## European call under geometric drift and arithmetic Brownian motion

Let $\mu,\alpha\in\mathbb{R}$ with $\alpha\geq0$. Consider a process following $$S\left(t\right)=S\left(0\right)+\int_{0}^{t}\mu S\left(s\right)ds+\int_{0}^{t}\alpha dW\left(s\right).$$ Note that the Ito integrand is constant (i.e. it is not the usual $\sigma S$ seen in financial applications). Ito's lemma applied to the pair $S$ and $f\left(t,x\right)\equiv e^{-\mu t}x$ yields $$S\left(t\right)=e^{\mu t}\left(S\left(0\right)+\int_{0}^{t}e^{-\mu s}\alpha dW\left(s\right)\right).$$ This suggests that the mean and variance of $S\left(t\right)$ are given by $$\mathbb{E}S\left(t\right)=e^{\mu t}S\left(0\right)$$ and $$\operatorname{Var}S\left(t\right)=e^{2\mu t}\int_{0}^{t}\left(e^{-\mu s}\alpha\right)^{2}ds=\frac{\alpha^{2}}{2\mu}\left(e^{2\mu t}-1\right)\equiv c^{2}.$$ We look to price a European call. That is, we seek $$\tilde{\mathbb{E}}\left[e^{-rt}\left(S\left(t\right)-K\right)^{+}\right]$$ where $\tilde{\mathbb{E}}$ denotes the risk-neutral measure. Let $Y\sim\mathcal{N}\left(0,1\right)$ and note that \begin{align*} \tilde{\mathbb{E}}\left[e^{-rT}\left(S\left(T\right)-K\right)^{+}\right] & =\tilde{\mathbb{E}}\left[e^{-rt}\left(e^{rt}S\left(0\right)-Yc-K\right)^{+}\right]\\ & =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-rt}\left(e^{rt}S\left(0\right)-yc-K\right)^{+}e^{-y^{2}/2}dy. \end{align*} The term $$e^{rt}S\left(0\right)-y\sqrt{\frac{\alpha^{2}}{2r}\left(e^{2rt}-1\right)}-K$$ is positive if and only if $$y<d\equiv\frac{e^{rt}S\left(0\right)-K}{c}.$$ Therefore, \begin{align*} \tilde{\mathbb{E}}\left[e^{-rT}\left(S\left(T\right)-K\right)^{+}\right] & =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{d}e^{-rt}\left(e^{rt}S\left(0\right)-yc-K\right)e^{-y^{2}/2}dy\\ & =S\left(0\right)N\left(d\right)-e^{-rt}\frac{c}{\sqrt{2\pi}}\int_{-\infty}^{d}ye^{-y^{2}/2}dy-e^{-rt}KN\left(d\right)\\ & =N\left(d\right)\left(S\left(0\right)-e^{-rt}K\right)+e^{-rt-d^{2}/2}\frac{c}{\sqrt{2\pi}}. \end{align*}