Nonzero-Sum Games of Optimal Stopping for Markov Processes

Nonzero-Sum Games of Optimal Stopping for Markov Processes
复制标题

马尔可夫过程最优停止的非零和博弈

DOI:
--
复制
发表时间:
2018
期刊:
影响因子:
--
通讯作者:
N. Attard
N. Attard
中科院分区:
--
文献类型:
--
作者:
N. Attard

文献摘要

被引文献

相似文献

Two players are observing a right-continuous and quasi-left-continuous strong Markov process X. We study the optimal stopping problem $$V^{1}_{sigma }(x)=sup _{ au } mathsf {M}_{x}^{1}( au ,sigma )$$Vσ1(x)=supτMx1(τ,σ) for a given stopping time $$sigma $$σ (resp. $$V^{2}_{ au }(x)=sup _{sigma } mathsf {M}_{x}^{2}( au ,sigma )$$Vτ2(x)=supσMx2(τ,σ) for given $$ au $$τ) where $$mathsf {M}_{x}^{1}( au ,sigma ) = mathsf {E}_{x} [G_{1}(X_{ au })I( au le sigma ) + H_{1}(X_{sigma })I(sigma < au )]$$Mx1(τ,σ)=Ex[G1(Xτ)I(τ≤σ)+H1(Xσ)I(σ<τ)] with $$G_1,H_1$$G1,H1 being continuous functions satisfying some mild integrability conditions (resp. $$mathsf {M}_{x}^{2}( au ,sigma ) = mathsf {E}_{x} [G_{2}(X_{sigma })I(sigma < au ) + H_{2}(X_{ au })I( au le sigma )]$$Mx2(τ,σ)=Ex[G2(Xσ)I(σ<τ)+H2(Xτ)I(τ≤σ)] with $$G_2,H_2$$G2,H2 being continuous functions satisfying some mild integrability conditions). We show that if $$sigma = sigma _{D_{2}} = inf {t ge 0: X_t in D_2}$$σ=σD2=inf{t≥0:Xt∈D2} (resp. $$ au = au _{D_{1}} = inf {t ge 0: X_t in D_1}$$τ=τD1=inf{t≥0:Xt∈D1}) where $$D_{2}$$D2 (resp. $$D_1$$D1) has a regular boundary, then $$V^{1}_{sigma _{D_{2}}}$$VσD21 (resp. $$V^{2}_{ au _{D_{1}}}$$VτD12) is finely continuous. If $$D_{2}$$D2 (resp. $$D_1$$D1) is also (finely) closed then $$ au _*^{sigma _{D_2}} = inf {t ge 0: X_{t} in D_{1}^{sigma _{D_{2}}}}$$τ∗σD2=inf{t≥0:Xt∈D1σD2} (resp. $$sigma _{*}^{ au _{D_1}} = inf {t ge 0: X_{t} in D_{2}^{ au _{D_{1}}}}$$σ∗τD1=inf{t≥0:Xt∈D2τD1}) where $$D_{1}^{sigma _{D_{2}}} = {V^{1}_{sigma _{D_{2}}} = G_{1}}$$D1σD2={VσD21=G1} (resp. $$D_{2}^{ au _{D_{1}}} = {V^{2}_{ au _{D_{1}}} = G_{2}}$$D2τD1={VτD12=G2}) is optimal for player one (resp. player two). We then derive a partial superharmonic characterisation for $$V^{1}_{sigma _{D_2}}$$VσD21 (resp. $$V^{2}_{ au _{D_1}}$$VτD12) which can be exploited in examples to construct a pair of first entry times that is a Nash equilibrium.
Two players are observing a right-continuous and quasi-left-continuous strong Markov process X. We study the optimal stopping problem $$V^{1}_{sigma }(x)=sup _{ au } mathsf {M}_{x}^{1}( au ,sigma )$$Vσ1(x)=supτMx1(τ,σ) for a given stopping time $$sigma $$σ (resp. $$V^{2}_{ au }(x)=sup _{sigma } mathsf {M}_{x}^{2}( au ,sigma )$$Vτ2(x)=supσMx2(τ,σ) for given $$ au $$τ) where $$mathsf {M}_{x}^{1}( au ,sigma ) = mathsf {E}_{x} [G_{1}(X_{ au })I( au le sigma ) + H_{1}(X_{sigma })I(sigma < au )]$$Mx1(τ,σ)=Ex[G1(Xτ)I(τ≤σ)+H1(Xσ)I(σ<τ)] with $$G_1,H_1$$G1,H1 being continuous functions satisfying some mild integrability conditions (resp. $$mathsf {M}_{x}^{2}( au ,sigma ) = mathsf {E}_{x} [G_{2}(X_{sigma })I(sigma < au ) + H_{2}(X_{ au })I( au le sigma )]$$Mx2(τ,σ)=Ex[G2(Xσ)I(σ<τ)+H2(Xτ)I(τ≤σ)] with $$G_2,H_2$$G2,H2 being continuous functions satisfying some mild integrability conditions). We show that if $$sigma = sigma _{D_{2}} = inf {t ge 0: X_t in D_2}$$σ=σD2=inf{t≥0:Xt∈D2} (resp. $$ au = au _{D_{1}} = inf {t ge 0: X_t in D_1}$$τ=τD1=inf{t≥0:Xt∈D1}) where $$D_{2}$$D2 (resp. $$D_1$$D1) has a regular boundary, then $$V^{1}_{sigma _{D_{2}}}$$VσD21 (resp. $$V^{2}_{ au _{D_{1}}}$$VτD12) is finely continuous. If $$D_{2}$$D2 (resp. $$D_1$$D1) is also (finely) closed then $$ au _*^{sigma _{D_2}} = inf {t ge 0: X_{t} in D_{1}^{sigma _{D_{2}}}}$$τ∗σD2=inf{t≥0:Xt∈D1σD2} (resp. $$sigma _{*}^{ au _{D_1}} = inf {t ge 0: X_{t} in D_{2}^{ au _{D_{1}}}}$$σ∗τD1=inf{t≥0:Xt∈D2τD1}) where $$D_{1}^{sigma _{D_{2}}} = {V^{1}_{sigma _{D_{2}}} = G_{1}}$$D1σD2={VσD21=G1} (resp. $$D_{2}^{ au _{D_{1}}} = {V^{2}_{ au _{D_{1}}} = G_{2}}$$D2τD1={VτD12=G2}) is optimal for player one (resp. player two). We then derive a partial superharmonic characterisation for $$V^{1}_{sigma _{D_2}}$$VσD21 (resp. $$V^{2}_{ au _{D_1}}$$VτD12) which can be exploited in examples to construct a pair of first entry times that is a Nash equilibrium.