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Consider, on a filtred probability space $ \left (\Omega, \mathcal F, \mathbb F , \mathbb P \right )$ where $ \mathbb F = \left(\mathcal F_ t \right )_ {t\geq 0}$ is filtration satisfying the habitual conditions, a non-negative continuous process $X = \left (X_t \right)_ {t\geq 0}$ satisfying $ \mathbb E \left \{ \bar X \right\}< \infty $ (where $ \bar X =\sup _{0\leq t \leq T} X_t $) and its Snell envelope

$$ \hat {X_\theta} = \underset {\tau \in \mathcal T _{\theta,T} } {\text{ess sup}} \ \mathbb E \left\{ X_\tau | \mathcal F_\theta \right \}$$

where $\mathcal T _ {\theta, T} := \{\tau \quad \mathbb F -\text{stopping time}: \tau \in [0, T] \quad \text{and} \quad \tau \geq \theta \quad \mathbb P -\text{a.s.} \}$ and $T \in \mathbb R_+$ is a deterministic constant.

I'd like to understand how justify the following inequality:

$$\mathbb E \left\{ \sup_{0\leq t \leq T} \hat X_t^p\right \} \leq \mathbb E \left\{ \sup_{0\leq t \leq T} \bar X_t^p\right \} $$

where $\bar X_t = \mathbb E \left\{ \bar X | \mathcal F_t \right \}$

Suplementary question Justify the following inequality: $$\mathbb E \left [ \sup_{0\leq t\leq T} \hat X_t^p \right] \leq \mathbb E \left [ \sup_{0\leq t\leq T} \mathbb E \left [ \bar X ^p| \mathcal F_t\right] ^p\right] $$

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    $\begingroup$ What is $\mathcal T_{\theta,T}$? $\endgroup$ Commented Feb 13, 2013 at 22:49
  • $\begingroup$ @DavideGiraudo : See edits, please. $\endgroup$
    – Paul
    Commented Feb 14, 2013 at 13:06
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    $\begingroup$ Cross-post on MO. I answered your question there $\endgroup$
    – SBF
    Commented Feb 14, 2013 at 14:10
  • $\begingroup$ Perfect answer! Thank you very much for your help. Please, answer here also just to help people in the future. $\endgroup$
    – Paul
    Commented Feb 14, 2013 at 14:20

1 Answer 1

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I think it is just a sequence of much stronger inequalities that goes as follows. All the inequalities below are $\Bbb P$-a.s.

  1. Let $\tau\in \mathcal T_{t,T}$ be an arbitrary stopping time, then $$ (X_\tau)(\omega) = (X_{\tau(\omega)})(\omega)\leq \sup_{t\leq s\leq T} X_s(\omega)\tag{1} $$ since $\Bbb P(\tau\in [0,T]) = 1$.

  2. As the latter term in the RHS of $(1)$ is smaller or equal to $\bar X$, we obtain $ X_\tau\leq \bar X$ for all $\tau\in \mathcal T_{t,T}$ and thus $\Bbb E[X_\tau|\mathcal F_t]\leq\Bbb E[\bar X|\mathcal F_t]$.

  3. As a result, we have that $\operatorname{esssup}\limits_{\tau\in \mathcal T_{t,T}}\;\Bbb E[X_\tau|\mathcal F_t]\leq \Bbb E[\bar X|\mathcal F_t]$ which in your notation is $\hat X_t\leq \bar X_t$.

  4. Applying to $\hat X_t\leq \bar X_t$ $\Bbb P$-a.s. first $\sup_{0\leq t\leq T}$ and then expectation shall make it.

P.S. I have not dealt with esssup over stopping times for a while, so I hope step $3$ is correct - but better if you double-check it.

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