Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

We are given a stochastic process $X$ and denote by $\mathbb{P}$ the set of all equivalent local martingale measure, that is the set of all equivalent measures $Q\approx P$, such that $X$ is a local $Q$ martingale. We can assume $\mathbb{P}\not=\emptyset$ and $\mathcal{F}_0$ is trivial. For $t\in \mathbb{R}_+$, $Q\in\mathbb{P} $ let $$A_t:=\{Z|Z \mbox{ is a density process with respect to } Q \mbox{ of some }R\in\mathbb{P}, Z_s=1,s\le t\}$$ Suppose we have $Z^1,Z^2\in A_t$ given and a set $F\in\mathcal{F}_t$. We define $Z^3:=\mathbf1_FZ^1+\mathbf1_{F^c}Z^2$. The claim is, that $Z^3$ is again an element of $A_t$. Clearly $Z^3_s=1$ for $s\le t$. However I'm unsure with other properties. How can I define such a $R$ with the desired properties?

$\mathbf{motivation:}$ The question arises when studying the so called hedging duality in mathemtical finance. There $X$ is a semimartingale and should model a risky asset. For a $H\in L_+^0(\mathcal{F}_T)$ we want to study the process

$$Y_t:=\underset{Q \in \mathbb{P}}{\text{esssup}} \ E_Q \left[ H | \mathcal{F}_t \right]$$ and prove that $(Y_t)$ is a supermartingale for every $Q\in\mathbb{P}$. The question is a part of the proof of this statement.

share|improve this question
    
are you sure you have your t's and s's all straight ? $Z^3_s$ is fine for $s > t$ but its not adapted for $s < t$. I think $F \in \mathcal F_s$ –  mike Apr 12 '13 at 13:32
    
Yes I just checked my notes, again. But maybe it could be a mistake there. For $s\le t$ we have $Z_s^3=1$, which is in $\mathcal{F}_0$ hence no problem. $\mathcal{F}_0$ is trivial. I will edid my question. Or am I wrong? –  user20869 Apr 12 '13 at 13:34
    
I think the questino should be migrated to quant.stackexchange. It would be appreciated if one of the moderator would migrate it. –  user20869 Apr 13 '13 at 7:57

1 Answer 1

up vote 1 down vote accepted

Let $Z$ in $A_t$ denote the density process with respect to $Q$ of some $R$ in $\mathbb P$. Let $G$ and $H$ in $\mathcal F_t$. Assume to simplify things that $X$ is a martingale for $Q$ and $R$ (and not only a local martingale).

Let $s\geqslant t$. Then, $E^Q[X_sZ_s\mathbf 1_G\mid\mathcal F_t]=\mathbf 1_GE^Q[X_sZ_s\mid\mathcal F_t]$. Furthermore, $$E^Q[X_sZ_s\mathbf 1_H]=E^R[X_s\mathbf 1_H]=E^R[E^R[X_s\mid \mathcal F_t]\mathbf 1_H]=E^R[X_t\mathbf 1_H]=E^Q[X_tZ_t\mathbf 1_H]. $$ This holds for every $H$ in $\mathcal F_t$ hence $E^Q[X_sZ_s\mid\mathcal F_t]=X_tZ_t$ and finally, $$ E^Q[X_sZ_s\mathbf 1_G\mid\mathcal F_t]=\mathbf 1_GX_tZ_t. $$ Applying this to $(G,Z)=(F,Z^1)$ and to $(G,Z)=(F^c,Z^2)$ for $Z^1$ and $Z^2$ in $A_t$ and summing the results, one sees that $Z^3=Z^1\mathbf 1_F+Z^2\mathbf 1_{F^c}$ is such that $E^Q[X_sZ^3_s\mid\mathcal F_t]=X_tZ^3_t$ for every $s\geqslant t$, hence $Z^3$ is in $A_t$.

share|improve this answer
    
I'm not quite sure, why this implies the result: For $Z^3$ of the given form, we must prove the existence of $R\in\mathbb{P}$ such that $Z^3$ is the density w.r.t to $Q$ of this $R$. Assume $Z^3$ is the density of such an $R$, then Bayes implies: $E^R[X_s|\mathcal{F}_t]=\frac{1}{Z_t}E^Q[X_sZ_s|\mathcal{F}_t]=X_t$, hence under $R$ $X$ is also a martingale. But why is $Z^3$ a density of an equivalent probability measure w.r.t $Q$? –  user20869 Apr 14 '13 at 9:14
    
I really want to accept your answer. Therefore it would be appreciated if you could explain the problems in my previous comment. Thank you for your help. –  user20869 Apr 20 '13 at 9:53
    
I think the only question which remains is the following: Why is $Z^3$ a $Q$ martingale? If it is, then I can define $dR^3:=Z^3_TdQ$, since $E^Q[Z^3_T]=E^Q[Z^3_0]=1$ –  user20869 Apr 20 '13 at 12:24
    
How would you extend this to $X$ a local martingale? Of course, for a localizing sequence $\{\tau_n\}$ we have $Z_rX_r=\lim_n Z_rX_{\tau_n\wedge r}=\lim E[Z_tX_{\tau_n\wedge t}|\mathcal{F}_r]$ for $r< t$. But why can I put the limit inside? –  user20869 Jun 27 '13 at 14:51

Your Answer

 
discard

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