**Appendix A**

*Appendix A.1. Proof of Theorem 2*

We denote

$$h\_n(\theta) = \frac{1}{n} \sum\_{i=1}^n \frac{f\_\theta(X\_i)^\pi}{C\_\pi(\theta)}.$$

Differentiating both sides of the equality, we have

$$\frac{\partial h\_n(\theta)}{\partial \theta} = \frac{\tau}{\mathbb{C}\_\tau(\theta)} \frac{1}{n} \sum\_{i=1}^n f\_\theta(X\_i)^\tau (\mu\_\theta(X\_i) - c\_\tau(\theta)).$$

Now we establish that

$$\left(\frac{\partial^2 h\_n(\boldsymbol{\theta})}{\partial \boldsymbol{\theta} \partial \boldsymbol{\theta}^T}\right)\_{\boldsymbol{\theta} = \boldsymbol{\theta}\_0} \overset{P}{\underset{\boldsymbol{\pi} \to \infty}{\longrightarrow}} -\frac{\boldsymbol{\pi}}{\mathbb{C}\_\boldsymbol{\pi}(\boldsymbol{\theta})} \mathbb{S}\_\boldsymbol{\pi}(\boldsymbol{\theta}\_0).$$

We have

*∂*2*hn*(*θ*) *<sup>∂</sup>θ∂θ<sup>T</sup>* <sup>=</sup> <sup>1</sup> *Cτ*(*θ***)** 2 & 1 *n n* ∑ *i*=1 *τ*<sup>2</sup> *<sup>f</sup>θ*(*Xi*)*τuθ*(*Xi*)*uθ*(*Xi*)*<sup>T</sup>* <sup>+</sup> *<sup>τ</sup> <sup>f</sup>θ*(*Xi*)*<sup>τ</sup> <sup>∂</sup>uθ*(*Xi*) *∂θ<sup>T</sup> Cτ*(*θ***)** <sup>−</sup>*τCτ*(*θ*)*cτ*(*θ*)*<sup>τ</sup> <sup>f</sup>θ*(*Xi*)*τuθ*(*Xi*)*<sup>T</sup>* ! <sup>−</sup> <sup>1</sup> *Cτ*(*θ***)** 2 & 1 *n n* ∑ *i*=1 *<sup>τ</sup> <sup>∂</sup>cτ*(*θ*) *<sup>∂</sup>θ<sup>T</sup> <sup>f</sup>θ*(*Xi*)*<sup>τ</sup>* <sup>+</sup> *<sup>τ</sup>*<sup>2</sup> *<sup>f</sup>θ*(*Xi*)*τcτ*(*θ*)*uθ*(*Xi*)*<sup>T</sup> Cτ*(*θ***)** <sup>−</sup>*τCτ*(*θ***)***cτ*(*θ*)*τcτ*(*θ*)*<sup>T</sup> <sup>f</sup>θ*(*Xi*)*<sup>τ</sup>* ! <sup>=</sup> <sup>1</sup> *Cτ*(*θ***)** & 1 *n n* ∑ *i*=1 *<sup>τ</sup>*<sup>2</sup> *<sup>f</sup>θ*(*Xi*)*τuθ*(*Xi*)*uθ*(*Xi*)*<sup>T</sup>* <sup>+</sup> *<sup>τ</sup> <sup>f</sup>θ*(*Xi*)*<sup>τ</sup> <sup>∂</sup>uθ*(*Xi*) *∂θ<sup>T</sup>* <sup>−</sup> *<sup>τ</sup>*2*cτ*(*θ*)*<sup>T</sup> <sup>f</sup>θ*(*Xi*)*τuθ*(*Xi*)*<sup>T</sup>* <sup>−</sup> *<sup>τ</sup> <sup>∂</sup>cτ*(*θ*) *<sup>∂</sup>θ<sup>T</sup> <sup>f</sup>θ*(*Xi*)*<sup>τ</sup>* <sup>−</sup>*τ*2*cτ*(*θ*)*fθ*(*Xi*)*τuθ*(*Xi*)*<sup>T</sup>* <sup>−</sup> *<sup>τ</sup>*2*cτ*(*θ*)*cτ*(*θ*)*<sup>T</sup> <sup>f</sup>θ*(*Xi*)*<sup>τ</sup>* !.

As *n* → ∞, we have

$$\left(\frac{\partial^2 h\_n(\theta)}{\partial \theta \partial \theta^T}\right)\_{\theta = \theta\_0} \stackrel{P}{\to}\_{n \to \infty} T(\theta\_0)$$

with *T*(*θ*0) being the matrix given by

$$\begin{split} T(\theta\_{0}) &= \ \frac{1}{\mathsf{C}\_{\mathsf{T}}(\theta)} \Big\{ \tau^{2} \int f\_{\theta}(\mathbf{x})^{\mathsf{T}+1} \mathsf{u}\_{\theta}(\mathbf{x}) \mathsf{u}\_{\theta}(\mathbf{x})^{\mathsf{T}} d\mathbf{x} + \tau \int f\_{\theta}(\mathbf{x})^{\mathsf{T}+1} \frac{\partial \mathsf{u}\_{\theta}(\mathbf{x})}{\partial \theta^{\mathsf{T}}} d\mathbf{x} \\ &\quad - \tau^{2} \mathsf{c}\_{\mathsf{T}}(\theta) \int f\_{\theta}(\mathbf{x})^{\mathsf{T}+1} \mathsf{u}\_{\theta}(\mathbf{x})^{\mathsf{T}} d\mathbf{x} - \tau \frac{\partial \mathsf{c}\_{\mathsf{T}}(\theta)}{\partial \theta^{\mathsf{T}}} \int f\_{\theta}(\mathbf{x})^{\mathsf{T}+1} d\mathbf{x} \\ &\quad + \tau^{2} \mathsf{c}\_{\mathsf{T}}(\theta) \int f\_{\theta}(\mathbf{x})^{\mathsf{T}+1} \mathsf{u}\_{\theta}(\mathbf{x})^{\mathsf{T}} d\mathbf{x} - \tau^{2} \mathsf{c}\_{\mathsf{T}}(\theta) \mathsf{c}\_{\mathsf{T}}(\theta)^{\mathsf{T}} \int f\_{\theta}(\mathbf{x})^{\mathsf{T}+1} d\mathbf{x} \Big\}. \end{split}$$

.

From the above, after some algebra, we obtain

$$\begin{split} T(\boldsymbol{\theta}\_{0}) &= \ \frac{1}{\mathbb{C}\_{\boldsymbol{\tau}}(\boldsymbol{\theta})} \Big\{ \boldsymbol{\tau}^{2} \int f\_{\boldsymbol{\theta}}(\boldsymbol{x})^{\boldsymbol{\tau}+1} \boldsymbol{u}\_{\boldsymbol{\theta}}(\boldsymbol{x}) \boldsymbol{u}\_{\boldsymbol{\theta}}(\boldsymbol{x})^{\boldsymbol{T}} d\boldsymbol{x} + \boldsymbol{\tau} \int f\_{\boldsymbol{\theta}}(\boldsymbol{x})^{\boldsymbol{\tau}+1} \frac{\partial \boldsymbol{u}\_{\boldsymbol{\theta}}(\boldsymbol{x})}{\partial \boldsymbol{\theta}^{\boldsymbol{T}}} d\boldsymbol{x} \\ &- \boldsymbol{\tau} \frac{\partial \boldsymbol{c}\_{\boldsymbol{\tau}}(\boldsymbol{\theta})}{\partial \boldsymbol{\theta}^{\boldsymbol{T}}} \int f\_{\boldsymbol{\theta}}(\boldsymbol{x})^{\boldsymbol{\tau}+1} d\boldsymbol{x} - \boldsymbol{\tau}^{2} \boldsymbol{c}\_{\boldsymbol{\tau}}(\boldsymbol{\theta})^{\boldsymbol{T}} \boldsymbol{c}\_{\boldsymbol{\tau}}(\boldsymbol{\theta}) \int f\_{\boldsymbol{\theta}}(\boldsymbol{x})^{\boldsymbol{\tau}+1} d\boldsymbol{x} \Big\}. \end{split}$$

On the other hand, it not difficult to establish that

$$\begin{array}{rcl} \frac{\partial \mathbf{c}\_{\tau}(\boldsymbol{\theta})}{\partial \boldsymbol{\theta}^{\mathsf{T}}} &=& (\tau+1) \frac{\int f\_{\boldsymbol{\theta}}(\boldsymbol{x})^{\tau+1} \mathsf{u}\_{\boldsymbol{\theta}}(\boldsymbol{x}) \mathsf{u}\_{\boldsymbol{\theta}}(\boldsymbol{x})^{\mathsf{T}} d\boldsymbol{x}}{\int f\_{\boldsymbol{\theta}}(\boldsymbol{x})^{\tau+1} d\boldsymbol{x}} + \frac{\int f\_{\boldsymbol{\theta}}(\boldsymbol{x})^{\tau+1} \frac{\partial \mathsf{u}\_{\boldsymbol{\theta}}(\boldsymbol{x})}{\partial \boldsymbol{\theta}^{\mathsf{T}}} d\boldsymbol{x}}{\int f\_{\boldsymbol{\theta}}(\boldsymbol{x})^{\tau+1} d\boldsymbol{x}} \\ & - (\tau+1) \frac{\int f\_{\boldsymbol{\theta}}(\boldsymbol{x})^{\tau+1} \mathsf{u}\_{\boldsymbol{\theta}}(\boldsymbol{x}) d\boldsymbol{x} \int f\_{\boldsymbol{\theta}}(\boldsymbol{x})^{\tau+1} \mathsf{u}\_{\boldsymbol{\theta}}(\boldsymbol{x})^{\mathsf{T}} d\boldsymbol{x}}{\left(\int f\_{\boldsymbol{\theta}}(\boldsymbol{x})^{\tau+1} d\boldsymbol{x}\right)^{2}}. \end{array}$$

Therefore we have

$$\begin{split} -\tau \frac{\partial c\_{\tau}(\boldsymbol{\theta})}{\partial \boldsymbol{\theta}^{\mathsf{T}}} \int f\_{\boldsymbol{\theta}}(\mathbf{x})^{\mathsf{T}+1} d\mathbf{x} &= -\tau(\tau+1) \int f\_{\boldsymbol{\theta}}(\mathbf{x})^{\mathsf{T}+1} \mathsf{u}\_{\boldsymbol{\theta}}(\mathbf{x}) \mathsf{u}\_{\boldsymbol{\theta}}(\mathbf{x})^{\mathsf{T}} d\mathbf{x} - \tau \int f\_{\boldsymbol{\theta}}(\mathbf{x})^{\mathsf{T}+1} \frac{\partial \mathsf{u}\_{\boldsymbol{\theta}}(\mathbf{x})}{\partial \boldsymbol{\theta}^{\mathsf{T}}} \mathsf{u}\_{\boldsymbol{\theta}}(\mathbf{x}) d\mathbf{x} \\ &+ \tau(\tau+1) \frac{\int f\_{\boldsymbol{\theta}}(\mathbf{x})^{\mathsf{T}+1} \mathsf{u}\_{\boldsymbol{\theta}}(\mathbf{x}) d\mathbf{x} \int f\_{\boldsymbol{\theta}}(\mathbf{x})^{\mathsf{T}+1} \mathsf{u}\_{\boldsymbol{\theta}}(\mathbf{x})^{\mathsf{T}} d\mathbf{x} . \end{split}$$

Finally,

*<sup>T</sup>*(*θ*0) <sup>=</sup> <sup>1</sup> *Cτ*(*θ***)** *τ*2 *fθ*(*x*)*τ*+1*uθ*(*x*)*uθ*(*x*)*Tdx* + *τ <sup>f</sup>θ*(*x*)*τ*+<sup>1</sup> *<sup>∂</sup>uθ*(*x*) *<sup>∂</sup>θ<sup>T</sup> dx* −*τ*(*τ* + 1) *<sup>f</sup>θ*(*x*)*τ*+1*uθ*(*x*)*uθ*(*x*)*Tdx* <sup>−</sup> *<sup>τ</sup> <sup>f</sup>θ*(*x*)*τ*+<sup>1</sup> *<sup>∂</sup>uθ*(*x*) *∂θ<sup>T</sup>* +*τ*(*τ* + 1) *fθ*(*x*)*τ*+1*uθ*(*x*)*dx fθ*(*x*)*τ*+1*uθ*(*x*)*Tdx <sup>f</sup>θ*(*x*)*τ*+1*dx* <sup>−</sup> *<sup>τ</sup>*2*cτ*(*θ*)*cτ*(*θ*)*<sup>T</sup> <sup>f</sup>θ*(*x*)*τ***+1***dx*' <sup>=</sup> <sup>1</sup> *Cτ*(*θ***)** & −*τ fθ*(*x*)*τ*+1*uθ*(*x*)*uθ*(*x*)*Tdx* + *τ fθ*(*x*)*τ*+1*uθ*(*x*)*dx fθ*(*x*)*τ*+1*uθ*(*x*)*Tdx <sup>f</sup>θ*(*x*)*τ*+1*dx* ' <sup>=</sup> <sup>−</sup> *<sup>τ</sup> Cτ*(*θ***)** *S*(*θ*0).

On the other hand,

$$\sqrt{n}\frac{\partial h\_{\mathfrak{n}}(\boldsymbol{\theta})}{\partial \boldsymbol{\theta}} = \frac{\boldsymbol{\tau}}{\mathbb{C}\_{\boldsymbol{\tau}}(\boldsymbol{\theta})}\frac{1}{\sqrt{n}}\sum\_{i=1}^{n}f\_{\mathfrak{k}}(\boldsymbol{X}\_{i})^{\boldsymbol{\tau}}(\boldsymbol{u}\_{\boldsymbol{\theta}}(\boldsymbol{X}\_{i}) - \boldsymbol{c}\_{\boldsymbol{\tau}}(\boldsymbol{\theta})) \underset{\boldsymbol{n}\to\infty}{\underset{\boldsymbol{n}\to\infty}{\mathrm{d}}}\mathcal{N}\left(\mathbf{0}\_{p\boldsymbol{\prime}}\left(\frac{\boldsymbol{\tau}}{\mathbb{C}\_{\boldsymbol{\tau}}(\boldsymbol{\theta})}\right)^{2}\mathbf{K}\_{\boldsymbol{\tau}}(\boldsymbol{\theta}\_{0})\right),$$

as

$$E\left[\left(\frac{\tau}{C\_{\tau}(\theta)}f\_{\theta}(X)^{\tau}(\mathfrak{u}\_{\theta}(X) - \mathfrak{c}\_{\tau}(\theta))\right)\_{\theta = \mathfrak{o}\_{\emptyset}}\right] = \mathfrak{o}\_{\mathcal{P}}$$

and

$$\text{Cov}\left[\left(\frac{\tau}{\mathbb{C}\_{\tau}(\boldsymbol{\theta})} f\_{\boldsymbol{\theta}}(\boldsymbol{X})^{\tau} (\boldsymbol{u}\_{\boldsymbol{\theta}}(\boldsymbol{X}) - \boldsymbol{c}\_{\tau}(\boldsymbol{\theta})) \right)\_{\boldsymbol{\theta} = \boldsymbol{\theta}\_{0}}\right] = \left(\frac{\tau}{\mathbb{C}\_{\tau}(\boldsymbol{\theta}\_{0})}\right)^{2} \mathbf{K}\_{\tau}(\boldsymbol{\theta}\_{0})^{\tau}$$

Then, the RMRPE estimator of *θ*, \* *θτ*, must satisfy

$$\begin{cases} \left. \frac{\partial}{\partial \boldsymbol{\theta}} \ln\_{\boldsymbol{n}} (\boldsymbol{\theta}) \right|\_{\boldsymbol{\theta} = \overline{\boldsymbol{\theta}}\_{\boldsymbol{\tau}}} + \mathbf{G} (\overline{\boldsymbol{\theta}}\_{\boldsymbol{\tau}}) \lambda\_{\boldsymbol{n}} = \mathbf{0}\_{\boldsymbol{p} \prime} \\ \mathbf{g} (\overline{\boldsymbol{\theta}}\_{\boldsymbol{\tau}}) = \mathbf{0}\_{\boldsymbol{r} \prime} \end{cases} \tag{A1}$$

where *λ<sup>n</sup>* is a vector of Lagrangian multipliers. Now we consider *θ<sup>n</sup>* = *θ*<sup>0</sup> + *mn*−1/2, with ||*m*|| < *k*, for 0 < *k* < ∞. We have,

$$\frac{\partial}{\partial \theta} h\_{\mathbb{H}}(\theta)|\_{\theta = \theta\_{\mathbb{H}}} = \frac{\partial}{\partial \theta} h\_{\mathbb{H}}(\theta)|\_{\theta = \theta\_0} + \frac{\partial}{\partial \theta^T} \frac{\partial}{\partial \theta} h\_{\mathbb{H}}(\theta)|\_{\theta = \theta\_0} (\theta\_{\mathbb{H}} - \theta\_0) + o(||\theta\_{\mathbb{H}} - \theta\_0||^2)$$

and

$$n^{1/2}\frac{\partial}{\partial\theta}h\_{\mathfrak{n}}(\theta)\Big|\_{\theta=\mathfrak{e}\_{\mathfrak{n}}} = n^{1/2}\frac{\partial}{\partial\theta}h\_{\mathfrak{n}}(\theta)|\_{\theta=\mathfrak{e}\_{\mathfrak{0}}} - \frac{\partial}{\partial\theta^T}\frac{\partial}{\partial\theta}h\_{\mathfrak{n}}(\theta)|\_{\theta=\mathfrak{e}\_{\mathfrak{0}}}n^{1/2}(\theta\_{\mathfrak{n}}-\theta\_{\mathfrak{0}}) + o(n^{1/2}||\theta\_{\mathfrak{n}}-\theta\_{\mathfrak{0}}||^2). \tag{A2}$$

However,

$$o(n^{1/2}||\theta\_n - \theta\_0||^2) = o(n^{1/2}||\mathfrak{m}||^2/n) = o(n^{-1/2}||\mathfrak{m}||^2) = o(O\_p(1)) = o\_p(1).$$

Since

$$\lim\_{n \to \infty} \frac{\partial}{\partial \theta^T} \frac{\partial}{\partial \theta} h\_{\mathbb{H}}(\theta)|\_{\theta = \theta\_0} = -\frac{\tau}{\mathbb{C}\_{\tau}(\theta)} \mathbb{S}\_{\tau}(\theta\_0)$$

we obtain

$$n^{1/2} \left. \frac{\partial}{\partial \theta} h\_n(\theta) \right|\_{\theta = \theta\_n} = n^{1/2} \frac{\partial}{\partial \theta} h\_n(\theta)|\_{\theta = \theta\_0} + \frac{\tau}{C\_\tau(\theta)} S\_\tau(\theta\_0) n^{1/2} (\theta\_n - \theta\_0) + o\_p(1). \tag{A3}$$

Now, we know that

$$n^{1/2}\mathbf{g}(\theta\_{\mathbb{H}}) = \mathbf{G}(\theta\_{\mathbb{O}})^T n^{1/2}(\theta\_{\mathbb{H}} - \theta\_{\mathbb{O}}) + o\_{\mathbb{P}}(1). \tag{A4}$$

Further, the RMRPE \* *θ<sup>τ</sup>* must satisfy the conditions in (A1), and in view of (A3) and (A4) we have

$$n^{1/2}\frac{\partial}{\partial\theta}h\_n(\theta)|\_{\theta=\theta\_0} + \frac{\tau}{C\_\tau(\theta)}\mathcal{S}\_\mathbf{r}(\theta\_0)n^{1/2}(\overline{\theta}\_\tau - \theta\_0) + \mathcal{G}(\theta\_0)n^{1/2}\lambda\_n + o\_p(1) = \mathbf{0}\_\overline{\rho}.\tag{A5}$$

From (A4) it follows that

$$G(\theta\_0)^T n^{1/2} (\tilde{\theta}\_7 - \theta\_0) + o\_p(1) = \mathbf{0}\_7. \tag{A6}$$

Now we can express equations (A5) and (A6) in matrix form as

$$
\begin{pmatrix}
\frac{\tau}{C\_{\tau}(\boldsymbol{\theta}\_{0})} \boldsymbol{\mathcal{S}}\_{\tau}(\boldsymbol{\theta}\_{0}) & \boldsymbol{\mathcal{G}}(\boldsymbol{\theta}\_{0}) \\
\boldsymbol{\mathcal{G}}(\boldsymbol{\theta}\_{0})^{\mathrm{T}} & \mathbf{0}\_{\tau \times \boldsymbol{\tau}}
\end{pmatrix}
\begin{pmatrix}
n^{1/2}(\boldsymbol{\tilde{\theta}}\_{\tau} - \boldsymbol{\theta}\_{0}) \\
n^{1/2}\boldsymbol{\lambda}\_{\boldsymbol{n}}
\end{pmatrix} = \begin{pmatrix}
\mathbf{0}\_{\boldsymbol{r}}
\end{pmatrix} + o\_{p}(1).
$$

Therefore

$$\begin{pmatrix} n^{1/2} (\tilde{\theta}\_{\mathsf{T}} - \theta\_{0}) \\ n^{1/2} \lambda\_{n} \end{pmatrix} = \begin{pmatrix} \frac{\mathsf{T}}{\mathsf{C}\_{\mathsf{T}}(\theta\_{0})} \mathsf{S}\_{\mathsf{T}}(\theta\_{0}) & \mathsf{G}(\theta\_{0}) \\ \mathsf{G}(\theta\_{0})^{T} & \mathsf{0}\_{\mathsf{T} \times \mathsf{r}} \end{pmatrix}^{-1} \begin{pmatrix} -n^{1/2} \frac{\partial}{\partial \mathsf{B}} h\_{n}(\theta)|\_{\theta = \theta\_{0}} \\ \mathsf{0}\_{\mathsf{r}} \end{pmatrix} + o\_{p}(1).$$

However,

$$
\begin{pmatrix}
\frac{\tau}{C\_{\tau}(\boldsymbol{\theta}\_{0})} \mathbf{S}\_{\tau}(\boldsymbol{\theta}\_{0}) & \mathbf{G}(\boldsymbol{\theta}\_{0}) \\
\mathbf{G}(\boldsymbol{\theta}\_{0})^{T} & \mathbf{0}
\end{pmatrix}^{-1} = \begin{pmatrix}
\left(\frac{\tau}{C\_{\tau}(\boldsymbol{\theta}\_{0})}\right)^{-1} \mathbf{P}\_{\tau}^{\*}(\boldsymbol{\theta}\_{0}) & \mathbf{Q}\_{\tau}(\boldsymbol{\theta}\_{0}) \\
\mathbf{Q}\_{\tau}(\boldsymbol{\theta}\_{0})^{T} & \mathbf{R}\_{\tau}(\boldsymbol{\theta}\_{0})
\end{pmatrix},
$$

where *P*∗ *<sup>τ</sup>*(*θ*0) and *Qτ*(*θ*0) are defined in (18) and (19), respectively. The matrix *Rτ*(*θ*0) is the quantity needed to make the right hand side of the above equation equal to the indicated inverse. Then,

$$n^{1/2}(\tilde{\theta}\_{\mathbb{T}} - \theta\_0) = -\left(\frac{\tau}{C\_{\tau}(\theta)}\right)^{-1} P\_{\tau}^\*(\theta\_0) n^{1/2} \frac{\partial}{\partial \theta} h\_{\mathbb{H}}(\theta)|\_{\theta = \theta\_0} + o\_{\mathbb{P}}(1),\tag{A7}$$

and we know

$$n^{1/2} \frac{\partial}{\partial \boldsymbol{\theta}} h\_n(\boldsymbol{\theta})|\_{\boldsymbol{\theta} = \boldsymbol{\theta}\_0} \xrightarrow[n \to \infty]{\mathcal{L}} \mathcal{N}\left(\mathbf{0}, \left(\frac{\boldsymbol{\tau}}{\mathbb{C}\_{\boldsymbol{\tau}}(\boldsymbol{\theta}\_0)}\right)^2 \mathbf{K}\_{\boldsymbol{\tau}}(\boldsymbol{\theta}\_0)\right). \tag{A8}$$

Now by (A7) and (A8), we have the desired result.
