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If the first logarithmic perturbation ${\chi }_1$ is a centered linear functional of a Gaussian random field, the complex Gaussian exponential moment gives the displayed mean of the first Rytov approximation ${\psi }_R={\psi }_i{e}^{\mu {\chi }_1}$. This is the mean of the truncated logarithmic model. A nonzero mean of the omitted second logarithmic perturbation can change the true field mean at order ${\mu }^2$.

Set $\psi =e^{ikx+\mu {\chi }_1+O({\mu }^2)}$. Substitution into the Helmholtz equation gives the first Rytov approximation equationFor the forward parabolic approximation, omit ${\partial }_x^2{\chi }_1$. With zero perturbation at $x=0$, the Fresnel propagator then givesA jointly Gaussian random field makes ${\chi }_1$ a centered complex Gaussian variable. The complex Gaussian exponential moment therefore gives the first-Rytov coherent meanThe second moment here has no complex conjugation. It cannot be replaced by the positive intensity variance $\mathbb{E}[|{\chi }_1{|}^2]$.

For example, let $\hat{C}(u,v;\mathbf{\kappa })$ be the transverse spectrum of the covariance between depths $u$ and $v$. The Fourier multiplier of the Fresnel propagator is $e^{-i(x-u)|\mathbf{\kappa }{|}^2/(2k)}$, soThis supplies the requested mean in terms of medium statistics. For the longitudinal white-noise model, $\hat{C}=\delta (u-v)\Phi$, which reduces the double longitudinal integral to one.

The exponential is the mean of the first-order logarithmic model. It is not a complete second-order perturbation expansion of the true mean: the omitted logarithmic term ${\mu }^2{\chi }_2$ can itself have a nonzero mean. Strictly at first order, $\mathbb{E}[\psi ]=e^{ikx}+O({\mu }^2)$. As an independent check, the closed white-noise parabolic model gives exactlyfor plane-wave input. A truncated first-Rytov covariance expression with transverse diffraction need not equal this exact mean; omitted logarithmic terms account for the difference.