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Let ${\psi }_R={\psi }_i{e}^{{\varphi }_1}$ with unit-modulus incidence and ${\varphi }_1={\int }_{D}K(\mathbf{r},{\mathbf{r}}^{\prime })V({\mathbf{r}}^{\prime })d{\mathbf{r}}^{\prime }$. For a centered real Gaussian random field $V$, the exponential moment of a Gaussian linear functional givesHere $C_V$ is the autocorrelation function of a random field. Both the covariance $\mathbb{E}|{\varphi }_1{|}^2$ and the generally nonzero term $\mathrm{Re}\mathbb{E}{\varphi }_1^2$ enter the exponent; dropping the latter is not justified for an arbitrary finite real random medium.

Use the leading far-field kernel from the previous part. DefineThen ${\varphi }_1={\int }_{D}KV$ and, because the incident intensity is one,To first order in weak contrast, $V=2\mu k_0^{2}W+O({\mu }^2)$. Its linearized part is a centered stationary Gaussian random field, with autocorrelation function of a random field $C_V(\mathbf{s})=4{\mu }^2{k}_0^4{C}_W(\mathbf{s})$. The exponential moment of a Gaussian linear functional therefore gives the usual Gaussian intensity in the first Rytov approximation:Equivalently the exponent is $\mathbb{E}|{\varphi }_1{|}^2+\mathrm{Re}\mathbb{E}{\varphi }_1^2$. The second term generally matters because a finite real random medium need not generate a circular complex Gaussian amplitude.

If the full stated potential $V=k_0^2(2\mu W+{\mu }^2{W}^2)$ is retained, it is not Gaussian and has mean $m_V={\mu }^2{k}_0^2$. To retain every term through order ${\mu }^2$ in the intensity of this first Rytov field, the cumulant expansion instead givesHere $C_V^{\mathrm{cov}}=\mathbb{E}[(V-m_V)(V^{\prime }-m_V)]$; the raw autocorrelation differs by $m_V^2=O({\mu }^4)$. The stated remainder uses Gaussian finite-dimensional statistics of $W$, for which odd total moments vanish. This mean term is absent if one first truncates the potential to $2\mu k_0^{2}W$. Both formulas concern the first Rytov approximation, and exponentiating it does not supply the omitted higher-order scattering corrections.