Expected Returns on Black–Scholes Call Options Under GBM
Summary
The document derives the expected value and holding-period return of a European call priced with the Black–Scholes formula when the underlying stock follows geometric Brownian motion. It assumes a constant drift and volatility for the stock, and initially sets the option’s implied volatility equal to that stock volatility. The future stock price is represented using a standard normal shock, which makes the option’s Black–Scholes value at the holding time a function of that shock.
The derivation evaluates the expected call value by integrating over the normal distribution. It gives closed-form expressions for the integrals using the normal cumulative distribution function, then defines expected return as expected future option value divided by its initial price, minus one. The treatment concerns expected value and return; although the question also asks about option volatility, the response does not derive it. Its results depend on the stated model assumptions, and it does not provide empirical validation or address departures from constant-volatility GBM.
Key ideas
- Assume the underlying stock follows geometric Brownian motion with constant drift and volatility.
- The option is repriced at the holding time using Black–Scholes and the assumed volatility.
- The expected option value is obtained by integrating its price over the stock’s normal shock.
- Expected holding-period return is calculated from expected future value relative to the initial option price.
- The answer does not derive the option’s volatility or test the assumptions empirically.
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# Expected return on Black-Scholes priced option?
# Expected return on Black-Scholes priced option?
Suppose we have a European-style call option on some stock, and it was priced according to Black-Scholes. Everybody agrees on the stock's volatility and expected return. What's the expected return (and volatility) of buying this option?
## Answer by RRL (score 6, accepted)
https://quant.stackexchange.com/a/60797
We can obtain a closed-form solution for the expected return over an arbitrary holding period under some typical assumptions.
Assuming geometric Brownian motion with drift $\mu$ and volatility $\sigma$, the stock price at time $t \geqslant 0$ is
$$S(t) = S(0)e^{(\mu - \frac{1}{2}\sigma^2)t}e^{\sigma \sqrt{t} z},$$
where $z \sim \mathcal{N}(0,1)$, a standard normal random variable.
To keep things simple, we will assume that the option implied volatility is $\sigma$, although relaxing this assumption presents no major difficulties. The Black-Scholes price of a call option with expiration at $T >t$, strike $K$, and risk-free interest rate $r$ is
$$C(t) = S(t)N(d) - Ke^{-r(T-t)}N(d - \sigma\sqrt{T-t}),$$
where $N(\cdot)$ is the standard normal CDF, and
$$d = \frac{\log \frac{S(t)}{Ke^{-r(T-t)}}+ \frac{1}{2}\sigma^2(T-t)}{\sigma\sqrt{T-t}} = \frac{\log \frac{S(0)e^{(\mu - \frac{1}{2}\sigma^2)t}}{Ke^{-r(T-t)}}+ \frac{1}{2}\sigma^2(T-t)}{\sigma\sqrt{T-t}} + \sqrt{\frac{t}{T-t}}z\\ = \alpha +\beta z$$
Hence, the expected value of the call price at time $t$ is
$$E(C(t)) = \int_{-\infty}^\infty [S(t)N(d) - Ke^{-r(T-t)}N(d - \sigma\sqrt{T-t})] \frac{e^{-z^2/2}}{\sqrt{2\pi}} \, dz \\ = S(0)e^{(\mu - \frac{1}{2}\sigma^2)t}\int_{-\infty}^\infty e^{\sigma\sqrt{t}z}N(\alpha+\beta z)\frac{e^{-z^2/2}}{{\sqrt{2\pi}}}\, dz + Ke^{-r(T-t)}\int_{-\infty}^\infty N(\alpha'+\beta z)\frac{e^{-z^2/2}}{{\sqrt{2\pi}}}\, dz,$$
where $\alpha' = \alpha - \sigma\sqrt{T-t}$.
Both integrals on the RHS can be evaluated in closed-form. With some effort it can be shown for the second integral that
$$\int_{-\infty}^\infty N(\alpha'+\beta z)\frac{e^{-z^2/2}}{{\sqrt{2\pi}}}\, dz = N(\alpha'/\sqrt{1+ \beta^2})$$
For the first integral we have
$$\sigma \sqrt{t} z - \frac{z^2}{2} = - \frac{1}{2}(z - \sigma\sqrt{t})^2 - \frac{1}{2}\sigma^2t,$$
and, thus,
$$\int_{-\infty}^\infty e^{\sigma\sqrt{t}z}N(\alpha+\beta z)\frac{e^{-z^2/2}}{{\sqrt{2\pi}}}\, dz = e^{-\sigma^2t/2}\int_{-\infty}^\infty N(\alpha+\beta z)\frac{e^{-(z- \sigma\sqrt{t})^2/2}}{{\sqrt{2\pi}}}\, dz \\ = e^{-\sigma^2t/2}\int_{-\infty}^\infty N((\alpha+\beta\sigma\sqrt{t})+\beta u)\frac{e^{-u^2/2}}{{\sqrt{2\pi}}}\, du \\ = e^{-\sigma^2t/2}N((\alpha + \beta\sigma\sqrt{t})/\sqrt{1+ \beta^2})$$
Finally, the expected return of the option over the period from time $0$ to $t$ is
$$\frac{E(C(t))}{C(0)}-1$$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.