 Solution

The diffusion generator is $\mathcal{L}=\mu {\partial }_x+\frac{1}{2}{\sigma }^2{\partial }_{xx}$. By the Feynman-Kac formula, the required conditional characteristic function solvesUse separation of variables in the form $F(t,x)=e^{iux}h(t)$. Since $F_x=iuF$ and $F_{xx}=-u^{2}F$, the temporal factor satisfies the ordinary differential equationThus $h(t)=\exp [(iu\mu -\frac{1}{2}{\sigma }^2{u}^2)(T-t)]$, andThis is bounded in absolute value by one, so it lies in the uniqueness class of the Feynman-Kac formula. As an independent check, integrating the stochastic differential equation gives $X_T=x+\mu (T-t)+\sigma (B_T-B_t)$. The independent Brownian increments have the normal distribution with mean zero and variance $T-t$, yielding exactly the same characteristic function. For $\sigma =0$, the conditional distribution is a point mass at $x+\mu (T-t)$, and the same boxed expression still applies.