Exercise: Black-Scholes Call Price as a Risk-Neutral Expectation

Prerequisites: Risk-Neutral Measure, Log-Normal Distribution, Derivation of the Black-Scholes Formula

Problem

Under the risk-neutral measure $\mathbb{Q}$, the stock price follows geometric Brownian motion:

so $S_T = S_0 \exp\bigl((r - \tfrac{1}{2}\sigma^2)T + \sigma W_T^{\mathbb{Q}}\bigr)$ with $W_T^{\mathbb{Q}} \sim \mathcal{N}(0, T)$.

The price of a European call with strike $K$ and maturity $T$ is

1. Split $\mathbb{E}^{\mathbb{Q}}[(S_T - K)^+] = \mathbb{E}^{\mathbb{Q}}[S_T \mathbf{1}_{\{S_T > K\}}] - K \,\mathbb{Q}(S_T > K)$. Compute $\mathbb{Q}(S_T > K)$ in closed form using the normal CDF $\Phi$. Express your answer in terms of

1. Compute $\mathbb{E}^{\mathbb{Q}}[S_T \mathbf{1}_{\{S_T > K\}}]$ by completing the square inside the Gaussian integral. Express your answer in terms of $S_0, r, T$, and

1. Combine to obtain the Black-Scholes call formula $C_0 = S_0 \Phi(d_1) - K e^{-rT}\Phi(d_2)$.
2. Verify that the $\mu$ (real-world drift) nowhere appears. Explain in one sentence why this is expected from risk-neutral pricing.

Hint

For part 2, write $S_T = S_0 e^{(r - \sigma^2/2)T} e^{\sigma W_T^{\mathbb{Q}}}$ and use the change of variables $z = W_T^{\mathbb{Q}}/\sqrt{T} \sim \mathcal{N}(0, 1)$. Inside the integrand $e^{\sigma\sqrt{T}z} e^{-z^2/2}$, complete the square: $\sigma\sqrt{T}z - z^2/2 = -(z - \sigma\sqrt{T})^2/2 + \sigma^2 T/2$.