Why Black–Scholes Requires No-Arbitrage, Not No Statistical Arbitrage
Summary
The discussion distinguishes the absence of risk-free arbitrage, which underpins Black–Scholes pricing, from the stronger claim that no strategy can have a positive expected payoff. A trade with a high probability of profit but a small probability of loss is not risk-free arbitrage, even if it might look attractive statistically. The answers explain that Black–Scholes does not specify the real-world drift and volatility values, and that its pricing result is conditional on the assumed price process.
The key pricing insight is that the model can determine an option price without knowing the asset’s expected return, so it need not rule out every favorable statistical opportunity. The discussion also notes that implausible parameter combinations could imply unusually attractive trades in the underlying itself, while real markets need not follow geometric Brownian motion exactly. This is a conceptual explanation rather than empirical evidence: it does not establish whether statistical arbitrage exists in practice, and its conclusions depend on the model assumptions and the meaning of arbitrage.
Key ideas
- Black–Scholes relies on the absence of risk-free arbitrage rather than the absence of every positive-expectation strategy.
- A strategy with likely gains and occasional losses is not risk-free arbitrage.
- The model’s option pricing result does not require specifying the asset’s expected return.
- The model is conditional on its assumptions and does not claim that real prices follow geometric Brownian motion.
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# Why doesn't Black-Scholes assume the absence of statistical arbitrage?
# Why doesn't Black-Scholes assume the absence of statistical arbitrage?
Both Black-Scholes and binomial model assume that there's no risk-free arbitrage in the market. But that sounds like a very weak condition.
If a trading scheme makes you gain 100 dollars with 99% probability and lose 5 dollars with 1% probability (starting from 0), this is not risk free but it will be surprising if such an arbitrage opportunity exists without being exploited.
So, either a) Black-Scholes does not make accurate predictions about the prices of derivatives, or b) statistical arbitrage of the type mentioned above does actually exist in the Black-Scholes Model. Which one of the two is correct?
## Answer by Mark Joshi (score 12, accepted)
https://quant.stackexchange.com/a/15600
Neither.
Black--Scholes says nothing about the parameter values: $\mu$ and $\sigma.$
A very large $\mu$ and very small $\sigma$ is very unlikely to actually occur in the market and if it did you could make money with high probability without using option contracts.
BS simply says that if the market follows a certain process then a certain option price is enforcable by no arbitrage. It says nothing about the existence or non-existence of very good deals.
In practice, no one knows $\mu$ and so the great strength of BS is that you don't need it to price.
Essentially, it doesn't outlaw silly parameter values because it doesn't need to. However, that doesn't mean that such parameters will occur in a real market. In any case, no one believes that stock prices follows geometric BM. The question is whether the model is good enough to be useful and it is.
## Answer by Ulysses (score 0)
https://quant.stackexchange.com/a/15604
Absence of statistical arbitrage is a stronger condition that usual NA condition $$ \nexists \varphi\in \Phi: V(\varphi)>0 \text{ and } \mathbb EV_T(\varphi)>0 \text{ for some } T\geq0 \tag{1} $$ since as soon as there exists an admissible trading strategy $\varphi\in \Phi$ satisfying $(1)$, it readily provides statistic arbitrage. Now, in BS the weaker condition $(1)$ is already sufficient to show that option prices can be uniquely determined, so why then impose stronger assumptions on the model if they are not necessary.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.