**Appendix A**

**Proof of Proposition 1.** Since (*θ*, *t*) → Ψ(*y*, *θ*, *t*) is continuous, by the uniform law of large numbers, Assumption 1 (a) implies

$$\sup\_{\theta \in V\_{\theta\_0, t \in V\_{\theta\_0}}} \parallel \int \Psi(y, \theta, t) dP\_n(y) - \int \Psi(y, \theta, t) dP\_0(y) \parallel \to 0,\tag{A1}$$

in probability. This result together with Assumption 1 (b) ensures the convergence in probability of the estimators *θ* 4*c <sup>ϕ</sup>* and 4*t c θ* 4*c ϕ* toward *θ*<sup>0</sup> and *tθ*<sup>0</sup> , respectively. The proof is the same as the one for Theorem 5.9 from [28], p. 46.

**Proof of Proposition 2.** By the definitions of *θ* 4*c <sup>ϕ</sup>* and 4*t c θ* 4*c ϕ* , they both satisfy

$$\int \psi(y, \hat{\theta}^{\hat{c}}\_{\hat{\theta}^{\prime}} \hat{t}^{\hat{c}}\_{\hat{\theta}^{\hat{c}}\_{\hat{\theta}}}) dP\_{\mathfrak{n}}(y) = 0 \quad (E1)$$

$$\int \frac{\partial}{\partial \theta} m(y, \hat{\theta}^{\hat{c}}\_{\hat{\theta}^{\prime}} \hat{t}^{\hat{c}}\_{\hat{\theta}^{\prime}\_{\hat{\theta}}}) dP\_{\mathfrak{n}}(y) = 0 \quad (E2)$$

Using a Taylor expansion in (*E*1), there exists (*θ* \**c <sup>ϕ</sup>*,\**t c <sup>ϕ</sup>*) inside the segment that links (*θ* 4*c <sup>ϕ</sup>*,4*t c θ* 4*c ϕ* ) and (*θ*0, *tθ*<sup>0</sup> ) such that

$$\begin{array}{l} 0 = \int \psi(y, \theta\_0, t\_{\theta\_0}) dP\_n(y) + \left[ \left( \int \frac{\partial}{\partial t} \psi(y, \theta\_0, t\_{\theta\_0}) dP\_n(y) \right)^\top \right] \\\ \left( \int \frac{\partial}{\partial \theta} \psi(y, \theta\_0, t\_{\theta\_0}) dP\_n(y) \right)^\top \right] \cdot a\_n + \frac{1}{2} a\_n^\top A\_n a\_{n\prime} \end{array} \tag{A2}$$

where

$$a\_{\boldsymbol{\theta}} := \left( (\widehat{\boldsymbol{t}}\_{\widehat{\boldsymbol{\theta}}^{c}\_{\boldsymbol{\theta}}} - \boldsymbol{t}\_{\boldsymbol{\theta}0})^{\top}, (\widehat{\boldsymbol{\theta}}^{c}\_{\boldsymbol{\theta}} - \boldsymbol{\theta}0)^{\top} \right)^{\top},\tag{A3}$$

and

$$A\_{\mathfrak{n}} := \begin{pmatrix} \int \frac{\partial^2}{\partial t^2} \psi(y, \overline{\theta}^c\_{q^c}, \overline{I}^c\_{\mathfrak{q}}) dP\_n(y) & \int \frac{\partial^2}{\partial t \partial \overline{\theta}} \psi(y, \overline{\theta}^c\_{q^c}, \overline{I}^c\_{\mathfrak{q}}) dP\_n(y) \\ \int \frac{\partial^2}{\partial t \partial \overline{\theta}} \psi(y, \overline{\theta}^c\_{q^c}, \overline{I}^c\_{\mathfrak{q}}) dP\_n(y) & \int \frac{\partial^2}{\partial t^2} \psi(y, \overline{\theta}^c\_{q^c}, \overline{I}^c\_{\mathfrak{q}}) dP\_n(y) \end{pmatrix}. \tag{A4}$$

By Assumption 2 (b), the law of large numbers implies that *An* = *OP*(1). Then, using Assumption 2 (a), the last term in (A2) can be written *oP*(1)*an*. On the other hand, by Assumption 2 (d), using the law of large numbers, we can write

$$\begin{array}{l} \left[ \left( \int \frac{\partial}{\partial t} \psi(\boldsymbol{y}, \boldsymbol{\theta}\_{0}, \boldsymbol{t}\_{\boldsymbol{\theta}\_{0}}) dP\_{n}(\boldsymbol{y}) \right)^{\top}, \left( \int \frac{\partial}{\partial \boldsymbol{\theta}} \psi(\boldsymbol{y}, \boldsymbol{\theta}\_{0}, \boldsymbol{t}\_{\boldsymbol{\theta}\_{0}}) dP\_{n}(\boldsymbol{y}) \right)^{\top} \right] \\ = \left[ \left( \int \frac{\partial}{\partial t} \psi(\boldsymbol{y}, \boldsymbol{\theta}\_{0}, \boldsymbol{t}\_{\boldsymbol{\theta}\_{0}}) dP\_{0}(\boldsymbol{y}) \right)^{\top}, \left( \int \frac{\partial}{\partial \boldsymbol{\theta}} \psi(\boldsymbol{y}, \boldsymbol{\theta}\_{0}, \boldsymbol{t}\_{\boldsymbol{\theta}\_{0}}) dP\_{0}(\boldsymbol{y}) \right)^{\top} \right] + o\_{P}(1). \end{array}$$

Consequently, (A2) becomes

$$\begin{array}{ll} \displaystyle -\int \psi(y,\theta\_0,t\_{\theta\_0})dP\_{\boldsymbol{n}}(\boldsymbol{y}) \\ \displaystyle = \left[ \left( \int \frac{\partial}{\partial t} \psi(y,\theta\_0,t\_{\theta\_0})dP\_{\boldsymbol{0}}(\boldsymbol{y}) \right)^{\top} + o\_P(1), \left( \int \frac{\partial}{\partial \boldsymbol{\theta}} \psi(y,\theta\_0,t\_{\theta\_0})dP\_{\boldsymbol{0}}(\boldsymbol{y}) \right)^{\top} + o\_P(1) \right] \cdot a\_n. \end{array} \tag{A5}$$

In the same way, using a Taylor expansion in (E2), there exists (*θ c <sup>ϕ</sup>*, *t c <sup>ϕ</sup>*) inside the segment that links (*θ* 4*c <sup>ϕ</sup>*,4*t c θ* 4*c ϕ* ) and (*θ*0, *tθ*<sup>0</sup> ) such that

$$\begin{array}{l} 0 = \int \frac{\partial}{\partial \boldsymbol{\theta}} m(\boldsymbol{y}, \boldsymbol{\theta}\_{0}, \boldsymbol{t}\_{\theta\_{0}}) dP\_{n}(\boldsymbol{y}) + \left[ \left( \int \frac{\partial^{2}}{\partial \boldsymbol{\theta} \partial \boldsymbol{t}} m(\boldsymbol{y}, \boldsymbol{\theta}\_{0}, \boldsymbol{t}\_{\theta\_{0}}) dP\_{n}(\boldsymbol{y}) \right)^{\top} \right. \\\ \left( \int \frac{\partial^{2}}{\partial^{2} \boldsymbol{\theta}} m(\boldsymbol{y}, \boldsymbol{\theta}\_{0}, \boldsymbol{t}\_{\theta\_{0}}) dP\_{n}(\boldsymbol{y}) \right)^{\top} \right] \cdot \boldsymbol{a}\_{n} + \frac{1}{2} \boldsymbol{a}\_{n}^{\top} \boldsymbol{B}\_{n} \boldsymbol{a}\_{n} . \end{array} \tag{A6}$$

where

$$B\_n := \begin{pmatrix} \int \frac{\partial^3}{\partial t \partial \overline{t}^t} m(y, \overline{\theta\_{q^\*}^\varepsilon} \overline{t}\_q^\varepsilon) dP\_n(y) & \int \frac{\partial^3}{\partial t^2 \partial t} m(y, \overline{\theta\_{q^\*}^\varepsilon} \overline{t}\_q^\varepsilon) dP\_n(y) \\ \int \frac{\partial^3}{\partial t \partial t \partial \theta} m(y, \overline{\theta\_{q^\*}^\varepsilon} \overline{t}\_q^\varepsilon) dP\_n(y) & \int \frac{\partial^3}{\partial t^3} m(y, \overline{\theta\_{q^\*}^\varepsilon} \overline{t}\_q^\varepsilon) dP\_n(y) \end{pmatrix}. \tag{A7}$$

Similarly, as in (A5), we obtain

$$\begin{array}{l} \displaystyle -\int \frac{\partial}{\partial \boldsymbol{\theta}} m(\boldsymbol{y}, \boldsymbol{\theta}\_{0}, \boldsymbol{t}\_{\boldsymbol{\theta}\_{0}}) dP\_{n}(\boldsymbol{y}) \\ = \left[ \left( \int \frac{\partial^{2}}{\partial \boldsymbol{\theta} \partial \boldsymbol{t}} m(\boldsymbol{y}, \boldsymbol{\theta}\_{0}, \boldsymbol{t}\_{\boldsymbol{\theta}\_{0}}) dP\_{0}(\boldsymbol{y}) \right)^{\top} + o\_{P}(1) , \int \frac{\partial^{2}}{\partial^{2} \boldsymbol{\theta}} m(\boldsymbol{y}, \boldsymbol{\theta}\_{0}, \boldsymbol{t}\_{\boldsymbol{\theta}\_{0}}) dP\_{0}(\boldsymbol{y}) + o\_{P}(1) \right] \cdot a\_{n} . \end{array} \tag{A8}$$

Using (A5) and (A8), we obtain

$$\begin{array}{ll}\sqrt{n}a\_{n} = \sqrt{n}\begin{pmatrix} \left(\int \frac{\partial}{\partial\theta}\varphi(y\_{\prime},\theta\_{0},t\_{\theta\_{0}})dP\_{0}(y)\right)^{\top} & \left(\int \frac{\partial}{\partial\theta}\varphi(y\_{\prime},\theta\_{0},t\_{\theta\_{0}})dP\_{0}(y)\right)^{\top} \\ \left(\int \frac{\partial^{2}}{\partial\theta\partial t}m(y,\theta\_{0},t\_{\theta\_{0}})dP\_{n}(y)\right)^{\top} & \int \frac{\partial^{2}}{\partial^{2}\theta}m(y,\theta\_{0},t\_{\theta\_{0}})dP\_{n}(y) \\ \times \begin{pmatrix} -\int \psi(y,\theta\_{0},t\_{\theta\_{0}})dP\_{n}(y) \\ -\int \frac{\partial}{\partial\theta}m(y,\theta\_{0},t\_{\theta\_{0}})dP\_{n}(y) \end{pmatrix} + o\_{P}(1). \end{array} \tag{A9}$$

Consider *S* the (*l* + 1 + *d*) × (*l* + 1 + *d*) matrix:

$$S := \begin{pmatrix} S\_{11} & S\_{12} \\ S\_{21} & S\_{22} \end{pmatrix} \\ \tag{A10}$$

with *<sup>S</sup>*<sup>11</sup> := ( *<sup>∂</sup> <sup>∂</sup>tψ*(*y*, *<sup>θ</sup>*0, *<sup>t</sup>θ*<sup>0</sup> )*dP*0(*y*)), *<sup>S</sup>*<sup>12</sup> := ( *<sup>∂</sup> ∂θ ψ*(*y*, *θ*0, *tθ*<sup>0</sup> )*dP*0(*y*)), *<sup>S</sup>*<sup>21</sup> := ( *<sup>∂</sup>*<sup>2</sup> *∂θ∂tm*(*y*, *<sup>θ</sup>*0, *<sup>t</sup>θ*<sup>0</sup> )*dP*0(*y*)), and *<sup>S</sup>*<sup>22</sup> :<sup>=</sup> *<sup>∂</sup>*<sup>2</sup> *<sup>∂</sup>*2*<sup>θ</sup> <sup>m</sup>*(*y*, *<sup>θ</sup>*0, *<sup>t</sup>θ*<sup>0</sup> )*dP*0(*y*). Through calculations, we have

$$S\_{21} = -[0\_{d\prime} \int \frac{\partial}{\partial \theta} g(y, \theta\_0) dP\_0(y)]\_{\prime} \tag{A11}$$

$$S\_{22} \quad = \ [0\_{d\prime}, \dots, 0\_d]. \tag{A12}$$

From (A9), we deduce that

$$\sqrt{n}\begin{pmatrix}\hat{\mathbf{f}}\_{\hat{\theta}\_{\Psi}^{c}}^{c} - t\_{\theta\_{0}}\\\hat{\theta}\_{\Psi}^{c} - \theta\_{0}\end{pmatrix} = \mathcal{S}^{-1}\sqrt{n}\begin{pmatrix}-\int \psi(y,\theta\_{0},t\_{\theta\_{0}})dP\_{n}(y)\\0\end{pmatrix} + o\_{P}(1). \tag{A13}$$

On the other hand, under assumption Assumption 2 (d), using the central limit theorem,

$$
\sqrt{n}\begin{pmatrix} -\int \psi(y,\theta\_{0\prime}t\_{\theta\_0})dP\_n(y) \\ 0\_d \end{pmatrix} \tag{A14}
$$

converges in distribution to a centred multivariate normal variable with covariance matrix:

$$M := \begin{pmatrix} M\_{11} & M\_{12} \\ M\_{21} & M\_{22} \end{pmatrix} \\ \text{\textsuperscript{0.0cm}} \tag{A15}$$

with

$$M\_{11} := \text{cov}[\psi(X, \theta\_0, t\_{\theta\_0})], \quad M\_{12} := \begin{pmatrix} 0\_d^\top \\ \vdots \\ 0\_d^\top \end{pmatrix}, \quad M\_{21} := \begin{pmatrix} 0 & 0\_l^\top \\ \vdots & \vdots \\ 0 & 0\_l^\top \end{pmatrix}, \quad M\_{22} := \begin{pmatrix} 0\_d^\top \\ \vdots \\ 0\_d^\top \end{pmatrix}.$$

Since *E*[*ψ*(*X*, *θ*0, *tθ*<sup>0</sup> )] = 0 by the construction of *ψ*, we obtain

$$M\_{11} = \text{cov}[\psi(X, \theta\_0, t\_{\theta\_0})] = \int \psi(y, \theta\_0, t\_{\theta\_0}) \psi(y, \theta\_0, t\_{\theta\_0})^\top dP\_0(y) = I\_{l+1} \tag{A16}$$

on the basis of (29) for *θ* = *θ*0.

Using then (A13) and the Slutsky theorem, we obtain that

$$\sqrt{n}\begin{pmatrix}\hat{\mathbf{f}}\_{\hat{\theta}^{c}\_{\varphi}}^{\varepsilon} - \mathbf{t}\_{\theta\_{0}}\\\hat{\theta}^{c}\_{\varphi} - \theta\_{0}\end{pmatrix}\tag{A17}$$

converges in distribution to a centred multivariate normal variable with the covariance matrix given by

> *C* = *S*−1*M*[*S*−1] . (A18)

If we denote

$$\mathbf{C} := \begin{pmatrix} \mathbf{C}\_{11} & \mathbf{C}\_{12} \\ \mathbf{C}\_{21} & \mathbf{C}\_{22} \end{pmatrix} \\ \text{\textquotedblleft} \\ \tag{A19}$$

through calculation, we obtain

$$\mathbf{C}\_{11} = \begin{bmatrix} \mathbf{S}\_{11}^{-1} - \mathbf{S}\_{11}^{-1} \mathbf{S}\_{12} [\mathbf{S}\_{21} \mathbf{S}\_{11}^{-1} \mathbf{S}\_{21}]^{-1} \mathbf{S}\_{21} \mathbf{S}\_{11}^{-1} \end{bmatrix} \times \begin{bmatrix} \mathbf{S}\_{11}^{-1} - \mathbf{S}\_{11}^{-1} \mathbf{S}\_{12} [\mathbf{S}\_{21} \mathbf{S}\_{11}^{-1} \mathbf{S}\_{12}]^{-1} \mathbf{S}\_{21} \mathbf{S}\_{11}^{-1} \end{bmatrix}^{\top},\tag{A20}$$

$$\mathbf{C}\_{12} = \begin{bmatrix} \mathbf{S}\_{11}^{-1} - \mathbf{S}\_{11}^{-1} \mathbf{S}\_{12} [\mathbf{S}\_{21} \mathbf{S}\_{11}^{-1} \mathbf{S}\_{12}]^{-1} \mathbf{S}\_{21} \mathbf{S}\_{11}^{-1} \end{bmatrix} \times \begin{bmatrix} [\mathbf{S}\_{21} \mathbf{S}\_{11}^{-1} \mathbf{S}\_{12}]^{-1} \mathbf{S}\_{21} \mathbf{S}\_{11}^{-1} \end{bmatrix}^{\top},\tag{A21}$$

$$\begin{array}{rcl} \mathbb{C}\_{21} &=& [[\mathbb{S}\_{21}\mathbb{S}\_{11}^{-1}\mathbb{S}\_{12}]^{-1}\mathbb{S}\_{21}\mathbb{S}\_{11}^{-1}] \times [\mathbb{S}\_{11}^{-1} - \mathbb{S}\_{11}^{-1}\mathbb{S}\_{12}[\mathbb{S}\_{21}\mathbb{S}\_{11}^{-1}\mathbb{S}\_{12}]^{-1}\mathbb{S}\_{21}\mathbb{S}\_{11}^{-1}]^{\top}, \\\\ &\ddots \end{array} \tag{A22}$$

$$\mathbf{C\_{22}} = \underbrace{[[\mathbf{S\_{21}S\_{11}^{-1}S\_{12}]^{-1}S\_{21}S\_{11}^{-1}]}}\_{\mathbf{CD}} \times [[\mathbf{S\_{21}S\_{11}^{-1}S\_{12}]^{-1}S\_{21}S\_{11}^{-1}]^{\top}}.\tag{A23}$$

**Proof of Proposition 3.** For the contaminated model *P*\**ε<sup>x</sup>* = (1 − *ε*)*P*<sup>0</sup> + *εδx*, whenever it exists, *t c <sup>θ</sup>* (*P*\**εx*) is defined as the solution of equation:

$$\int H\_{\varepsilon}(A\_{\theta}(t\_{\theta}^{\varepsilon}(\overline{P}\_{\varepsilon x}))) [\frac{\partial}{\partial t} m\_{\theta}(y, t\_{\theta}^{\varepsilon}(\overline{P}\_{\varepsilon x})) - x\_{\theta}(t\_{\theta}^{\varepsilon}(\overline{P}\_{\varepsilon x}))]) d\overline{P}\_{\varepsilon x}(y) = 0. \tag{A24}$$

It follows that

$$\begin{cases} (1-\varepsilon) \int H\_{\varepsilon}(A\_{\theta}(t^{\varepsilon}\_{\theta}(\overline{P}\_{\varepsilon x})) [\frac{\partial}{\partial \overline{\theta}} m\_{\theta}(y, t^{\varepsilon}\_{\theta}(\overline{P}\_{\varepsilon x})) - \pi\_{\theta}(t^{\varepsilon}\_{\theta}(\overline{P}\_{\varepsilon x}))]) dP\_{0}(y) \\ + \varepsilon H\_{\varepsilon}(A\_{\theta}(t^{\varepsilon}\_{\theta}(\overline{P}\_{\varepsilon x})) [\frac{\partial}{\partial \overline{\theta}} m\_{\theta}(x, t^{\varepsilon}\_{\theta}(\overline{P}\_{\varepsilon x})) - \pi\_{\theta}(t^{\varepsilon}\_{\theta}(\overline{P}\_{\varepsilon x}))]) = 0. \end{cases} \tag{A25}$$

Derivation with respect to *ε* in (A25) yields

$$\begin{split} & -\int H\_{\mathsf{C}}(A\_{\theta}(t\_{\theta}(P\_{0}))[\frac{\partial}{\partial t}m\_{\theta}(y, t\_{\theta}(P\_{0})) - \pi\_{\theta}(t\_{\theta}(P\_{0}))] \, dP\_{0}(y) \\ & + \int \frac{\partial}{\partial t} [H\_{\mathsf{C}}(A\_{\theta}(t)[\frac{\partial}{\partial t}m\_{\theta}(y, t) - \pi\_{\theta}(t)])]\_{t = t\_{\theta}(P\_{0})} \, dP\_{0}(y) \mathrm{IF}(\mathsf{x}; t\_{\theta}^{c}, P\_{0}) \\ & + H\_{\mathsf{C}}(A\_{\theta}(t\_{\theta}(P\_{0}))[\frac{\partial}{\partial t}m\_{\theta}(\mathsf{x}, t\_{\theta}(P\_{0})) - \pi\_{\theta}(t\_{\theta}(P\_{0}))]). \end{split} \tag{A26}$$

Since the first integral in (A26) equals zero, we obtain

$$\begin{split} \text{IF}(\mathbf{x}; t\_{\theta}^{c}, P\_{0}) &= -\left\{ \int \frac{\partial}{\partial t} [H\_{\varepsilon}(A\_{\theta}(t))[\frac{\partial}{\partial t}m\_{\theta}(y, t) - \mathbf{r}\_{\theta}(t)]] \right]\_{t = t\_{\theta}(P\_{0})} dP\_{0}(y) \right\}^{-1} \\ &\cdot H\_{\varepsilon}(A\_{\theta}(t\_{\theta}(P\_{0}))[\frac{\partial}{\partial t}m\_{\theta}(\mathbf{x}, t\_{\theta}(P\_{0})) - \mathbf{r}\_{\theta}(t\_{\theta}(P\_{0}))]) \\ &= -\left\{ \int \frac{\partial}{\partial t} \Psi\_{\theta}(y, t\_{\theta}(P\_{0})) dP\_{0}(y) \right\}^{-1} \cdot \Psi\_{\theta}(\mathbf{x}, t\_{\theta}(P\_{0})) . \end{split} \tag{A27}$$

For each *θ*, the influence function (55) is bounded with respect to *x*; therefore, the estimators 4*t c <sup>θ</sup>* are B-robust.

**Proof of Proposition 4.** For the contaminated model *<sup>P</sup>*\**ε<sup>x</sup>* = (<sup>1</sup> <sup>−</sup> *<sup>ε</sup>*)*P*<sup>0</sup> <sup>+</sup> *εδx*, *<sup>T</sup>c*(*P*\**εx*) is defined as the solution of equation:

$$\int \frac{\partial}{\partial \theta} m(y, T^c(\overline{P}\_{\varepsilon x}), t^c(T^c(\overline{P}\_{\varepsilon x}), \overline{P}\_{\varepsilon x})) d\overline{P}\_{\varepsilon x}(y) = 0,\tag{A28}$$

whenever this solution exists. Then,

$$(1 - \varepsilon) \int \frac{\partial}{\partial \theta} m(\underline{y}, T^{\varepsilon}(\tilde{P}\_{\varepsilon x}), t^{\varepsilon}(T^{\varepsilon}(\tilde{P}\_{\varepsilon x}), \tilde{P}\_{\varepsilon x})) dP\_0(\underline{y}) + \varepsilon \frac{\partial}{\partial \theta} m(\underline{x}, T^{\varepsilon}(\tilde{P}\_{\varepsilon x}), t^{\varepsilon}(T^{\varepsilon}(\tilde{P}\_{\varepsilon x}), \tilde{P}\_{\varepsilon x})) = 0. \tag{A29}$$

Derivation with respect to *ε* in (A29) yields

$$\begin{cases} -\int \frac{\partial}{\partial \theta} m(y, \theta\_0, t^c(\theta\_0, P\_0)) dP\_0(y) + \int \frac{\partial^2}{\partial^2 \theta} m[y, \theta, t]\_{\theta = \theta\_0, t = t\_{\theta\_0}(P\_0)} dP\_0(y) \text{IF}(\mathbf{x}; T^c, P\_0) + \int \frac{\partial}{\partial \theta} m[y, \theta, t]\_{\theta = \theta\_0, t = t\_{\theta\_0}(P\_0)} dP\_0(y) \text{IF}(\mathbf{x}; T^c, P\_0) + \int \frac{\partial}{\partial \theta} m[y, \theta, t]\_{\theta = \theta\_0, t = t\_{\theta\_0}(P\_0)} dP\_0(y) \text{IF}(\mathbf{x}; T^c, P\_0) \\ + \int \mathbf{F}(\mathbf{x}; t\_{\theta\_0, t}^c, P\_0) \Big\{ + \frac{\partial}{\partial \theta} [m(\mathbf{x}, \theta, t)]\_{\theta = \theta\_0, t = t\_{\theta\_0}(P\_0)} = 0. \end{cases} \tag{A30}$$

Some calculations show that

$$\frac{\partial}{\partial \theta} [m(\mathbf{x}, \theta, t)]\_{\theta = \theta\_0, t = t\_{\theta\_0}(P\_0)} = 0 \quad \text{and} \quad \frac{\partial^2}{\partial^2 \theta} [m(\mathbf{x}, \theta, t)]\_{\theta = \theta\_0, t = t\_{\theta\_0}(P\_0)} = 0,\tag{A31}$$

for any *x*; therefore, (A30) reduces to

$$\int \frac{\partial^2}{\partial \theta \partial t} m[y, \theta\_r, t]\_{\theta = \theta\_0, t = t\_{\theta\_0}(P\_0)} dP\_0(y) \cdot \left\{ \frac{\partial}{\partial \theta} t^\varepsilon (\theta\_0, P\_0) \text{IF}(\mathbf{x}; T^\varepsilon, P\_0) + \text{IF}(\mathbf{x}; t\_{\theta\_0}^\varepsilon, P\_0) \right\} = 0. \tag{A32}$$

On the other hand,

$$\begin{split} \int \frac{\partial^2}{\partial \theta \partial t} m[\underline{y}, \theta, t]\_{\theta = \theta\_0, t = t\_{\theta\_0}(P\_0)} dP\_0(\underline{y}) &= -\psi'(\underline{\rho}'(\mathbf{1})) \int \frac{\partial}{\partial \theta} \overline{\xi}(\underline{y}, \theta\_0) dP\_0(\underline{y}) \\ &= -\int \frac{\partial}{\partial \theta} \overline{\xi}(\underline{y}, \theta\_0) dP\_0(\underline{y}) \end{split}$$

since *ψ* (*u*) = *ϕ*−1(*u*).

Taking into account that *t <sup>c</sup>*(*θ*, *P*0) = *t*(*θ*, *P*0) and *t*(*θ*, *P*0) verifies

$$\int \frac{\partial}{\partial t} m(y, \theta, t(\theta, P\_0)) dP\_0(y) = 0,\tag{A33}$$

and the derivation with respect to *θ* yields

$$\int \frac{\partial^2}{\partial t \partial \theta} m(y, \theta, t(\theta, P\_0)) dP\_0(y) + \int \frac{\partial^2}{\partial^2 t} m(y, \theta, t(\theta, P\_0)) dP\_0(y) \cdot \frac{\partial}{\partial \theta} t(\theta, P\_0) = 0,\tag{A34}$$

which implies

$$\begin{split} \frac{\partial}{\partial\theta}t^{\varepsilon}(\theta\_{0},P\_{0}) &= \frac{\partial}{\partial\theta}t(\theta\_{0},P\_{0}) = -\left\{\int \frac{\partial^{2}}{\partial^{2}t}m(y,\theta\_{0},t(\theta\_{0},P\_{0}))dP\_{0}(y)\right\}^{-1}\int \frac{\partial^{2}}{\partial t\partial\theta}m(y,\theta\_{0},t(\theta\_{0},P\_{0}))dP\_{0}(y) \\ &= -\varrho''(1)\left\{\int \overline{\chi}(y,\theta\_{0})\overline{\chi}(y,\theta\_{0})^{\top}dP\_{0}(y)\right\}^{-1}\int \frac{\partial}{\partial\theta}\overline{\chi}(y,\theta\_{0})dP\_{0}(y), \end{split}$$

because

$$\int \frac{\partial^2}{\partial^2 t} m(y, \theta\_0, t(\theta\_0, P\_0)) dP\_0(y) = -\frac{1}{q^{\prime\prime}(1)} \int \widetilde{\g}(y, \theta\_0) \widetilde{\g}(y, \theta\_0)^\top dP\_0(y). \tag{A35}$$

Then, (A32) becomes

$$\begin{cases} \int \frac{\partial}{\partial \boldsymbol{\theta}} \overline{\boldsymbol{\mathfrak{g}}}(\boldsymbol{y}, \boldsymbol{\theta}\_{0}) dP\_{0}(\boldsymbol{y}) \Big[ \int -\boldsymbol{\eta}^{\prime\prime}(\boldsymbol{1}) \left\{ \int \underline{\boldsymbol{\mathfrak{g}}}(\boldsymbol{y}, \boldsymbol{\theta}\_{0}) \overline{\boldsymbol{\mathfrak{g}}}(\boldsymbol{y}, \boldsymbol{\theta}\_{0})^{\top} dP\_{0}(\boldsymbol{y}) \right\}^{-1} \int \frac{\partial}{\partial \boldsymbol{\theta}} \overline{\boldsymbol{\mathfrak{g}}}(\boldsymbol{y}, \boldsymbol{\theta}\_{0}) dP\_{0}(\boldsymbol{y}) \mathrm{IF}(\boldsymbol{x}; \boldsymbol{T}^{c}, \boldsymbol{P}\_{0}) \\ \quad + \mathrm{IF}(\boldsymbol{x}; \boldsymbol{t}^{c}\_{\boldsymbol{\theta}\_{0}}, \boldsymbol{P}\_{0}) \Big] = 0, \end{cases}$$

and consequently,

$$\begin{split} \text{IIF}(\mathbf{x};T^{c},P\_{0}) &= \left\{ \left[\frac{\partial}{\partial\boldsymbol{\theta}}\overline{\mathbf{g}}(\boldsymbol{y},\boldsymbol{\theta}\_{0})dP\_{0}(\boldsymbol{y})\right]^{\top} \left[\int \overline{\mathbf{g}}(\boldsymbol{y},\boldsymbol{\theta}\_{0})\overline{\mathbf{g}}(\boldsymbol{y},\boldsymbol{\theta}\_{0})^{\top}dP\_{0}(\boldsymbol{y})\right]^{-1} \left[\frac{\partial}{\partial\boldsymbol{\theta}}\overline{\mathbf{g}}(\boldsymbol{y},\boldsymbol{\theta}\_{0})dP\_{0}(\boldsymbol{y})\right] \right\}^{-1} \\ &\cdot \left[\frac{\partial}{\partial\boldsymbol{\theta}}\overline{\mathbf{g}}(\boldsymbol{y},\boldsymbol{\theta}\_{0})dP\_{0}(\boldsymbol{y})\right]^{\top} \frac{1}{\overline{\mathbf{g}}^{\prime}(\boldsymbol{1})} \text{IF}(\mathbf{x};t\_{\theta\_{0}}^{c},P\_{0}). \end{split} \tag{A.36}$$
