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The following problem comes up reading some notes on utility maximization in mathematical finance. Since it is a purely stochastic process problem, I place it in this forum. The time index is covering $[0,T]$. We are given a process $X$, non negative and a set of processes $\mathcal{V}$. It is not important what condition are made on the processes from $\mathcal{V}$. The process $XV$ is a supermartingale for every $V\in\mathcal{V}$ over the dense subset $\mathbb{Q}\cap [0,T]$. A standard construction from martingale theory, e.g. Dellacherie Meyer yields the existence of a $RCLL$ supermartingale $Z$, that coincides on the rationals with $X$. For general reals, one takes $Z_t:=\lim_{q\downarrow t}X_q$, where $q\in\mathbb{Q}$. Moreover $Z_T:=X_T$. How can I extend the supermartingale property of $XV$ to $ZV$ over $[0,T]$? So I want to prove that $ZV$ is a supermartingale over $[0,T]$ for every $V\in\mathcal{V}$

We have to establish the supermartingale property $E[Z_tV_t|\mathcal{F}_s]\le Z_sV_s$. If $s$ is rational everything is clear. So we just have to consider the case where $s$ is not rational.

The problem is about the $\mathcal{F}_s$. Here I do not see how I can simplify things to the rational case. Thanks in advance for your help.

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@ math let me sum up: you have XV is a SM (shorthand for supermartingale) over rationals for every V then (Dellacherie Meyer theorem) there exists a process Z that coincides with X over rationals and with the SM property over reals. Now you want to know if ZV is SM over reals correct ? –  TheBridge Aug 4 '13 at 7:03
    
@TheBridge yes exactly –  math Aug 4 '13 at 10:18
    
What is your definition of $\mathcal{F}_s$ for $s\in[0,T]\cap \mathbb{Q}^c$? Is it the generated filtration for the rational $s$? –  Dima McGreen Aug 7 '13 at 12:19
    
@DimaMcGreen Sorry if this was not clear: We have given a filtration $\{\mathcal{F}_t:t\in[0,T]\}$, and $XV$ is then just a supermartingale w.r.t to $\{\mathcal{F}_s:s\in[0,T]\cap \mathbb{Q}\}$ –  math Aug 8 '13 at 6:34
    
Since the bounty does not help to solve this problem, I also asked it on QSE. –  math Aug 15 '13 at 6:15

1 Answer 1

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Maybe I'm overlooking something, but can't you just use this theorem again for $ZV$? It is easy to show that $ZV$ is again a supermartingale over $\mathbb{Q}\cap [0,T]$. Now we could apply the theorem of Dellacherie Meyer again on this pair of processes to get a RCLL supermartingale over the entire filtration. I think you will need for the theorem from Dellacherie Meyer right continuity of the filtration.

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