How to show $σ =inf(t ≥ 0 : |B_t| =sqrt{3})$ is optimal for $M_t = B_t^4 − 6tB^2_t +e^{2B_t−2t}+3t^2$












-1














Given $σ =inf(t ≥ 0 : |B_t| =sqrt{3})$, show $sigma$ is optimal for $M_t = B_t^4 − 6tB^2_t +e^{2B_t−2t}+3t^2$ for $t ≥ 0$.

Struggling to show that it is bounded, my attempt so far is:
$$|M_{t wedge sigma}|=|B_{t wedge sigma}^4 − 6({t wedge sigma})B^2_{t wedge sigma} +e^{2B_{t wedge sigma}−2({t wedge sigma})}+3({t wedge sigma})^2|leq|B_{t wedge sigma}^4+e^{2B_{t wedge sigma}−2({t wedge sigma})}+3({t wedge sigma})^2|$$
Unsure how to proceed.










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    What do you mean by "$sigma$ is optimal" ...? When do you call a stopping time optimal?
    – saz
    Jan 1 at 16:55
















-1














Given $σ =inf(t ≥ 0 : |B_t| =sqrt{3})$, show $sigma$ is optimal for $M_t = B_t^4 − 6tB^2_t +e^{2B_t−2t}+3t^2$ for $t ≥ 0$.

Struggling to show that it is bounded, my attempt so far is:
$$|M_{t wedge sigma}|=|B_{t wedge sigma}^4 − 6({t wedge sigma})B^2_{t wedge sigma} +e^{2B_{t wedge sigma}−2({t wedge sigma})}+3({t wedge sigma})^2|leq|B_{t wedge sigma}^4+e^{2B_{t wedge sigma}−2({t wedge sigma})}+3({t wedge sigma})^2|$$
Unsure how to proceed.










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  • 2




    What do you mean by "$sigma$ is optimal" ...? When do you call a stopping time optimal?
    – saz
    Jan 1 at 16:55














-1












-1








-1







Given $σ =inf(t ≥ 0 : |B_t| =sqrt{3})$, show $sigma$ is optimal for $M_t = B_t^4 − 6tB^2_t +e^{2B_t−2t}+3t^2$ for $t ≥ 0$.

Struggling to show that it is bounded, my attempt so far is:
$$|M_{t wedge sigma}|=|B_{t wedge sigma}^4 − 6({t wedge sigma})B^2_{t wedge sigma} +e^{2B_{t wedge sigma}−2({t wedge sigma})}+3({t wedge sigma})^2|leq|B_{t wedge sigma}^4+e^{2B_{t wedge sigma}−2({t wedge sigma})}+3({t wedge sigma})^2|$$
Unsure how to proceed.










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Zugzwangerz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Given $σ =inf(t ≥ 0 : |B_t| =sqrt{3})$, show $sigma$ is optimal for $M_t = B_t^4 − 6tB^2_t +e^{2B_t−2t}+3t^2$ for $t ≥ 0$.

Struggling to show that it is bounded, my attempt so far is:
$$|M_{t wedge sigma}|=|B_{t wedge sigma}^4 − 6({t wedge sigma})B^2_{t wedge sigma} +e^{2B_{t wedge sigma}−2({t wedge sigma})}+3({t wedge sigma})^2|leq|B_{t wedge sigma}^4+e^{2B_{t wedge sigma}−2({t wedge sigma})}+3({t wedge sigma})^2|$$
Unsure how to proceed.







probability-theory brownian-motion martingales






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edited 17 hours ago









Did

246k23221455




246k23221455






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asked Jan 1 at 11:38









Zugzwangerz

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Zugzwangerz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.








  • 2




    What do you mean by "$sigma$ is optimal" ...? When do you call a stopping time optimal?
    – saz
    Jan 1 at 16:55














  • 2




    What do you mean by "$sigma$ is optimal" ...? When do you call a stopping time optimal?
    – saz
    Jan 1 at 16:55








2




2




What do you mean by "$sigma$ is optimal" ...? When do you call a stopping time optimal?
– saz
Jan 1 at 16:55




What do you mean by "$sigma$ is optimal" ...? When do you call a stopping time optimal?
– saz
Jan 1 at 16:55










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