# 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?

• Black-Scholes is known to be inaccurate, but not for this reason. Can you show an example where Black-Scholes behaves poorly for statistical arbitrage? Lack of statistical arbitrage is implicit in Black-Scholes and in the market. – user59 Nov 22 '14 at 19:43
• Not sure exactly what an example would look like, but say you are taking a short position on an option and you want to hedge it by borrowing money and buying stocks. You might be able to get lower interest rates on your loan if you're willing to take a slight risk; a lower interest rate will imply a lower price for the option. The Binomial model assumes you don't want to take any risk whatsoever, which seems unrealistic, and hence it should make incorrect predictions. – Vinayak Pathak Nov 22 '14 at 23:40

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.

• 'In any case, no one believes that stock prices follows geometric BM' is a very strong statement considering BS is based on GBM, isn't it? – wildbunny Apr 1 '19 at 10:59
• @wildbunny Download some daily stock prices for S&P500 from yahoo finance and try to fit a normal distribution to the log daily returns and run any normality test on the data. I hope that convinces you of Mark Joshi is saying. – baibo Dec 28 '19 at 14:33

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.

• Because the aim is not to just find a formula, but to find one that represents the real world to a close approximation. – Vinayak Pathak Nov 25 '14 at 18:29