equation , at the converging point and as $N$ tends to infinity, as $$\label{eqlimit}
    \overline{E}\{\hat{\varphi}_f(t_k,\bar{\theta})\varphi_f^\top(t_k,\bar{\theta})\}^{-1} \overline{E}\{\hat{\varphi}_f(t_k,\bar{\theta})\varepsilon(t_k,\bar{\theta})\}=0,$$ where $\varepsilon(t_k,\bar{\theta})$ is the GEE evaluated at the converging point. Since the matrix inverse in is assumed to be non-singular, the second expectation in must be zero, i.e., $$\label{eqcon2}
    \overline{E}\{\hat{\varphi}_f(t_k,\bar{\theta})\varepsilon(t_k,\bar{\theta})\}=0.$$

Let $\bar{A}(p)B^*(p)-\bar{B}(p)A^*(p)=h_0p^r+h_1 p^{r-1}+\dots + h_r$, where $r = \max(n+m^*,n^*+m)=n+m$. Then, the GEE in can be rearranged as $$\begin{aligned}
    \varepsilon(t_k,\bar{\theta})&=\left\{\frac{\bar{A}(p)B^*(p)-\bar{B}(p)A^*(p)}{\bar{A}(p)A^*(p)}u(t)\right\}_{t_k}+v(t_k) \notag \\
    &= \left\{\frac{U_{du}^\top(t)}{\bar{A}(p)A^*(p)}H \right\}_{t_k} + v(t_k), \notag
    \end{aligned}$$ where $H = \begin{bmatrix}
    h_0, & h_1, & \hspace{0.1cm}\dots, & h_{n+m}\end{bmatrix}^\top$. Now, note that the instrument vector $\hat{\varphi}_f(t_k)$ can be written as $$\hat{\varphi}_f(t_k) = S(-\bar{B},\bar{A})\left[\frac{U_{du}(t)}{\bar{A}^2(p)}\right]_{t_k}, \notag$$ where $S(-\bar{B},\bar{A})$ is a Sylvester matrix associated with the polynomials $\bar{B}(p)$ and $\bar{A}(p)$, which again is non-singular. So, we can express as $$\begin{aligned}
    &0 = S(-\bar{B},\bar{A}) \underbrace{\overline{E}\left\{\left[\frac{U_{du}(t)}{\bar{A}^2(p)}\right]_{t_k} \left[\frac{U_{du}(t)}{\bar{A}(p)A^*(p)}\right]_{t_k}^\top\right\}}_{:=\bar{\Phi}} H \notag \\
    \label{eqcon3}
    &\hspace{0.2cm}+ S(-\bar{B},\bar{A})\underbrace{\overline{E}\left\{ \left[\frac{U_{du}(t)}{\bar{A}^2(p)}\right]_{t_k}v(t_k) \right\}}_{:=\bar{\Psi}}.
    \end{aligned}$$ Following a similar approach as in Lemma \[lemma1\], we conclude that $\bar{\Psi}=0$, and by Lemma \[lemma2\], $\bar{\Phi}$ is generically non-singular. Thus, for to hold we need $H=0$, which implies that $$\begin{aligned}
    \bar{A}(p)B^*(p)&-\bar{B}(p)A^*(p)=0 \notag \\
    \frac{\bar{B}(p)}{\bar{A}(p)}&= \frac{B^*(p)}{A^*(p)}, \notag 
    \end{aligned}$$ i.e., $\theta^*$ is the unique limiting point.

*Statement 3*. The proof follows from the analysis made for proving Statement 3 of Theorem 1 in [@pan2020consistency].

