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The quasibinomial regression retains the logit link and mean function $p_i=(1+e^{-x_i^T\beta }{)}^{-1}$ but uses the working variance function $\mathrm{Var}(Y_i)=\varphi p_i(1-p_i)$. Its quasi-score equation isso multiplying by the common dispersion parameter does not change the coefficient estimates. Iteratively reweighted least squares therefore gives the same fitted means and coefficients as the binomial model. The usual Pearson dispersion estimator isThe estimated coefficient covariance matrix is $\hat{\varphi }(X^{T}WX{)}^{-1}$, with $W_{ii}={\hat{p}}_i(1-{\hat{p}}_i)$. From the intercept standard errors, $(1.08921/1.13031{)}^2\simeq 0.92860$; the rounded slope errors give approximately the same factor. The display uses approximate Student's t-tests with 118 residual degrees of freedom instead of fixed-dispersion normal tests.

For a genuinely individual binary response, the Bernoulli distribution identity $Y_i^2=Y_i$ forces $\mathrm{Var}(Y_i)=p_i(1-p_i)$. Thus $\varphi \ne 1$ is a working quasi-likelihood specification, not a different independent binary distribution with that mean. The mild estimated underdispersion does not establish an improved model, and the mean predictions are unchanged. In particular, a scalar rescaling does not model correlation among users, and usual likelihood-based Akaike information criterion comparisons are unavailable for a family without a specified probability likelihood. The output gives no convincing reason to prefer model3 to model2.

There is a genuine syntax error in the printed command: random=list(...) must be a separate argument, following a comma rather than a plus. The intended call isWith the default Gaussian family, this generalized additive mixed model isHere $(b_{0j},b_{1j}{)}^T\stackrel{\mathrm{iid}}{\sim }{N}_2(0,D)$ across doctors, $D$ is an estimated covariance matrix, and the Gaussian errors have variance ${\sigma }^2$ and are independent of the random effects. The formula ~age includes a random intercept as well as a random slope; it does not force these two effects to be uncorrelated.

The random intercept allows different prescribing baselines, and the random slope allows doctor-specific linear modifications to the population age curve. Conditional on these effects the observations are independent under the model, whereas two patients of the same doctor have additional covariance $(1,a_{ij})D(1,a_{kj}{)}^T$. This accounts for clustered data and avoids treating doctor variation as independent patient-level noise. The population smooth and doctor-specific predictions are different targets. With only four doctors, estimates of the random-effect covariance can be imprecise; the Gaussian support issue for counts also remains. The new model accounts for doctor-specific baselines and age slopes.

Write $z=(Z_{s_1},\dots ,Z_{s_n}{)}^T$, let ${\Sigma }_{ij}=C(\Vert s_i-s_j\Vert )$, let $c_i=C(\Vert s_i-s_0\Vert )$, and put $v_0=C(0)$. Assume the covariance matrix $\Sigma$ is invertible. Because the mean is known to be zero, every linear predictor $w^{T}z$ is unbiased. Its mean squared prediction error isCompleting the square yieldsThe positive-definite matrix property makes the last term nonnegative, proving the simple kriging formulasThe residual is uncorrelated with $z$. Since the joint law is multivariate normal, it is independent of $z$, so the same predictor and variance describe the conditional distribution. Consequently a 95-percent prediction interval isThere is no constraint that the weights sum to one: that is required for an unknown constant mean in ordinary kriging, not for the stated known-zero-mean problem. If $\Sigma$ is singular, use a Moore-Penrose inverse with the corresponding covariance compatibility condition. At an already observed location prediction of that very same field value has zero error; predicting an independent noisy replicate is a different target. Estimated covariance parameters make this a plug-in interval and generally add uncertainty that the formula does not include.

Let $X$ have rows $(1,y_i)$ and let ${\Sigma }_{\vartheta }$ be the spatial covariance matrix with $\vartheta =({\tau }^2,{\sigma }^2,a)$. Conditional on chosen covariance parameters, the best linear unbiased estimator of the drift coefficients is the generalized least squares estimatorThe printed procedure first estimates the linear drift from depth ~ y, constructs a binned residual empirical semivariogram, fits a Gaussian semivariogram to it, and then uses the fitted covariance in generalized least squares trend estimation and universal kriging. The missing fit could be supplied aswhere those arguments are initial values, not the reported final estimates. Such semivariogram fitting typically minimizes a weighted sum of squared discrepancies between binned empirical values and model values; it is not automatically a joint Gaussian likelihood maximization. One can iterate: form residuals using the current drift, refit the covariance model, recompute the generalized least squares drift, and repeat until changes are small. The displayed commands themselves show a single sequence, not evidence that such iteration occurred.

A likelihood-based alternative estimates the two blocks coherently fromSubstitute $\hat{\beta }(\vartheta )$ to obtain a profile log-likelihood, maximize over ${\tau }^2,{\sigma }^2\ge 0$ and $a>0$ with a positive-definite matrix ${\Sigma }_{\vartheta }$, and recover the drift coefficients from the optimizing covariance. Restricted maximum likelihood adds the design determinant term $\log \det (X^T{\Sigma }_{\vartheta }^{-1}X)$ and uses $n-2$ in place of $n$ in the normalizing term, up to a fixed-design constant. It can reduce bias from estimating the drift before the covariance. A zero fitted nugget is a legitimate boundary estimate, not proof that measurement error is absent. Estimate drift using covariance-weighted regression, and spatial dependence from drift-adjusted variation, with the two stages or likelihood explicitly distinguished.

Use one joint hypothesis test for $H_0:{\beta }_{\lambda }={\beta }_{\mu }=0$. Fit the unrestricted model with statistical parameters $({\lambda }_0,{\mu }_0,{\beta }_{\lambda },{\beta }_{\mu })$, where ${\lambda }_i={\lambda }_0{e}^{{\beta }_{\lambda }{z}_i}$ and ${\mu }_i={\mu }_0{e}^{{\beta }_{\mu }{z}_i}$ for treatment indicator $z_i$. Refit under $H_0$, estimating the two common baseline intensities; do not fix them at their unrestricted estimates. The likelihood-ratio test statistic isUnder regular identifiable positive-rate models, sufficient independent patient trajectories, and the null hypothesis, Wilks theorem gives two degrees of freedom because two coefficients are constrained. Reject if $D$ exceeds the chosen upper chi-squared distribution quantile. Alternatively a two-dimensional Wald test uses ${\hat{\beta }}^T\widehat{\mathrm{Cov}}(\hat{\beta }{)}^{-1}\hat{\beta }$, including the off-diagonal covariance. Two separate tests do not provide this single calibrated assessment. The point estimates alone cannot produce a numerical p-value without fitted likelihoods or a coefficient covariance matrix.

The most plausible cause is strong dependence between the latent variables and covariance hyperparameters in their joint posterior distribution. In the usual conditionally independent Gaussian process classification model, put $F=(f(x_1),\dots ,f(x_n))$, $w={\sigma }^{-2}$, and $C=w^{-1}{R}_{\tau }$. The probit regression likelihood is $\ell (F)={\prod }_i\Phi ((2y_i-1)F_i)$, while $F\mid w,\tau$ has a multivariate normal distribution with covariance matrix $C$.

For distinct inputs and positive length scales, $R_{\tau }$ is positive definite. Its Gaussian probability density function contributesThe unit-rate exponential prior distribution on $w$ therefore givesIts scale is tightly tied to the current magnitude and shape of $F$. Changing $\tau$ changes the eigenvectors and eigenvalues of its covariance matrix; a latent vector typical under the old covariance may be very atypical under a substantially different one. Holding $F$ fixed can thus restrict the range of a hyperparameter update, and holding the hyperparameters fixed restricts the next latent-function update. Even exact sampling of each full conditional distribution can move slowly along a narrow ridge of the joint posterior distribution.

Binary observations provide limited information about the absolute size of large correctly signed latent values. This can leave substantial scale uncertainty and strengthen the dependence. Long correlation lengths can also make the covariance matrix nearly singular and the latent coordinates highly dependent. These are plausible explanations for a large mixing time, not proofs that every data set causes slow convergence. A blocked Gibbs sampler removes dependence within its chosen blocks, but not dependence between them. Repeated identical inputs should be represented by one shared latent value; otherwise the nominal Gaussian matrix is singular and the displayed inverse formula must be replaced by a representation on its support.

Let $s={\sigma }^2$ and $\vartheta =(s,{\tau }_1,\dots ,{\tau }_m)$. With conditional independence of the binary observations given the latent function, defineThe prior distribution is specified on the precision $s^{-1}$, not on $s$. Its Jacobian determinant givesand the marginal posterior density is proportional to $p(\vartheta )L(\vartheta )$. Working instead with precision or logarithmic coordinates is possible, provided all corresponding Jacobian determinants are included.

Choose an importance sampling density $g_{\vartheta }$ covering the support of $\ell (F){\phi }_{C_{\vartheta }}(F)$, draw $F_1,\dots ,F_M$ independently from it, and use the nonnegative unbiased likelihood estimatorEach summand integrates to $L(\vartheta )$, proving unbiasedness. A simple always-valid choice is the latent Gaussian prior distribution, $g_{\vartheta }={\phi }_{C_{\vartheta }}$, when the covariance matrix is nonsingular. Then the estimator is just the average of the bounded values $\ell (F_j)\in (0,1)$. For duplicate inputs, sample only the distinct latent values and multiply all likelihood contributions at each common input. A Gaussian approximation to the latent posterior distribution, or its mixture with the Gaussian prior distribution, can reduce relative variance while keeping complete support. Its approximation does not replace the exact weights.

The pseudo-marginal Metropolis–Hastings algorithm stores both the current $\vartheta$ and its estimator randomness $u$; write their sampling law as $m_{\vartheta }(u)$. Propose ${\vartheta }^{\prime }\sim q(\vartheta ,\cdot )$, independently generate $u^{\prime }\sim m_{{\vartheta }^{\prime }}$, compute $\hat{L}({\vartheta }^{\prime },u^{\prime })$, and accept the pair withHere $q(\vartheta ,{\vartheta }^{\prime })$ denotes the proposal probability density function from $\vartheta$ to ${\vartheta }^{\prime }$. On rejection retain both the old parameter and the old likelihood estimate. Initialize with a positive estimate. One must not independently refresh the denominator after rejection.

Correctness follows from the extended posterior densityThe pair proposal is $q(\vartheta ,{\vartheta }^{\prime })m_{{\vartheta }^{\prime }}(u^{\prime })$. Its Metropolis–Hastings acceptance probability simplifies to the displayed ratio, since the auxiliary densities cancel. Hence detailed balance holds for $\tilde{\pi }$. Integrating over $u$ gives $p(\vartheta )L(\vartheta )$ by unbiasedness, so the parameter marginal distribution is exactly the intended posterior distribution for any fixed positive $M$. This is exactness of the stationary target, not independent exact samples or a guarantee of rapid convergence. Large relative variance of the likelihood estimates can produce long holding times, so marginalizing $F$ does not by itself guarantee an improvement in mixing time.

Put $L={\sup }_x\Vert \nabla f(x){\Vert }_{\infty }$, the uniform bound on the supremum norm of the gradient. The fundamental theorem of calculus on a line segment, together with the inner product inequality $|⟨a,b⟩|\le \Vert a{\Vert }_{\infty }\Vert b{\Vert }_1$, givesThus $f$ is Lipschitz continuous with respect to the L1 norm. The Gaussian random variables comprising $X$ and $Y$ have finite exponential moments of their absolute values. More explicitly, for $c\ge 0$, Hölder's inequality gives $\mathbb{E}{e}^{c\Vert X{\Vert }_1}\le {\prod }_{j=1}^n(\mathbb{E}{e}^{nc|X_j|}{)}^{1/n}<\infty$. Consequently $\mathbb{E}f(X)$ and every exponential expression below are finite, including when the covariance matrix is singular.

Use the convex function $\Psi (v)=e^{\lambda |v|}$. Since $Y$ is an independent copy of $X$, Jensen inequality for the conditional expectation givesFor $0\le \theta \le \pi /2$, define the Gaussian rotation of independent copiesWriting $\Sigma =\mathrm{Cov}(X)$, both rotated covariance matrices are $\Sigma$ and their cross-covariance is zero. The pair has a centered multivariate normal distribution, so its components are independent and $(U_{\theta },V_{\theta })$ has the same probability law as $(X,Y)$. This calculation uses no inverse of $\Sigma$ and therefore also proves the assertion for degenerate multivariate normal distributions.

The path $\theta \mapsto U_{\theta }$ runs from $Y$ to $X$. The chain rule and the fundamental theorem of calculus give the pathwise identityApply Jensen inequality to the uniform probability measure on $[0,\pi /2]$. Then use Tonelli theorem and the Gaussian rotation of independent copies to obtainThis proves the Gaussian rotation interpolation inequality in the required form:The stated almost-sure domination now gives an exponential moment of an absolute standard normal variable bound. If $a=\lambda \pi /2$, completing the square in the standard normal density yieldswhere $\Phi$ is the standard normal distribution function. Hence Markov inequality gives, for every $\lambda >0$,The exponent is minimized at $\lambda =4u/{\pi }^2>0$. Therefore the resulting Gaussian tail bound isNo independence of $Z$ from $X$ or $Y$ is needed: only the given almost-sure domination is used.

The equal-variance form of Slepian's lemma is the following. If $U,V$ are centered vectors with multivariate normal distributions, $\mathbb{E}{U}_i^2=\mathbb{E}{V}_i^2$ for every $i$, and $\mathbb{E}{U}_i{U}_j\ge \mathbb{E}{V}_i{V}_j$ whenever $i\ne j$, thenfor every threshold vector $(u_i)$. In particular $\mathbb{P}({\max }_i{U}_i\le u)\ge \mathbb{P}({\max }_i{V}_i\le u)$. No nonsingularity of the covariance matrices is required.

Here the pointwise variances $2t^{2H}$ and $2t^{2H^{\prime }}$ differ. To make the comparison rigorous, use the increment form, namely the Sudakov-Fernique inequality: if centered multivariate normal distributions satisfy $\mathbb{E}(U_i-U_j{)}^2\le \mathbb{E}(V_i-V_j{)}^2$ for every pair, then $\mathbb{E}{\max }_i{U}_i\le \mathbb{E}{\max }_i{V}_i$. Unlike the preceding form of Slepian's lemma, this statement has no equal-pointwise-variance hypothesis.

For completeness, the increment form follows by Gaussian integration by parts. Make $U,V$ independent, define $Z_r=\sqrt{1-r}U+\sqrt{r}V$ for $0<r<1$, and letThe $p_i$ are softmax function weights, with $p_i\ge 0$ and ${\sum }_i{p}_i=1$. Direct differentiation gives ${\partial }_i{F}_{\tau }=p_i$ and ${\partial }_{ij}{F}_{\tau }=\tau (p_i{1}_{i=j}-p_i{p}_j)$. The Gaussian integration by parts identity therefore yieldsThe last summands are nonnegative because $D_{ii}+D_{jj}-2D_{ij}=\mathbb{E}(V_i-V_j{)}^2-\mathbb{E}(U_i-U_j{)}^2$. Sinceintegration in $r$ and the limit $\tau \to \infty$ prove the Sudakov-Fernique inequality. All derivatives needed here are bounded for a fixed $\tau$. Singular covariance matrices can be handled by adding small independent centered normal noise with the same variance to both vectors, proving the identity for the nonsingular vectors, and letting the noise variance decrease to zero.

Use the printed normalization of fractional Brownian motion, which omits the conventional factor $1/2$. Its increment variance isFor $s,t\in [0,1]$ and $H>H^{\prime }$, $|t-s{|}^{2H}\le |t-s{|}^{2H^{\prime }}$. Enumerate $T$ by increasing finite subsets $T_n$ whose union is $T$, with $0\in T_n$. The Sudakov-Fernique inequality givesBoth Gaussian processes vanish at zero almost surely. Thus these maxima are nonnegative and increase with $n$; monotone convergence theorem givesTheir continuous modifications agree with the original Gaussian processes simultaneously on the countable set $T$, after discarding a single null event. Continuity and density of the rational parameters make each rational supremum equal to the supremum over $[0,1]$ for these modifications.

The expected suprema are finite. Indeed, the canonical distance of $Y_{H^{\prime }}$ is $d(s,t)=\sqrt{2}|t-s{|}^{H^{\prime }}$. A Euclidean grid supplies a closed-ball metric covering number bound $N(\epsilon )\le 1+\lceil (\sqrt{2}/\epsilon {)}^{1/H^{\prime }}\rceil$ for $0<\epsilon \le \sqrt{2}$. The Dudley entropy integral bound for an anchored centered separable Gaussian process isThe integral is finite since its integrand grows at most like a constant times $\sqrt{1+\log (1/\epsilon )}$ near zero. The same argument applies to $X_H$.

For any real path vanishing at zero,A centered Gaussian process has the same probability law as its negative. Combining this symmetry with the increment comparison gives the stronger boundIn particular the requested comparison isThe common normalization factor $\sqrt{2}$ affects both sides equally. The comparison is valid for every $0<H^{\prime }<H<1$, including the equality cases $s=t$ or $|s-t|=1$ in the increment bounds.

In factor analysis, the independent multivariate normal distributions of $Z_i$ and ${\xi }_i$ imply that their linear combination is also multivariate normal. Its expected value is zero and its covariance matrix isThereforeThe noise variance is assumed positive. The factor contribution is positive semidefinite and has rank at most $k$, while the noise makes the marginal covariance nonsingular.

All moments below are under the proper posterior distribution of part (b). Write the known covariance matrix in block formThe score covariance $S$ is positive definite in this model. Indeed, a further integration by parts gives $\mathbb{E}[g(\beta ){\beta }^T]=-I_p$. If $a^{T}Sa=0$, then $a^{T}g=0$ almost surely, because $\mathbb{E}[g]=0$. Multiplying by $a^T\beta$ would give $0=-\Vert a{\Vert }^2$, so $a=0$.

For a fixed coefficient vector $a$, the control variate summand has varianceThus the unique minimum is attained at $a_{\ast }=S^{-1}c$, givingThe coefficient is fixed because the covariance is assumed known, and the posterior score control variate has mean zero, so this estimator remains unbiased.

We can identify exactly when the improvement is strict. Integration by parts with the bounded smooth function $h$ givesThe scalar expectation is strictly positive. Therefore $c\ne 0$ whenever $x_{\mathrm{test}}\ne 0$, and positive definiteness gives a strictly smaller variance in that case. If $x_{\mathrm{test}}=0$, $h\equiv 1/2$ and both estimators already have variance zero. The printed request for a smaller variance therefore needs this nondegeneracy qualification; a non-increasing variance always holds.

Let $Z_{ir}=1_{\{\text{observation }i\text{ belongs to rat }r\}}$. After integrating out the random intercepts,The fixed-effect design matrix has four independent columns, so the given $A$ has dimensions $176\times 172$, with $A^{T}A=I_{172}$ and $A^{T}X=0$. Consequently the error contrasts $U=A^{T}Y$ have a multivariate normal distribution with mean zero and covariance matrix $W=A^{T}VA$.

The restricted likelihood from orthogonal error contrasts is their density. Equivalently, the estimates maximizeThis is restricted maximum likelihood, not maximization over individual rat effects. The unknown $\beta$ disappears because $A^{T}X=0$. Replacing $A$ by $AQ$ for an orthogonal $Q$ rotates $U$ and conjugates $W$, preserving the determinant and quadratic form; hence the objective is independent of the chosen orthonormal basis. The software's integrated REML log-likelihood can differ by an additive constant depending only on $X$, which has no effect on these variance estimates.

Under a correctly specified Poisson regression with adequate expected counts, the residual Poisson deviance is approximately chi-squared distribution with $n-p=107-5=102$ residual degrees of freedom. The observed $D=319.86$ givesThus the ordinary Poisson model gives a very poor fit under this goodness-of-fit approximation. This is consistent with overdispersion, although a wrong mean model or dependence can also cause lack of fit; quasi-Poisson variance inflation alone does not establish that the mean model is adequate.

For Quasi-Poisson regression, $\mathrm{Var}(Y_i)=\varphi {\mu }_i$. The Pearson dispersion estimator isIt uses squared Pearson residuals, not the residual deviance divided by 102. The fitted coefficients are unchanged, and their estimated covariance matrix is multiplied by $\hat{\varphi }$. With $\hat{\varphi }=3.24$, standard errors increase by $1.8$, so the main treatment coefficient hasThe main Progabide coefficient is not significant at 5% after this dispersion adjustment. R uses a Student's t-distribution when dispersion is estimated, as described in its GLM summary documentation; the large-sample normal approximation also gives the same decision.

Because an interaction term is present, this tests the treatment contrast at baseline zero. An overall absence of treatment influence requires the joint null ${\beta }_T={\beta }_{TB}=0$. The printed marginal standard errors do not give their covariance, so they are insufficient to calculate that joint Wald test, or the treatment contrast test at an arbitrary nonzero baseline.

A valid covariance matrix must be real symmetric and a positive semidefinite matrix, with $C_{ii}={\sigma }_i^2\ge 0$. Conversely, every such matrix is the covariance of a possibly degenerate multivariate normal distribution. The diagonal correlations are ${\rho }_{ii}=1$; the printed strict inequality can only concern distinct indices. Pairwise bounds do not suffice: the $3$ by $3$ correlation matrix with diagonal $1$ and every off-diagonal entry $-0.9$ has eigenvalue $-0.8$ along $1$ and is invalid. Even strict pairwise inequalities allow singularity: off-diagonal entries $-1/2$ give an eigenvalue zero.

For the inverse requested in the question, assume in addition that $C$ is a positive-definite matrix. Write $x=({\hat{\mu }}_i)$, $\Lambda =C^{-1}$ and $S=1^T\Lambda 1>0$. Up to a constant, the log-likelihood is $-\frac{1}{2}(x-\mu 1{)}^T\Lambda (x-\mu 1)$. Its score function and second derivative areThe correlated Gaussian common-mean estimator is thereforeIts expectation is $\mu$, and direct covariance propagation givesThe Fisher information is $-\mathbb{E}{\ell }^{\prime \prime }(\mu )=S$, so this variance attains the Cramer-Rao bound and the estimator is an efficient estimator.

If $C$ is singular, an ordinary Lebesgue density and $C^{-1}$ are unavailable. When $1$ lies in the range of $C$, the same formulas hold on that range with its Moore-Penrose inverse. Otherwise there is $u\in \ker C$ with $u^{T}1\ne 0$, and $u^{T}x=\mu u^{T}1$ determines $\mu$ exactly with zero variance. The usual nonsingular Cramér–Rao calculation then does not apply.

Use the time-harmonic wave convention $\psi e^{-i\omega t}$ and write $\mathbf{z}=(y,z)$, $E(0,\mathbf{z})=E_0(\mathbf{z})$, with a deterministic incident wave envelope. The printed speed ratio is inconsistent with $k=k_{0}n$: the refractive index used below is $n=c_0/c$. Substituting $\psi =e^{i{k}_{0}x}E$ in the Helmholtz equation givesThe paraxial approximation discards ${\partial }_x^{2}E$. The usual weak-fluctuation model also linearizes $n^2-1=2\mu W+O({\mu }^2{W}^2)$, givingDropping the quadratic contrast is an additional weak-fluctuation assumption, not a consequence of small propagation angles. Although the linearized random potential generates attenuation of order ${\mu }^2$, it does not retain every effect of that order in the literal finite-correlation index: the discarded quadratic contrast can also produce a mean wave phase shift. This linearization must precede a Gaussian white noise limit: the square of ideal white noise has no ordinary pointwise meaning.

The split-step Fourier method alternates free-space diffraction, $U_h=e^{hD}$, with a random phase screen,For jointly Gaussian random fields, Gaussian phase averaging gives $\mathbb{E}{e}^{i{\varphi }_h}=e^{-\mathrm{Var}({\varphi }_h)/2}$. More generally, a product of $p$ fields and $q$ conjugate fields picks up $\exp (i{\sum }_a{ϵ}_a{\varphi }_h({\mathbf{z}}_a))$, with signs ${ϵ}_a=+1$ or $-1$. Its screen average is determined entirely by the wave phase covariance matrix. The deterministic diffraction step acts on each coordinate, with opposite signs on factors formed by complex conjugation. This is the basis of the field-moment equations.

There is an important closure qualification. A stationary Gaussian random field need not have independent longitudinal increments. For finite-correlation fluctuations, the unlinearized parabolic wave equation instead givesIn the weak-fluctuation model, the exact first-moment equation isThe last term cannot in general be replaced by a constant times $m$. An exact generic solution of the linearized model iswhere $\mathcal{T}$ denotes time ordering of the propagation operators. For the unlinearized model, add $i{k}_0{\mu }^2{W}^2/2$ inside the propagation generator. The covariance function of the medium is needed to evaluate this expression; One-point Gaussian distributions alone would not even determine the joint wave phase statistics.

For example, omit diffraction and take a longitudinal autocovariance function $R(s)=\mathbb{E}[W(s,\mathbf{z})W(0,\mathbf{z})]$. Direct Gaussian phase averaging givesEven here, unit-variance stationary Gaussian random fields with different $R$ give different answers. In the general problem the diffraction and multiplication operators do not commute, so this scalar attenuation cannot simply be multiplied by $U_x$ without an additional approximation.

The standard closed answer uses the Markov approximation for a random medium, made explicit in part ii. Let the longitudinally integrated covariance kernel beReplace the medium by longitudinal Gaussian white noise with this strength. A screen of thickness $h$ then has $\mathbb{E}[{\varphi }_h({\mathbf{z}}_a){\varphi }_h({\mathbf{z}}_b)]=k_0^2{\mu }^{2}hB({\mathbf{z}}_a-{\mathbf{z}}_b)$ and is independent of the incoming field. For the moment $M=\mathbb{E}{\prod }_a{E}_a^{({ϵ}_a)}$, expanding both steps to order $h$ givesIn particular, the coherent attenuation in a white-noise random medium and its Fresnel propagator solution areEquivalently, write $Wdx=d{\beta }_x$, where ${\beta }_x(\mathbf{z})$ is Brownian motion in $x$ with transverse covariance kernel $\mathbb{E}[d{\beta }_x(\mathbf{z})d{\beta }_x({\mathbf{z}}^{\prime })]=B(\mathbf{z}-{\mathbf{z}}^{\prime })dx$. The Stratonovich integral formulation is $dE=DEdx+i{k}_0\mu E\circ d{\beta }_x$. Its Itô integral form isThe mean of the Itô integral vanishes, independently confirming the attenuation drift. For $x>0$ in two transverse dimensions,Thus a unit plane wave has $m=e^{-\gamma x}$. For a general incident wave envelope, the Fresnel propagator supplies its spreading. The attenuation is redistribution between the coherent and diffuse wave fields, rather than wave absorption: for a field and its conjugate at the same point, the screen contribution in the second-moment equation cancels.

The covariance matrix isIf $\mathrm{Cov}(X_1,X_2)=0$, the moment-generating function in part (ii) factors into the product of the two marginal moment-generating functions. Thus $X_1$ and $X_2$ are independent random variables; this is the independence of uncorrelated jointly normal variables.

By bilinearity of covariance,ChooseThen $(X_1,Y)$ is bivariate normal by part (i) and has zero covariance, so $Y$ is a normal random variable independent of $X_1$. Hence $X_2=a{X}_1+Y$.