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Motivation: I want to prove that the existence of a $\sigma$-martingale implies NFLVR (No Free Lunch With Vanishing Risk). This comes from arbitrage theory in mathematical finance and was proved by Delbaen/Schachermayer.

We call $S$ a $\sigma$-martingale if there exists a local martingale $X$ and a predictable, $X$-integrable strictly positive process $\psi$ such that $S=\int\psi dX$. Then you can also define an equivalent $\sigma$-martingale measure in the obvious way.

Suppose $S$ is a semimartingale (under $P$) and it admits an equivalent $\sigma$-martingale measure, i.e. there exists a probability measure $Q\approx P$, s.t. $S$ is a $\sigma$-martingale (under $Q$). I know that there is predictable and $S$-integrable process $\vartheta$ such that $\int_0^t\vartheta_udS_u\ge -a$ for all $t\ge 0$ $P-$a.s. and an $a\ge 0$.

Clearly, by the equivalence of $Q$ and $P$, we have $\int_0^t\vartheta_udS_u\ge -a$ for all $t\ge 0$ $Q-$a.s. Since under $Q$ $S$ admits a $\sigma$-martingale we have:

$$-a\le\int_0^t\vartheta_udS_u=\int_0^t\vartheta_u\psi_udX_u$$

The thing which bothers me is: Is the integral $\int_0^t\vartheta_u\psi_udX_u$ well-defined? Obviously $\psi\vartheta$ is again predictable and $\psi$ is by definition $X$ integrable. But why is the product $\psi\vartheta$ $X$-integrable?

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The tricky thing here seems to be to pin down exactly what is meant by integrability. We can take the statement "$\vartheta$ is $S$-integrable" to mean that $\vartheta$ is predictable and that for each $t\geq 0$

$$ \int_0^t\vartheta_u^2d[S]_u < \infty \text{ a.s.,} $$

since we can readily define the stochastic integral in this case.

By Kunita-Watanabe

$$[S]_u = \int_0^u\psi_s^2d[X]_s,$$

so, by a property of Lebesgue integrals, which follows from a Monotone class argument,

$$\begin{align} \int_0^t\vartheta_u^2\psi_u^2d[X]_u &= \int_0^t\vartheta_u^2d\left(\int_0^u\psi_s^2d[X]_s\right)\\ &= \int_0^t\vartheta_u^2d[S]_u\\ &< \infty, \end{align} $$

since $\vartheta$ is $S$-integrable, so $\vartheta\psi$ is $X$-integrable.

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