No-Arbitrage Conditions for Drift and Volatility in a Diffusion Market
Summary
The document considers a single risky asset whose price follows a diffusion with deterministic drift and volatility, in a market with a zero interest rate. It asks what absence of arbitrage implies when volatility vanishes at some times, and whether the drift measure must be dominated by the volatility measure so that drift can be written as volatility multiplied by a measurable function.
The proposed reasoning is that a non-negligible interval or set with zero volatility and nonzero drift would permit a position aligned with the drift to earn a gain without exposure to randomness, creating an arbitrage. The answer notes that admissibility of such a trading strategy still needs justification. Once the zero-volatility, nonzero-drift set is negligible, it invokes Lebesgue decomposition to establish the stated absolute-continuity relationship and representation. The response is brief and does not supply the full measure-theoretic proof or formal admissibility conditions.
Key ideas
- With zero interest, nonzero drift during periods of zero volatility can create an arbitrage opportunity.
- The set where volatility is zero and drift is nonzero must have zero time measure under the stated no-arbitrage premise.
- The answer relies on admissibility of the proposed trading strategy, but leaves its details open.
- Lebesgue decomposition is used to motivate expressing drift as volatility times a measurable function.
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Full text
# Non-arbitrage theory and existence of a risk premium
# Non-arbitrage theory and existence of a risk premium
Consider a probability filtred space $(\Omega, \mathcal F, \mathbb F, \mathbb P)$, where $\mathbb F = (\mathcal F_t)_{0\leq t\leq T}$ satisfing the habitual conditions and isgenerated by $1 d $- Brownian Motion (with $\mathcal F_T = \mathcal F$).
Also, consider a finantial market where the interest rate is nul, $r=0$, and the dynamics of the risky asset $S$ is given by $$S_t= S_0 + \int_0^t \mu_s ~ds +\int_0^t \sigma_s ~dW_s \quad , t \geq 0$$
where $t \in [0,T] \mapsto \mu_t$ and $t \in [0,T] \mapsto \sigma_t \geq 0$ are deterministic and continuous functions.
Show that:
- If the absence of arbitrage opportunity hypothesis is verified, then $B:=\{t \in [0,T] : \sigma_t=0 \ \text{and} \ \mu_t \neq 0\}$ is a Lebesgue nul-measure set (ie, $\int_0^T\mathbf1_{t \in B} dt=0$).
- $\nu_\sigma(O):= \int_0^T\mathbf1_{t \in O}\sigma_t ~dt$ dominates $\nu_\mu (O):= \int_0^T\mathbf1_{t \in O} \mu_t ~dt$, where $O$ is a borelian of $[0,T]$ ( ie, $\nu_\mu \ll \nu_\sigma$) and deduce from it that there is a measurable function $\lambda$ such that $\mu = \sigma \lambda$.
## Answer by TheBridge (score 4, accepted)
https://quant.stackexchange.com/a/7701
For the first one absurd reasoning allows you to construct an arbitrage (as r=0) by investing (or short selling according to the sign of $\mu$) at the time where $\sigma$ is null, or if you prefer as soon as $t$ is in $B$ (which is not a Lebesgue negligible set by hypothesis) which is absurd as no-arbitrage holds. The details that remain to be proved is that such a strategy is an admissible one.
For the second question, as soon as the first part is done, it is only the application of Lebesgue's decomposition theorem.
For the first part as any borelian set such that $\nu_\sigma(O):= \int_0^T\mathbf1_{t \in O}\sigma_t dt $ doesn't dominate $\nu_\mu (O):= \int_0^T\mathbf1_{t \in O} \mu_t dt$ is included in $B$ and as $B$ is of null probability the conclusion holds true.
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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.