=  {{\boldsymbol{U}}}^{\dagger}(t) a({{\boldsymbol{U}}}(t) y, t)  \\
    B_j(y, t) = {{\boldsymbol{U}}}^{\dagger}(t) b_j({{\boldsymbol{U}}}(t) y, t) \\
    {{\boldsymbol{G}}}(t) = (-i) {\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} \biggl( {\mathbb{E}}_{\theta_t} \bigl[a({{\boldsymbol{U}}}(t) y, t) ({{\boldsymbol{U}}}(t) y)^{\dagger}\bigr] \\ \hspace{10em} + \sum_{j=1}^{N} {\mathbb{E}}_{\theta_t}\bigl[b_j({{\boldsymbol{U}}}(t) y, t) b_j^{\dagger}({{\boldsymbol{U}}}(t) y, t) \bigr] \biggr) {{\boldsymbol{U}}}(t) {\mathbb{E}}_{\theta_t}[yy^{\dagger}]^{-1} {{\boldsymbol{U}}}^{\dagger}(t) + {\text{h.c.}}\end{cases}$$

(ii) The solution is optimal in the sense that, for any given ${{\boldsymbol{U}}}(t)$ and $\theta_t$, $$\begin{aligned}
          \min_{{\widetilde{{\mathcal{A}}}}_{{LR},t}}\; d_{{\mathsf{F}}}\bigl({\widetilde{{\mathcal{A}}}}_{{LR},t} \mu_{{LR},t},\
          {{\mathcal{A}}}_{t} \mu_{{LR},t} \bigr) & =
          d_{{\mathsf{F}}}\bigl( {{\mathcal{A}}}_{{LR},t}
          \mu_{{LR},t},\ {\mathcal{A}}_{t} \mu_{{LR},t} \bigr)\\
          & = {\left\lVert\sum_{j=1}^{N} {\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} {\mathbb{E}}_{\theta_t}
            \bigl[b_j({{\boldsymbol{U}}}(t)y,t) b_j^{\dagger}({{\boldsymbol{U}}}(t)y,t)\bigr]
            {\mathcal{Q}}_{{{\boldsymbol{U}}}(t)}\right\rVert}_{{HS}},
        \end{aligned}$$ where the adjoint of ${{\mathcal{A}}}_{{LR},t}$ ([*i.e.*]{}, ${{\mathcal{A}}}^{*}_{{LR},t}$) is given by [Eq. ]{}, ${\widetilde{{\mathcal{A}}}}_{{LR},t}$ is any operator acting on probability measures whose corresponding infinitesimal generator ${\widetilde{{\mathcal{A}}}}_{{LR},t}^{*}$ is parametrized by $A$, $B_j$ and ${{\boldsymbol{G}}}$ as in Lemma \[lemma::generator\].

(iii) Time-evolution equation of ${{\boldsymbol{U}}}(t)$ and $\theta_t$ is given by the following system: $$\left\{
          \label{eqn::low_rank_xx_v2}
          \begin{split}
            {\,\mathrm{d}}Y &= {{\boldsymbol{U}}}^{\dagger}(t) a({{\boldsymbol{U}}}(t) Y, t)\ {\,\mathrm{d}}t 
            + \sum_{j=1}^{N} {{\boldsymbol{U}}}^{\dagger}(t) b_j({{\boldsymbol{U}}}(t) Y, t)\ {\,\mathrm{d}}W_j\,,\\
            \frac{{\,\mathrm{d}}{{\boldsymbol{U}}}}{{\,\mathrm{d}}t} &= {\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} \left({\mathbb{E}}\bigl[a({{\boldsymbol{U}}}(t) Y, t) Y^{\dagger}\bigr] + \sum_{j=1}^{N} {\mathbb{E}}\bigl[b_j({{\boldsymbol{U}}}(t) Y, t) b_j^{\dagger}({{\boldsymbol{U}}}(t) Y, t)\bigr] {{\boldsymbol{U}}}(t) \right) {\mathbb{E}}\bigl[Y Y^{\dagger}\bigr]^{-1}\,. \\
          \end{split}\right.$$ If rank $r = n$, then ${\mathcal{Q}}_{{{\boldsymbol{U}}}(t)} = 0$ and ${\mathcal{P}}_{{{\boldsymbol{U}}}(t)} = {\mathrm{Id}}$. Consequently, $\frac{{\,\mathrm{d}}{{\boldsymbol{U}}}}{{\,\mathrm{d}}t} = 0$ and the infinitesimal generator for SDE of $X_{{LR},t}$ is the same as that for $X_t$. 