= Solution
{c}
The notation $P_0=\mathcal N(0,I)$ denotes generalized <Gaussian white noise> over $H=L^2(\mathbb R^2)$, not a <Gaussian measure> supported on $H$: the identity is not a <trace-class operator> there. For example, realize the observation on a suitable space of <tempered distributions>. Its <Cameron-Martin space of a Gaussian measure> is $H$. For deterministic $h\in H$, the stochastic pairing $W_m(h)$ has <normal distribution> $\mathcal N(0,\|h\|_H^2)$ under $P_0$ and is the <isonormal Gaussian process> indexed by $H$.
Interpret the stated maps between <Sobolev spaces> as bounded <linear operators>, as usual. In particular $K:H\to H^2(\mathbb R^2)\subset H$ is bounded, so its <adjoint operator> $K^*$ is bounded on $H$. Hence $A=K^*K$ maps $H$ into $H$, which is the property needed here. The stronger smoothing follows by <Sobolev duality>: boundedness of $K:H^{-4}\to H^{-2}$ gives $K^*:H^2\to H^4$, and therefore $A:H\to H^4\subset H$. As $\Pi(H)=1$, the shift $Au$ belongs to the noise <Cameron-Martin space of a Gaussian measure> for $\Pi$-almost every $u$.
The <white-noise likelihood for a square-integrable shift>, obtained from the <Cameron-Martin theorem for a Gaussian measure> in its white-noise form, is
$$
L(u,m)=\frac{dP_u}{dP_0}(m)
=\exp\left(W_m(Au)-\tfrac12\|Au\|_H^2\right)
=e^{-\Phi(u;m)},\qquad
\Phi(u;m)=\tfrac12\|Au\|_H^2-W_m(Au).
$$
Here $P_u$ is the law of $Au+\eta$. In an <orthonormal basis> $(e_j)$ of $H$, with white-noise coordinates $m_j$, the pairing under $P_0$ is
$$
W_m(Au)=\lim_{N\to\infty}\sum_{j=1}^Nm_j\langle Au,e_j\rangle_H.
$$
For fixed $u$ this has <mean-square convergence> and converges <almost surely>, since $\sum_j|\langle Au,e_j\rangle|^2=\|Au\|_H^2$. The stipulated joint measurability allows the likelihood to be used under $\nu_0(du,dm)=\Pi(du)P_0(dm)$. One must not replace this expression by a finite $\|m-Au\|_H^2$, because white noise is not $H$-valued. On the unbounded domain $\mathbb R^2$, membership of white noise in a global unweighted negative <Sobolev space> must not be assumed either; the stochastic pairing avoids that issue.
Let $\nu(du,dm)=\Pi(du)P_u(dm)$ be the actual joint law. The shift formula gives $\nu\ll\nu_0$ with <Radon-Nikodym derivative> $L$. Define
$$
Z(m)=\int_HL(u,m)\,\Pi(du).
$$
For each fixed admissible $u$, the <moment-generating function of a normal distribution> gives
$$
\int L(u,m)\,P_0(dm)=
e^{-\|Au\|_H^2/2}\mathbb E^{P_0}e^{W_m(Au)}=1.
$$
By <Tonelli theorem>, $\int Z(m)P_0(dm)=1$, so $Z(m)<\infty$ for $P_0$-almost every $m$. The likelihood is finite and strictly positive for $\nu_0$-almost every $(u,m)$; <Fubini's theorem> then gives $Z(m)>0$ for $P_0$-almost every $m$. The marginal observation law is $Z(m)P_0(dm)$ and is an <equivalent probability measure> to $P_0$.
The <Bayes formula for a dominated observation model> therefore defines
$$
\boxed{\frac{d\Pi^m}{d\Pi}(u)=\frac1{Z(m)}
\exp\left(W_m(Au)-\tfrac12\|Au\|_{L^2}^2\right).}
$$
To verify that it is the <conditional distribution>, for measurable sets $B$ of unknowns and $C$ of data,
$$
\int_C\Pi^m(B)Z(m)\,P_0(dm)
=\int_{B\times C}L(u,m)\,\nu_0(du,dm)=\nu(B\times C).
$$
Thus the <posterior distribution> is well defined for almost every observation under the actual data law. This is the appropriate almost-everywhere assertion supplied by <Bayes' theorem>; the stated assumptions do not prescribe a canonical posterior at every exceptional generalized datum.
Back to article page
