Black–Scholes Pricing and the Expected Return on Options

# Assuming Black-Scholes assumptions are correct, would the expected return from buying/selling options be 0?


# Assuming Black-Scholes assumptions are correct, would the expected return from buying/selling options be 0?












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.

## Answer by Mark Joshi (score 3)

https://quant.stackexchange.com/a/21563

BS does not require this.

The real-world drift of the stock can be greater than the risk-free rate so on average you make money.

If this seems weird, forget the option and consider the stock. If you buy it and hold it for a week then you ought to make money on average about the risk-free rate since you get a risk compensation for the fact you can lose money.

## Answer by Louis. B (score 1)

https://quant.stackexchange.com/a/21564

This is entirely true. The basic pricing formula that is intended to work for all assets (including options) is \begin{equation} P=E[m*X] \end{equation}

where $P$ is the price, $m$ the discount factor, and $X$ the payoff. This can also be rewrite \begin{equation} 1=E[m*R] \end{equation}

with $R$ the return. This is know as the Euler Equation.

In the Black and Scholes world the Stochastic Discount Factor is the risk-neutral probability. So, in the Black and Scholes world, this is true that if you buy any option (paying the Black and Scholes price), you will not make any money on average. Of course, the relation holds on average, this means that you can make (or loose) money for a given option, but if you repeat your strategy an infinite number of time, you will make exactly nothing.

You can indeed think about that by saying that realizations in which you win exactly balance the realizations in which you loose (loose the price you paid for the option).

In real life, volatility is not constant (unlike in Black and Scholes world) so the global payoff of following your strategy will depend on the difference between the realized volatility and the volatility used to price your option, that is the implied volatility.

Keep in mind that this is true only if you pay the Black Scholes price, in general there is a cost of trading option, that is the implied volatility used to price the option you're going to buy/sell is different from the market's one (sort of bid/ask spread). So in this case, if we assume no time-varying volatility, you will basically loose money on average with your strategy.

## Answer by Frakel.X (score 1)

https://quant.stackexchange.com/a/21568

To get the answer, you should know the difference between forward and futures. If all options in your strategy will not be really settled, instead, just an P&L is marked, then you will find in the long run your return is 0. This is similar to a forward contract. However, if these options are settled, you will get a realized P&L which is MtM weekly. This is like a futures contract. The difference should be your realized return.

## Answer by Kiwiakos (score 1)

https://quant.stackexchange.com/a/21584

No. Under BS you get rewarded for the risk you bear, and the risk premium (ie the instantaneous excess return over the risk free rate) is $\frac{\mu-r}{\sigma}\ \sigma^*_t\ dt$ where $\sigma^*_t$ is the instantaneous vol of your position.

For a simple stock position the instantaneous vol is constant, $\sigma^*_t=\sigma$ therefore you have a constant excess return of $(\mu-r)\ dt$ at all times.

For an option the instanteneous vol is not constant, as the option moves in and out of the money, therefore you need to integrate over all paths/ensemble. However it is always positive.