= Solution
The price process is a <geometric Brownian motion> with
$$
\frac{dS_t}{S_t}=
\left(\mu+\frac12\sigma^2\right)dt+\sigma dB_t.
$$
Put
$$
\vartheta=\frac{\mu+\sigma^2/2-r}{\sigma}
$$
and define the equivalent measure $Q$ on $\mathcal F_T$ by the <exponential Brownian martingale>
$$
\frac{dQ}{d\mathbb P}
=\exp\left(-\vartheta B_T-\frac12\vartheta^2T\right).
$$
The <Cameron-Martin-Girsanov theorem> says that $W_t^Q=B_t+\vartheta t$ is Brownian motion under $Q$. Therefore
$$
S_t=S_0\exp\left(\sigma W_t^Q+
\left(r-\frac12\sigma^2\right)t\right),
$$
so $e^{-rt}S_t$ is a $Q$-martingale. This is the unique <equivalent martingale measure>, since cancellation of the single Brownian risk forces the displayed drift shift.
By <risk-neutral valuation>,
$$
\pi_C=e^{-rT}\mathbb E_Qf(S_T).
$$
Writing $W_T^Q=\sqrt T Y$ with $Y\sim N(0,1)$ gives
$$
\boxed{
\pi_C=e^{-rT}\int_{-\infty}^{\infty}
f\left(S_0e^{\sigma\sqrt T y+(r-\sigma^2/2)T}
\right)
\frac{e^{-y^2/2}}{\sqrt{2\pi}} dy.}
$$
Back to article page
