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If we are going to have the form \begin{align*} dr = A dt + BdW_t, \end{align*} Then both A and B are functions of $t$ and $r_t$, otherwise, $r_t$ is normal. However, note that \begin{align*} r_t = \exp\Bigg(\frac{1}{\sigma(t)}\bigg(\int_0^t \theta(s)\sigma(s) ds +\sigma(0)\ln r_0 + \int_0^t\sigma^2(s) dW_s\bigg)\Bigg). \end{align*} That is, $r_t$ is ...