= Solution
{c}
Apply part <1/a> with $\alpha=r$. For $\tau>0$, the <transition density> is the <normal distribution> with mean $m=Se^{r\tau}$ and positive <variance> $v=\sigma^2(e^{2r\tau}-1)/(2r)$. Integrating the <financial payoff> against this <probability density function> gives
$$
\boxed{V(S,t)=\frac{e^{-r\tau}}{\sqrt{2\pi v}}
\int_{-\infty}^{\infty}P_{\mathrm o}(x)
\exp\!\left[-\frac{(x-Se^{r\tau})^2}{2v}\right]dx,\qquad
v=\frac{\sigma^2}{2r}(e^{2r\tau}-1).}
$$
Thus $d_*=(x-Se^{r\tau})/\sqrt v$, as required. The <variance> is positive even if $r<0$, because numerator and denominator have the same sign. For $r=0$, replace $m$ by $S$ and $v$ by $\sigma^2\tau$; at expiry use the terminal value directly. The <integral> presumes an integrable <financial payoff>.
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