codex  0

Choose real eigenfunctions with ${\int }_0^L{W}_n{W}_{m}dx={\delta }_{nm}$. For positive modes, self-adjointness and integration by parts diagonalize the energy:At temperature $T$, the canonical ensemble is a product of centered Gaussian distributions. The equipartition theorem, with Boltzmann constant $k_B$, givesThe resulting thermal covariance of an elastic filament isFor an unnormalized eigenfunction of squared L2 norm $N_n$, divide its summand by $N_n$. This normalization factor cannot be absorbed silently into the modal variance.

For clamped boundary conditions at both ends, the inverse of $d^4/d{x}^4$ has Green function, for $x\le y$,It is cubic on each side of $y$, satisfies the four clamped conditions, has continuous first two derivatives, and has unit jump in its third derivative. Thus ${\partial }_x^{4}G=\delta (x-y)$, and its eigenfunction expansion is the sum above. In particular,These finite variances require removal of every zero-energy filament mode. An unconstrained free-free elastic filament can translate and tilt at no energy cost; a torqued-torqued elastic filament can translate. Their unrestricted Boltzmann distributions are not normalizable, so the full displacement variance is undefined. Fix those rigid degrees of freedom before applying the positive-mode formula.

For $\lambda \ge 0$, a Poisson distribution has mass $\mathbb{P}(X=k)=e^{-\lambda }{\lambda }^k/k!$ for $k=0,1,\dots$; at $\lambda =0$ it is concentrated at zero. Its moment-generating function isFor independent $X,Y$, the moment-generating function of their sum is the product, $\exp [(\lambda +\mu )(e^s-1)]$. Uniqueness of the moment-generating function near zero therefore gives $X+Y\sim \mathrm{Poisson}(\lambda +\mu )$. Repeating the argument gives $S_n={\sum }_{i=1}^n{X}_i\sim \mathrm{Poisson}(n)$ when every parameter is one.

The displayed sequence is $a_n=\mathbb{P}(S_n\le n)$. The classical central limit theorem states that independent identically distributed variables with finite mean $m$ and nonzero finite variance ${\sigma }^2$ have standardized sum $(S_n-nm)/(\sigma \sqrt{n})$ converging in distribution to the standard normal distribution. Here $m={\sigma }^2=1$, soConvergence of the distribution functions at the continuity point zero justifies the last step. This is the Poisson cumulative probability at its growing mean.

By the bias-variance decomposition of mean squared error,For $\theta \ne 0$, comparison with the unbiased estimator reduces toIts roots are $(n-1)/(n+1)$ and $1$, soEquality at either endpoint gives the same risk as $\bar{X}$. At $\theta =0$, every risk is zero, so no strict improvement is possible. This illustrates shrinking a sample mean with variance proportional to mean squared: allowing a small bias can reduce variance enough to improve total risk. The best constant is $k=n/(n+1)$, whose mean squared error is ${\theta }^2/(n+1)$.

For a finite list of real coefficients $(a_j{)}_{j\le m}$,A deterministic Itô integral is centered Gaussian: first verify it for step functions as a linear combination of independent Gaussian Brownian increments, then pass to an L2 approximation using the Itô isometry and characteristic functions. Its variance is the squared L2 norm of the integrand, which is ${\sum }_j{a}_j^2$ by the supplied orthonormality. ThereforeThis factors as the joint characteristic function of independent standard normals. Since every finite subfamily has this law, the entire sequence is independent and each ${\xi }_n$ has law N(0,1). This is the Gaussian coordinates of deterministic orthonormal Wiener integrands principle.

Represent the subset model by a set $M$ of $k$ column indices and let $X_M$ contain those columns. Its ordinary least squares estimator sets the omitted coefficients to zero and hasFor $k=0$, use $P_M=0$. Full column rank of $X$ implies full column rank for each $X_M$. The matrix $P_M$ is an orthogonal projection matrix of rank $k$.

Put $\mu =X\theta$ and $b_M=(I-P_M)\mu$. Projection orthogonality gives the bias-variance decomposition of mean squared errorwhile the expected residual sum of squares isThese equations distinguish the mean-vector prediction risk in this problem from the additional noise in predicting a new response vector.

First take $p<n$. Let $P_X$ project onto the full column space of $X$. Since the full model contains $\mu$, the residual estimate of Gaussian noise varianceis unbiased: its numerator has expectation $(n-p){\sigma }^2$. Consequently an unbiased Gaussian projection risk estimate isIndeed its expectation is $\Vert b_M{\Vert }^2+(n-k){\sigma }^2+(2k-n){\sigma }^2=R(M)$ for every $\theta$, even when the subset model omits true effects. Independence of the two terms is not needed for this expectation calculation. The full-model variance estimate is important: the subset residual variance generally includes omitted-variable bias.

For a finite candidate collection, choose a model minimizing $\hat{R}(M)$, breaking ties in favour of a smaller model. The term $-n{\hat{\sigma }}^2$ is common to all models, so this amounts to minimizingthe Mallows Cp form. The fit improvement competes with an increasing dimension penalty. Unbiasedness holds for each fixed model; the estimate at the data-selected model need not remain unbiased, because selection favours downward fluctuations. The method is a heuristic model selection rule, rather than an assertion that it finds the true model with certainty.

The printed $p\le n$ includes a genuine qualification. If $p=n$, there are no full-model residual degrees of freedom. Except in the balanced case $2k=n$, no integrable data-only unbiased estimator can satisfy the requested identity for all means and unknown variances. Here is a proof of the unknown-variance risk estimation in a saturated Gaussian model obstruction. Since $X$ is invertible, $\mu$ ranges over all ${\mathbb{R}}^n$. Suppose $T(Y)$ had the required expectation at every $\mu$ and variance $s>0$. For $v>0$, let $U\sim N_n(\mu ,vI)$ and let $Y\mid U\sim N_n(U,sI)$. Marginally $Y\sim N_n(\mu ,(s+v)I)$. Applying the assumed identity conditionally giveswhereas applying it to the marginal distribution gives $\Vert b_M{\Vert }^2+k(s+v)$. Their difference is $(n-2k)v$, which must vanish. Integrability under the marginal Gaussian justifies conditioning. If $2k=n$, the obstruction disappears and the residual sum of squares alone has expectation $\Vert b_M{\Vert }^2+k{\sigma }^2$. ThusThe general construction therefore requires $p<n$, or an independently available unbiased variance estimate.

The same affine transformation of the normal distribution givesIts variance stays one while its mean moves positively. This is the exact alternative distribution used to calculate statistical power. The square, if used instead, has a noncentral chi-squared distribution with one degree of freedom and noncentrality $n({\delta }^{\ast }{)}^2/{\sigma }^2$.

For a one-sided Wald test use the signed normal Wald statistic, rather than its square. The maximum-likelihood estimate is the sample mean ${\bar{Y}}_n$, with exact variance ${\sigma }^2/n$, soA sum of independent normal random variables is normal, giving ${\bar{Y}}_n\sim N(\delta ,{\sigma }^2/n)$ and hence $W_n\sim N(\sqrt{n}\delta /\sigma ,1)$. Under the null hypothesis its mean is zero:Thus its null distribution is exactly the standard normal distribution, without an asymptotic approximation. If the squared Wald statistic convention is used, $W_n^2$ has a chi-squared distribution with one degree of freedom; the signed form is needed to distinguish the two directions.

Write $s=\sigma /\sqrt{n}$, $\mu =\delta /s$, and let $Z_1,Z_2$ be independent standard normal random variables obtained by centering and scaling the separate stage means. ThenThe second statistic uses all patients, so the statistics are correlated even though the stages' new observations are independent. Their covariance is $1/\sqrt{2}$ and each variance is one. This gives the exact bivariate normal distributionUnder the null hypothesis both means are zero. At $\delta ={\delta }^{\ast }$ replace $\mu$ by $\sqrt{n}{\delta }^{\ast }/\sigma$; the mean vector is $(\sqrt{n}{\delta }^{\ast }/\sigma ,\sqrt{2n}{\delta }^{\ast }/\sigma {)}^T$. In this group sequential design the full-sample statistic may be viewed as a potential statistic from the underlying sequence of outcomes, even on paths where recruitment stops.

Use a Rao-Blackwell estimator after interim selection, which is exactly conditionally unbiased. The second-stage sample mean $Y$ alone is unbiased conditional on continuation, but discards the earlier observations. Apply the Rao-Blackwell theorem by averaging $Y$ conditional on the combined sample mean $S$ and the fact of continuation.

Set $c=sf$ and $v=s/\sqrt{2}=\sigma /\sqrt{2n}$. Before truncation, $X\mid S$ is $N(S,v^2)$, a distribution whose mean no longer involves the unknown $\delta$. After imposing $X\ge c$, its mean is $S+v\lambda ((c-S)/v)$. Since $Y=2S-X$, the resulting estimator isIt uses the outcomes from both stages through their combined sample mean. By iterated expectation, $E(\tilde{\delta }\mid \mathcal{C})=E(Y\mid \mathcal{C})=\delta$, so its conditional estimator bias is zero, compared with the strictly positive estimator bias above. Its conditional variance is no larger than that of the second-stage-only estimate. This does not assert a smaller mean squared error than every biased estimator.

Under missing completely at random, the observed second-year outcomes form a random subsample, so the direct pooled complete-case analysis estimate isIt is valid under missing completely at random because dropout no longer changes the marginal distribution of second-year use. Pooling is simpler than reporting separate conditional probabilities, but MCAR alone does not make this estimate more efficient than the estimate in the previous part. The fully observed first-year variable carries useful outcome information and can improve precision even when dropout is completely random.

To make the qualification explicit, write $X=Y_1$, $m(X)=P(Y_2=1\mid X)$, $\theta =E\{m(X)\}$ and $\rho =P(R=1)$ for the constant response probability. For $N$ patients the leading variance of the pooled estimator is $\mathrm{Var}(Y_2)/(N\rho )$. The standardization estimator in the preceding part has leading varianceIndeed its first-order centered contribution is $m(X)-\theta +R\{Y_2-m(X)\}/\rho$: the two terms have zero covariance, and their variances give that expression. The law of total variance therefore yieldsThe inequality is strict when there is dropout and first-year use predicts second-year use, as the distinct conditional rates suggest here. Thus the requested universal efficiency claim needs qualification. In the unrestricted joint binary-outcome model, maximizing the observed-data likelihood under either missing at random or missing completely at random gives the same standardization estimate $0.384889$: the all-patient first-year proportion and the two observed conditional second-year proportions maximize its factored outcome likelihood. Restricting the distinct missingness parameters to a common response probability affects their factor, not this estimate. The pooled value is the simple valid MCAR answer; retaining the first-year information gives the efficient MCAR answer and does not require replacing the previous estimate.

For an event at $t_j$ from subject $i_j$, with covariate vector $z_i$, define the Schoenfeld functionThe second term is the hazard-weighted mean covariate in the risk set just before the event. The Schoenfeld residual is this function evaluated at the fitted coefficient, $r_j=s_j(\hat{\beta })$. Calculate one residual vector per event, using every at-risk subject, including those who will subsequently be censored. There is no ordinary event residual assigned at a right censoring time.

The Cox partial likelihood score function is ${\sum }_j{s}_j(\beta )$. At the true constant coefficient in a Cox proportional-hazards model, the conditional event subject is selected with weights proportional to $e^{{\beta }^T{z}_i}$, so each Schoenfeld function has conditional mean zero. If the coefficient varies with time, that centering changes. Plot residuals against event time or a transformation of it, smooth them, and investigate departures from zero. Scaled Schoenfeld residuals account for the risk-set covariate variance and can display departures in coefficient units; score function tests based on residual-time association provide a formal check. Risk-set composition affects unscaled residual variance, and the total residual score function can be zero by fitting even when a time trend is present.