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Differentiate the product of Dirac delta functions defining $\tilde{P}$, regarding $B$ and $\sigma$ as independent density arguments. The random continuity equation isFor $0<s<t$, the causal functional derivatives follow from the explicit strain solution and $\tilde{B}(t)=B_0\exp [{\int }_0^t\tilde{\sigma }(u)du]$:Consequently,In the Furutsu–Novikov formula, the covariance $\kappa \delta (t-s)$ samples the upper endpoint of the causal time integral. Symmetric regularization of the white noise gives half the mass there. The $B$ response vanishes as $s\uparrow t$, while the strain response tends to one, so $\mathbb{E}[f(t)\tilde{P}(t)]=-(\kappa /2){\partial }_{\sigma }P$. Averaging the continuity equation yields the Fokker-Planck equationThere is no direct diffusion term in $B$: its equation contains the time integral of colored noise, whereas the strain equation receives Gaussian white noise. The same result follows from the Itô diffusion with drift $(\sigma B,-\sigma /\tau )$ and diffusion vector $(0,\sqrt{\kappa })$.

For a general colored medium and unit plane-wave input $E(0,\mathbf{s})=1$, let $U_W(x,0)$ be the propagator of ${\partial }_x=i{\Delta }_{\perp }/(2k)+ik\mu W(x,\mathbf{s})$. Then the solution isA time-ordered operator exponential defines $U_W$. Its expectation depends on the full longitudinal covariance; the initial Gaussian and transverse-stationarity assumptions do not specify a universal closed expression.

With the stated longitudinal Gaussian white noise and transverse statistical homogeneity, the moment depends only on $\mathbf{\rho }={\mathbf{s}}_1-{\mathbf{s}}_2$. Hence $({\Delta }_1-{\Delta }_2)\Gamma =0$. The closed equation in the previous part and initial value $\Gamma (0)=1$ giveIn particular $\Gamma (x,0)=1$: the ensemble mean intensity remains constant for plane-wave input, even though coherent attenuation in a white-noise random medium reduces the mean complex field.

Use a consistent transverse power spectrum convention,For an isotropic spectrum, angular integration gives the isotropic transverse covariance spectrum, $C(\rho )=(2\pi {)}^{-1}{\int }_0^{\infty }\Phi (\kappa )J_0(\kappa \rho )\kappa d\kappa$. ThereforeThe supplied two-dimensional Fourier transform definition has angular factor $2\pi$, whereas the displayed radial transform pair omits it. The latter is a normalized Hankel transform pair. If its radial spectrum is denoted ${\Phi }_H=\Phi /(2\pi )$, use ${\Phi }_H$ in the integral and omit the prefactor $1/(2\pi )$. These are equivalent conventions; mixing them changes the attenuation rate.

In the Gaussian white noise model, observe the entire pathwhere $W$ is standard Brownian motion and the deterministic drift belongs to L2 space on $[0,1]$. Equivalently, $dY(t)=f(t)dt+n^{-1/2}dW(t)$. For every deterministic $h\in L^2[0,1]$, the observed stochastic integral satisfiesThe noise $W(h)=\int hdW$ is an isonormal Gaussian process. In particular its variance is $\Vert h{\Vert }_2^2$. Gaussian white noise is interpreted through these integrals, rather than as an ordinary random function with a pointwise value at every time. No smoothness of $f$ is needed.

Here is the Gaussian maximum bound without independence. Put $M={\max }_{1\le i\le N}|g_i|$. It is integrable since it is bounded by ${\sum }_i|g_i|$. For every $\lambda >0$,The second inequality uses only the moment-generating function of the standard normal distribution, so independence is unnecessary. Jensen inequality gives $e^{\lambda EM}\le E{e}^{\lambda M}$, henceMinimizing at $\lambda =\sqrt{2\log (2N)}$ proves

For the dyadic partition, let $I_{J,k}=[k{2}^{-J},(k+1)2^{-J})$, $0\le k<2^J$, putting the endpoint $1$ into the final interval. The Haar scaling functionsform an orthonormal basis of the space $V_J$ of functions constant on each interval. The Haar wavelets can be written asFor $J\ge 1$, the constant function ${\varphi }_{0,0}=1$ together with ${\psi }_{j,k}$, $0\le j<J$, is another orthonormal basis of $V_J$. Indeed, each successive level splits a cell's two constants into their sum and difference; the number of basis elements is $1+{\sum }_{j=0}^{J-1}{2}^j=2^J$. Thus the Haar projection is the orthogonal projectionOn cell $I_{J,k}$ its value is $2^J{\int }_{I_{J,k}}f(t)dt$. The first formula also applies to $J=0$.

Estimate each coefficient by its observed stochastic integral:This is the Haar projection estimator in Gaussian white noise. The noise integrals over disjoint intervals form a Gaussian vector with zero off-diagonal covariance. The principle that uncorrelated jointly Gaussian variables are independent then makes these coefficients independent. ThereforeTaking expectations shows that ${\hat{\Pi }}_{V_J}f$ is an unbiased estimator of ${\Pi }_{V_J}f$, both coefficientwise and pointwise for the stated step-function representatives.

Only one scaling function is nonzero in each cell, so the supremum norm of the error has the exact formThe endpoint convention ensures the same identity at $1$. Apply the Gaussian maximum bound without independence with $N=2^J$ to obtain the supremum norm risk of a Haar projection estimator:The factor $2^{J/2}$ comes from cellwise scaling, while the extra square root of $J+1$ comes from taking the maximum over $2^J$ Gaussian errors.

Write $m=⟨E⟩$ and $L_0=i{\partial }_z^2/(2k)$. Substituting $n=1+\mu W$ into part (a), without prematurely replacing $n^2$ by its linear part, givesTaking the expectation yields the exact mean equation within the paraxial approximation:The correlation term cannot be discarded just because $⟨W⟩=0$: the field depends on the same random medium as $W$. Nor can $⟨W^{2}E⟩$ be factored exactly from the given variance alone. The printed statistics do not specify the autocorrelation function of a random field, so they do not determine a unique local closed equation for $m$.

A useful covariance-dependent equation follows to second order in $\mu$. Defineand letbe the Fresnel propagator kernel for one transverse coordinate. The Duhamel principle applied to the linear fluctuation term giveswhere $E^{(0)}=E_{\mathrm{free}}$. Multiply by $W(x,z)$ and average. Also $⟨W(x,z{)}^{2}E(x,z)⟩=E^{(0)}(x,z)+O(\mu )$. Since $m=E^{(0)}+O({\mu }^2)$, replacing $E^{(0)}$ by $m$ in terms already multiplied by ${\mu }^2$ preserves second-order accuracy. We obtain the weak-fluctuation memory equation for the coherent fieldThis perturbation statement is for fixed propagation intervals with sufficient covariance regularity and moment bounds. The kernel is understood as an oscillatory integral. If the field is jointly Gaussian, odd perturbative contributions vanish and the remainder can be improved under the corresponding expansion hypotheses. Joint Gaussianity itself still leaves the covariance unspecified. The term $ik{\mu }^{2}m/2$ is the quadratic refractive-index shift in the coherent field; it comes from the mean of $W^2$.

For the usual local answer, impose the additional Markov approximation for a random medium. Its longitudinal correlation length must be much shorter than the distance on which the envelope varies, and diffraction across that correlation distance must not resolve substantial transverse covariance variation. Away from the entry layer, replace the slowly varying factors in the memory integral by their local values and extend the longitudinal integral to infinity. Define the integrated covarianceFor an integrable real stationary covariance, $C(s,0)$ is even, and $B(0)\ge 0$ is its zero-longitudinal-frequency spectral density. The conventional leading weak-index model keeps only the linear random potential $ik\mu W$; in that model the local first-moment equation isThe factor $1/2$ can be checked independently in the white-noise model. If ${\mathcal{B}}_x(z)$ is a Brownian field with $⟨d{\mathcal{B}}_x(z)d{\mathcal{B}}_x(z^{\prime })⟩=B(z-z^{\prime })dx$, writeConverting the Stratonovich integral to an Itô integral adds $-k^2{\mu }^{2}B(0)Edx/2$. The remaining stochastic integral has zero mean, giving the boxed equation andIf the finite-correlation model retains $n^2$ consistently through order ${\mu }^2$, the local second-order expansion also has $+ik{\mu }^{2}m/2$, adding the phase factor $e^{ik{\mu }^2(x-x_0)/2}$. This phase is computed before the white-noise idealization: squaring ideal Gaussian white noise is not the finite-variance operation $W^2$.

The need for a correlation assumption can be seen without any closure argument. The stationary Gaussian random field $W(x,z)=Z$, with a single standard normal variable $Z$, has all the printed one-point statistics. In the linear weak-index model it givesIts attenuation is quadratic in propagation distance, whereas the Markov model gives linear-distance exponential attenuation. The local attenuation equation requires the additional correlation/Markov assumption; it does not follow from statistical homogeneity and unit variance alone.

The printed strong white noise assumption does not imply a normal distribution or even a finite fourth moment. Thus it does not, by itself, determine the covariance of a quadratic transform. We give the intended Gaussian calculation, and then the general finite-fourth-moment answer.

If $\eta$ is Gaussian, $X$ is a centered Gaussian process. Put $v={\gamma }_X(0)=136/15$ and $U_t=X_t/\sqrt{v}$. Since $H_1(u)=u$ and $H_2(u)=u^2-1$, the centered transform isThe supplied Hermite polynomial identity makes the cross terms vanish and givesEquivalently,The constant $a$ has no effect on covariance.

For a general iid noise with $E{\eta }^4<\infty$, put $m_3=E{\eta }^3$ and ${\kappa }_4=E{\eta }^4-3$, its fourth cumulant. Independence and expansion of third and fourth moments give, for $k=|h|$,whereSumming the geometric series explicitly givesThe covariance of quadratic transforms of a linear process follows from a fourth-moment expansion consisting of the three Isserlis theorem pairings, plus the fourth-cumulant contribution when all four noise indices coincide. The third-moment contribution similarly requires three coincident indices. This proves the general formula without assuming a normal distribution. For Gaussian white noise $m_3={\kappa }_4=0$, recovering the simpler answer. If $c\ne 0$ and the noise has infinite fourth moment, $Z_t$ need not have finite variance, so an autocovariance function may not exist.

Put $c_t=\cos (2\pi t/S)$ and assume ${\sigma }^2>0$. ThenThus $Y$ generally has a periodic mean, and $Z$ generally has a periodic variance. The edge cases allowed by $S\in {\mathbb{N}}^{\ast }$ matter: $Y$ is stationary when $S=1$, because the mean is then constant, and $Z$ is stationary for $S=1$ or $S=2$. In the latter case multiplying iid centered Gaussian white noise by $(-1{)}^t$ leaves its iid distribution unchanged. For every $S\ge 3$, $c_0^2=1$ but $c_1^2<1$, so $Z$ is nonstationary. For $S>1$, $Y$ is nonstationary because $c_t$ is not constant.

Use the seasonal difference operator ${\Delta }_S=1-B^S$. Since $c_t=c_{t-S}$,This is a stationary moving-average model of iid Gaussian white noise. Its covariance is $2{\sigma }^2$ at lag $0$, $-{\sigma }^2$ at lags $\pm S$, and zero otherwise.

For the variance-modulated process the same operation givesIts variance is $2{\sigma }^2{c}_t^2$ and its covariance at lag $S$ is $-{\sigma }^2{c}_t^2$. Therefore it remains nonstationary for $S\ge 3$, and the operation doubles the marginal variance at nonzero seasons. Seasonal differencing does not remove periodic variance: it removes a periodic mean, but the periodic variance generally remains. Practical alternatives are a periodic model or seasonal variance standardization; at seasons with $c_t=0$ the observations are deterministically zero, so division by $c_t$ is not possible there. For the already stationary special cases $S=1,2$, filtering preserves stationarity.

A periodically correlated process with period $S$ has its mean and two-time covariance unchanged when both times are shifted by $S$. For $Z$, the mean is zero andHence $Z$ is a periodically correlated process. Its covariance period can be smaller than $S$: when $S$ is even, $c_{t+S/2}^2=c_t^2$, so $S/2$ is already a covariance period.

Here $\Gamma >0$ is a kinetic coefficient, and the additive Gaussian white noise must be centered. A rigorous starting point is a finite spatial discretization or finite set of modes of the coarse-grained field. For each field coordinate, the Fokker-Planck probability current has the continuum notationThe equilibrium Boltzmann distribution ${\mathcal{P}}_{\mathrm{eq}}\propto e^{-\beta F}$ has $\delta {\mathcal{P}}_{\mathrm{eq}}/\delta \varphi =-\beta (\delta F/\delta \varphi ){\mathcal{P}}_{\mathrm{eq}}$. Detailed balance for this time-even field requires zero configuration-space current. ThereforeThis is the Model A fluctuation-dissipation relation. Conversely, that noise strength makes the equilibrium current vanish and gives reversible relaxational dynamics when the equilibrium measure is normalizable.

Alternatively, the Onsager--Machlup path probability for Model A dynamics compares the squared required noise histories $\dot{\varphi }+\Gamma \delta F/\delta \varphi$ and $-\dot{\varphi }+\Gamma \delta F/\delta \varphi$. Their difference gives $\log ({\mathbb{P}}_F/{\mathbb{P}}_B)=-2\Gamma \Delta F/{\sigma }^2$, using the functional chain rule. The midpoint convention and time-reversal invariance of a path Jacobian justify cancellation of the common Jacobian. Comparing with the conditional path ratio gives the same result. The spatial white-noise continuum is a coarse-grained idealization: retain a microscopic cutoff when pointwise fluctuations or ultraviolet divergences matter.

Interpret the longitudinal Dirac delta function as the Markov approximation for a random medium. The two arguments on the left of the printed covariance must be $x_1$ and $x_2$. The transverse covariance kernel is $B(|{\mathbf{z}}_1-{\mathbf{z}}_2|)$. Here statistical isotropy means transverse isotropy: a medium with a distinguished longitudinal white noise direction and a smooth transverse covariance is not literally isotropic in all three directions. Also, ideal Gaussian white noise replaces the original finite-variance field; it cannot simultaneously satisfy a pointwise normalization $\mathbb{E}{W}^2=1$.

Write $Q(r)={\mu }^{2}B(r)$ for the covariance strength of the refractive index fluctuations. Define the transverse power spectrum using the unnormalized forward Fourier transform printed in equation 3:This is the spectrum of the fluctuations, excluding the deterministic mean index. Equivalently it is the zero-longitudinal-frequency slice of the three-dimensional fluctuation spectrum before the Markov approximation. The angular integral is $2\pi J_0(\nu r)$, where $J_0$ is the Bessel function of the first kind. Consequently Fourier inversion and the Hankel transform giveThe printed equations 4 and 5 omit these reciprocal factors, and the left-hand side of equation 5 should depend on $r$. They are instead a consistent order-zero Hankel transform pair if their $F$ denotes $S_H=S_F/(2\pi )$.

Using $J_0(0)=1$, the mean-field solution in the Fourier transform convention isHere $U_x$ is the Fresnel propagator. Finiteness of the integral ensures a finite screen wave phase variance. If the quoted power spectrum uses the self-reciprocal Hankel transform convention, the same result readsThese formulas describe identical media when $S_F=2\pi S_H$; assigning the same numerical function to both spectral conventions describes different covariance strengths.

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.

Write $E(t)=\delta A/\delta x(t)$ for the functional derivative of the action. Under time reversal in classical mechanics, $\dot{x}$ changes sign whereas $E$ is even by assumption. Thus the forward and reversed Gaussian white noise histories associated with the same geometric path areThe Onsager--Machlup path probability therefore givesHere the path probabilities are densities conditional on their respective initial states; they are not separately normalized bridges conditioned on both endpoints. The noise-to-path Jacobian determinant must be treated with a consistent discretization: it cancels when it is invariant under time reversal in classical mechanics. This is the usual additive-noise Langevin dynamics convention, including a constant-mass Underdamped Langevin dynamics or the time-reversal invariance of a path Jacobian in midpoint overdamped Langevin dynamics. A completely arbitrary velocity-dependent Lagrangian would require specifying that measure rather than deducing its cancellation solely from the parity of $E$.

For an autonomous Lagrangian, the Euler-Lagrange equation usesApply the chain rule to the Hamiltonian along an arbitrary smooth path, without assuming the unforced Euler-Lagrange equation:Consequently the energy balance for an autonomous Lagrangian isWith explicit time dependence, an additional $-{\partial }_{t}L$ enters $dH/dt$. For ideal Gaussian white noise, the identity is understood through smooth-noise regularization or the Stratonovich chain rule. The original PDF correctly differentiates $L$ with respect to $\dot{x}$ in $H$; the supplied TeX's derivative with respect to $x$, its endpoint $t^2$, and its ordinary derivative of $A$ are transcription errors.

At thermal equilibrium, microscopic reversibility equates the probabilities of a path and its reversed path when both include their Boltzmann distribution initial weights. Denote their endpoint states by $z_1,z_2$, including velocity if needed. Since the Hamiltonian is even under time reversal in classical mechanics,This detailed balance condition and the energy balance for an autonomous Lagrangian yieldThis is the fluctuation-dissipation relation for a Langevin particle: the strength of Gaussian white noise is fixed by the damping and temperature, with $k_B$ the Boltzmann constant.

For an equilibrium coarse-grained variable, the unresolved microscopic states contribute entropy; their statistical weight is encoded in the Helmholtz free energy, rather than in a single microscopic energy. Relative to the same reference measure, ${\rho }_{\mathrm{eq}}(x)\propto e^{-\beta F(x)}$, so microscopic reversibility becomesThis extension assumes an equilibrium coarse-grained description with reversible path statistics; externally driven dynamics need not obey this relation.

The Neumann boundary conditions give Fourier modes $q=n\pi /L$, including the spatially uniform mode $q=0$. In the frame translating with the mean deposition height, that mode has no restoring force and obeysHere $W_t$ is Brownian motion, independent of the initial height. More generally $⟨h_0(t{)}^2⟩=⟨h_0(0{)}^2⟩+{\sigma }^{2}t$ when these second moments exist. Removing the mean deposition drift does not remove the Gaussian white noise.

The Brownian zero mode of a fluctuating interface has no stationary distribution on the real height axis for ${\sigma }^2>0$. This conclusion does not depend on assuming finite variance: a stationary characteristic function would satisfy $\hat{\mu }(k)=\hat{\mu }(k)e^{-{\sigma }^2{k}^{2}t/2}$, hence vanish for every $k\ne 0$, contradicting its continuity at zero and $\hat{\mu }(0)=1$. Any stationary joint distribution of the whole height would have a stationary zero-mode marginal, which is impossible. Thus the full unpinned height has no Boltzmann equilibrium. Pinning the mean height, or retaining only the nonzero Fourier modes, removes this obstruction. The PDF has ${\dot{h}}_q$; the TeX's ${\hslash }_q$ is a transcription error.

Write ${\mu }_j=\delta F/\delta p_j$ and $\int ={\int }_{t_1}^{t_2}dt\int d\mathbf{r}$. Eliminating the noise gives $f=\dot{p}+\Gamma \mu$. The Onsager--Machlup path probability is thereforeThese are conditional path densities given the corresponding starting configuration. For a time-even vector order parameter, reversal reads the configurations backward, changes $\dot{p}$ to $-\dot{p}$, and keeps $\mu$ unchanged on each corresponding configuration.

The two Gaussian white noise measures have the same normalization. With a common midpoint discretization, the noise-to-path Jacobian also agrees under reversal by time-reversal invariance of a path Jacobian. The common factor $\mathcal{K}[p]$ may include that path-dependent Jacobian; it need not be a universal constant. This makes the shared-factor assertion precise while leaving the requested ratio unaffected.