# 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?

• 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 Commented Jan 28, 2021 at 19:41

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

$$\frac{E(C(t))}{C(0)}-1$$