Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 usuual 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:

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

where $\bar X_t = \mathbb E \left [ \bar X | \mathcal F_t\right]$

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

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
    
It's perfect! Thank you very much for your help. –  Paul Feb 14 '13 at 14:18
    
@Ilya: Could you please check the suplementary question I've just added in the text, please? –  Paul Feb 14 '13 at 14:47
    
@Paul: are you sure you have to take $p$ power in the RHS two times? Or you are just asking the question in the math.stackexchange version? –  Ilya Feb 14 '13 at 16:37
    
@Ilya: Inside the article I was reading, it's writen exactelly like that with p power taken two times. Nonetheless, I'm afraid it's a mistake. But, it's a published article so it's difficult to take this conclusion. Obvioslly, by experience most of time when we think there is a mistake in article writen by good researchers,actually there are something we misunderstood. So I assume this posture of humility –  Paul Feb 14 '13 at 17:01
    
@Paul: I'm not sure whether it's true. Let $X\equiv \frac 12$ and let $p = 2$, then LHS is $\frac14$ and RHS is $\frac1{16}$, so your philosophy does not apply at least in such case. –  Ilya Feb 14 '13 at 17:36
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.