Stopping with expectation constraints: 3 points suffice

Bayesian sequential testing with expectation constraints

“…We assume that the agent can accept H 0 only if the posterior probability Y is smaller than or equal to a given threshold α ∈ 0, 1 2 . Similarly, the agent can accept H 1 only if Y is greater than or equal to an upper threshold.…”

“…In Section 3 and Section 4 we show that if V (T, y) > V 2 (T, y), then there exists an optimal stopping time τ * for (1.2) such that the law of Y τ * under P y is a weighted sum of Dirac measures in α, b * , 1 − α, where b * ∈ α, 1 2 . The mass point b * is independent of T and y.…”

“…If we impose that both hypotheses have probability 1 2 at the beginning of the observation, i.e. y = 1 2 , then the function q 1 2 simplifies to…”

“…Let (F Y t ) be the filtration generated by (Y t ) and denote by T (T ) the set of all (F Y t )-stopping times satisfying the constraint E[τ ] ≤ T ∈ [0, ∞). Let α ∈ 0, 1 2 , β ≥ 1 and suppose that the agent faces the stopping problem maximize E 1 (0,α] (Y τ ) + β 1 [1−α,1) (Y τ ) subject to τ ∈ T (T ). (0.1)…”

“…The results from [1] on general stopping problems with expectation constraints imply that for solving (0.1) it is sufficient to consider only stopping times such that the law of the posterior probability process Y at the stopping time is a weighted sum of at most 3 Dirac measures (cf. Theorem 1.4 in [1]). We show that a reduction to 2 mass points is not possible in the specific stopping problem (0.1).…”