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No: the first moment generally depends on the transverse coordinate. At entry to the random medium, the initial data are deterministic, sowhich is a nonconstant Gaussian beam profile. Statistical homogeneity of the medium says that shifting both the medium and the incident data gives correspondingly shifted field statistics. It does not make the response to fixed, localized incident data translation invariant.

In particular, Gaussian coherent-field propagation in the Markov approximation, derived in part (d), givesfor the linear weak-index model. A homogeneous attenuation factor multiplies the varying beam profile. Thus spatially homogeneous medium statistics can coexist with a transversely inhomogeneous coherent field. The first moment is not the mean wave intensity, and disappearance of the coherent component is not itself disappearance of the total wave energy.

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.

Let the medium's autocorrelation function of a random field and power spectrum use the explicit conventionSince $n^2-1=2\mu W+{\mu }^2{W}^2$, only the linear term in $\mu$ is needed to obtain the cross-section through second order:Squaring and taking the expectation givesThis is the finite-volume formula in the Born approximation for scalar wave scattering; the zero mean of $W$ does not make its squared scattered amplitude vanish. For a jointly centered Gaussian random field, the third-order contribution is zero, so the next perturbative contribution is fourth order when that expansion is valid.

Introduce the overlap volume $H_V(\mathbf{s})=|V\cap (V-\mathbf{s})|$. Changing variables rewrites the double integral asWhen the linear dimensions of a regular scattering volume are large compared with the correlation length, and $C$ is integrable, $H_V(\mathbf{s})\simeq |V|$ over the significant covariance range. This gives the finite-volume random scattering spectrum resultThere are two distinct approximations here: the weak-scattering expansion in $\mu$ and the negligible-boundary-overlap approximation in the volume-to-correlation-scale ratio. If $S$ instead denotes the power spectrum of the index fluctuation $n-1=\mu W$, then $S={\mu }^2{S}_W$ and the external factor ${\mu }^2$ is absorbed into $S$.

Statistical homogeneity alone allows an anisotropic spectrum, so its argument is generally the vector ${\mathbf{k}}_s$. If statistical isotropy is additionally assumed, $S_W$ depends only onThe scalar notation $S(k_s)$ is valid in that isotropic case. For the alternative convention $S_{\mathrm{norm}}=(2\pi {)}^{-3}{S}_W$, the bulk formula is $2\pi {\mu }^2{k}^4|V|S_{\mathrm{norm}}({\mathbf{k}}_s)$. Changing the sign of the scattering vector does not change the spectrum of a real stationary scalar field; changing the Fourier transform normalization does change the displayed $2\pi$ factors. Both conventions have therefore been stated explicitly.

For noisy data, exact equality $Ax=y$ may be impossible: the data can have a component outside the operator range. A least-squares solution of a linear inverse problem instead minimizesFor any variation $h\in X$, the adjoint operator givesThus statistical homogeneity in every direction is the normal equation for a linear inverse problem, $A^{\ast }Ax=A^{\ast }y$. It is also sufficient for a global minimum: when it holds,The residual is orthogonal to the closure of the range, since it lies in $\ker A^{\ast }$. A normal-equation solution therefore fits the component of the data representable by the forward map, while leaving the perpendicular component unfitted. If there is more than one such solution, select the minimum-norm least-squares solution, which lies in $(\ker A{)}^{\perp }$.

These facts explain the formulation, but do not make the inverse problem well posed. A compact operator of infinite rank typically has nonclosed operator range, so even the least-squares infimum need not be attained. A least-squares solution exists exactly when the orthogonal projection of $y$ onto $\overline{\mathrm{Ran}A}$ belongs to $\mathrm{Ran}A$. In a singular system of a compact operator, this becomes the Picard criterion. Also, the nonzero eigenvalues of $A^{\ast }A$ are the squares of the singular values of $A$: solving the operator normal equation directly can worsen numerical conditioning. The operator normal equation expresses least squares; it does not itself regularize the inverse problem. The subsequent finite Landweber iteration and its stopping rule provide the inverse-problem regularization.

Statistical homogeneity forces a scalar two-point function to have momenta $(\mathbf{k},-\mathbf{k})$. Statistical isotropy then makes it a function only of $|\mathbf{k}|$, so it is parity even without using perturbation theory.

For a scalar three-point function, momentum conservation gives ${\mathbf{k}}_1+{\mathbf{k}}_2+{\mathbf{k}}_3=0$, so all three vectors lie in one plane. A rotation by $\pi$ around the normal to that plane sends every ${\mathbf{k}}_a$ to $-{\mathbf{k}}_a$. Rotational invariance therefore identifies a triangle with its parity reverse, proving nonperturbatively that the scalar primordial bispectrum is parity even.

A rotationally invariant local parity-odd three-scalar vertex must contain a Levi-Civita symbol contracted with three spatial momenta. Its momentum-space factor is proportional tobecause momentum conservation makes the momenta linearly dependent. The equivalent position-space expression is a total divergence; its spatial integral vanishes under the stated vanishing boundary condition. Thus a parity-odd interaction of three scalar fields contributes nothing to the action.