Tests for the explanatory power of latent factors

Abstract

We propose herein a factor-augmented semi-varying coefficient model and discuss whether the extracted factors have significant explanatory power. We first use Principal Component Analysis (PCA) to estimate the model and then develop the PCA-based Wald test. We find that the PCA-based Wald test statistic is asymptotically chi-squared distributed with degrees of freedom equal to the unknown number of factors. To avoid estimating the degrees of freedom, we then use Common Correlated Effects (CCE) to estimate the model and develop the CCE-based Wald test. However, as opposed to the PCA-based estimator, the CCE-based estimator of loadings is ambiguous in the sense that the estimation depends on the dimensions of factors and predictors and the estimator can even be inconsistent. If we construct the Wald test based on the CCE-based estimator, the test lacks power. We overcome these difficulties and construct a powerful CCE-based Wald test that is immune to factor-number uncertainty. In addition to the two Wald tests, a new CCE-based goodness-of-fit test is also proposed. The test is irrelevant to the unknown number of factors and spares us the work of estimating asymptotic covariance matrix. Finally, three empirical examples are provided to demonstrate the usefulness of the tests.

Appendix: Proofs of theorems

To avoid unnecessary repetitions, we first state a unified result in Lemma 1, which asserts that, for any matrix ${\mathbf {Q}}_{t}$ satisfying some regularities, ${\mathbf {W}}_{j}(u)$ in (7.1) converges to its expectation in probability. Here, ${\mathbf {Q}}_{t}$ could be, for example, ${\mathbf {X}}_{t}{\mathbf {X}}^{\top }_{t}$ and ${\mathbf {X}}_{t}\mathbf {{\overline{Z}}}^{\top }_{t}$ that will appear in Lemma 2.

Lemma 1

Let $U_{t}$ be univariate and ${\mathbf {Q}}_{t}$ be any matrix of finite dimensions. They satisfy following conditions:

• 1. Each (l, k)th element $Q_{t,lk}$ of ${\mathbf {Q}}_{t}$ and $U_{t}$ are strictly stationary; $U_{t}$ has bounded density f(u) with compact support.

• 2. The $\alpha$-mixing coefficients $\alpha (m)$ of $\{\left( Q_{t,lk},U_{t}\right) : t\ge 1\}$ satisfy $\sum _{m=1}^{\infty }m^{b}\left[ \alpha (m)\right] ^{\delta /(2+\delta )}<\infty$ where $b>2+\delta ,\,\delta >0$.

• 3. $\mathrm {E}(\left| Q_{t,lk}\right| ^{2+\delta })<\infty$, $\mathrm {E}(\vert Q_{t,lk}Q_{t+m,lk}\vert )<\infty$ for any t and any m.

• 4. For kernel function k(u), $\mu _{j}=\int u^{j}k(u)\mathrm {d}u$ and $\nu _{j}=\int u^{j}k^{2}(u)\mathrm {d}u$ exist for $j\le 4$.

Define

$$\begin{aligned} {\mathbf {W}}_{j}(u)=\frac{1}{T}\sum \limits _{t=1}^{T} {\mathbf {Q}}_{t}\left( \frac{U_{t}-u}{h}\right) ^{j}k_{h}(U_{t}-u) \end{aligned}$$

(7.1)

then we have

$$\begin{aligned}&\mathrm {E}({\mathbf {W}}_{j}(u)) =f(u)\mu _{j}\mathrm {E}({\mathbf {Q}}_{t}|U_{t}=u)(1+O(h^\delta )) \end{aligned}$$

(7.2)

$$\begin{aligned}&Th{{\mathrm {Var}}}\left[ {\mathbf {W}}_{j}(u)\right] = O(1) \end{aligned}$$

(7.3)

where $\delta =2$ if $j=0,2$; $\delta =1$ if $j=1$.

Proof Firstly, by change of variables and Taylor expansion, we have

$$\begin{aligned} \mathrm {E}({\mathbf {W}}_{j}(u))= & {} \mathrm {E}\left( \mathrm {E}({\mathbf {Q}}_{t}|U_{t})\left( \frac{U_{t}-u}{h}\right) ^{j}k_{h}(U_{t}-u)\right) \nonumber \\= & {} \int \mathrm {E}({\mathbf {Q}}_{t}|s)\left( \frac{s-u}{h}\right) ^{j}\frac{1}{h}k\left( \frac{s-u}{h}\right) f(s)\mathrm {d}s\nonumber \\= & {} \int \mathrm {E}({\mathbf {Q}}_{t}|u)(1+O(h))x^{j}k(x)[f(u)+f^\prime (u)xh+O(h^2)]\mathrm {d}x\nonumber \\= & {} f(u)\mu _{j}\mathrm {E}({\mathbf {Q}}_{t}|U_{t}=u)(1+O(h^\delta )) \end{aligned}$$

(7.4)

since k(x) is symmetric, $\delta =2$ if $j=0,2$; $\delta =1$ if $j=1$.

Secondly, we derive the variance of $({\mathbf {W}}_{j}(u))_{lk}$, the (l, k)th element of ${\mathbf {W}}_{j}(u)$. To that end, let $w_{t}(u)=Q_{t,lk}\left( \frac{U_{t}-u}{h}\right) ^{j}k_{h}(U_{t}-u)$. Then we have

$$\begin{aligned} Th{{\mathrm {Var}}}\left[ ({\mathbf {W}}_{j}(u))_{lk}\right]= & {} h {\mathrm {Var}}(w_{t}(u))+2h\sum _{m=1}^{T-1}\left( 1-\frac{m}{T}\right) {\mathrm {Cov}}\left( w_{1}(u),w_{m+1}(u)\right) \\\equiv & {} {\mathbb {I}}_{1}+{\mathbb {I}}_{2} \end{aligned}$$

For ${\mathbb {I}}_{1}$, we have

$$\begin{aligned} {\mathbb {I}}_{1}= & {} h{\mathrm {Var}}(w_{t}(u)) =h\mathrm {E}\left( Q^{2}_{t,lk}\left( \frac{U_{t}-u}{h}\right) ^{2j}k^{2}_{h}(U_{t}-u)\right) -h\mathrm {E}^2(w_{t}(u))\nonumber \\= & {} h\mathrm {E}\left( \mathrm {E}(Q^{2}_{t,lk}|U_{t})\left( \frac{U_{t}-u}{h}\right) ^{2j}k^{2}_{h}(U_{t}-u)\right) +O(1)\nonumber \\= & {} h\int \mathrm {E}(Q^{2}_{t,lk}|s)\left( \frac{s-u}{h}\right) ^{2j}\frac{1}{h^2}k^2\left( \frac{s-u}{h}\right) f(s)\mathrm {d}s+O(1)\nonumber \\= & {} O(1) \end{aligned}$$

(7.5)

For ${\mathbb {I}}_{2}$, we can first get

$$\begin{aligned}&{\mathrm {Cov}}\left( w_{t}(u),w_{t+m}(u)\right) \nonumber \\&\quad = \mathrm {E}\Bigg ( \mathrm {E}(Q_{t,lk}Q_{t+m,lk}|U_{t},U_{t+m})\Bigg [\left( \frac{U_{t}-u}{h}\right) ^{j}k_{h}(U_{t}-u)\nonumber \\&\quad \quad \times \,\left( \frac{U_{t+m}-u}{h}\right) ^{j}k_{h}(U_{t+m}-u)\Bigg ]\Bigg )+O(1)\nonumber \\&\quad = \iint \mathrm {E}(Q_{t,lk}Q_{t+m,lk}|x,y)\left( \frac{x-u}{h}\right) ^{j}k\left( \frac{x-u}{h}\right) \nonumber \\&\quad \quad \times \,\left( \frac{y-u}{h}\right) ^{j}k\left( \frac{y-u}{h}\right) f_{U_{t},U_{t+m}}(x,y)\mathrm {d}\left( \frac{x-u}{h}\right) \mathrm {d}\left( \frac{y-u}{h}\right) +O(1)\nonumber \\&\quad =O(1) \end{aligned}$$

(7.6)

where $f_{U_{t},U_{t+m}}(x,y)$ is the joint density function of $(U_{t},U_{t+m})$. Hence, $\sum _{m=1}^{d_{T}-1}{\mathrm {Cov}}\left( w_{1}(u),w_{1+m}(u)\right) =O(d_{T})$.

Analogous to the derivation of ${\mathbb {I}}_{1}$, we can also derive

$$\begin{aligned} \mathrm {E}\left| w_{t}(u)\right| ^{2+\delta }= & {} \mathrm {E}\left| Q_{t,lk}\left( \frac{U_{t}-u}{h}\right) ^{j}k_{h}(U_{t}-u) \right| ^{2+\delta } \le \frac{C}{h^{1+\delta }} \end{aligned}$$

(7.7)

Therefore if $d^{b}_{T}h^{\delta /(2+\delta )}=O(1)$, by the Davydov inequality and assumption $\sum _{m=1}^{\infty }m^{b}\left[ \alpha (m)\right] ^{\delta /(2+\delta )}<\infty$ $(b>2+\delta ,\,\delta >0)$, we have

$$\begin{aligned} \sum _{m=d_{T}}^{T-1}\left| \mathrm {{\mathrm {Cov}}}\left( w_{1}(u),w_{m+1}(u)\right) \right|\le & {} \sum _{m=d_{T}}^{T-1}8\left[ \alpha (m)\right] ^{\delta /(2+\delta )}\left[ \mathrm {E}\left| w_{1}\right| ^{2+\delta }\right] ^{2/(2+\delta )}\nonumber \\\le & {} 8h^{-2(1+\delta )/(2+\delta )}\sum _{m=d_{T}}^{T-1}\left[ \alpha (m)\right] ^{\delta /(2+\delta )}\nonumber \\\le & {} 8h^{-2(1+\delta )/(2+\delta )}d_{T}^{-b} \sum _{m=d_{T}}^{T-1}m^{b}\left[ \alpha (m)\right] ^{\delta /(2+\delta )}\nonumber \\= & {} O(h^{-1}) \end{aligned}$$

(7.8)

To sum up, if $d_{T}\rightarrow \infty$, $d_{T}h=O(1)$ and $d^{b}_{T}h^{\delta /(2+\delta )}=O(1)$, then ${\mathbb {I}}_{2}\le h\sum _{m=1}^{T-1}\left| {{\mathrm {Cov}}}\left( w_{1}(u),w_{m+1}(u)\right) \right| =O(d_{T}h)+O(1)=O(1)$. Hence, we have $Th{\mathrm {Var}}\left[ ({\mathbf {W}}_{j}(u))_{lk}\right] = O(1)$. $\blacksquare$

To handle the divergence problem of $(T^{-1}\mathbf {{\overline{Z}}}^{\top }\mathbf {{\overline{Z}}})^+$ when $d>R$, we follow Karabiyik et al. (2017) and left multiple $\mathbf {{\overline{Z}}}_{t}$ with ${\mathbf {A}}^{\top }_{N}{\mathbf {B}}^{\top }$, where ${\mathbf {B}}$ and ${\mathbf {A}}_{N}$ are specified in the proof of (7.10). Denote $\mathbf {{\overline{Z}}}{\mathbf {B}}{\mathbf {A}}_{N}$ as $\mathbf {{\overline{Z}}}^{0}$, $\mathbf {{\overline{V}}}{\mathbf {B}}{\mathbf {A}}_{N}$ as $\mathbf {{\overline{V}}}^{0}$, $\left( {\mathbf {F}},{\mathbf {0}}_{T\times (d-R)}\right)$ as ${\mathbf {F}}^{0}$, then $\mathbf {{\overline{Z}}}^{0}={\mathbf {F}}^{0}+\mathbf {{\overline{V}}}^{0}$. Using this device, we could guarantee the inverse of the left side of (7.10) also converges to the inverse of the right side of (7.10).

Lemma 2

Let $c_{NTh1}=(Th)^{-1/2}+h+N^{-1/2}$, $c_{Th1}=(Th)^{-1/2}+h$, while $c_{NTh}=(Th)^{-1/2}+h^2+N^{-1/2}$, $c_{Th}=(Th)^{-1/2}+h^2$, we have

$$\begin{aligned}&(a)\quad T^{-1}\mathbf {{\overline{Z}}}^{\top }\mathbf {{\overline{Z}}} ={\mathbf {D}}^{\top }\varvec{\varSigma }_{F}{\mathbf {D}}+O_{p}(1/\sqrt{NT}+1/N) \end{aligned}$$

(7.9)

$$\begin{aligned}&(b)\quad T^{-1}\mathbf {{\overline{Z}}}^{0\top }\mathbf {{\overline{Z}}}^{0}=\left( \begin{array}{cc} \varvec{\varSigma }_{F} &{} {\mathbf {0}}_{R\times (d-R)} \\ {\mathbf {0}}_{(d-R)\times R} &{} {\mathbf {B}}^{\top }_{-R}\varvec{\varSigma }_{v} {\mathbf {B}}_{-R} \\ \end{array} \right) +O_{p}(1/\sqrt{N}+1/\sqrt{T}) \end{aligned}$$

(7.10)

$$\begin{aligned}&(c)\quad T^{-1}\mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u)\mathbf {{\overline{Z}}}^{0}=f(u)\left( \begin{array}{cc} {\mathbf {L}}^{\top }\varvec{\varSigma }_{F}+O_{p}(c_{NTh}) &{} {\mathbf {0}}_{p\times (d-R)}+O_{p}(c_{NTh}) \\ {\mathbf {0}}_{p\times R}+O_{p}(c_{NTh1}) &{} {\mathbf {0}}_{p\times (d-R)}+O_{p}(c_{NTh1}) \\ \end{array} \right) \end{aligned}$$

(7.11)

$$\begin{aligned}&(d)\quad T^{-1}\mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u)\mathbf {{\widetilde{X}}}(u)=f(u)\left( \begin{array}{cc} \varvec{\varOmega }_{xx}(u)+O_{p}(c_{Th}) &{} {\mathbf {0}}_{p\times p}+O_{p}(c_{Th1}) \\ {\mathbf {0}}_{p\times p}+O_{p}(c_{Th1}) &{} \mu _{2}\varvec{\varOmega }_{xx}(u)+O_{p}(c_{Th}) \\ \end{array} \right) \end{aligned}$$

(7.12)

$$\begin{aligned}&(e)\quad T^{-1}\mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u)\mathbf {{\widetilde{A}}}(u)=f(u)\mu _{2}\left( \begin{array}{c} {\mathbf {L}}^{\top }\varvec{\varSigma }_{F}{\mathbf {L}}+\varvec{\varSigma }_{e} + O_{p}(c_{Th}) \\ {\mathbf {0}}_{p\times p}+ O_{p}(c_{Th1})\\ \end{array} \right) . \end{aligned}$$

(7.13)

where $\mathbf {{\widetilde{A}}}(u)^{\top }=\left( \begin{array}{ccc} {\mathbf {X}}_{1}(\frac{U_{1}-u}{h})^2 &{} \cdots &{} {\mathbf {X}}_{T}(\frac{U_{T}-u}{h})^2 \\ \end{array} \right)$ and $\varvec{\varOmega }_{xx}(u) = \mathrm {E}({\mathbf {X}}_{t}{\mathbf {X}}^{\top }_{t}\vert U_{t}=u)$.

Proof

(a) $T^{-1}\mathbf {{\overline{Z}}}^{\top }\mathbf {{\overline{Z}}}=T^{-1}\mathbf {{\overline{D}}}^{\top }{\mathbf {F}}^{\top }{\mathbf {F}}\mathbf {{\overline{D}}} +T^{-1}\mathbf {{\overline{D}}}^{\top }{\mathbf {F}}^{\top }\mathbf {{\overline{V}}} +T^{-1}\mathbf {{\overline{V}}}^{\top }{\mathbf {F}}\mathbf {{\overline{D}}}+T^{-1}\mathbf {{\overline{V}}}^{\top }\mathbf {{\overline{V}}} ={\mathbf {D}}^{\top }\varvec{\varSigma }_{F}{\mathbf {D}}+O_{p}(1/\sqrt{T})+O_{p}(1/\sqrt{NT})+O_{p}(1/N)$. But ${\mathbf {D}}^{\top }\varvec{\varSigma }_{F}{\mathbf {D}}$ is of deficient rank. Hence $(T^{-1}\mathbf {{\overline{Z}}}^{\top }\mathbf {{\overline{Z}}})^{+} {\mathop {\nrightarrow }\limits ^{\mathrm {P}}} ({\mathbf {D}}^{\top }\varvec{\varSigma }_{F}{\mathbf {D}})^{+}$.

(b) To counteract the degenerating matrix ${\mathbf {D}}^{\top }\varvec{\varSigma }_{F}{\mathbf {D}}$, we premultiple $\mathbf {{\overline{Z}}}_{t}$ with ${\mathbf {A}}^{\top }_{N}{\mathbf {B}}^{\top }$ where ${\mathbf {B}}=({\mathbf {B}}_{R},{\mathbf {B}}_{-R})=\left( \begin{array}{cc} {\mathbf {D}}_{R}^{-1} &{} -{\mathbf {D}}_{R}^{-1}{\mathbf {D}}_{-R} \\ {\mathbf {0}}_{(d-R)\times R}&{} \varvec{\mathrm {I}}_{d-R} \\ \end{array} \right)$ is of full rank, ${\mathbf {D}}_{R}$ is an $R\times R$ submatrix of ${\mathbf {D}}$ with full rank, ${\mathbf {D}}_{-R}$ is the remaining $R\times (d-R)$ submatrix of ${\mathbf {D}}$ and ${\mathbf {A}}_{N}=\text {diag}\left( \varvec{\mathrm {I}}_{R},\sqrt{N}\varvec{\mathrm {I}}_{d-R}\right)$. Then $T^{-1}\mathbf {{\overline{Z}}}^{0\top }\mathbf {{\overline{Z}}}^{0}=\left( \begin{array}{cc} \varvec{\varSigma }_{F} &{} {\mathbf {0}} \\ {\mathbf {0}} &{} {\mathbf {B}}^{\top }_{-R}\varvec{\varSigma }_{v} {\mathbf {B}}_{-R} \\ \end{array} \right)$ $+O_{p}(1/\sqrt{N}+1/\sqrt{T})$. This limiting matrix is now of full rank.

(c) Since ${\mathbf {e}}_{t}$ is of finite dimension, even if its components come from some of ${\mathbf {V}}_{jt}$’s, the number of these ${\mathbf {V}}_{jt}$’s is finite. We denote the set of these ${\mathbf {V}}_{jt}$’s as ${\mathfrak {M}}$. Then

$$\begin{aligned}&T^{-1}\sum _{t=1}^{T}{\mathbf {e}}_{t}\overline{{\mathbf {V}}}^{\top }_{t}k_{h}(U_{t}-u)\nonumber \\&\quad = (NT)^{-1}\sum _{j\notin {\mathfrak {M}}}\sum _{t=1}^{T}{\mathbf {e}}_{t}{\mathbf {V}}^{\top }_{jt}k_{h}(U_{t}-u) +(NT)^{-1}\sum _{j\in {\mathfrak {M}}}\sum _{t=1}^{T}{\mathbf {e}}_{t}{\mathbf {V}}^{\top }_{jt}k_{h}(U_{t}-u) \nonumber \\&\quad = O_{p}(1/\sqrt{NTh})+O_{p}(1/N) \end{aligned}$$

(7.14)

Now we are in a good position to analyze

$$\begin{aligned} T^{-1}\mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u)\mathbf {{\overline{Z}}}^{0}= & {} \left( \begin{array}{c} T^{-1}\sum _{t=1}^{T}{\mathbf {X}}_{t}\mathbf {{\overline{Z}}}^{\top }_{t}{\mathbf {B}}{\mathbf {A}}_{N} k_{h}(U_{t}-u)\\ T^{-1}\sum _{t=1}^{T}{\mathbf {X}}_{t}\mathbf {{\overline{Z}}}^{\top }_{t}{\mathbf {B}}{\mathbf {A}}_{N} h^{-1}(U_{t}-u)k_{h}(U_{t}-u)\\ \end{array} \right) \nonumber \\ \end{aligned}$$

(7.15)

By Lemma 1, the upper block matrix

$$\begin{aligned}&T^{-1}\sum _{t=1}^{T}{\mathbf {X}}_{t}\mathbf {{\overline{Z}}}^{\top }_{t}{\mathbf {B}}{\mathbf {A}}_{N} k_{h}(U_{t}-u)\nonumber \\&\quad = T^{-1}\sum _{t=1}^{T} \left[ {\mathbf {L}}^{\top }{\mathbf {f}}_{t}{\mathbf {f}}_{t}^{\top }\overline{{\mathbf {D}}} +{\mathbf {L}}^{\top }{\mathbf {f}}_{t}\overline{{\mathbf {V}}}_{t}^{\top } +{\mathbf {e}}_{t}{\mathbf {f}}_{t}^{\top }\overline{{\mathbf {D}}} +{\mathbf {e}}_{t}\overline{{\mathbf {V}}}_{t}^{\top }\right] {\mathbf {B}}{\mathbf {A}}_{N}k_{h}(U_{t}-u) \nonumber \\&\quad =f(u){\mathbf {L}}^{\top }\varvec{\varSigma }_{F}\left( \varvec{\mathrm {I}}_{R}, {\mathbf {0}}_{R\times (d-R)}\right) +O_{p}(c_{NTh}) \end{aligned}$$

(7.16)

Analogously, for the lower block matrix, it is of order $O_{p}(h+1/\sqrt{Th}+1/\sqrt{N})$.

(d) and (e) are the straightforward results of Lemma 1. $\square$

By $\frac{1}{dNT}{\mathbf {Z}}{\mathbf {Z}}^{\top }\widehat{{\mathbf {F}}}=\widehat{{\mathbf {F}}}\widehat{\varvec{\varXi }}_{NT}$, we have

$$\begin{aligned} (\widehat{{\mathbf {F}}}-{\mathbf {F}}{\mathbf {H}}) \widehat{\varvec{\varXi }}_{NT}= & {} \frac{1}{dNT}{\mathbf {F}}{\mathfrak {D}}{\mathbf {V}}^{\top } \widehat{{\mathbf {F}}}+\frac{1}{dNT}{\mathbf {V}}{\mathfrak {D}}^{\top }{\mathbf {F}}^{\top }\widehat{{\mathbf {F}}} +\frac{1}{dNT}{\mathbf {V}}{\mathbf {V}}^{\top }\widehat{{\mathbf {F}}} \end{aligned}$$

(7.17)

where ${\mathbf {H}}=\frac{1}{dNT}{\mathfrak {D}}{\mathfrak {D}}^{\top }{\mathbf {F}}^{\top }\widehat{{\mathbf {F}}}\widehat{\varvec{\varXi }}^{-1}_{NT}$ is invertible and both ${\mathbf {H}}$ and $\widehat{\varvec{\varXi }}_{NT}$ are shown to be $O_{p}(1)$ in Lemma A.3 of Bai (2003).

Lemma 3

Denote ${\mathbf {E}}=({\mathbf {e}}_{1},\cdots ,{\mathbf {e}}_{T})^{\top }$. Then we have

$$\begin{aligned}&(a)\quad \frac{1}{T}{\mathbf {F}}^{\top }K_{h}(u)(\widehat{{\mathbf {F}}}-{\mathbf {F}}{\mathbf {H}})=O_{p}\left( \frac{1}{N}+\frac{1}{\sqrt{NTh}}+\frac{1}{T\sqrt{h}}\right) \end{aligned}$$

(7.18)

$$\begin{aligned}&(b)\quad \frac{1}{T}{\mathbf {E}}^{\top }K_{h}(u)(\widehat{{\mathbf {F}}}-{\mathbf {F}}{\mathbf {H}})=O_{p}\left( \frac{1}{N}+\frac{1}{\sqrt{NTh}}\right) \end{aligned}$$

(7.19)

$$\begin{aligned}&(c)\quad \frac{1}{T}{\mathbf {X}}^{\top }K_{h}(u)(\widehat{{\mathbf {F}}}-{\mathbf {F}}{\mathbf {H}})=O_{p}\left( \frac{1}{N}+\frac{1}{\sqrt{NTh}}+\frac{1}{T\sqrt{h}}\right) \end{aligned}$$

(7.20)

Proof

(a) We left-multiply $\frac{1}{T}{\mathbf {F}}^{\top }K_{h}(u)$ to (7.17), then the first term of the right side of (7.17) becomes

$$\begin{aligned} \frac{1}{T}{\mathbf {F}}^{\top }{\mathbf {K}}_{h}(u){\mathbf {F}}\frac{1}{dNT}{\mathfrak {D}}{\mathbf {V}}^{\top }\widehat{{\mathbf {F}}}= & {} O_{p}(1)\frac{1}{dNT}{\mathfrak {D}}{\mathbf {V}}^{\top }(\widehat{{\mathbf {F}}}-{\mathbf {F}}{\mathbf {H}})+ O_{p}(1)\frac{1}{dNT}{\mathfrak {D}}{\mathbf {V}}^{\top }{\mathbf {F}}{\mathbf {H}}\nonumber \\= & {} O_{p}\left( \frac{1}{N}+\frac{1}{\sqrt{NT}}\right) +O_{p}\left( \frac{1}{\sqrt{NT}}\right) \end{aligned}$$

(7.21)

where $\frac{1}{dN\sqrt{T}}{\mathfrak {D}}{\mathbf {V}}^{\top }=O_{p}(1/\sqrt{N})$, $\frac{1}{\sqrt{T}}(\widehat{{\mathbf {F}}}-{\mathbf {F}}{\mathbf {H}})=O_{p}(1/\sqrt{N}+1/\sqrt{T})$ is the direct result of Theorem 1 of Bai and Ng (2002) and $\frac{1}{dNT}{\mathfrak {D}}{\mathbf {V}}^{\top }{\mathbf {F}}=\frac{1}{dNT}\sum _{t=1}^{T}\sum _{j=1}^{N}{\mathbf {D}}_{j}{\mathbf {V}}_{jt}{\mathbf {f}}^{\top }_{t} =O_{p}(1/\sqrt{NT})$.

In a similar manner, the resulted new 2nd term plus the new 3rd term is $\frac{1}{dNT}{\mathbf {F}}^{\top }{\mathbf {K}}_{h}(u){\mathbf {V}}{\mathfrak {D}}^{\top }\frac{1}{T}{\mathbf {F}}^{\top }\widehat{{\mathbf {F}}} +\frac{1}{dNT}{\mathbf {F}}^{\top }{\mathbf {K}}_{h}(u){\mathbf {V}}\frac{1}{T}{\mathbf {V}}^{\top }\widehat{{\mathbf {F}}} =O_{p}(\frac{1}{\sqrt{NTh}})+O_{p}(\frac{1}{T\sqrt{h}})$. So $\frac{1}{T}{\mathbf {F}}^{\top }K_{h}(u)(\widehat{{\mathbf {F}}}-{\mathbf {F}}{\mathbf {H}})=O_{p}(\frac{1}{N}+\frac{1}{\sqrt{NTh}}+\frac{1}{T\sqrt{h}})$.

(b) It can be obtained analogously to (a).

(c) This is the immediate result of (a) and (b). $\square$

Proof of Theorem 1

$$\begin{aligned} \varvec{{\widehat{\lambda }}}_{P}= & {} \left[ \widehat{{\mathbf {F}}}^{\top }(\varvec{\mathrm {I}}_{T} -{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}_{T} -{\mathbf {S}})\widehat{{\mathbf {F}}}\right] ^{-1} \widehat{{\mathbf {F}}}^{\top }(\varvec{\mathrm {I}}_{T} -{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}}) \left[ {\mathbf {M}}+{\mathbf {F}}\varvec{\lambda } +\varvec{\epsilon }\right] \nonumber \\ \end{aligned}$$

(7.22)

To derive the asymptotic distribution of (7.22), we follow several steps. In the 1st step, we find the limit of the denominator in (7.22). According to (7.12) and (7.20), we have

$$\begin{aligned}&({\mathbf {X}}_{t}^{\top },{\mathbf {0}}_{1\times p})\left[ \mathbf {{\widetilde{X}}}^{\top }(U_{t}){\mathbf {K}}_{h}(U_{t}) \mathbf {{\widetilde{X}}}(U_{t})\right] ^{+}\mathbf {{\widetilde{X}}}^{\top }(U_{t}){\mathbf {K}}_{h}(U_{t})\widehat{{\mathbf {F}}}\nonumber \\&\quad ={\mathbf {X}}_{t}^{\top } \varvec{\varOmega }^{-1}_{xx}(U_{t}){\mathbf {L}}^{\top }\varvec{\varSigma }_{F}{\mathbf {H}}+O_{p}(c_{NTh}) \end{aligned}$$

(7.23)

then the denominator

$$\begin{aligned}&\frac{1}{T}\left[ \widehat{{\mathbf {F}}}^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})\widehat{{\mathbf {F}}}\right] +O_{p}(c_{NTh})\nonumber \\&\quad = \frac{1}{T}\sum _{t=1}^{T}\left( \widehat{{\mathbf {f}}}_{t}-{\mathbf {H}}^{\top }\varvec{\varSigma }_{F} {\mathbf {L}}\varvec{\varOmega }^{-1}_{xx}(U_{t}){\mathbf {X}}_{t} \right) \left( \widehat{{\mathbf {f}}}^{\top }_{t}-{\mathbf {X}}_{t}^{\top } \varvec{\varOmega }^{-1}_{xx}(U_{t}){\mathbf {L}}^{\top }\varvec{\varSigma }_{F}{\mathbf {H}}\right) \nonumber \\&\quad = {\mathbf {H}}^{\top }\varvec{\varSigma }_{F}{\mathbf {H}}-{\mathbf {H}}^{\top }\varvec{\varSigma }_{F}{\mathbf {L}}({\mathbf {L}}^{\top }\varvec{\varSigma }_{F}{\mathbf {L}}+\varvec{\varSigma }_{e})^{+}{\mathbf {L}}^{\top }\varvec{\varSigma }_{F}{\mathbf {H}} \equiv \varvec{\varSigma }^{-1}_{P} \end{aligned}$$

(7.24)

where we use the fact that $\frac{1}{T}\sum _{t=1}^{T}\widehat{{\mathbf {f}}}_{t}\widehat{{\mathbf {f}}}^{\top }_{t}={\mathbf {H}}^{\top }\varvec{\varSigma }_{F}{\mathbf {H}}+O_{p}(1/\sqrt{T}+1/\sqrt{N})$.

In the 2nd step, we find the limit of the term containing ${\mathbf {M}}$. Simply substituting $\widehat{{\mathbf {F}}}$ with ${\mathbf {M}}$ in (7.23), we have

$$\begin{aligned}&\frac{1}{T}\left[ \widehat{{\mathbf {F}}}^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}}){\mathbf {M}}\right] \nonumber \\&\quad = \frac{1}{T}\sum _{t=1}^{T}\left( \widehat{{\mathbf {f}}}_{t}-{\mathbf {H}}^{\top }\varvec{\varSigma }_{F} {\mathbf {L}}\varvec{\varOmega }^{-1}_{xx}(U_{t}){\mathbf {X}}_{t} \right) {\mathbf {X}}_{t}^{\top }\varvec{\beta }(U_{t})[1+O_{p}(c_{NTh})] O_{p}(c_{NTh})\nonumber \\&\quad =O_{p}(c_{NTh}^2) \end{aligned}$$

(7.25)

In the 3rd step, we will show the limit of the term containing ${\mathbf {F}}\varvec{\lambda }$. First, $({\mathbf {X}}_{t}^{\top },{\mathbf {0}}_{1\times p})$$\left[ \mathbf {{\widetilde{X}}}^{\top }(U_{t}){\mathbf {K}}_{h}(U_{t}) \mathbf {{\widetilde{X}}}(U_{t})\right] ^{+}\mathbf {{\widetilde{X}}}^{\top }(U_{t}){\mathbf {K}}_{h}(U_{t})({\mathbf {F}}-\widehat{{\mathbf {F}}}{\mathbf {H}}^{-1}) ={\mathbf {X}}_{t}^{\top } \varvec{\varOmega }^{-1}_{xx}(U_{t})O_{p}(\frac{1}{N}+\frac{1}{\sqrt{NTh}}+\frac{1}{T\sqrt{h}})$ according to (7.12) and (7.20). Then

$$\begin{aligned}&\frac{1}{T}\left[ \widehat{{\mathbf {F}}}^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})^{\top }{\mathbf {S}}({\mathbf {F}}-\widehat{{\mathbf {F}}}{\mathbf {H}}^{-1})\right] \nonumber \\&\quad = \frac{1}{T}\sum _{t=1}^{T}\left( \widehat{{\mathbf {f}}}_{t}-{\mathbf {H}}^{\top }\varvec{\varSigma }_{F} {\mathbf {L}}\varvec{\varOmega }^{-1}_{xx}(U_{t}){\mathbf {X}}_{t}+O_{p}(c_{NTh}) \right) \nonumber \\&\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {\mathbf {X}}_{t}^{\top } \varvec{\varOmega }^{-1}_{xx}(U_{t})O_{p}\left( \frac{1}{N}+\frac{1}{\sqrt{NTh}}+\frac{1}{T\sqrt{h}}\right) \nonumber \\&\quad = O_{p}(c_{NTh})O_{p}\left( \frac{1}{N}+\frac{1}{\sqrt{NTh}}+\frac{1}{T\sqrt{h}}\right) \end{aligned}$$

(7.26)

And next

$$\begin{aligned}&\frac{1}{T}\left[ \widehat{{\mathbf {F}}}^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})^{\top }({\mathbf {F}}-\widehat{{\mathbf {F}}}{\mathbf {H}}^{-1})\right] \nonumber \\&\quad = \frac{1}{T}\sum _{t=1}^{T}\left( \widehat{{\mathbf {f}}}_{t}-{\mathbf {H}}^{\top }\varvec{\varSigma }_{F} {\mathbf {L}}\varvec{\varOmega }^{-1}_{xx}(U_{t}){\mathbf {X}}_{t}+O_{p}(c_{NTh}) \right) ({\mathbf {f}}^{\top }_{t}-\widehat{{\mathbf {f}}}^{\top }_{t}{\mathbf {H}}^{-1})\nonumber \\&\quad = O_{p}(1/N+1/T)+ O_{p}(1/N+1/T)+O_{p}(c_{NTh})O_{p}(1/T+1/\sqrt{NT})\nonumber \\ \end{aligned}$$

(7.27)

Combining (7.26) and (7.27), we have

$$\begin{aligned}&\sqrt{T}\left[ \widehat{{\mathbf {F}}}^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})\widehat{{\mathbf {F}}}\right] ^{+}\widehat{{\mathbf {F}}}^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}}){\mathbf {F}}\varvec{\lambda }\nonumber \\&\quad ={\mathbf {H}}^{-1}\varvec{\lambda }+O_{p}\left( \frac{\sqrt{T}}{N}+\frac{1}{\sqrt{T}}+\frac{1}{N\sqrt{h}}+\frac{1}{h\sqrt{NT}}+ \frac{1}{Th}\right) \end{aligned}$$

(7.28)

So if $\sqrt{T}/N \rightarrow \tau$ and $\varvec{\lambda }\ne {\mathbf {0}}$, a bias term will show up.

In the last step, we show the limit of the term having $\varvec{\varvec{\epsilon }}$. We first observe that $\widehat{{\mathbf {F}}}^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})\varvec{\varvec{\epsilon }} = \frac{1}{T}\sum _{t=1}^{T}\left( \widehat{{\mathbf {f}}}_{t}-{\mathbf {H}}^{\top }\varvec{\varSigma }_{F} {\mathbf {L}}\varvec{\varOmega }^{-1}_{xx}(U_{t}){\mathbf {X}}_{t} \right) \epsilon _{t} [1+O_{p}(c_{NTh})]$, then $\frac{\sqrt{T}}{T}\left[ \widehat{{\mathbf {F}}}^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})\varvec{\epsilon }\right] {\mathop {\rightarrow }\limits ^{\mathrm {d}}} \text {N}(0,\sigma ^{2}\varvec{\varSigma }^{-1}_{P})$. All in all,

$$\begin{aligned} \sqrt{T}(\widehat{\varvec{\lambda }}_{P}-{\mathbf {H}}^{-1}\varvec{\lambda })&{\mathop {\rightarrow }\limits ^{\mathrm {d}}}&\text {N}({\mathbf {0}},\sigma ^2\varvec{\varSigma }_{P}) \end{aligned}$$

(7.29)

$\square$

Proof of Theorem 2

According to

$$\begin{aligned} \widehat{\varvec{\beta }}_{P}(u) = (\varvec{\mathrm {I}}_{p}, {\mathbf {0}}_{p\times p}) \left[ \mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u) \mathbf {{\widetilde{X}}}(u)\right] ^{+}\mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u)({\mathbf {Y}}-\widehat{{\mathbf {F}}}\widehat{\varvec{\lambda }}_{P}) \end{aligned}$$

we first have

$$\begin{aligned} \widehat{\varvec{\beta }}_{P}(u)-\varvec{\beta }(u)= & {} \,\frac{h^{2}}{2} (\varvec{\mathrm {I}}_{p}, {\mathbf {0}}_{p\times p}) \left[ \mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u) \mathbf {{\widetilde{X}}}(u)\right] ^{+}\mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u)\mathbf {{\widetilde{A}}}(u)\varvec{\beta }^{\prime \prime }(\xi )\nonumber \\&\quad +\, (\varvec{\mathrm {I}}_{p}, {\mathbf {0}}_{p\times p}) \left[ \mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u) \mathbf {{\widetilde{X}}}(u)\right] ^{+}\mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u)({\mathbf {F}}\varvec{\lambda }-\widehat{{\mathbf {F}}}\widehat{\varvec{\lambda }}_{P})\nonumber \\&\quad +\, (\varvec{\mathrm {I}}_{p}, {\mathbf {0}}_{p\times p}) \left[ \mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u) \mathbf {{\widetilde{X}}}(u)\right] ^{+}\mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u)\varvec{\varvec{\epsilon }}\nonumber \\\equiv & {} J_{1}+J_{2}+J_{3} \end{aligned}$$

(7.30)

where $\xi$ lies between $U_{t}$ and u.

For $J_{1}$, according to (7.13) and (7.12), we have $J_{1}=\frac{h^{2}}{2}\mu _{2}\varvec{\beta }^{\prime \prime }(\mu )+o_{p}(h^2)$.

For $J_{2}$, we focus on

$$\begin{aligned}&\frac{\sqrt{Th}}{T}{\mathbf {X}}^{\top }{\mathbf {K}}_{h}(u)({\mathbf {F}}\varvec{\lambda }-\widehat{{\mathbf {F}}}\widehat{\varvec{\lambda }}_{P}) =\frac{\sqrt{Th}}{T}{\mathbf {X}}^{\top }{\mathbf {K}}_{h}(u)({\mathbf {F}}-\widehat{{\mathbf {F}}}{\mathbf {H}}^{-1})\varvec{\lambda }\nonumber \\&\quad +\frac{\sqrt{Th}}{T}{\mathbf {X}}^{\top }{\mathbf {K}}_{h}(u)(\widehat{{\mathbf {F}}}-{\mathbf {F}}{\mathbf {H}})({\mathbf {H}}^{-1}\varvec{\lambda }-\widehat{\varvec{\lambda }}_{P}) +\frac{\sqrt{Th}}{T}{\mathbf {X}}^{\top }{\mathbf {K}}_{h}(u){\mathbf {F}}{\mathbf {H}}({\mathbf {H}}^{-1}\varvec{\lambda }-\widehat{\varvec{\lambda }}_{P})\nonumber \\&=O_{p}\left( \frac{\sqrt{Th}}{N}+\frac{1}{\sqrt{N}}+\frac{1}{\sqrt{T}}\right) +O_{p}(\sqrt{h}) \end{aligned}$$

(7.31)

For $J_{3}$, we have

$$\begin{aligned} \frac{\sqrt{Th}}{T}{\mathbf {X}}^{\top }{\mathbf {K}}_{h}(u)\varvec{\epsilon }= & {} \frac{\sqrt{Th}}{T}\sum _{t=1}^{T}{\mathbf {X}}_{t}\epsilon _{t}k_{h}(U_{t}-u){\mathop {\rightarrow }\limits ^{\mathrm {d}}}\text {N}({\mathbf {0}}, \nu _{0}\sigma ^2 f(u)({\mathbf {L}}^{\top }\varvec{\varSigma }_{F}{\mathbf {L}}+\varvec{\varSigma }_{e})) \end{aligned}$$

Combing these terms together yields

$$\begin{aligned} \sqrt{Th}(\widehat{\varvec{\beta }}_{P}(u)-\varvec{\beta }(u)-\frac{h^{2}}{2}\mu _{2}\varvec{\beta }^{\prime \prime }(\mu )){\mathop {\rightarrow }\limits ^{\mathrm {d}}}\text {N}\left( {\mathbf {0}}, \nu _{0}\frac{\sigma ^2}{f(u)}\left[ {\mathbf {L}}^{\top }\varvec{\varSigma }_{F}{\mathbf {L}}+\varvec{\varSigma }_{e}\right] ^{-1}\right) \end{aligned}$$

$\square$

Proof of Theorem 3

According to (7.29), it is easy to see

$$\begin{aligned} T\sigma ^{-2}\widehat{\varvec{\lambda }}_{P}^{\top }\varvec{\varSigma }^{-1}_{P}\widehat{\varvec{\lambda }}_{P}&{\mathop {\rightarrow }\limits ^{\mathrm {d}}}&\chi ^2(R,\psi _{\lambda }) \end{aligned}$$

(7.32)

where $\psi _{\lambda }=\lim _{N,T\rightarrow \infty } T \sigma ^2 \varvec{\lambda }^{\top }{\mathbf {H}}^{-1\top }\varvec{\varSigma }^{-1}_{P}{\mathbf {H}}^{-1}\varvec{\lambda }$. But $\varvec{\varSigma }^{-1}_{P}$ is unknown, it can be estimated by (7.24) and ${\widehat{\sigma }}_{P}^2=T^{-1}\widehat{\varvec{\varvec{\epsilon }}}^{T}\widehat{\varvec{\varvec{\epsilon }}}{\mathop {\rightarrow }\limits ^{\mathrm {P}}}\sigma ^2$ is obvious. $\square$

Proof of Theorem 4

The proof is similar to, even simpler than the one of Theorem 5. Hence, we omit it. $\square$

Proof of Theorem 5

$$\begin{aligned} \varvec{{\widehat{\lambda }}}_{C}^{0}= & {} \left[ \mathbf {{\overline{Z}}}^{0\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})\mathbf {{\overline{Z}}}^{0}\right] ^{-1}\mathbf {{\overline{Z}}}^{0\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})\left[ {\mathbf {M}}+{\mathbf {F}}\varvec{\lambda }+\varvec{\epsilon }\right] \nonumber \\ \end{aligned}$$

(7.33)

Part (a) To derive the asymptotical distribution of (7.33), we follow several steps. In the 1st step, we find the limit of the denominator in (7.33). According to (7.12) and (7.11), we have

$$\begin{aligned}&({\mathbf {X}}_{t}^{\top },{\mathbf {0}}_{1\times p})\left[ \mathbf {{\widetilde{X}}}^{\top }(U_{t}){\mathbf {K}}_{h}(U_{t}) \mathbf {{\widetilde{X}}}(U_{t})\right] ^{+}\mathbf {{\widetilde{X}}}^{\top }(U_{t}){\mathbf {K}}_{h}(U_{t})\mathbf {{\overline{Z}}}^{0}\nonumber \\&\quad = ({\mathbf {X}}_{t}^{\top } \varvec{\varOmega }^{-1}_{xx}(U_{t}){\mathbf {L}}^{\top }\varvec{\varSigma }_{F},{\mathbf {0}}_{1\times (d-R)})+O_{p}(c_{NTh}) \end{aligned}$$

(7.34)

then the denominator

$$\begin{aligned}&\frac{1}{T}\left[ \mathbf {{\overline{Z}}}^{0\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})\mathbf {{\overline{Z}}}^{0}\right] +O_{p}(c_{NTh})\nonumber \\&\quad = \frac{1}{T}\sum _{t=1}^{T}\left( \mathbf {{\overline{Z}}}^{0}_{t}-\left( \begin{array}{c} \varvec{\varSigma }_{F} {\mathbf {L}}\varvec{\varOmega }^{-1}_{xx}(U_{t}){\mathbf {X}}_{t}\\ {\mathbf {0}}_{(d-R)\times 1} \\ \end{array} \right) \right) \left( \mathbf {{\overline{Z}}}^{0\top }_{t}-({\mathbf {X}}_{t}^{\top } \varvec{\varOmega }^{-1}_{xx}(U_{t}){\mathbf {L}}^{\top }\varvec{\varSigma }_{F},{\mathbf {0}}_{1\times (d-R)})\right) \nonumber \\&\quad =\left( \begin{array}{cc} \varvec{\varSigma }_{F}-\varvec{\varSigma }_{F}{\mathbf {L}}({\mathbf {L}}^{\top }\varvec{\varSigma }_{F}{\mathbf {L}}+\varvec{\varSigma }_{e})^{+}{\mathbf {L}}^{\top }\varvec{\varSigma }_{F} &{} {\mathbf {0}}_{R\times (d-R)} \\ {\mathbf {0}}_{(d-R)\times R} &{} {\mathbf {B}}^{\top }_{-R}\varvec{\varSigma }_{v} {\mathbf {B}}_{-R} \\ \end{array} \right) \nonumber \\&\quad \equiv \varvec{\varSigma } \end{aligned}$$

(7.35)

In the 2nd step, we find the limit of the term containing ${\mathbf {M}}$ in (7.33). Simply substituting $\mathbf {{\overline{Z}}}^{0}$ with ${\mathbf {M}}$ in (7.34), we have

$$\begin{aligned}&\frac{1}{T}\left[ \mathbf {{\overline{Z}}}^{0\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}}){\mathbf {M}}\right] \nonumber \\&\quad = \frac{1}{T}\sum _{t=1}^{T}\left( \mathbf {{\overline{Z}}}^{0}_{t}-\left( \begin{array}{c} \varvec{\varSigma }_{F} {\mathbf {L}}\varvec{\varOmega }^{-1}_{xx}(U_{t}){\mathbf {X}}_{t}\\ {\mathbf {0}}_{(d-R)\times 1} \\ \end{array} \right) \right) {\mathbf {X}}_{t}^{\top }\varvec{\beta }(U_{t})[1+O_{p}(c_{NTh})] O_{p}(c_{NTh})\nonumber \\&\quad = O_{p}(c_{NTh}^2) \end{aligned}$$

(7.36)

In the 3rd step, we will show the limit of the term involving ${\mathbf {F}}\varvec{\lambda }$.

$$\begin{aligned}&\left[ \mathbf {{\overline{Z}}}^{0\top } (\varvec{\mathrm {I}}_{T}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})\mathbf {{\overline{Z}}}^{0}\right] ^{-1}\mathbf {{\overline{Z}}}^{0\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})\left[ {\mathbf {F}}\varvec{\lambda }-\mathbf {{\overline{Z}}}^{0}\left( \begin{array}{c} \varvec{\mathrm {I}}_{R\times R}\\ {\mathbf {0}}_{(d-R)\times R} \\ \end{array} \right) \varvec{\lambda }\right] \nonumber \\&\quad =-\left[ \mathbf {{\overline{Z}}}^{0\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})\mathbf {{\overline{Z}}}^{0}\right] ^{-1}\mathbf {{\overline{Z}}}^{0\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})\mathbf {{\overline{V}}}_{R}{\mathbf {D}}^{-1}_{R}\varvec{\lambda }\nonumber \\&\quad \equiv -\left[ \mathbf {{\overline{Z}}}^{0\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})\mathbf {{\overline{Z}}}^{0}\right] ^{-1}\left( {\mathbf {a}}+{\mathbf {b}}\right) \end{aligned}$$

(7.37)

For ${\mathbf {a}}$, we have

$$\begin{aligned} \frac{\sqrt{T}}{T}{\mathbf {a}}= & {} \frac{\sqrt{T}}{T}\mathbf {{\overline{Z}}}^{0\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})^{\top }\mathbf {{\overline{V}}}_{R}{\mathbf {D}}^{-1}_{R}\varvec{\lambda }\nonumber \\= & {} \frac{\sqrt{T}}{T}\sum _{t=1}^{T}\left( \mathbf {{\overline{Z}}}^{0}_{t}-\left( \begin{array}{c} \varvec{\varSigma }_{F} {\mathbf {L}}\varvec{\varOmega }^{-1}_{xx}(U_{t}){\mathbf {X}}_{t}\\ {\mathbf {0}}_{(d-R)\times 1} \\ \end{array} \right) \right) \mathbf {{\overline{V}}}_{R}{\mathbf {D}}^{-1}_{R}\varvec{\lambda } [1+O_{p}(c_{NTh})]\nonumber \\= & {} \left( \begin{array}{c} \sqrt{T}/T\mathbf {F^{\top }}\mathbf {{\overline{V}}}_{R} +\sqrt{T}/T{\mathbf {D}}_{R}^{-1\top }\mathbf {{\overline{V}}}^{\top }_{R}\mathbf {{\overline{V}}}_{R}\\ \sqrt{NT}/T\mathbf {{\overline{V}}}^{\top }_{-R}\mathbf {{\overline{V}}}_{R} -\sqrt{NT}/T({\mathbf {D}}_{R}^{-1}{\mathbf {D}}_{-R})^{\top }\mathbf {{\overline{V}}}^{\top }_{R}\mathbf {{\overline{V}}}_{R}\\ \end{array} \right) {\mathbf {D}}^{-1}_{R}\varvec{\lambda } [1+O_{p}(c_{NTh})]\nonumber \\= & {} \left( \begin{array}{c} {\mathbf {0}}_{R\times R}+O_{p}(1/\sqrt{N})+O_{p}(\sqrt{T}/N)\\ {\mathbf {0}}_{(d-R)\times R}+O_{p}(\sqrt{NT}/N)\\ \end{array} \right) {\mathbf {D}}^{-1}_{R}\varvec{\lambda } [1+O_{p}(c_{NTh})] \end{aligned}$$

(7.38)

The implication of (7.38) is that if $\sqrt{NT}/N \nrightarrow 0$ (which is true if $N/T\rightarrow \tau$ (a constant)), then $\sqrt{T}/{T}{\mathbf {a}}$ is not $o_{p}(1)$.

For ${\mathbf {b}}$, we first have $T^{-1}\mathbf {{\widetilde{X}}}^{\top }(U_{t}){\mathbf {K}}_{h}(U_{t})\mathbf {{\overline{V}}}_{R}=O_{p}(1/N+1/\sqrt{NTh})$, then we have $\sqrt{T}({\mathbf {X}}_{t}^{\top },{\mathbf {0}}_{1\times p})\left[ \mathbf {{\widetilde{X}}}^{\top }(U_{t}){\mathbf {K}}_{h}(U_{t}) \mathbf {{\widetilde{X}}}(U_{t})\right] ^{+}\mathbf {{\widetilde{X}}}^{\top }(U_{t}){\mathbf {K}}_{h}(U_{t})\mathbf {{\overline{V}}}_{R} = {\mathbf {X}}_{t}^{\top } \varvec{\varOmega }^{-1}_{xx}(U_{t})O_{p}(1/\sqrt{Nh}+\sqrt{T}/N)$. So

$$\begin{aligned} \frac{\sqrt{T}}{T}{\mathbf {b}}= & {} \frac{\sqrt{T}}{T}\mathbf {{\overline{Z}}}^{0\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})^{\top }{\mathbf {S}}\mathbf {{\overline{V}}}_{R}{\mathbf {D}}^{-1}_{R}\varvec{\lambda }\nonumber \\= & {} \frac{1}{T}\sum _{t=1}^{T}\left( \mathbf {{\overline{Z}}}^{0}_{t}-\left( \begin{array}{c} \varvec{\varSigma }_{F} {\mathbf {L}}\varvec{\varOmega }^{-1}_{xx}{\mathbf {X}}_{t}\\ {\mathbf {0}}_{(d-R)\times 1} \\ \end{array} \right) \right) {\mathbf {X}}_{t}^{\top } \varvec{\varOmega }^{-1}_{xx}{\mathbf {D}}^{-1}_{R}\varvec{\lambda }\nonumber \\&\times O_{p}\left( \frac{1}{\sqrt{Nh}}+\frac{\sqrt{T}}{N}\right) [1+O_{p}(c_{NTh})]\nonumber \\= & {} \left[ \left( \begin{array}{c} \varvec{\varSigma }_{F}{\mathbf {L}}\\ {\mathbf {0}}\\ \end{array} \right) -\left( \begin{array}{c} \varvec{\varSigma }_{F}{\mathbf {L}}\\ {\mathbf {0}}\\ \end{array} \right) +O_{p}\left( \frac{1}{\sqrt{T}}+\frac{1}{\sqrt{N}}\right) \right] \varvec{\varOmega }^{-1}_{xx}{\mathbf {D}}^{-1}_{R}\varvec{\lambda }\nonumber \\&\times O_{p}\left( \frac{1}{\sqrt{Nh}}+\frac{\sqrt{T}}{N}\right) \nonumber \\= & {} O_{p}\left( \frac{1}{\sqrt{NTh}}+\frac{1}{N}+\frac{1}{N\sqrt{h}}+\frac{\sqrt{T}}{N\sqrt{N}}\right) \end{aligned}$$

(7.39)

The implication of (7.39) is that $\sqrt{T}/T {\mathbf {b}}$ is negligible if $N^2 h\rightarrow \infty$ and $T/N^{3}\rightarrow 0$ (both are true if $\sqrt{T}/N\rightarrow 0$ and $\sqrt{T}h\rightarrow \infty$).

In the last step, we show the limit of the term having $\varvec{\varvec{\epsilon }}$. We first observe that

$$\begin{aligned} \mathbf {{\overline{Z}}}^{0\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})\varvec{\varvec{\epsilon }} = \sum _{t=1}^{T}\left( \mathbf {{\overline{Z}}}^{0}_{t}-\left( \begin{array}{c} \varvec{\varSigma }_{F} {\mathbf {L}}\varvec{\varOmega }^{-1}_{xx}(U_{t}){\mathbf {X}}_{t}\\ {\mathbf {0}}_{(d-R)\times 1} \\ \end{array} \right) \right) \epsilon _{t} [1+O_{p}(c_{NTh})]\nonumber \\ \end{aligned}$$

(7.40)

then

$$\begin{aligned} \frac{\sqrt{T}}{T}\mathbf {{\overline{Z}}}^{0\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})\varvec{\epsilon }&{\mathop {\rightarrow }\limits ^{\mathrm {d}}}&\text {N}({\mathbf {0}},\sigma ^{2}\varvec{\varSigma }) \end{aligned}$$

(7.41)

where $\varvec{\varSigma }$ is expressed in (7.35). Combining these terms, we have

$$\begin{aligned}&\varvec{{\widehat{\lambda }}}_{C}^{0}-\left( \begin{array}{c} \varvec{\mathrm {I}}_{R\times R}\\ {\mathbf {0}}_{(d-R)\times R} \\ \end{array} \right) \varvec{\lambda }=O_{p}(c^2_{NTh})+O_{p}(1/\sqrt{T})\nonumber \\&\quad - \lim _{N,T\rightarrow \infty } \varvec{\varSigma }^{-1}\frac{1}{\sqrt{T}}\left( \begin{array}{c} \sqrt{T}/T[\mathbf {F^{\top }}\mathbf {{\overline{V}}}_{R}+{\mathbf {D}}_{R}^{-1\top }\mathbf {{\overline{V}}}^{\top }_{R}\mathbf {{\overline{V}}}_{R}]\\ \frac{\sqrt{NT}}{T}\left[ \mathbf {{\overline{V}}}^{\top }_{-R}\mathbf {{\overline{V}}}_{R} -({\mathbf {D}}_{R}^{-1}{\mathbf {D}}_{-R})^{\top }\mathbf {{\overline{V}}}^{\top }_{R}\mathbf {{\overline{V}}}_{R}\right] \\ \end{array} \right) {\mathbf {D}}^{-1}_{R}\varvec{\lambda }\nonumber \\ \end{aligned}$$

(7.42)

$\square$

Part (b) Due to $\varvec{{\widehat{\lambda }}}_{C}={\mathbf {B}}{\mathbf {A}}_{N} \varvec{{\widehat{\lambda }}}_{C}^{0}$, we pre-multiply ${\mathbf {B}}{\mathbf {A}}_{N}$ to (7.42). First we observe that ${\mathbf {B}}{\mathbf {A}}_{N}\varvec{\varSigma }^{-1}=\left( \begin{array}{cc} {\mathbf {D}}_{R}^{-1}\varvec{\varSigma }^{-1}_{11} &{} -\sqrt{N}{\mathbf {D}}_{R}^{-1}{\mathbf {D}}_{-R} \varvec{\varSigma }^{-1}_{22} \\ {\mathbf {0}}_{(d-R)\times R}&{} \sqrt{N}\varvec{\varSigma }^{-1}_{22} \\ \end{array} \right)$ where $\varvec{\varSigma }_{11}$ and $\varvec{\varSigma }_{22}$ refer to the two diagonal block matrices in $\varvec{\varSigma }$ respectively. So if $\sqrt{T}/N\rightarrow 0$, $Th^8\rightarrow 0$, then $\frac{\sqrt{T}}{\sqrt{N}}\Vert {\mathbf {B}}{\mathbf {A}}_{N}\varvec{\varSigma }^{-1}\Vert O_{p}(c^2_{NTh})=o_{p}(1)$.

Next, pre-multiplying $\frac{\sqrt{T}}{\sqrt{N}}{\mathbf {B}}{\mathbf {A}}_{N}$ to the 3rd term of the right hand side of (7.42) yields

$$\begin{aligned}&\lim _{N,T\rightarrow \infty }\frac{1}{\sqrt{N}}{\mathbf {B}}{\mathbf {A}}_{N}\varvec{\varSigma }^{-1}\left( \begin{array}{c} \sqrt{T}/T[\mathbf {F^{\top }}\mathbf {{\overline{V}}}_{R}+{\mathbf {D}}_{R}^{-1\top }\mathbf {{\overline{V}}}^{\top }_{R}\mathbf {{\overline{V}}}_{R}]\\ \frac{\sqrt{NT}}{T}\left[ \mathbf {{\overline{V}}}^{\top }_{-R}\mathbf {{\overline{V}}}_{R} -({\mathbf {D}}_{R}^{-1}{\mathbf {D}}_{-R})^{\top }\mathbf {{\overline{V}}}^{\top }_{R}\mathbf {{\overline{V}}}_{R}\right] \\ \end{array} \right) {\mathbf {D}}^{-1}_{R}\varvec{\lambda }\\&\quad = \lim _{N,T\rightarrow \infty } \left( \begin{array}{c} -{\mathbf {D}}^{-1}_{R}{\mathbf {D}}_{-R} \varvec{\varSigma }^{-1}_{22}\frac{\sqrt{NT}}{T}\left[ \mathbf {{\overline{V}}}^{\top }_{-R}\mathbf {{\overline{V}}}_{R} -({\mathbf {D}}_{R}^{-1}{\mathbf {D}}_{-R})^{\top }\mathbf {{\overline{V}}}^{\top }_{R}\mathbf {{\overline{V}}}_{R}\right] \\ \frac{\sqrt{NT}}{T}\varvec{\varSigma }^{-1}_{22}\left[ \mathbf {{\overline{V}}}^{\top }_{-R}\mathbf {{\overline{V}}}_{R} -({\mathbf {D}}_{R}^{-1}{\mathbf {D}}_{-R})^{\top }\mathbf {{\overline{V}}}^{\top }_{R}\mathbf {{\overline{V}}}_{R}\right] \\ \end{array} \right) {\mathbf {D}}^{-1}_{R}\varvec{\lambda } \end{aligned}$$

At last $\varvec{\varSigma }_{C}$ can be derived from $\lim _{N,T\rightarrow \infty }\frac{T}{N}\frac{1}{T} {\mathbf {B}}{\mathbf {A}}_{N}\varvec{\varSigma }^{-1}A_{N}^{\top }{\mathbf {B}}^{\top }$. $\blacksquare$

Proof of Theorem 6

According to (13)

$$\begin{aligned} \widehat{\varvec{\beta }}_{C}(u) = (\varvec{\mathrm {I}}_{p}, {\mathbf {0}}_{p\times p}) \left[ \mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u) \mathbf {{\widetilde{X}}}(u)\right] ^{+}\mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u)({\mathbf {Y}}-\mathbf {{\overline{Z}}}\varvec{{\widehat{\lambda }}}_{C}) \end{aligned}$$

we first have

$$\begin{aligned} \widehat{\varvec{\beta }}_{C}(u)-\varvec{\beta }(u)= & {} \frac{h^{2}}{2} (\varvec{\mathrm {I}}_{p}, {\mathbf {0}}_{p\times p}) \left[ \mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u) \mathbf {{\widetilde{X}}}(u)\right] ^{+}\mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u)\mathbf {{\widetilde{A}}}(u)\varvec{\beta }^{\prime \prime }(\xi )\nonumber \\&\quad + (\varvec{\mathrm {I}}_{p}, {\mathbf {0}}_{p\times p}) \left[ \mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u) \mathbf {{\widetilde{X}}}(u)\right] ^{+}\mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u)({\mathbf {F}}\varvec{\lambda }-\mathbf {{\overline{Z}}} \varvec{{\widehat{\lambda }}}_{C})\nonumber \\&\quad + (\varvec{\mathrm {I}}_{p}, {\mathbf {0}}_{p\times p}) \left[ \mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u) \mathbf {{\widetilde{X}}}(u)\right] ^{+}\mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u)\varvec{\varvec{\epsilon }}\nonumber \\\equiv & {} J_{1}+J_{2}+J_{3} \end{aligned}$$

(7.43)

where $\xi$ lies between $U_{t}$ and u.

For $J_{2}$, we first decompose it into

$$\begin{aligned} J_{2}= & {} (\varvec{\mathrm {I}}_{p}, {\mathbf {0}}_{p\times p}) \left[ \mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u) \mathbf {{\widetilde{X}}}(u)\right] ^{+}\mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u)\left[ {\mathbf {F}}\varvec{\lambda }-\mathbf {{\overline{Z}}}^{0}\left( \begin{array}{c} \varvec{\mathrm {I}}_{R\times R}\\ {\mathbf {0}}_{(d-R)\times R} \\ \end{array} \right) \varvec{\lambda }\right] \nonumber \\&\quad +\, (\varvec{\mathrm {I}}_{p}, {\mathbf {0}}_{p\times p}) \left[ \mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u) \mathbf {{\widetilde{X}}}(u)\right] ^{+}\mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u)\nonumber \\&\quad \times \left[ \mathbf {{\overline{Z}}}^{0}\left( \begin{array}{c} \varvec{\mathrm {I}}_{R\times R}\\ {\mathbf {0}}_{(d-R)\times R} \\ \end{array} \right) \varvec{\lambda }-\mathbf {{\overline{Z}}}^{0} \varvec{{\widehat{\lambda }}}_{C}^{0}\right] \nonumber \\\equiv & {} \, J_{21}+J_{22} \end{aligned}$$

(7.44)

We can first observe that $T^{-1} {\mathbf {X}}^{\top }{\mathbf {K}}_{h}(u)\mathbf {{\overline{V}}}=T^{-1} {\mathbf {L}}^{\top }{\mathbf {F}}^{\top }{\mathbf {K}}_{h}(u)\overline{{\mathbf {V}}} +T^{-1}\sum _{t=1}^{T}{\mathbf {e}}_{t}\overline{{\mathbf {V}}}^{\top }_{t}k_{h}(U_{t}-u) =O_{p}(1/\sqrt{NTh})+N^{-1}f(u)\varvec{\varSigma }_{ev}+O_{p}(h^2/N)+O_{p}(1/(N\sqrt{Th}))$ where $\varvec{\varSigma }_{ev}=\mathrm {E}({\mathbf {e}}_{t}{\mathbf {V}}_{jt}^{\top })$. Then $J_{21}$ can be rewritten as

$$\begin{aligned} \sqrt{Th}J_{21}= & {} \sqrt{Th}(\varvec{\mathrm {I}}_{p}, {\mathbf {0}}_{p\times p}) \left[ \mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u) \mathbf {{\widetilde{X}}}(u)\right] ^{+}\mathbf {{\widetilde{X}}}^{\top }(u){\mathbf {K}}_{h}(u)\mathbf {{\overline{V}}}_{R}{\mathbf {D}}^{-1}_{R}\varvec{\lambda }\\= & {} \frac{\sqrt{Th}}{N}[{\mathbf {L}}^{\top }\varvec{\varSigma }_{F}{\mathbf {L}}+\varvec{\varSigma }_{e}]^{-1}\varvec{\varSigma }_{ev,R}{\mathbf {D}}^{-1}_{R}\varvec{\lambda }+O_{p}\left( \frac{1}{\sqrt{N}}\right) \end{aligned}$$

where $\varvec{\varSigma }_{ev,R}$ is the left $p\times R$ block matrix of $\varvec{\varSigma }_{ev}$.

For $J_{22}$, according to (7.12) and (7.11), it can be rewritten as

$$\begin{aligned}&\sqrt{Th}J_{22}\nonumber \\&\quad = (\varvec{\mathrm {I}}_{p}, {\mathbf {0}}_{p\times p})\left( \begin{array}{cc} \varvec{\varOmega }_{xx}(u) &{} {\mathbf {0}}_{p\times p} \\ {\mathbf {0}}_{p\times p} &{} \mu _{2}\varvec{\varOmega }_{xx}(u) \\ \end{array} \right) ^{+} \left( \begin{array}{cc} {\mathbf {L}}^{\top }\varvec{\varSigma }_{F} &{} {\mathbf {0}}_{p\times (d-R)} \\ {\mathbf {0}}_{p\times R} &{} {\mathbf {0}}_{p\times (d-R)} \\ \end{array} \right) [1+O_{p}(c_{NTh})]\nonumber \\&\qquad \times \,\sqrt{Th}\left[ \left( \begin{array}{c} \varvec{\mathrm {I}}_{R\times R}\\ {\mathbf {0}}_{(d-R)\times R} \\ \end{array} \right) \varvec{\lambda }- \varvec{{\widehat{\lambda }}}_{C}^{0}\right] \nonumber \\&\quad = \left( \varvec{\varOmega }_{xx}(u)^{+}{\mathbf {L}}^{\top }\varvec{\varSigma }_{F},{\mathbf {0}}_{p\times (d-R)}\right) \sqrt{Th}\left[ \left( \begin{array}{c} \varvec{\mathrm {I}}_{R\times R}\\ {\mathbf {0}}_{(d-R)\times R} \\ \end{array} \right) \varvec{\lambda }- \varvec{{\widehat{\lambda }}}_{C}^{0}\right] [1+O_{p}(c_{NTh})]\nonumber \\ \end{aligned}$$

(7.45)

Therefore the asymptotic properties of $\sqrt{Th}J_{22}$ depend on (7.42). We first look at the order $O_{p}(c^2_{NTh})$ which comes from (7.36). Thus multiplying $\sqrt{Th}$ to (7.36) and selecting the upper R rows, we have

$$\begin{aligned}&\frac{\sqrt{Th}}{T}(\varvec{\mathrm {I}}_{R}, {\mathbf {0}}_{R\times (d-R)})\left[ \mathbf {{\overline{Z}}}^{0\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}_{T}-{\mathbf {S}}){\mathbf {M}}\right] =(\varvec{\mathrm {I}}_{R}, {\mathbf {0}}_{R\times (d-R)})\nonumber \\&\frac{\sqrt{Th}}{T}\sum _{t=1}^{T}\left( \mathbf {{\overline{Z}}}^{0}_{t}-\left( \begin{array}{c} \varvec{\varSigma }_{F} {\mathbf {L}}\varvec{\varOmega }^{-1}_{xx}(U_{t}){\mathbf {X}}_{t}\\ {\mathbf {0}}_{(d-R)\times 1} \\ \end{array} \right) \right) {\mathbf {X}}_{t}^{\top }\varvec{\beta }(U_{t})[1+O_{p}(c_{NTh})] O_{p}(c_{NTh})\nonumber \\&\quad = O_{p}(c_{NTh})\frac{\sqrt{Th}}{T}\sum _{t=1}^{T}[{\mathbf {f}}_{t}{\mathbf {X}}_{t}^{\top }\varvec{\beta }(U_{t})+\mathbf {{\overline{V}}}_{t}{\mathbf {X}}_{t}^{\top }\varvec{\beta }(U_{t})]\nonumber \\&\qquad -O_{p}(c_{NTh})\frac{\sqrt{Th}}{T}\sum _{t=1}^{T} \varvec{\varSigma }_{F} {\mathbf {L}}\varvec{\varOmega }^{-1}_{xx}(U_{t}){\mathbf {X}}_{t}{\mathbf {X}}_{t}^{\top }\varvec{\beta }(U_{t}) \nonumber \\&\quad = O_{p}(\sqrt{Th}/N)O_{p}(c_{NTh}) \end{aligned}$$

(7.46)

For the 2nd term of the right hand side of (7.42), multiplying $\sqrt{Th}$ to it yields $O_{p}(\sqrt{h})$. Left-multiplying $\sqrt{Th}\left( \varvec{\varOmega }_{xx}(u)^{+}{\mathbf {L}}^{\top }\varvec{\varSigma }_{F},{\mathbf {0}}_{p\times (d-R)}\right)$ to the 3rd term yields $\varvec{\varOmega }_{xx}(u)^{+}{\mathbf {L}}^{\top }\varvec{\varSigma }_{F}\varvec{\varSigma }_{11} ^{-1}\frac{\sqrt{Th}}{T}[\mathbf {F^{\top }}\mathbf {{\overline{V}}}_{R}+{\mathbf {D}}_{R}^{-1\top }\mathbf {{\overline{V}}}^{\top }_{R}\mathbf {{\overline{V}}}_{R}] =O_{p}(\sqrt{h}/\sqrt{N})+\frac{\sqrt{Th}}{T}\varvec{\varOmega }_{xx}(u)^{+}{\mathbf {L}}^{\top }\varvec{\varSigma }_{F}\varvec{\varSigma }_{11} ^{-1}{\mathbf {D}}_{R}^{-1\top }\mathbf {{\overline{V}}}^{\top }_{R}\mathbf {{\overline{V}}}_{R}$ where the latter term is of order $O_{p}(\sqrt{Th}/N)$.

Therefore, $\sqrt{Th}J_{2}=O_{p}(\frac{\sqrt{Th}}{N})+O_{p}(\frac{1}{\sqrt{N}})+O_{p}(\sqrt{Th}/N)O_{p}(c_{NTh})$. $\blacksquare$

Proofs of Theorems 7 and 8. We left multiply $(\varvec{\mathrm {I}}_{R\times R},{\mathbf {0}}_{R\times (d-R)})$ to (7.42), then we have

$$\begin{aligned} \sqrt{T}\left( {\mathbf {D}} \varvec{{\widehat{\lambda }}}_{C}-\varvec{\lambda }\right)&{\mathop {\rightarrow }\limits ^{\mathrm {d}}}&\text {N}({\mathbf {0}},\sigma ^{2}\varvec{\varSigma }_{11} ^{-1}) \end{aligned}$$

(7.47)

where $\varvec{\varSigma }_{11}$ is the upper left block matrix $\varvec{\varSigma }_{F}-\varvec{\varSigma }_{F}{\mathbf {L}}({\mathbf {L}}^{\top }\varvec{\varSigma }_{F}{\mathbf {L}}+\varvec{\varSigma }_{v})^{+}{\mathbf {L}}^{\top }\varvec{\varSigma }_{F}$ of $\varvec{\varSigma }$. Based upon (7.47), a Wald test statistic can be constructed

$$\begin{aligned} T \varvec{{\widehat{\lambda }}}_{C}^{\top }{\mathbf {D}}^{\top }(\sigma ^{-2}\varvec{\varSigma }_{11} ){\mathbf {D}} \varvec{{\widehat{\lambda }}}_{C} {\mathop {\rightarrow }\limits ^{\mathrm {d}}}\chi ^{2}(R,\psi _{D}) \end{aligned}$$

To estimate ${\mathbf {D}}^{\top }\varvec{\varSigma }_{11}{\mathbf {D}}$, we use $T^{-1}\overline{{\mathbf {Z}}}^{\top }\overline{{\mathbf {Z}}}-T^{-1}\overline{{\mathbf {Z}}}^{\top }{\mathbf {X}} [T^{-1}{\mathbf {X}}^{\top }{\mathbf {X}}]^{-1}T^{-1}{\mathbf {X}}^{\top }\overline{{\mathbf {Z}}}$. But note that the rank is d rather than R. Hence replacing ${\mathbf {D}}^{\top }\varvec{\varSigma }_{11}{\mathbf {D}}$ with the estimator, the chi-squared distribution has d degrees of freedom. To see this,

$$\begin{aligned}&T \sigma ^{-2} \varvec{{\widehat{\lambda }}}_{C}^{\top }\left[ T^{-1}\overline{{\mathbf {Z}}}^{\top }\overline{{\mathbf {Z}}}-T^{-1}\overline{{\mathbf {Z}}}^{\top }{\mathbf {X}} (T^{-1}{\mathbf {X}}^{\top }{\mathbf {X}})^{-1}T^{-1}{\mathbf {X}}^{\top }\overline{{\mathbf {Z}}}\right] \varvec{{\widehat{\lambda }}}_{C}\nonumber \\&\quad = T \sigma ^{-2} \varvec{{\widehat{\lambda }}}_{C}^{0\top }\left[ T^{-1}\overline{{\mathbf {Z}}}^{0\top }\overline{{\mathbf {Z}}}^{0}-T^{-1}\overline{{\mathbf {Z}}}^{0\top }{\mathbf {X}} (T^{-1}{\mathbf {X}}^{\top }{\mathbf {X}})^{-1}T^{-1}{\mathbf {X}}^{\top }\overline{{\mathbf {Z}}}^{0}\right] \varvec{{\widehat{\lambda }}}^{0}_{C}\nonumber \\&\quad {\mathop {\rightarrow }\limits ^{\mathrm {P}}} T\sigma ^{-2} \varvec{{\widehat{\lambda }}}_{C} ^{0\top }\varvec{\varSigma } \varvec{{\widehat{\lambda }}}_{C}^{0} \end{aligned}$$

(7.48)

Based on (7.42), we can get the desired result. The convergency of the estimator ${\widehat{\sigma }}_{C}^2$ will be dealt with in the following theorem. $\blacksquare$

Proof of Theorem 9. In the first place, we will show that $T^{-1}{\mathrm {RSS}}_{1}=\sigma ^{2}(1+o_{p}(1))$. Let us start by decomposing

$$\begin{aligned} {\mathrm {RSS}}_{1}&= [{\mathbf {Y}}-\mathbf {{\widetilde{M}}}-\mathbf {{\overline{Z}}} \varvec{{\widehat{\lambda }}}_{C}]^{\top }[{\mathbf {Y}}-\mathbf {{\widetilde{M}}}-\mathbf {{\overline{Z}}} \varvec{{\widehat{\lambda }}}_{C}]\nonumber \\&\quad = [{\mathbf {F}}\varvec{\lambda }-\mathbf {{\overline{Z}}} \varvec{{\widehat{\lambda }}}_{C}+{\mathbf {M}}+\varvec{\epsilon }]^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})[{\mathbf {F}}\varvec{\lambda }-\mathbf {{\overline{Z}}} \varvec{{\widehat{\lambda }}}_{C}+{\mathbf {M}}+\varvec{\epsilon }]\nonumber \\&= \varvec{\epsilon }^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})\varvec{\epsilon } +[{\mathbf {F}}\varvec{\lambda }-\mathbf {{\overline{Z}}} \varvec{{\widehat{\lambda }}}_{C}]^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})[{\mathbf {F}}\varvec{\lambda }-\mathbf {{\overline{Z}}} \varvec{{\widehat{\lambda }}}_{C}]\nonumber \\&\quad +{\mathbf {M}}^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}}){\mathbf {M}} +\varvec{\epsilon }^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}}){\mathbf {M}}+{\mathbf {M}}^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})\varvec{\epsilon }\nonumber \\&\quad +\,{\mathbf {M}}^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})[{\mathbf {F}}\varvec{\lambda }-\mathbf {{\overline{Z}}} \varvec{{\widehat{\lambda }}}_{C}] +[{\mathbf {F}}\varvec{\lambda }-\mathbf {{\overline{Z}}} \varvec{{\widehat{\lambda }}}_{C}]^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}}){\mathbf {M}}\nonumber \\&\quad +\,\varvec{\epsilon }^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})[{\mathbf {F}}\varvec{\lambda }-\mathbf {{\overline{Z}}} \varvec{{\widehat{\lambda }}}_{C}] +[{\mathbf {F}}\varvec{\lambda }-\mathbf {{\overline{Z}}} \varvec{{\widehat{\lambda }}}_{C}]^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})\varvec{\epsilon }\nonumber \\&\equiv J_{1}+\cdots +J_{9} \end{aligned}$$

(7.49)

where $T^{-1}J_{1}=\sigma ^{2}(1+o_{p}(1))$, $T^{-1}J_{3}=O_{p}(c_{NTh}^{2})$, $T^{-1}J_{4}=T^{-1}J_{5}=O_{p}(c_{NTh})$ are easy to derive. Now we only focus on

$$\begin{aligned} T^{-1}J_{2}= & {} T^{-1} [{\mathbf {F}}(\varvec{\lambda }-{\mathbf {D}} \varvec{{\widehat{\lambda }}}_{C})-\mathbf {{\overline{V}}} \varvec{{\widehat{\lambda }}}_{C}] ^{\top } (\varvec{\mathrm {I}}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})[{\mathbf {F}}(\varvec{\lambda }-{\mathbf {D}} \varvec{{\widehat{\lambda }}}_{C})-\mathbf {{\overline{V}}} \varvec{{\widehat{\lambda }}}_{C}]\nonumber \\\equiv & {} J_{21}+J_{22}+J_{22}^{\top }+J_{24}=O_{p}\left( 1/T+1/N\right) \end{aligned}$$

where

$$\begin{aligned}&J_{21} = T^{-1} (\varvec{\lambda }-{\mathbf {D}} \varvec{{\widehat{\lambda }}}_{C}) ^{\top } \mathbf {F^{\top }} (\varvec{\mathrm {I}}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}}){\mathbf {F}}(\varvec{\lambda }-{\mathbf {D}} \varvec{{\widehat{\lambda }}}_{C}) =(\varvec{\lambda }-{\mathbf {D}} \varvec{{\widehat{\lambda }}}_{C}) ^{\top }\\&\quad \frac{1}{T}\sum _{t=1}^{T}\left( {\mathbf {f}}_{t}^{\top }-{\mathbf {X}}_{t}^{\top } \varvec{\varOmega }^{-1}_{xx}{\mathbf {L}}^{\top }\varvec{\varSigma }_{F}\right) ^{\top }\left( {\mathbf {f}}_{t}^{\top }-{\mathbf {X}}_{t}^{\top } \varvec{\varOmega }^{-1}_{xx}{\mathbf {L}}^{\top }\varvec{\varSigma }_{F}\right) (\varvec{\lambda }-{\mathbf {D}} \varvec{{\widehat{\lambda }}}_{C})[1+o_{p}(1)]\\&\quad =O_{p}(1/T)\\&\quad J_{22}= T^{-1} \varvec{{\widehat{\lambda }}}_{C} ^{\top }\mathbf {{\overline{V}}} ^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}}){\mathbf {F}}(\varvec{\lambda }-{\mathbf {D}} \varvec{{\widehat{\lambda }}}_{C})\\&\quad = \varvec{{\widehat{\lambda }}}_{C} ^{\top }\frac{1}{T}\sum _{t=1}^{T}\mathbf {{\overline{V}}}_{t}\left( {\mathbf {f}}_{t}^{\top }-{\mathbf {X}}_{t}^{\top } \varvec{\varOmega }^{-1}_{xx}{\mathbf {L}}^{\top }\varvec{\varSigma }_{F}\right) (\varvec{\lambda }-{\mathbf {D}} \varvec{{\widehat{\lambda }}}_{C})[1+o_{p}(1)]=O_{p}(1/(T\sqrt{N}))\\&J_{24}= T^{-1} \varvec{{\widehat{\lambda }}}_{C} ^{\top }\mathbf {{\overline{V}}} ^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})\mathbf {{\overline{V}}} \varvec{{\widehat{\lambda }}}_{C}= \varvec{{\widehat{\lambda }}}_{C} ^{\top }\frac{1}{T}\sum _{t=1}^{T}\mathbf {{\overline{V}}}_{t}\mathbf {{\overline{V}}}^{\top }_{t} \varvec{{\widehat{\lambda }}}_{C}[1+o_{p}(1)]=O_{p}(1/N) \end{aligned}$$

In a similar way, we have

$$\begin{aligned}&T^{-1}J_{6} =T^{-1}J_{7}= T^{-1} {\mathbf {M}}^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})[{\mathbf {F}}(\varvec{\lambda }-{\mathbf {D}} \varvec{{\widehat{\lambda }}}_{C})-\mathbf {{\overline{V}}} \varvec{{\widehat{\lambda }}}_{C}]\\&\quad =O_{p}\left( \frac{1}{\sqrt{T}}c_{NTh}\right) \\&T^{-1}J_{8} =T^{-1}J_{9}=T^{-1}\varvec{\epsilon }^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})[{\mathbf {F}}(\varvec{\lambda }-{\mathbf {D}} \varvec{{\widehat{\lambda }}}_{C})-\mathbf {{\overline{V}}} \varvec{{\widehat{\lambda }}}_{C}]\\&\quad =O_{p}\left( \frac{1}{T}+\frac{1}{\sqrt{NT}}\right) \end{aligned}$$

If the loadings in Y equations are zero, we do not need to apply CCE technique. Then

$$\begin{aligned}&{\mathrm {RSS}}_{0} = [{\mathbf {Y}}-\mathbf {{\widetilde{M}}}-\mathbf {{\overline{Z}}} \varvec{{\widehat{\lambda }}}_{C}+(\varvec{\mathrm {I}}-{\mathbf {S}})\mathbf {{\overline{Z}}} \varvec{{\widehat{\lambda }}}_{C}]^{\top }[{\mathbf {Y}}-\mathbf {{\widetilde{M}}}-\mathbf {{\overline{Z}}} \varvec{{\widehat{\lambda }}}_{C}+(\varvec{\mathrm {I}}-{\mathbf {S}})\mathbf {{\overline{Z}}} \varvec{{\widehat{\lambda }}}_{C}]\nonumber \\&\quad = {\mathrm {RSS}}_{1}+J_{1}+[{\mathbf {Y}}-\mathbf {{\widetilde{M}}}-\mathbf {{\overline{Z}}} \varvec{{\widehat{\lambda }}}_{C}]^{\top }[(\varvec{\mathrm {I}}-{\mathbf {S}})\mathbf {{\overline{Z}}} \varvec{{\widehat{\lambda }}}_{C}]+[(\varvec{\mathrm {I}}-{\mathbf {S}})\mathbf {{\overline{Z}}} \varvec{{\widehat{\lambda }}}_{C}]^{\top }[{\mathbf {Y}}-\mathbf {{\widetilde{M}}}-\mathbf {{\overline{Z}}} \varvec{{\widehat{\lambda }}}_{C}]\nonumber \\ \end{aligned}$$

(7.50)

where the last two terms converge to 0 while $T^{-1} J_{1} = T^{-1} [(\varvec{\mathrm {I}}-{\mathbf {S}})\mathbf {{\overline{Z}}} \varvec{{\widehat{\lambda }}}_{C}]^{\top }[(\varvec{\mathrm {I}}-{\mathbf {S}})\mathbf {{\overline{Z}}} \varvec{{\widehat{\lambda }}}_{C}] =T^{-1} \varvec{{\widehat{\lambda }}}_{C} ^{0\top }\mathbf {{\overline{Z}}} ^{0\top }(\varvec{\mathrm {I}}-{\mathbf {S}})^{\top }(\varvec{\mathrm {I}}-{\mathbf {S}})\mathbf {{\overline{Z}}}^{0} \varvec{{\widehat{\lambda }}}_{C}^{0}{\mathop {\rightarrow }\limits ^{\mathrm {P}}} \varvec{{\widehat{\lambda }}}_{C} ^{0\top }\varvec{\varSigma } \varvec{{\widehat{\lambda }}}_{C}^{0}$. Then

$$\begin{aligned} T^{-1}({\mathrm {RSS}}_{0}-{\mathrm {RSS}}_{1})=T^{-1}J_{1}&{\mathop {\rightarrow }\limits ^{\mathrm {P}}}&\varvec{{\widehat{\lambda }}}_{C} ^{0\top }\varvec{\varSigma } \varvec{{\widehat{\lambda }}}_{C}^{0} \end{aligned}$$

(7.51)

Based on (7.42), it is easy to obtain the result. $\square$