= Solution
Represent the thin layer by its <wave phase> jump at the exit $x=0$ and take the incident envelope to have amplitude one. Then $E(0,z)=e^{i\phi(z)}$. <Stationarity> makes the <wave phase> mean a constant $\mu$; it need not be zero, since the question specifies only the <variance>. For a <normal distribution> marginal, write $\phi=\mu+\sigma Z$ with $Z\sim N(0,1)$. If $F(t)=\mathbb E[e^{itZ}]$, integration by parts in the Gaussian density gives $F'(t)=-tF(t)$ and $F(0)=1$, hence $F(t)=e^{-t^2/2}$. Thus its <characteristic function> gives
$$
\langle E(0,z)\rangle=\langle e^{i\phi(z)}\rangle=e^{i\mu-\sigma^2/2}\equiv M.
$$
The <Fourier transform> of this constant is $2\pi M\delta(\nu)$, interpreted distributionally for the infinite <plane wave>. Applying the propagator from part (a),
$$
\boxed{\langle\widehat E(x,\nu)\rangle
=2\pi e^{i\mu-\sigma^2/2}\delta(\nu)e^{-i\nu^2x/(2k)}
=2\pi e^{i\mu-\sigma^2/2}\delta(\nu)}.
$$
If the <wave phase> has zero mean, omit $e^{i\mu}$. A nonunit incident amplitude simply multiplies the expression. Only the one-point <Gaussian distribution> is needed for this <expected value>; no assumption about joint <Gaussian distributions> at separated points is needed.
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