No-Arbitrage Constraints for Assets with Shared Brownian Risk

Summary

The document questions whether two assets can follow geometric Brownian motions with the same volatility and Brownian driver but different expected drifts in an arbitrage-free market. Because their random shocks are identical, a portfolio holding offsetting exposures can remove the shared source of uncertainty. If the drifts differ, the remaining relative movement raises an arbitrage concern under the stated assumptions.

The question also examines the risk-neutral measure: one change of measure must make both discounted asset prices martingales, but the required Brownian drift adjustment differs when the assets have different physical drifts. That inconsistency indicates that the assumptions cannot all hold in the standard setting. The document presents the puzzle rather than a full resolution; conclusions depend on market completeness, tradability, financing, and whether the assets truly share the same risk exposure.

Key ideas

• A shared Brownian driver and equal volatility give the two assets the same source of instantaneous risk.
• Different physical drifts imply different changes of measure would be needed to make discounted prices martingales.
• Under the stated tradability assumptions, that inconsistency signals a violation of no-arbitrage conditions.
• The conclusion depends on the assets’ actual risk exposures and market assumptions.

Tags

# Stocks with same volatility but different drifts


# Stocks with same volatility but different drifts












In the book Quant Job Interview Questions & Answers, in section 2, question 2.4 says suppose two assets in a Black-Scholes world have the same volatility but different drifts. How will the price of call options on them compare? In the answer, it's assumed that the two asset prices follow $$dS_t^1=\mu_1S^1_tdt+\sigma S_t^1dW_t$$ $$dS_t^2=\mu_2S_t^2dt+\sigma S_t^2dW_t$$ where $\mu_1\neq \mu_2$. My question is whether this is even an arbitrage-free model? If we know that $\mu_1<\mu_2$, can't we have a strategy of buying $S^2$ and shorting $S^1$ to make a riskless profit? In particular, if I would like to find a risk neutral measure, then I need the discounted prices $e^{-rt}S_t^1$ and $e^{-rt}S_t^2$ to be martingales. Since $$d(e^{-rt}S_t^1)=e^{-rt}S_t^1((\mu_1-r)dt+\sigma dW_t)$$ $$d(e^{-rt}S_t^2)=e^{-rt}S_t^2((\mu_2-r)dt+\sigma dW_t),$$ this would mean the Brownian motion under the risk neutral measure would have $$d\tilde{W_t}=dW_t+\frac{\mu_1-r}{\sigma}dt=dW_5+\frac{\mu_2-r}{\sigma}dt$$ which implied $\mu_1=\mu_2$. Am I missing something here? How can this model be valid?

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.