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Let $\{X_i\}$ be a sequence sequence of nonnegative r.v. which has the lattice property. This implies that there is a sequence of indices $\{i_n\}$ so that $\{X_{i_n}\}$ is nondecreasing and $\operatorname{esssup}_{i\in I}X_i= \sup_n{X_{i_n}}= \lim_nX_{i_n} $ a.s. Note that the set $I$ is general an therefore it could be uncountable.

I would like to show $E[\operatorname{esssup}_{i\in I}X_i]=\sup_{i\in I}E[X_i]$.

I tried to apply monotone convergence together with the existence of a sequence $\{i_n\}$:

$$E[\operatorname{esssup}_{i\in I}X_i] = E[\lim_nX_{i_n}]= \lim_nE[X_{i_n}]\le \sup_{i\in I}E[X_i]$$.

However, I'm struggling with the other inequality mainly for two reason: a priori $\sup_{i\in I}X_i \ge \sup_n X_{i_n}$ and how to show $ \lim_nE[X_{i_n}]\ge \sup_{i\in I}E[X_i]$

math
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  • The main problem is that the sequence $(i_n)$ depends on $\omega$ - and this means that your argumentation doesn't work. – saz May 02 '15 at 12:55
  • @saz so how can one proof this? its from this lecture note, proposition 1.1.14 http://www.fmf.uni-lj.si/finmath09/ShortCourseAmericanOptions.pdf – math May 02 '15 at 15:54
  • @ saz : Lattice property is stronger than an $\omega$ by $\omega$ property it is an a.s. property from what I get from the lecture notes, so for almost all $\omega$ $i_n$ has the same value, so that the line of argument of user8 is ok. Finishing the job for the reverse equality is quite simple and result form the property of integration. You have $\forall i\in I, essup_I X_i \ge X_i$ almost surely taking expectation on both sides (which makes sense as we work with positive random variables) gives plainly : $E[essup_I X_i] \ge E[X_i]$ for all $i\in I$ which is the desired claim. Best regards. – TheBridge May 03 '15 at 14:07
  • @TheBridge ah stupid me! many thanks! – math May 03 '15 at 14:56

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