We see that, after integration with respect to $\mathbb{P}_n - P$, the i-th component of the first term on the right hand side of (B.61) can be written as a finite summation of summands

$$ \int \left( u(d(\mathbf{s}, \boldsymbol{\xi}_n)) - u(d(\mathbf{s}, \boldsymbol{\xi}_P)) \right) x_{ji} x_{st} d(\mathbb{P}_n - P)(\mathbf{s}) m_{n,js} \boldsymbol{\beta}_{n,t}, $$

for $i = 1, \dots, q$ and $s, t, j \in \{1, \dots, k\}$, where $m_{n,js} \to m_{P,js}$ and $\boldsymbol{\beta}_{n,t} \to \boldsymbol{\beta}_{P,t}$, in probability. All functions $u(d(\mathbf{s}, \boldsymbol{\xi}))x_{ji}y_s$, for $\boldsymbol{\xi} \in \mathbb{R}^q \times \mathbb{R}^l$, are members of the class

$$ \mathcal{F}_{ab} = \{ u(\|\mathbf{V}^{-1/2}(\mathbf{y} - \mathbf{X}\boldsymbol{\beta})\|)s_a s_b : \boldsymbol{\beta} \in \mathbb{R}^q, \mathbf{V} \in \text{PDS}(k) \}, $$

and $u$ is of bounded variation. According to Lemma B.7, the corresponding class of subgraphs has polynomial discrimination, so similar to (B.55) it follows that

$$ \int \left( u(d(\mathbf{s}, \boldsymbol{\xi}_n)) - u(d(\mathbf{s}, \boldsymbol{\xi}_P)) \right) x_{ji} x_{st} d(\mathbb{P}_n - P)(\mathbf{s}) = o_P(n^{-1/2}), $$

which means that

$$ \int \left( u(d(\mathbf{s}, \boldsymbol{\xi}_n)) - u(d(\mathbf{s}, \boldsymbol{\xi}_P)) \right) \mathbf{X}^T \mathbf{M}_n \mathbf{X} \boldsymbol{\beta}_n d(\mathbb{P}_n - P)(\mathbf{s}) = o_P(n^{-1/2}). \quad (B.62) $$

Next, consider the second term on the right hand side of (B.61). We then have to deal with summands of the form

$$ \int u(d(\mathbf{s}, \boldsymbol{\xi}_P)) x_{ji} x_{st} d(\mathbb{P}_n - P)(\mathbf{s}) (m_{n,js} \boldsymbol{\beta}_{n,t} - m_{P,js} \boldsymbol{\beta}_{P,t}), $$

for $i = 1, \dots, q$, and $s, t, j \in \{1, \dots, k\}$, where $m_{n,js} \to m_{P,js}$ and $\boldsymbol{\beta}_{n,t} \to \boldsymbol{\beta}_{P,t}$, in probability. According to the central limit theorem, the integral is of the order $O_P(n^{-1/2})$. Because, $m_{n,js} \boldsymbol{\beta}_{n,t} \to m_{P,js} \boldsymbol{\beta}_{P,t}$, in probability, it follows that the product is of the order $o_P(n^{-1/2})$. We conclude,

$$ \int u(d(\mathbf{s}, \boldsymbol{\xi}_P)) (\mathbf{X}^T \mathbf{M}_n \mathbf{X} \boldsymbol{\beta}_n - \mathbf{X}^T \mathbf{M}_P \mathbf{X} \boldsymbol{\beta}_P) d(\mathbb{P}_n - P)(\mathbf{s}) = o_P(n^{-1/2}). \quad (B.63) $$

Putting together (B.60) and (B.63), proves (B.49)

Finally, consider $\Psi_{2,j}$ in (B.48), with $\mathbf{H}_j$ defined (7.3). Write

$$ \begin{aligned} \mathbf{M}_n &= \mathbf{V}(\boldsymbol{\theta}_n)^{-1} \mathbf{H}_j(\boldsymbol{\theta}_n) \mathbf{V}(\boldsymbol{\theta}_n)^{-1} \\ \mathbf{M}_P &= \mathbf{V}(\boldsymbol{\theta}_P)^{-1} \mathbf{H}_j(\boldsymbol{\theta}_P) \mathbf{V}(\boldsymbol{\theta}_P)^{-1}, \end{aligned} $$

so that $\mathbf{M}_n \to \mathbf{M}_P$, in probability, according to condition (V4). Decompose $\Psi_{2,j}(\mathbf{s}, \boldsymbol{\xi}_n) - \Psi_{2,j}(\mathbf{s}, \boldsymbol{\xi}_P)$ as follows

$$ \left\{ u(d(\mathbf{s}, \boldsymbol{\xi}_n)) - u(d(\mathbf{s}, \boldsymbol{\xi}_P)) \right\} (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_n)^T \mathbf{M}_n (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_n) + u(d(\mathbf{s}, \boldsymbol{\xi}_P)) \left\{ (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_n)^T \mathbf{M}_n (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_n) - (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_P)^T \mathbf{M}_P (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_P) \right\}. \quad (B.64) $$

The first term in (B.64), can be written as the trace of the matrix

$$ \left\{ u(d(\mathbf{s}, \boldsymbol{\xi}_n)) - u(d(\mathbf{s}, \boldsymbol{\xi}_P)) \right\} (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_n)(\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_n)^T \mathbf{M}_n, $$

where $\boldsymbol{\beta}_n \to \boldsymbol{\beta}_P$ and $\mathbf{M}_n \to \mathbf{M}_P$, in probability. As before, we consider each single entry of this $k \times k$ matrix. The $(i, j)$-th element of $(\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_n)(\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_n)^T$ is equal to

$$ \begin{aligned} (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_n)_i (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_n)_j = y_i y_j & - \sum_{s=1}^{k} y_j x_{is} \boldsymbol{\beta}_{n,s} - \sum_{t=1}^{k} y_i x_{jt} \boldsymbol{\beta}_{n,t} \\ & + \sum_{s=1}^{k} \sum_{t=1}^{k} x_{is} x_{jt} \boldsymbol{\beta}_{n,s} \boldsymbol{\beta}_{n,t}. \end{aligned} \quad (B.65) $$
We see that the $(i, j)$-th entry of

$$ \{u(d(\mathbf{s}, \boldsymbol{\xi}_n)) - u(d(\mathbf{s}, \boldsymbol{\xi}_P))\}(\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_n)(\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_n)^T $$

is a combination of four summations. The last of these summations arising from (B.65), after integration with respect to $\mathbb{P}_n - P$, has summands

$$ \int (u(d(\mathbf{s}, \boldsymbol{\xi}_n)) - u(d(\mathbf{s}, \boldsymbol{\xi}_P))) x_{is} x_{jt} \, d(\mathbb{P}_n - P)(\mathbf{s}) \beta_{n,s} \beta_{n,t}, $$

where $\beta_{n,s}\beta_{n,t} \to \beta_{P,s}\beta_{P,t}$, in probability. All functions $u(d(\mathbf{s}, \boldsymbol{\xi}))x_{ji}y_s$, for $\boldsymbol{\xi} \in \mathbb{R}^q \times \mathbb{R}^l$, are members of the class

$$ \mathcal{F}_{ab} = \{u(\|\mathbf{V}^{-1/2}(\mathbf{y} - \mathbf{X}\boldsymbol{\beta})\|)s_a s_b : \boldsymbol{\beta} \in \mathbb{R}^q, \mathbf{V} \in \text{PDS}(k)\}, $$

where $u$ is of bounded variation. According to Lemma B.7, the corresponding class of subgraphs has polynomial discrimination, so similar to (B.55) it follows that

$$ \int \{u(d(\mathbf{s}, \boldsymbol{\xi}_n)) - u(d(\mathbf{s}, \boldsymbol{\xi}_P))\} x_{is} x_{jt} \, d(\mathbb{P}_n - P)(\mathbf{s}) = o_P(n^{-1/2}). $$

The other three summations that arise from the right hand side of (B.65) can be handled in the same way. It follows, that for each $i, j \in \{1, \dots, k\}$ fixed,

$$ \int \{u(d(\mathbf{s}, \boldsymbol{\xi}_n)) - u(d(\mathbf{s}, \boldsymbol{\xi}_P))\} (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_n)_i (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_n)_j \, d(\mathbb{P}_n - P)(\mathbf{s}) = o_P(n^{-1/2}), $$

which means that

$$ \int \{u(d(\mathbf{s}, \boldsymbol{\xi}_n)) - u(d(\mathbf{s}, \boldsymbol{\xi}_P))\} (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_n)(\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_n)^T \mathbf{M}_n \, d(\mathbb{P}_n - P)(\mathbf{s}) = o_P(n^{-1/2}). $$

After taking traces, we conclude that

$$ \int \{u(d(\mathbf{s}, \boldsymbol{\xi}_n)) - u(d(\mathbf{s}, \boldsymbol{\xi}_P))\} (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_n)^T \mathbf{M}_n (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_n) \, d(\mathbb{P}_n - P)(\mathbf{s}) = o_P(n^{-1/2}). \quad (B.66) $$

Next, consider the second term on the right hand side of (B.64). First, note that

$$ (\mathbf{y} - \mathbf{X}\boldsymbol{\beta})^T \mathbf{M}(\mathbf{y} - \mathbf{X}\boldsymbol{\beta}) = \sum_{i=1}^{k} \sum_{j=1}^{k} y_i y_j m_{ij} + \sum_{i=1}^{k} \sum_{j=1}^{k} y_i \sum_{s=1}^{k} x_{js} \beta_s m_{ij} \\ + \sum_{i=1}^{k} \sum_{j=1}^{k} y_j \sum_{t=1}^{k} x_{it} \beta_t m_{ij} + \sum_{i=1}^{k} \sum_{j=1}^{k} \sum_{s=1}^{k} \sum_{t=1}^{k} x_{is} x_{jt} \beta_s \beta_t m_{ij}. $$

This means that

$$ \begin{aligned} & (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_n)^T \mathbf{M}_n (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_n) - (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_P)^T \mathbf{M}_P (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_P) \\ & = \sum_{i=1}^{k} \sum_{j=1}^{k} y_i y_j (m_{n,ij} - m_{P,ij}) + \sum_{i=1}^{k} \sum_{j=1}^{k} \sum_{s=1}^{k} y_i x_{js} (\beta_{n,s} m_{n,ij} - \beta_{P,s} m_{P,ij}) \\ & \quad + \sum_{i=1}^{k} \sum_{j=1}^{k} \sum_{t=1}^{k} y_j x_{it} (\beta_{n,t} m_{n,ij} - \beta_{P,t} m_{P,ij}) \\ & \quad + \sum_{i=1}^{k} \sum_{j=1}^{k} \sum_{s=1}^{k} \sum_{t=1}^{k} x_{is} x_{jt} (\beta_{n,s} \beta_{n,t} m_{n,ij} - \beta_{P,s} \beta_{P,t} m_{P,ij}). \end{aligned} \quad (B.67) $$
We see that the (i, j)-th entry of

$$u(d(\mathbf{s}, \boldsymbol{\xi}_P)) \left\{ (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_n)^T \mathbf{M}_n (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_n) - (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_P)^T \mathbf{M}_P (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_P) \right\}$$

can be written as the combination of four summations. The last of the summations arising from (B.67), after integration with respect to $\mathbb{P}_n - P$, has summands

$$\int u(d(\mathbf{s}, \boldsymbol{\xi}_P))x_{is}x_{jt} d(\mathbb{P}_n - P)(\mathbf{s}) (\beta_{n,s}\beta_{n,t}m_{n,ij} - \beta_{P,s}\beta_{P,t}m_{P,ij}),$$

where $\beta_{n,s}\beta_{n,t}m_{n,ij} \to \beta_{P,s}\beta_{P,t}m_{P,ij}$, in probability. According to the central limit theorem, the integral is of the order $O_P(n^{-1/2})$, whereas the second term tends to zero. All functions $u(d(\mathbf{s}, \boldsymbol{\xi}))x_{is}x_{jt}$, for $\boldsymbol{\xi} \in \mathbb{R}^q \times \mathbb{R}^l$, are members of the class

$$\mathcal{F}_{ab} = \{u(\|\mathbf{V}^{-1/2}(\mathbf{y} - \mathbf{X}\boldsymbol{\beta})\|)s_a s_b : \boldsymbol{\beta} \in \mathbb{R}^q, \mathbf{V} \in \text{PDS}(k)\},$$

where $u$ is of bounded variation. According to Lemma B.7, the corresponding class of subgraphs has polynomial discrimination, so similar to (B.55) it follows that

$$\int u(d(\mathbf{s}, \boldsymbol{\xi}_P))x_{is}x_{jt} d(\mathbb{P}_n - P)(\mathbf{s}) (\beta_{n,s}\beta_{n,t}m_{n,ij} - \beta_{P,s}\beta_{P,t}m_{P,ij}) = o_P(n^{-1/2}).$$

The other three summations that arise from the right hand side of (B.67) can be handled in the same way, so that

$$\int u(d(\mathbf{s}, \boldsymbol{\xi}_P)) \left\{ (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_n)^T \mathbf{M}_n (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_n) - (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_P)^T \mathbf{M}_P (\mathbf{y} - \mathbf{X}\boldsymbol{\beta}_P) \right\} d(\mathbb{P}_n - P)(\mathbf{s}) \quad (B.68) \\ = o_P(n^{-1/2}).$$

Putting together (B.66) and (B.68), proves (B.50) for each $j = 1, \dots, l$. This finishes the proof of Lemma B.8. $\square$

## Proof of Corollary 6

*Proof.* As in the proof of Corollary 5, it follows that $\partial\Lambda/\partial\boldsymbol{\xi}$ is continuously differentiable with a non-singular derivative at $\boldsymbol{\xi}(P)$, so that Theorem 6 applies. According to Theorem 6 and Lemma 2, it follows that $\sqrt{n}(\boldsymbol{\beta}_n - \boldsymbol{\beta}(P))$ is asymptotically normal with mean zero and covariance matrix

$$\frac{1}{\alpha^2} (\mathbb{E}[\mathbf{X}^T \boldsymbol{\Sigma}^{-1} \mathbf{X}])^{-1} \mathbb{E}[\Psi_{\boldsymbol{\beta}}(\mathbf{s}, \boldsymbol{\xi}_P) \Psi_{\boldsymbol{\beta}}(\mathbf{s}, \boldsymbol{\xi}_P)^T] (\mathbb{E}[\mathbf{X}^T \boldsymbol{\Sigma}^{-1} \mathbf{X}])^{-1}$$

where $\Psi_{\boldsymbol{\beta}}$ is defined in (7.10). We find that

$$\mathbb{E}[\Psi_{\boldsymbol{\beta}}(\mathbf{s}, \boldsymbol{\xi}_P) \Psi_{\boldsymbol{\beta}}(\mathbf{s}, \boldsymbol{\xi}_P)^T] = \mathbb{E}[\mathbf{X}^T \mathbb{E}[u(d)^2 \boldsymbol{\Sigma}^{-1}(\mathbf{y} - \boldsymbol{\mu})(\mathbf{y} - \boldsymbol{\mu})^T \boldsymbol{\Sigma}^{-1} | \mathbf{X}] \mathbf{X}],$$

where $d^2 = (\mathbf{y} - \boldsymbol{\mu})^T \boldsymbol{\Sigma}^{-1} (\mathbf{y} - \boldsymbol{\mu})$ and $u(s) = \rho'(s)/s$. As before, with $\mathbf{z} = \boldsymbol{\Sigma}^{-1/2}(\mathbf{y} - \boldsymbol{\mu})$ and $\mathbf{u} = \mathbf{z}/\|\mathbf{z}\|$, according to Lemma B.4, the inner conditional expectation can be written as

$$\boldsymbol{\Sigma}^{-1/2} \mathbb{E}_{\mathbf{0}, \mathbf{I}_k} [u(\|\mathbf{z}\|)^2 \|\mathbf{z}\|^2] \mathbb{E}_{\mathbf{0}, \mathbf{I}_k} [\mathbf{u}\mathbf{u}^T] \boldsymbol{\Sigma}^{-1/2} = \frac{\mathbb{E}_{\mathbf{0}, \mathbf{I}_k} [u(\|\mathbf{z}\|)^2 \|\mathbf{z}\|^2]}{k} \boldsymbol{\Sigma}^{-1}.$$

This implies that the asymptotic covariance of $\sqrt{n}(\boldsymbol{\beta}_n - \boldsymbol{\beta}(P))$ is given by

$$\frac{\mathbb{E}_{\mathbf{0}, \mathbf{I}_k} [\rho'(\|\mathbf{z}\|)^2]}{k\alpha^2} (\mathbb{E}[\mathbf{X}^T \boldsymbol{\Sigma}^{-1} \mathbf{X}])^{-1}.$$
Again, according to Theorem 6 and Lemma 2, it follows that $\sqrt{n}(\theta_n - \theta(P))$ is asymptotically normal with mean zero and covariance matrix

$$ \left( \frac{\partial \Lambda_{\theta}(\xi(P))}{\partial \theta} \right)^{-1} \mathbb{E} \left[ \mathrm{vec}(\Psi_{\theta}(\mathbf{s}, \xi_P)) \mathrm{vec}(\Psi_{\theta}(\mathbf{s}, \xi_P))^T \right] \left( \frac{\partial \Lambda_{\theta}(\xi(P))}{\partial \theta} \right)^{-1} $$

where $\Psi_{\theta}$ is defined in (7.10). We have

$$ \mathbb{E} \left[ \mathrm{vec}(\Psi_{\theta}(\mathbf{s}, \xi_P)) \mathrm{vec}(\Psi_{\theta}(\mathbf{s}, \xi_P))^T \right] \\ = \mathbf{E}^T \mathbb{E} \left[ \mathrm{vec} \left( \Sigma^{-1/2} \Psi_{\mathbf{V}}(\mathbf{s}, \xi_P) \Sigma^{-1/2} \right) \mathrm{vec} \left( \Sigma^{-1/2} \Psi_{\mathbf{V}}(\mathbf{s}, \xi_P) \Sigma^{-1/2} \right)^T \right] \mathbf{E} $$

where $\Psi_{\mathbf{V}}$ is defined in (7.8) and $\mathbf{E} = (\Sigma^{-1/2} \otimes \Sigma^{-1/2}) \mathbf{L}$ and

$$ \begin{aligned} & \mathbb{E} \left[ \mathrm{vec} \left( \Sigma^{-1/2} \Psi_{\theta}(\mathbf{s}, \xi_P) \Sigma^{-1/2} \right) \mathrm{vec} \left( \Sigma^{-1/2} \Psi_{\theta}(\mathbf{s}, \xi_P) \Sigma^{-1/2} \right)^T \right] \\ & = k^2 \mathbb{E}_{\mathbf{0}, \mathbf{I}_k} \left[ u(\|\mathbf{z}\|)^2 \|\mathbf{z}\|^4 \right] \mathbb{E}_{\mathbf{0}, \mathbf{I}_k} \left[ \mathrm{vec}(\mathbf{u}\mathbf{u}^T) \mathrm{vec}(\mathbf{u}\mathbf{u}^T)^T \right] \\ & \quad - k \mathbb{E}_{\mathbf{0}, \mathbf{I}_k} \left[ u(\|\mathbf{z}\|) v(\|\mathbf{z}\|) \|\mathbf{z}\|^2 \right] \mathbb{E}_{\mathbf{0}, \mathbf{I}_k} \left[ \mathrm{vec}(\mathbf{u}\mathbf{u}^T) \mathrm{vec}(\mathbf{I}_k)^T \right] \\ & \quad - k \mathbb{E}_{\mathbf{0}, \mathbf{I}_k} \left[ u(\|\mathbf{z}\|) v(\|\mathbf{z}\|) \|\mathbf{z}\|^2 \right] \mathbb{E}_{\mathbf{0}, \mathbf{I}_k} \left[ \mathrm{vec}(\mathbf{I}_k) \mathrm{vec}(\mathbf{u}\mathbf{u}^T)^T \right] \\ & \quad + \mathbb{E}_{\mathbf{0}, \mathbf{I}_k} \left[ v(\|\mathbf{z}\|)^2 \right] \mathbb{E}_{\mathbf{0}, \mathbf{I}_k} \left[ \mathrm{vec}(\mathbf{I}_k) \mathrm{vec}(\mathbf{I}_k)^T \right]. \end{aligned} $$

From Lemma B.4, the first term on the right hand side is equal to

$$ \frac{k \mathbb{E}_{\mathbf{0}, \mathbf{I}_k} \left[ u(\|\mathbf{z}\|)^2 \|\mathbf{z}\|^4 \right]}{k+2} \left( \mathbf{I}_{k^2} + \mathbf{K}_{k,k} + \mathrm{vec}(\mathbf{I}_k) \mathrm{vec}(\mathbf{I}_k)^T \right). $$

This leads to one term $\mathbf{I}_{k^2} + \mathbf{K}_{k,k}$ with coefficient

$$ \frac{k \mathbb{E}_{\mathbf{0}, \mathbf{I}_k} \left[ u(\|\mathbf{z}\|)^2 \|\mathbf{z}\|^4 \right]}{k+2} $$

and using that, according to Lemma B.4, $\mathbb{E}_{\mathbf{0}, \mathbf{I}_k} [\mathbf{u}\mathbf{u}^T] = (1/k)\mathbf{I}_k$, we find a second term $\mathrm{vec}(\mathbf{I}_k)\mathrm{vec}(\mathbf{I}_k)^T$ with coefficient

$$ \frac{k \mathbb{E}_{\mathbf{0}, \mathbf{I}_k} \left[ u(\|\mathbf{z}\|)^2 \|\mathbf{z}\|^4 \right]}{k+2} - 2 \mathbb{E}_{\mathbf{0}, \mathbf{I}_k} \left[ u(\|\mathbf{z}\|) v(\|\mathbf{z}\|) \|\mathbf{z}\|^2 \right] + \mathbb{E}_{\mathbf{0}, \mathbf{I}_k} \left[ v(\|\mathbf{z}\|)^2 \right]. $$

Since $v(s) = u(s)s^2 - \rho(s) + b_0$, we have that

$$ \frac{k}{k+2} u(s)^2 s^4 - 2u(s)v(s)s^2 + v(s)^2 = -\frac{2}{k+2} u(s)^2 s^4 + (\rho(s) - b_0)^2. $$

This means that

$$ \begin{aligned} & \mathbb{E} \left[ \mathrm{vec} \left( \Sigma^{-1/2} \Psi_{\theta}(\mathbf{s}, \xi_P) \Sigma^{-1/2} \right) \mathrm{vec} \left( \Sigma^{-1/2} \Psi_{\theta}(\mathbf{s}, \xi_P) \Sigma^{-1/2} \right)^T \right] \\ & = \delta_1 (\mathbf{I}_{k^2} + \mathbf{K}_{k,k}) + \delta_2 \mathrm{vec}(\mathbf{I}_k) \mathrm{vec}(\mathbf{I}_k)^T \end{aligned} $$

where

$$ \begin{aligned} \delta_1 &= \frac{k \mathbb{E}_{\mathbf{0}, \mathbf{I}_k} \left[ u(\|\mathbf{z}\|)^2 \|\mathbf{z}\|^4 \right]}{k+2} \\ \delta_2 &= -\frac{2}{k+2} \mathbb{E}_{\mathbf{0}, \mathbf{I}_k} \left[ u(\|\mathbf{z}\|)^2 \|\mathbf{z}\|^4 \right] + \mathbb{E}_{\mathbf{0}, \mathbf{I}_k} \left[ (\rho(\|\mathbf{z}\|) - b_0)^2 \right] \quad (B.69) \\ &= -\frac{2}{k} \delta_1 + \mathbb{E}_{\mathbf{0}, \mathbf{I}_k} \left[ (\rho(\|\mathbf{z}\|) - b_0)^2 \right] \end{aligned} $$
Since, $\mathbf{K}_{k,k}(\mathbf{\Sigma}^{-1/2} \otimes \mathbf{\Sigma}^{-1/2}) = (\mathbf{\Sigma}^{-1/2} \otimes \mathbf{\Sigma}^{-1/2})\mathbf{K}_{k,k}$, together with (B.32), this implies that

$$ 
\begin{aligned}
\mathbb{E} \left[ \text{vec}(\Psi_{\theta}(\mathbf{s}, \boldsymbol{\xi}_P)) \text{vec}(\Psi_{\theta}(\mathbf{s}, \boldsymbol{\xi}_P))^T \right] &= \mathbf{E}^T (\delta_1 (\mathbf{I}_{k^2} + \mathbf{K}_{k,k}) + \delta_2 \text{vec}(\mathbf{I}_k)\text{vec}(\mathbf{I}_k)^T) \mathbf{E} \\
&= 2\delta_1 \mathbf{E}^T \mathbf{E} + \delta_2 \mathbf{E}^T \text{vec}(\mathbf{I}_k)\text{vec}(\mathbf{I}_k)^T \mathbf{E}.
\end{aligned}
$$

Furthermore, according to Lemma B.5,

$$ \left( \frac{\partial \Lambda_{\theta}(\xi(P))}{\partial \theta} \right)^{-1} = a(\mathbf{E}^T \mathbf{E})^{-1} + b(\mathbf{E}^T \mathbf{E})^{-1} \mathbf{E}^T \text{vec}(\mathbf{I}_k) \text{vec}(\mathbf{I}_k)^T \mathbf{E} (\mathbf{E}^T \mathbf{E})^{-1}, $$

with a and b defined in (B.44). By application of (B.39), (B.40), and (B.42), we find that

$$ 
\begin{aligned}
\left( \frac{\partial \Lambda_{\theta}(\xi(P))}{\partial \theta} \right)^{-1} \mathbb{E} \left[ \text{vec}(\Psi_{\theta}(\mathbf{s}, \boldsymbol{\xi}_P)) \text{vec}(\Psi_{\theta}(\mathbf{s}, \boldsymbol{\xi}_P))^T \right] \\
= 2a\delta_1 \mathbf{I}_k + (a\delta_2 + 2b\delta_1 + bk\delta_2)(\mathbf{E}^T \mathbf{E})^{-1} \mathbf{E}^T \text{vec}(\mathbf{I}_k) \text{vec}(\mathbf{I}_k)^T \mathbf{E}
\end{aligned}
$$

and

$$ 
\begin{aligned}
\left( \frac{\partial \Lambda_{\theta}(\xi(P))}{\partial \theta} \right)^{-1} \mathbb{E} \left[ \text{vec}(\Psi_{\theta}(\mathbf{s}, \boldsymbol{\xi}_P)) \text{vec}(\Psi_{\theta}(\mathbf{s}, \boldsymbol{\xi}_P))^T \right] \left( \frac{\partial \Lambda_{\theta}(\xi(P))}{\partial \theta} \right)^{-1} \\
&= 2\sigma_1(\mathbf{E}^T \mathbf{E})^{-1} + \sigma_2(\mathbf{E}^T \mathbf{E})^{-1} \mathbf{E}^T \text{vec}(\mathbf{I}_k) \text{vec}(\mathbf{I}_k)^T \mathbf{E} (\mathbf{E}^T \mathbf{E})^{-1} \\
&= 2\sigma_1(\mathbf{E}^T \mathbf{E})^{-1} + \sigma_2 \theta(P) \theta(P)^T
\end{aligned}
$$

where

$$ 
\begin{aligned}
\sigma_1 &= a^2 \delta_1 \\
\sigma_2 &= 2b(2a + kb)\delta_1 + (a + kb)^2 \delta_2.
\end{aligned}
$$

When we insert the expressions for $\delta_1$, $\delta_2$, a, and b given in (B.69) and (B.44), then we find

$$ 
\begin{aligned}
\sigma_1 &= \frac{k(k+2)\mathbb{E}_{\mathbf{0},\mathbf{I}_k} [u(\|\mathbf{z}\|)^2 \|\mathbf{z}\|^4]}{(\mathbb{E}_{\mathbf{0},\mathbf{I}_k} [\rho''(\|\mathbf{z}\|) \|\mathbf{z}\|^2 + (k+1)\rho'(\|\mathbf{z}\|) \|\mathbf{z}\|])^2} \\
\sigma_2 &= -\frac{2}{k}\sigma_1 + \frac{4\mathbb{E}_{\mathbf{0},\mathbf{I}_k} [(\rho(\|\mathbf{z}\|) - b_0)^2]}{(\mathbb{E}_{\mathbf{0},\mathbf{I}_k} [\rho'(\|\mathbf{z}\|) \|\mathbf{z}\|])^2}
\end{aligned}
$$

By substituting $\mathbf{E} = (\mathbf{\Sigma}^{-1/2} \otimes \mathbf{\Sigma}^{-1/2}) \mathbf{L}$, we find that the limiting covariance of $\sqrt{n}(\boldsymbol{\theta}_n - \boldsymbol{\theta}(P))$ is given by

$$ 2\sigma_1 (\mathbf{L}^T (\mathbf{\Sigma}^{-1} \otimes \mathbf{\Sigma}^{-1}) \mathbf{L})^{-1} + \sigma_2 \theta(P) \theta(P)^T $$

This finishes the proof. □