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Suppose we have continuous local martingal $L$ given. We define $Z=\mathcal{E}(L)$, the stochastic exponential of $L$. I am interested in finding some condition such that $Z$ defines a density, i.e. I can define an equivalent measure $Q$ by setting $dQ:=Z_\infty dP$.

First of all: $Z$ is again a continuous local martingale, positive hence a supermartingale and $>0$ on $[0,\infty)$. Hence $Z_\infty$ exists. Now I need two assumption, such that everything works out nicely

  1. $E[Z_\infty]=1$, which is equivalent to the property that $Z$ is a $P$-martingale on $[0,\infty]$, hence uniformly integrable.
  2. I have to guarantee that $Z_\infty>0$ $P$-a.s.

For $2.$ we have $Z_\infty =e^{L_\infty -\frac{1}{2}\langle L\rangle_\infty}$. Therefore $L_\infty -\frac{1}{2}\langle L\rangle_\infty \not= -\infty$. Since the bracket process is increasing, we must have $\langle L\rangle_\infty <\infty$. The question is:

Can it happen that $\langle L\rangle_\infty <\infty$ but $L_\infty = -\infty$. If so, are there conditions which excludes $L_\infty=-\infty$ and $\langle L\rangle_\infty =\infty$?

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You're shooting questions with a constant speed :) this is by no means forbidden, but would you think of spending more time also on previous question of yours? –  Ilya Jan 18 '13 at 12:09

1 Answer 1

up vote 0 down vote accepted

Let $L$ be a continuous local martingale with initial value zero. Define

$$ F = \{\omega\in\Omega\mid L_t \textrm{ converges as }t\to\infty\}\\ G = \{\omega\in\Omega\mid \langle L\rangle_t \textrm{ converges as }t\to\infty\}. $$

It then holds that $F$ and $G$ are $P$-almost surely equal, in the sense that $P(F\setminus G)=P(G\setminus F)=0$. This implies that almost surely, it cannot happen that $\langle L\rangle_\infty<\infty$ but $L_\infty=-\infty.$

The proof of the above claim is not completely straightforward. You can find a detailed proof as the solution to Exercise 2.3 (in the appendix) in the following:

http://alexandersokol.dk/StochInt/ContinuousStochasticIntegration.pdf

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