By Martin Simon

This monograph is worried with the research and numerical answer of a stochastic inverse anomaly detection challenge in electric impedance tomography (EIT). Martin Simon stories the matter of detecting a parameterized anomaly in an isotropic, desk bound and ergodic conductivity random box whose realizations are swiftly oscillating. For this objective, he derives Feynman-Kac formulae to carefully justify stochastic homogenization on the subject of the underlying stochastic boundary worth challenge. the writer combines recommendations from the speculation of partial differential equations and useful research with probabilistic rules, paving tips to new mathematical theorems that could be fruitfully utilized in the therapy of the matter handy. additionally, the writer proposes an effective numerical technique within the framework of Bayesian inversion for the sensible answer of the stochastic inverse anomaly detection challenge.

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**Additional info for Anomaly Detection in Random Heterogeneous Media: Feynman-Kac Formulae, Stochastic Homogenization and Statistical Inversion**

**Sample text**

Then by the Chapman-Kolmogorov equation p(t, x, x) − |D|−1 p(t/2, x, y)p(t/2, y, x) dy − |D|−1 = D (p(t/2, x, y))2 dy − |D|−1 = D (p(t/2, x, y) − |D|−1 )2 dy ≥ 0, = D where we have used that D p(t, x, y) dy = 1 for all t ≥ 0 and x ∈ D. Moreover, we have by the analyticity of the mapping t → p(t, x, x), cf. 4) and the Poincaré inequality φ − |D|−1 φ(x) dx D 2 ≤ cD ||∇φ||2 for all φ ∈ H 1 (D). , there exist positive constants c1 and c2 such that 0 ≤ p(t, x, x) − |D|−1 ≤ c1 exp(−c2 t) for every t ≥ 0.

S. e. x ∈ D. 35) holds for arbitrary v ∈ H 1 (D). 12) we have that E g (u, v) = f, v for all v ∈ H 1 (D) ∩ C(D). s. e. x ∈ D. s. e. 35) and the Markov property of X. 15. 26). s. 26) is well-deﬁned. s. e. x ∈ D. Note that the second term on the right-hand side is a local Px -martingale and that eg is continuous, adapted to {Ft , t ≥ 0} and of bounded variation. Multiplication by such functions leaves the class of semimartingales invariant. e. , where the second summand on the right-hand side is a local Px martingale.

18 that the right-hand side in the last equality is continuous up to the boundary, the assertion holds for every x ∈ D. 20. 15 fails for the Neumann problem corresponding to the continuum model. 36) becomes inﬁnite. 2 from [34], specialized to a zero lower-order term, does not yield the desired Feynman-Kac formula for the continuum model either. 3 of the previous chapter. Recall that in this setting ∂D consists of two disjoint parts ∂1 D and ∂2 D and that measurements can be taken only on the accessible boundary ∂1 D while the electric potential vanishes on the inaccessible boundary ∂2 D.