Why Trading Gains from Local Martingales Remain Local Martingales

# 61576


# Why does it hold true that $\theta_{t} d\overline{X}_{t}$ is a local $Q$ martingale if $\overline{X}$ is a local $Q$ martingale












I am learning from Bernt Oksendal's Stochastic Differential Equations and on page 276 Lemma 12.1.6, it is stated that:

The existence of an equivalent martingale measure $Q$ on the discounted price process $\overline{X}$ implies that the market is arbitrage-free.

A statement in the proof has come up, that I cannot justify (I will only look at dimension $1$ for simplicity of notation):

Let $\theta$ be an admissible trading strategy, then we have $d\overline{V}_{t}^{\theta}=\theta_{t}d\overline{X}_{t}$ which is clear by the self-financing property. Next, since $\overline{X}$ is a local $Q$ martingale, it must follow that $\overline{V_{t}}=\overline{V_{0}}+\int_{0}^{t}\theta_{s}d\overline{X}_{s}$ is also a local $Q$ martingale $(*)$.

This last sentence is the part I do not understand. The assumptions that could be relevant are:

$dX_{t}=\mu_{t} dt+\sigma _{t}dB_{t}$

$d\overline{X_{t}}=\overline{X}_{t}((\mu_{t}-r_{t})dt+\sigma_{t}dB_{t})$

and $\theta$ is admissible if it is self-financing, lower bounded and

$\int_{0}^{T}\sigma^{2}_{t}\theta_{t}^{2}dt< \infty \; a.s. $ where $\sigma$ is fixed, progressive and satisfies $\int_{0}^{T}\sigma^{2}_{t}dt< \infty\; a.s.$

Now I have

What I know thus far: If $\Phi$ is a progressive process such that $\int^{T}_{0}(\Phi_{t})^{2}dt<\infty\; a.s.$, then for a brownian motion $B$, the stochastic integral:

$\int_{0}^{t}\Phi_{s}dB_{s}$ is a local martingale on $[0,T]$.

Now to prove $(*)$ I think I need to show $\int^{T}_{0}(\Phi_{t})^{2}dt<\infty\; a.s.$ , which I have been unable to do thus far. And even then, $\overline{X}$ is certainly not necessarily a Brownian motion, so I am lost as to how I can justify why $\theta_{t}d\overline{X}_{t}$ is indeed a local $Q$ martingale. Any ideas that I am missing?