# 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?

• – ir7 Aug 6 '20 at 2:12
• Sorry I don't think this answers the question. The model in the question is such that the two assets have the same volatility but different drifts. – Xiaohuolong Jan 22 at 5:36