I'm trying to solidify my understanding of options pricing and risk neutral distributions.
If the assumptions of the Black-Scholes option pricing were true for an underlying (namely that the future stock price would be described by a lognormal distribution) and one were always able to find an option where the underlying price, the IV, and strike were exactly the same, would the expected return of purchasing (or selling) this option over and over again, be equal to zero? (Maybe if not exactly zero, it is the forward price of the option adjusted by the risk-free rate.)
For example, let's say I buy a weekly call one week to expiration. Every week, I can find some stock that is trading at 100 and has an IV of 15%. Every week I buy the 100 strike option. If the risk free rate is zero, we can say that 50% of the time the call expires OTM and 50% ITM. What is the expected returned (within the assumptions of BS) of this strategy? I think the answer is zero. Is that correct? In other words, the times we make money exactly balances the times the call expires worthless.
EDIT: So I thought I'd try to approach the question with a simulation. What I did was perform a Monte Carlo simulation. My understanding is that BS assumes the future price will follow: $$ S = S_0 \exp\left(\mu t - \frac{\sigma^2}{2} t + \sigma \sqrt{t} u \right) $$ where $u$ is a draw from a standard normal distribution. If I set $\mu=0$ and calculate $S$ for 50,000 different $u$'s, I can then calculate the value of the call at expiry by $\max(S-K,0)$.
Following the hypothetical above:
- Initial price S0 = 100
- Strike K = 100
- Risk free rate r = 0
- Time to expiry T = 7/365
The BS equations will value this call at 0.829. The Monte Carlo simulation agrees to within two decimals.
What about the risk free rate? If we set it to 3%, BS value will increase to 0.857. I can make the Monte Carlo match if I set $\mu$ in the equation above to $r$.
Still interested in comments as to whether I'm approaching this correctly.