A numerical approach to infinite-dimensional linear programming in $L_1$ spaces

Ito, Satoshi; Wu, Soon‐Yi; Shiu, Ting-Jang; Teo, Kok Lay

doi:10.3934/jimo.2010.6.15

Cited by 4 publications

(4 citation statements)

References 21 publications

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“…However, since it involves the optimization over a continuous function, it falls in the class of infinite-dimensional optimization problems. The solution methods for this kind of problem range from classic calculus of variations (Gelfand, 2003) to general numerical approaches (Schochetman, 2001;Devolder, 2010) and approaches customized for linear problems (Ito, 2009). However, a detailed analysis of infinite-dimensional optimization techniques is beyond the scope of this work.…”

Section: Discussionmentioning

confidence: 99%

A Linear Programming Approach to Sequential Hypothesis Testing

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2015

Sequential Analysis

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Under some mild Markov assumptions it is shown that the problem of designing optimal sequential tests for two simple hypotheses can be formulated as a linear program. This result is derived by investigating the Lagrangian dual of the sequential testing problem, which is an unconstrained optimal stopping problem depending on two unknown Lagrangian multipliers. It is shown that the derivative of the optimal cost function, with respect to these multipliers, coincides with the error probabilities of the corresponding sequential test. This property is used to formulate an optimization problem that is jointly linear in the cost function and the Lagrangian multipliers and can be solved for both with off-the-shelf algorithms. To illustrate the procedure, optimal sequential tests for Gaussian random sequences with different dependency structures are derived, including the Gaussian AR(1) process.

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Section: Discussionmentioning

confidence: 99%

A Linear Programming Approach to Sequential Hypothesis Testing

Fauß

Zoubir

2015

Sequential Analysis

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“…Since the closure of N 1 is N, (7) is also held for all (y, r) ∈ N, which implies that ρ 0 ≥ 0. Indeed, N 1 contains pairs with arbitrary small negative values of r. Therefore, if ρ 0 < 0 we can increase the right side of (7) as much as we wish and get a contradiction with (7).…”

Section: Proof By Lemma 2 the Setmentioning

confidence: 93%

“…Clearly, (0, I(x 0 ) ∈ S. Thus, for each z ≤ 0 we have (z, I(x 0 )) ∈ S. On the other hand (0, I(x 0 )) ∈ N. Then for each z ≤ 0 by (7) ρ 0 I(x 0 ) + (y * 0 , z) ≥ ρ 0 I(x 0 ) Consequently, for all z ≤ 0 we get (y * 0 , z) ≥ 0. Therefore, y * 0 ≤ 0.…”

Section: Proof By Lemma 2 the Setmentioning

confidence: 98%