The Gaussian conditional affine map is quadratic-cost optimal

Since ${\sigma }_t=(1-t)+t{\sigma }_{\min }>0$, every affine map ${\psi }_t$ is invertible. For each $v\in {\mathbb{R}}^d$, the projection of ${\psi }_t(Z\mid x_1)$ is $v^T{\mu }_t(x_1)+{\sigma }_t{v}^{T}Z$. By [F1], this is normal with mean $v^T{\mu }_t(x_1)$ and variance ${\sigma }_t^2\mathrm{\parallel }v{\mathrm{\parallel }}_2^2$. Thus $P_t(\cdot \mid x_1)$ is multivariate normal with mean ${\mu }_t(x_1)$ and covariance ${\sigma }_t^2{I}_d$, which is positive definite.

By [F2], the Lebesgue density of $P_t(\cdot \mid x_1)$ is ${\rho }_t(x\mid x_1)=(2\pi {\sigma }_t^2{)}^{-d\mathrm{/}2}\exp \text{\hspace{0.17em}⁣}(-\frac{\mathrm{\parallel }x-{\mu }_t(x_1){\mathrm{\parallel }}_2^2}{2{\sigma }_t^2})\mathrm{.}$ It integrates to one for every $x_1$ and is continuous in $(t,x,x_1)$ because ${\sigma }_t>0$. Therefore for any endpoint probability law $\pi$ it is product-measurable and satisfies the conditional-density requirement in Probability paths and velocity fields.

We have ${\dot{\mu }}_t(x_1)=x_1$ and ${\dot{\sigma }}_t={\sigma }_{\min }-1$. Applying [F3] gives $u_t(x\mid x_1)=x_1+\frac{{\sigma }_{\min }-1}{{\sigma }_t}(x-t{x}_1)$. The density formula in step 1.2 and this affine field are continuously differentiable in $(t,x)$ by the algebra, chain, and exponential derivative rules [F6, F7, F8]. Thus ${\rho }_t$ and each component of ${\rho }_t{u}_t$ meet the regularity clauses of the conditional probability-path definition.

At $t=0$ the parameters are ${\mu }_0=0,{\sigma }_0=1$; at $t=1$ they are ${\mu }_1=x_1,{\sigma }_1={\sigma }_{\min }$. Thus the endpoints are the standard Gaussian law and the law of $x_1+{\sigma }_{\min }Z$.

Put $r=x-t{x}_1$ and write $\dot{\sigma }={\sigma }_{\min }-1$. We have ${\mathrm{\partial }}_{t}r=-x_1$ and ${\mathrm{\partial }}_t({\sigma }_t^{-d})=-d\dot{\sigma }{\sigma }_t^{-d-1}$ by the product and quotient rules [F6]. Applying the chain rule [F7] to $-\mathrm{\parallel }r{\mathrm{\parallel }}_2^2\mathrm{/}(2{\sigma }_t^2)$ and then the exponential derivative [F8] gives ${\mathrm{\partial }}_t{\rho }_t={\rho }_t(-d\frac{\dot{\sigma }}{{\sigma }_t}+\frac{⟨r,x_1⟩}{{\sigma }_t^2}+\frac{\dot{\sigma }\mathrm{\parallel }r{\mathrm{\parallel }}_2^2}{{\sigma }_t^3}),{\mathrm{\nabla }}_x{\rho }_t=-{\rho }_t\frac{r}{{\sigma }_t^2}\mathrm{.}$ The field in step 2.1 has ${\mathrm{div}}_x{u}_t=d\dot{\sigma }\mathrm{/}{\sigma }_t$.

The product rule gives ${\mathrm{div}}_x({\rho }_t{u}_t)=⟨{\mathrm{\nabla }}_x{\rho }_t,u_t⟩+{\rho }_t{\mathrm{div}}_x{u}_t={\rho }_t(-\frac{⟨r,x_1⟩}{{\sigma }_t^2}-\frac{\dot{\sigma }\mathrm{\parallel }r{\mathrm{\parallel }}_2^2}{{\sigma }_t^3}+d\frac{\dot{\sigma }}{{\sigma }_t})\mathrm{.}$ Adding this to ${\mathrm{\partial }}_t{\rho }_t$ in step 3.1 cancels each term, so the conditional continuity equation holds.

The standard Gaussian vector has mean $0$ and covariance $I_d$ by [F1]. Summing its $d$ coordinate second moments gives $\int \mathrm{\parallel }z{\mathrm{\parallel }}_2^2{N}_d(0,I_d)(dz)=d$. The endpoint law is the affine image under $T(z)=x_1+{\sigma }_{\min }z$. Applying [F4] with $a=x_1$, $s={\sigma }_{\min }$, and $m_2=d$ shows that every coupling has cost at least $\mathrm{\parallel }{x}_1{\mathrm{\parallel }}_2^2+d(1-{\sigma }_{\min }{)}^2$, and that the law of $(Z,T(Z))$ attains equality.