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?

  • 4
    $\begingroup$ If the underlying stock has a market beta of $\beta_S$ and a volatility of $\sigma_S$, then you only need to compute the option's elasticity (or lambda), i.e. $\Omega=\frac{\partial V}{\partial S}\frac{S}{V}$. The option's beta is then $\Omega\beta_S$ and the volatility of the option's return is $|\Omega|\sigma_S$. $\Omega$ is greater than one for call options (they're riskier than their underlying) but less than zero for put options (which act as insurances). See also this answer $\endgroup$
    – Kevin
    Commented Jan 28, 2021 at 19:41

1 Answer 1


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



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