From Geometric Brownian Motion to the Black-Scholes Call Formula

From Stock Dynamics to a Pricing Expectation

Assume a stock has constant volatility and is modelled under a physical probability measure \(\mathbb{P}\) by geometric Brownian motion. Pricing a European call requires a change of perspective: under the risk-neutral measure \(\mathbb{Q}\), the stock's drift is the constant interest rate \(r\), not its physical expected return. The option price is the discounted \(\mathbb{Q}\)-expectation of its terminal payoff. We derive that expectation, evaluate it, and verify the resulting function solves the pricing PDE.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Suppose under \(\mathbb{P}\) the stock satisfies \(dS_t=\mu S_t\,dt+\sigma S_t\,dB_t\), where \(S_0>0\), \(\sigma>0\), and \(\mu,r\) are constants. For each finite horizon \(T\), the integrand \(\sigma S_t\) is progressively measurable and belongs to \(\mathcal{L}^2\): the explicit solution below gives \(\mathbb{E}\int_0^T \sigma^2S_t^2\,dt<\infty\). Thus its Ito integral is defined in the usual square-integrable sense.

Set \(\theta=(\mu-r)/\sigma\). The exponential density \(Z_T=\exp(-\theta B_T-\tfrac12\theta^2T)\) has expectation one. In fact, Novikov's condition \(\mathbb{E}\exp(\tfrac12\int_0^T\theta^2\,dt)<\infty\) holds because its left side is \(e^{\theta^2T/2}<\infty\). Define \(\mathbb{Q}\) by \(d\mathbb{Q}=Z_T\,d\mathbb{P}\). The constant-kernel case of Girsanov's theorem gives a \(\mathbb{Q}\)-Brownian motion \(B_t^{\mathbb{Q}}=B_t+\theta t\), and substitution in the stock equation gives \(dS_t=rS_t\,dt+\sigma S_t\,dB_t^{\mathbb{Q}}\). The measures are equivalent on this finite horizon. If the model is specified directly under \(\mathbb{Q}\), this risk-neutral stock equation is the starting assumption.

Under \(\mathbb{Q}\), solving the stock equation gives \(S_T=S_t\exp((r-\tfrac12\sigma^2)\tau+\sigma(B_T^{\mathbb{Q}}-B_t^{\mathbb{Q}}))\), where \(\tau=T-t\). The \(-\tfrac12\sigma^2\tau\) term is Ito's correction. Conditional on \(\mathcal{F}_t\), the Brownian increment is independent of the past and has law \(N(0,\tau)\).

For completeness, applying Ito's formula to \(f(x)=\log x\) gives \(f'(x)=1/x\), \(f''(x)=-1/x^2\), and the correction \(\tfrac12 f''(S_t)(\sigma S_t)^2dt=-\tfrac12\sigma^2dt\). Hence \(d\log S_t=(r-\tfrac12\sigma^2)dt+\sigma dB_t^{\mathbb{Q}}\), which yields the displayed solution. The stochastic integral has an integrand in \(\mathcal{L}^2\) on finite intervals: under \(\mathbb{Q}\), \(\mathbb{E}^{\mathbb{Q}}[S_u^2\mid\mathcal{F}_t]=S_t^2e^{(2r+\sigma^2)(u-t)}\), finite for finite \(u-t\).

Evaluating the Lognormal Payoff

For \(t<T\), write the conditional price as an ordinary normal integral. Define \(d_1=\dfrac{\log(s/K)+(r+\tfrac12\sigma^2)\tau}{\sigma\sqrt{\tau}}\) and \(d_2=d_1-\sigma\sqrt{\tau}=\dfrac{\log(s/K)+(r-\tfrac12\sigma^2)\tau}{\sigma\sqrt{\tau}}\), where \(s=S_t>0\). Let \(\phi(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\) be the standard normal density and \(\Phi(x)=\int_{-\infty}^x\phi(z)\,dz\) its CDF. The event \(S_T>K\) is the event \(Z>-d_2\), with \(Z=(B_T^{\mathbb{Q}}-B_t^{\mathbb{Q}})/\sqrt{\tau}\sim N(0,1)\).

Proof. Split the positive-part payoff according to whether \(S_T>K\). The conditional expectation is \(e^{-r\tau}\mathbb{E}^{\mathbb{Q}}[S_T\mathbf{1}_{\{S_T>K\}}\mid\mathcal{F}_t]-Ke^{-r\tau}\mathbb{Q}(S_T>K\mid\mathcal{F}_t)\). The probability in the second term is \(\mathbb{P}(Z>-d_2)=\Phi(d_2)\), using the symmetry \(\phi(z)=\phi(-z)\). In the first term, insert the explicit stock solution. Its discounted conditional expectation becomes \(s\int_{-d_2}^{\infty}e^{-\sigma^2\tau/2+\sigma\sqrt{\tau}z}\phi(z)\,dz\). Complete the square: \(e^{-\sigma^2\tau/2+\sigma\sqrt{\tau}z}\phi(z)=\phi(z-\sigma\sqrt{\tau})\). Changing variables to \(y=z-\sigma\sqrt{\tau}\), the integral is \(s\int_{-d_2-\sigma\sqrt{\tau}}^\infty\phi(y)\,dy=s\Phi(d_2+\sigma\sqrt{\tau})=s\Phi(d_1)\). Combining the two terms proves the formula. At \(t=T\), the conditional payoff is exactly \((s-K)^+\), proving the terminal statement as well. \(\square\)

Why the Formula Solves the Black–Scholes PDE

The formula also satisfies a differential equation in time and current stock price. The following verification is for \(s>0\) and \(t<T\), so \(\tau>0\) and all the displayed derivatives exist. A key identity is \(s\phi(d_1)=Ke^{-r\tau}\phi(d_2)\). To check it, calculate \(d_1^2-d_2^2=(d_1-d_2)(d_1+d_2)=2\sigma\sqrt{\tau}\,d_2+\sigma^2\tau=2\log(s/K)+2r\tau\). Consequently \(\phi(d_1)/\phi(d_2)=e^{-(d_1^2-d_2^2)/2}=(K/s)e^{-r\tau}\), which gives the identity.

Proof. Differentiating the formula with respect to \(s\), using \(\partial_s d_1=\partial_s d_2=1/(s\sigma\sqrt{\tau})\) and \(s\phi(d_1)=Ke^{-r\tau}\phi(d_2)\), gives \(C_s=\Phi(d_1)\): the two density terms cancel. Differentiating again gives \(C_{ss}=\phi(d_1)/(s\sigma\sqrt{\tau})\). For the time derivative, first differentiate in \(\tau\). The terms involving derivatives of \(d_1,d_2\) combine using the density identity and \(\partial_\tau(d_1-d_2)=\sigma/(2\sqrt{\tau})\). Hence \(C_\tau=\tfrac12\sigma s\phi(d_1)/\sqrt{\tau}+rKe^{-r\tau}\Phi(d_2)\), and \(C_t=-C_\tau\). Now compute the PDE's spatial and discount terms: \(rsC_s+\tfrac12\sigma^2s^2C_{ss}-rC =rs\Phi(d_1)+\tfrac12\sigma s\phi(d_1)/\sqrt{\tau}-r[s\Phi(d_1)-Ke^{-r\tau}\Phi(d_2)]\). The \(rs\Phi(d_1)\) terms cancel, leaving exactly \(C_\tau\). Since \(C_t=-C_\tau\), the PDE follows. The terminal condition was established from the payoff at \(\tau=0\). \(\square\)

This PDE verification is also consistent with Ito's formula. Under \(\mathbb{Q}\), apply Ito's formula to \(e^{-rt}C(t,S_t)\). For the function \(f(t,s)=e^{-rt}C(t,s)\), its derivatives are \(f_t=e^{-rt}(C_t-rC)\), \(f_s=e^{-rt}C_s\), and \(f_{ss}=e^{-rt}C_{ss}\). The second-order correction is explicitly \(\tfrac12 f_{ss}(\sigma S_t)^2dt=\tfrac12e^{-rt}\sigma^2S_t^2C_{ss}dt\). Thus \(d(e^{-rt}C(t,S_t))=e^{-rt}\sigma S_tC_s(t,S_t)\,dB_t^{\mathbb{Q}}\), because the drift is the PDE's left side multiplied by \(e^{-rt}\), hence zero. The integrand is in \(\mathcal{L}^2\) on any finite interval before maturity: \(0\le C_s=\Phi(d_1)\le1\) and \(\mathbb{E}^{\mathbb{Q}}\int_0^T S_t^2dt<\infty\). The Ito integral is therefore a true martingale, not merely a local martingale. Its conditional expectation property gives the discounted risk-neutral expectation used for pricing.

Common Pitfall: Keeping the Physical Drift

A frequent error is to use \(\mu\) in the risk-neutral call formula simply because the stock's physical SDE has drift \(\mu\). For example, with \(s=K=1\), \(\tau=1\), \(\sigma=1\), \(r=0\), and \(\mu=1\), the incorrect substitution would set \(d_2=(\mu-\tfrac12)=\tfrac12\). But under \(\mathbb{Q}\), the drift is \(r=0\), so the correct value is \(d_2=-\tfrac12\), as in the first worked example. Using the wrong drift prices the payoff under \(\mathbb{P}\), not the discounted risk-neutral expectation. The correction is not to adjust the payoff; it is to use the risk-neutral dynamics and measure consistently.

Another common error is to omit Ito's \(-\tfrac12\sigma^2\tau\) term from the exponential solution. That changes the threshold for exercise and the lognormal integral. The independent PDE check is useful: a formula derived with the wrong correction will generally fail the displayed Black–Scholes equation or its risk-neutral conditional-expectation interpretation.