Typo in Paper? (Simple Calculation)

Question

I have a question about obtaining Eq. (17) of this paper.

Context: In Theorem 3, they show that for a Gaussian probability path $p_{t}(x\vert x_{1}) = \mathcal N(x\vert \mu_{t}(x_{1}), \sigma^2_{t}(x_{1})I)$, the conditional vector field (VF) generating it is given by $$u_{t}(x\vert x_{1}) = \frac{\sigma'_{t}(x_{1})}{\sigma_{t}(x_{1})}(x - \mu_{t}(x_{1})) + \mu_{t}'(x_{1})$$ I checked their proof in the App. A, and it is correct.

Before Eq. (17), they specify $\sigma_{t}(x_{1}) = \sigma_{1-t}$ and $\mu_{t}(x_{1}) = x_{1}$, so IMO, the conditional VF should now be $$u_{t}(x\vert x_{1}) = \frac{\sigma'_{1-t}}{\sigma_{1-t}}(x - x_{1}),$$ but the authors have an additional $-$ sign, and I just don't see why.

Why does this matter? Usually, I'd think this is just a typo, but in App. D, they show that another method can be obtained through their method, and while proving it, they obtain the conditional VF as in Eq. (17) of their paper, so it's important to understand the $-$ sign issue.

Answer 1

The minus comes from the derivative respect to "t" in the definition $\sigma_t(x_1)=\sigma_{1-t}$, because you differentiate respect to $t$ this quantity: \begin{eqnarray} \sigma'_t(x_1)=\frac{d}{dt}\sigma_t(x_1)=\frac{d}{dt}\sigma_{1-t}=(-1)\sigma'_{1-t}, \end{eqnarray} due to the chain rule.