*Appendix A.2. Proof of Theorem 3*

Consider the expression *<sup>R</sup>γ*(*fθ*, *<sup>f</sup>*\* *θτ* ). A Taylor expansion for an arbitrary *θ* ∈ Θ, around \* *θ<sup>τ</sup>* leads to the relation

$$\begin{split} R\_{\gamma}(f\_{\boldsymbol{\theta}\prime}f\_{\overline{\boldsymbol{\theta}}\_{\boldsymbol{\tau}}}) &= R\_{\gamma}(f\_{\overline{\boldsymbol{\theta}}\_{\boldsymbol{\tau}}\prime}f\_{\overline{\boldsymbol{\theta}}\_{\boldsymbol{\tau}}}) + \left(\frac{\partial R\_{\gamma}(f\_{\boldsymbol{\theta}\prime}f\_{\overline{\boldsymbol{\theta}}\_{\boldsymbol{\tau}}})}{\partial\boldsymbol{\theta}}\right)\_{\boldsymbol{\theta}=\overline{\boldsymbol{\theta}}\_{\boldsymbol{\tau}}} \left(\boldsymbol{\theta}-\overline{\boldsymbol{\theta}}\_{\boldsymbol{\tau}}\right) \\ &+ \frac{1}{2} \Big(\boldsymbol{\theta}-\overline{\boldsymbol{\theta}}\_{\boldsymbol{\tau}}\Big) \left(\frac{\partial^{2}R\_{\gamma}(f\_{\boldsymbol{\theta}\prime}f\_{\overline{\boldsymbol{\theta}}\_{\boldsymbol{\tau}}})}{\partial\boldsymbol{\theta}\partial\boldsymbol{\theta}^{\overline{\boldsymbol{\theta}}\_{\boldsymbol{\tau}}}}\right)\_{\boldsymbol{\theta}=\overline{\boldsymbol{\theta}}\_{\boldsymbol{\tau}}} \left(\boldsymbol{\theta}-\overline{\boldsymbol{\theta}}\_{\boldsymbol{\tau}}\right)^{T} + o\left(\left\|\boldsymbol{\theta}-\overline{\boldsymbol{\theta}}\_{\boldsymbol{\tau}}\right\|^{2}\right). \end{split}$$

It is clear that *<sup>R</sup>γ*(*f*\* *θτ* , *f*\* *θτ* ) = 0 and

$$\frac{\partial R\_{\gamma}(f\_{\mathfrak{G}'}f\_{\overline{\mathfrak{G}}\_{\mathfrak{r}}})}{\partial \theta} = \frac{\partial L^1\_{\gamma}(\theta)}{\partial \theta} - \frac{\partial L^2\_{\gamma}(\theta)}{\partial \theta}.$$

being

and

$$L^1\_\gamma(\theta) = \frac{1}{\gamma + 1} \log \left( \int f\_\theta(x)^{\gamma + 1} dx \right)$$

$$L^2\_\gamma(\theta) = \frac{1}{\gamma} \log \left( \int f\_\theta(x)^\gamma f\_{\overline{\theta}\_\Gamma}(x) dx \right).$$

Then,

$$\frac{\partial L^1\_\gamma(\theta)}{\partial \theta} = \frac{\int f\_\theta(\mathbf{x})^{\gamma+1} \mathsf{u}\_\theta(\mathbf{x}) d\mathbf{x}}{\int f\_\theta(\mathbf{x})^{\gamma+1} d\mathbf{x}} \text{ and } \frac{\partial L^2\_\gamma(\theta)}{\partial \theta} = \frac{\int f\_\theta(\mathbf{x})^\gamma \mathsf{u}\_\theta(\mathbf{x}) f\_{\overline{\theta}\_\mathbf{r}}(\mathbf{x}) d\mathbf{x}}{\int f\_\theta(\mathbf{x})^\gamma f\_{\overline{\theta}\_\mathbf{r}}(\mathbf{x}) d\mathbf{x}}.$$

Therefore,

$$\left(\frac{\partial R\_{\gamma}(f\_{\theta\prime}f\_{\bar{\theta}\_{\mathsf{f}}})}{\partial \theta}\right)\_{\mathfrak{o}=\bar{\theta}\_{\mathsf{f}}} = 0.$$

Regarding the second derivatives, we have

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

and

$$\begin{split} \frac{\partial^{2}L\_{\gamma}^{2}(\boldsymbol{\theta})}{\partial\boldsymbol{\theta}\partial\boldsymbol{\theta}^{T}} &= \quad \gamma \frac{\int f\_{\boldsymbol{\theta}}(\boldsymbol{x})^{\gamma}\boldsymbol{u}\_{\boldsymbol{\theta}}(\boldsymbol{x})\boldsymbol{u}\_{\boldsymbol{\theta}}(\boldsymbol{x})^{T}f\_{\overline{\boldsymbol{\theta}}\_{\boldsymbol{r}}}(\boldsymbol{x})d\boldsymbol{x}}{\int f\_{\boldsymbol{\theta}}(\boldsymbol{x})^{\gamma}f\_{\overline{\boldsymbol{\theta}}\_{\boldsymbol{r}}}(\boldsymbol{x})d\boldsymbol{x}} + \frac{\int f\_{\boldsymbol{\theta}}(\boldsymbol{x})^{\gamma}\frac{\partial\boldsymbol{u}\_{\boldsymbol{\theta}}(\boldsymbol{x})}{\partial\boldsymbol{\theta}^{T}}f\_{\overline{\boldsymbol{\theta}}\_{\boldsymbol{r}}}(\boldsymbol{x})d\boldsymbol{x}}{\int f\_{\boldsymbol{\theta}}(\boldsymbol{x})^{\gamma}f\_{\overline{\boldsymbol{\theta}}\_{\boldsymbol{r}}}(\boldsymbol{x})d\boldsymbol{x}} \\ &- \gamma \frac{\int f\_{\boldsymbol{\theta}}(\boldsymbol{x})^{\gamma}f\_{\overline{\boldsymbol{\theta}}\_{\boldsymbol{r}}}(\boldsymbol{x})\boldsymbol{u}\_{\boldsymbol{\theta}}(\boldsymbol{x})d\boldsymbol{x} \int f\_{\boldsymbol{\theta}}(\boldsymbol{x})^{\gamma}f\_{\overline{\boldsymbol{\theta}}\_{\boldsymbol{r}}}(\boldsymbol{x})\boldsymbol{u}\_{\boldsymbol{\theta}}(\boldsymbol{x})^{T}d\boldsymbol{x}}{\left(\int f\_{\boldsymbol{\theta}}(\boldsymbol{x})^{\gamma}f\_{\overline{\boldsymbol{\theta}}\_{\boldsymbol{r}}}(\boldsymbol{x})d\boldsymbol{x}\right)^{2}}. \end{split}$$

and so

$$\left(\frac{\partial^2 R\_\gamma(f\_{\mathfrak{G}\_\mathbf{r}} f\_{\overline{\mathfrak{G}}\_\mathbf{r}})}{\partial \theta \partial \theta^T}\right)\_{\mathfrak{G} = \overline{\mathfrak{g}}\_\mathbf{r}} = \frac{\mathbf{S}\_\gamma(\overline{\mathfrak{G}}\_\mathbf{r})}{\kappa\_\gamma(\overline{\mathfrak{G}}\_\mathbf{r})}$$

Therefore,

$$T\_{\gamma}(\widehat{\boldsymbol{\theta}}\_{\mathsf{T}},\widetilde{\boldsymbol{\theta}}\_{\mathsf{T}}) = 2n\mathbb{R}\_{\boldsymbol{\gamma}}(f\_{\widetilde{\boldsymbol{\theta}}\_{\mathsf{T}}}f\_{\widetilde{\boldsymbol{\theta}}\_{\mathsf{T}}}) = n^{1/2}(\widehat{\boldsymbol{\theta}}\_{\mathsf{T}}-\widetilde{\boldsymbol{\theta}}\_{\mathsf{T}})^{T}\frac{\mathbf{S}\_{\gamma}(\widehat{\boldsymbol{\theta}}\_{\mathsf{T}})}{\kappa\_{\boldsymbol{\gamma}}(\widetilde{\boldsymbol{\theta}}\_{\mathsf{T}})}n^{1/2}(\widehat{\boldsymbol{\theta}}\_{\mathsf{T}}-\widetilde{\boldsymbol{\theta}}\_{\mathsf{T}}) + n \times o\left(\left\|\widehat{\boldsymbol{\theta}}\_{\mathsf{T}}-\widetilde{\boldsymbol{\theta}}\_{\mathsf{T}}\right\|^{2}\right).$$

Under *θ*<sup>0</sup> ∈ Θ0,

$$\frac{\mathbf{S}\_{\gamma}(\boldsymbol{\theta}\_{\mathsf{T}})}{\kappa\_{\gamma}(\boldsymbol{\tilde{\theta}}\_{\mathsf{T}})} \underset{\boldsymbol{n} \longrightarrow \boldsymbol{\infty}}{\longrightarrow} \frac{\mathbf{S}\_{\gamma}(\boldsymbol{\theta}\_{0})}{\kappa\_{\tau}(\boldsymbol{\theta}\_{0})} .$$

Based on 4 *θ<sup>τ</sup>* and using by (A4) and (A5), we have that

$$n^{1/2} \, \_{\widehat{\partial}\theta} \, h\_{\pi}(\theta)|\_{\theta = \theta\_0} = -\frac{\pi}{C\_{\pi}(\theta\_0)} n^{1/2} \, \mathbf{S}\_{\pi}(\theta\_0)(\widehat{\theta}\_{\pi} - \theta\_0) + o\_{\mathbb{P}}(1),$$

and using (A7), we obtain

$$\begin{split} n^{1/2} (\tilde{\theta}\_{\mathsf{T}} - \theta\_{0}) &= P\_{\mathsf{T}}^{\*} (\theta\_{0}) n^{1/2} \mathcal{S}\_{\mathsf{T}} (\theta\_{0}) (\hat{\theta}\_{\mathsf{T}} - \theta\_{0}) + o\_{p}(1) \\ &= n^{1/2} (\hat{\theta}\_{\mathsf{T}} - \theta\_{0}) - \mathcal{Q}\_{\mathsf{T}} (\theta\_{0}) G (\theta\_{0})^{T} n^{1/2} (\hat{\theta}\_{\mathsf{T}} - \theta\_{0}) + o\_{p}(1) .\end{split}$$

Therefore,

$$n^{1/2}(\hat{\boldsymbol{\theta}}\_{\tau} - \tilde{\boldsymbol{\theta}}\_{\tau}) = \mathbf{Q}\_{\tau}(\boldsymbol{\theta}\_{0})\mathbf{G}(\boldsymbol{\theta}\_{0})^{T}n^{1/2}(\hat{\boldsymbol{\theta}}\_{\tau} - \boldsymbol{\theta}\_{0}) + o\_{p}(1). \tag{A9}$$

On the other hand, we know that

$$n^{1/2}(\hat{\boldsymbol{\theta}}\_{\mathsf{T}} - \boldsymbol{\theta}\_{0}) \xrightarrow[n \longrightarrow \infty]{\mathcal{L}} \mathcal{N}(\boldsymbol{0}, \mathbf{S}\_{\mathsf{T}}(\boldsymbol{\theta}\_{0})^{-1}\mathbf{K}\_{\mathsf{T}}(\boldsymbol{\theta}\_{0})\mathbf{S}\_{\mathsf{T}}(\boldsymbol{\theta}\_{0})^{-1}).$$

From equations (19) and (25), we can establish that

$$\begin{split} \mathcal{B}\_{\tau}(\boldsymbol{\theta}\_{0}) &= \quad \mathcal{S}\_{\tau}(\boldsymbol{\theta}\_{0})^{-1} \mathcal{G}(\boldsymbol{\theta}\_{0}) \Big[ \mathcal{G}(\boldsymbol{\theta}\_{0})^{T} \mathcal{S}\_{\tau}(\boldsymbol{\theta}\_{0})^{-1} \mathcal{G}(\boldsymbol{\theta}\_{0}) \Big]^{-1} \mathcal{G}(\boldsymbol{\theta}\_{0})^{T} \mathcal{S}\_{\tau}(\boldsymbol{\theta}\_{0})^{-1} \\ &= \quad \mathcal{Q}\_{\tau}(\boldsymbol{\theta}\_{0}) \mathcal{G}(\boldsymbol{\theta}\_{0})^{-1} \mathcal{S}\_{\tau}(\boldsymbol{\theta}\_{0})^{-1} . \end{split}$$

Therefore, it follows that

$$m^{1/2}(\hat{\boldsymbol{\theta}}\_{\tau} - \tilde{\boldsymbol{\theta}}\_{\tau}) \underset{n \to \infty}{\longrightarrow} \mathcal{N}(0, \boldsymbol{\mathcal{B}}\_{\tau}(\boldsymbol{\theta}\_{0}) \boldsymbol{\mathcal{K}}\_{\tau}(\boldsymbol{\theta}\_{0}) \boldsymbol{\mathcal{B}}\_{\tau}(\boldsymbol{\theta}\_{0})^{T}).$$

Now, observe from the definition that *Bτ*(*θ*0) = *Bτ*(*θ*0) *T*. Then, the asymptotic distribution of the random variables

$$T\_{\gamma}(\tilde{\boldsymbol{\theta}}\_{\tau}, \tilde{\boldsymbol{\theta}}\_{\tau}) = 2n \boldsymbol{R}\_{\gamma}(f\_{\tilde{\boldsymbol{\theta}}\_{\tau} \prime} f\_{\tilde{\boldsymbol{\theta}}\_{\tau}})$$

and

$$n^{1/2} (\widehat{\boldsymbol{\theta}}\_{\tau} - \overline{\boldsymbol{\theta}}\_{\tau})^T \frac{\mathbf{S}\_{\gamma}(\boldsymbol{\theta}\_0)}{\kappa\_{\gamma}(\boldsymbol{\theta}\_0)} n^{1/2} (\widehat{\boldsymbol{\theta}}\_{\tau} - \overline{\boldsymbol{\theta}}\_{\tau})^T$$

are the same, as we have established that

$$m \times o\left(\left\|\left\|\widehat{\theta}\_{\tau} - \overline{\theta}\_{\tau}\right\|\right\|^2\right) = o\_p(1).$$

Next, we apply Corollary 2.1 in Dik and Gunst [30], which states: "Let *X* be a *q*-variate normal random variable with mean vector **0** and variance-covariance matrix **Σ**. Let *M* be a real symmetric matrix of order *q*. Let *k* = rank(**Σ***M***Σ)**, *k* ≥ 1 and let *λ*1, ... , *λk*, be the nonzero eigenvalues of *M***Σ**. Then, the distribution of the quadratic form *XTMX* coincides with the distribution of the random variable *k* ∑ *i*=1 *λiZ*<sup>2</sup> *<sup>i</sup>* , where *Z*1, ... , *Zk* are independent, each having a standard normal variable". In our case, the asymptotic distribution of *Tγ*(4 *θτ*, \* *θτ*) coincides with the distribution of the random variable *k* ∑ *i*=1 *λτ*,*<sup>γ</sup> <sup>i</sup>* (*θ*0)*Z*<sup>2</sup> *<sup>i</sup>* where *λτ*,*<sup>γ</sup>* <sup>1</sup> (*θ*0),..., *<sup>λ</sup>τ*,*<sup>γ</sup> <sup>k</sup>* (*θ*0), are the nonzero eigenvalues of *Aγ*(*θ*0)*Bτ*(*θ*0)*Kτ*(*θ*0)*Bτ*(*θ*0) and

$$k = \min\{r, rank(\mathcal{B}\_{\mathsf{T}}(\theta\_0)\mathcal{K}\_{\mathsf{T}}(\theta\_0)\mathcal{B}\_{\mathsf{T}}(\theta\_0)A\_{\mathcal{I}}(\theta\_0)\mathcal{B}\_{\mathsf{T}}(\theta\_0)\mathcal{K}\_{\mathsf{T}}(\theta\_0)\mathcal{B}\_{\mathsf{T}}(\theta\_0))\}.\tag{A10}$$

We now establish that *k* = *r*. The matrix,

$$N\_{\mathbb{T}}(\theta\_0) = B\_{\mathbb{T}}(\theta\_0) K\_{\mathbb{T}}(\theta\_0) B\_{\mathbb{T}}(\theta\_0)$$

is given by

$$\begin{split} \mathcal{N}\_{\tau}(\boldsymbol{\theta}\_{0}) &= \quad \mathcal{S}\_{\tau}(\boldsymbol{\theta}\_{0})^{-1} \mathcal{G}(\boldsymbol{\theta}\_{0}) \Big[ \mathcal{G}(\boldsymbol{\theta}\_{0})^{\mathrm{T}} \mathcal{S}\_{\tau}(\boldsymbol{\theta}\_{0})^{-1} \mathcal{G}(\boldsymbol{\theta}\_{0}) \Big]^{-1} \mathcal{G}(\boldsymbol{\theta}\_{0})^{\mathrm{T}} \mathcal{S}\_{\tau}(\boldsymbol{\theta}\_{0})^{-1} \\ & \quad \quad \quad \quad \mathcal{K}\_{\tau}(\boldsymbol{\theta}\_{0}) \mathcal{S}\_{\tau}(\boldsymbol{\theta}\_{0})^{-1} \mathcal{G}(\boldsymbol{\theta}\_{0}) \Big[ \mathcal{G}(\boldsymbol{\theta}\_{0})^{\mathrm{T}} \mathcal{S}\_{\tau}(\boldsymbol{\theta}\_{0})^{-1} \mathcal{G}(\boldsymbol{\theta}\_{0}) \Big]^{-1} \mathcal{G}(\boldsymbol{\theta}\_{0})^{\mathrm{T}} \mathcal{S}\_{\tau}(\boldsymbol{\theta}\_{0})^{-1} . \end{split}$$

Corollary 14.11.3 in Harville [31] (p. 259) establishes the following: "For any *m* × *n* matrix *<sup>A</sup>* and any *<sup>m</sup>* <sup>×</sup> *<sup>m</sup>* symmetric positive definite matrix *<sup>W</sup>*, *rank*(*ATW A*) = *rank*(*A*)". Based on this Corollary we have that*rank*(*Nτ*(*θ*0)) coincides with *rank*(*Nτ*(*θ*0)*Sγ*(*θ*0)*Nτ*(*θ*0)). On the other hand, we know the following additional properties:

(a) *rank*(*AB***) =** *rank***(***A***)** if *B* is full rank (Corollary b.3.3 in Harville [31] (p. 83)).

(b) *rank*(*AB***) =** *rank*(*BA***)** if dimension of *A* coincides with dimension of *BT*.

Matrix *Kτ*(*θ*0) should be "full rank"; in fact, if *Kτ*(*θ*0) were not full rank, the variance– covariance matrix of 4 *θ<sup>β</sup>* and \* *θ<sup>β</sup>* would not be full rank (there were redundant components in *θ* and this is not true).

Therefore, we have

*rank*(*Nτ*(*θ*0)) <sup>=</sup> (*a*)*rank <sup>S</sup>τ*(*θ*0)<sup>−</sup> <sup>1</sup> <sup>2</sup> *G*(*θ*0) *G*(*θ*0) *<sup>T</sup>Sτ*(*θ*0)−1*G*(*θ*0) −<sup>1</sup> *G*(*θ*0) *<sup>T</sup>Sτ*(*θ*0)−1*Kτ*(*θ*0)*Sτ*(*θ*0)−1*G*(*θ*0) *G*(*θ*0) −1 *Sτ*(*θ*0)−1*G*(*θ*0) −<sup>1</sup> *G*(*θ*0) *<sup>T</sup>Sτ*(*θ*0)<sup>−</sup> <sup>1</sup> 2 <sup>=</sup> (*b*)*rank G*(*θ*0) *<sup>T</sup>Sτ*(*θ*0)−1*Kτ*(*θ*0)*Sτ*(*θ*0)−1*G*(*θ*0) *G*(*θ*0) *<sup>T</sup>Sτ*(*θ*0)−1*G*(*θ*0) −<sup>1</sup> <sup>=</sup> (*a*)*ran*- *G*(*θ*0) *<sup>T</sup>Sτ*(*θ*0)−1*Kτ*(*θ*0)*Sτ*(*θ*0)−1*G*(*θ*0) <sup>=</sup> *Corollary*14.11.3*rank*- *Sτ*(*θ*0)−1*G*(*θ*0) = (*a*)*rank*(*G*(*θ*0)) = *r*.

### *Appendix A.3. Rényi's Pseudodistance between Normal Populations*

Here, we compute the expression of the RP between densities belonging to the normal model with parameters (*μ*1, *σ*1) and (*μ*2, *σ*2), respectively. The RP between N (*μ*1, *σ*1) and N (*μ*2, *σ*2) is given by

$$\begin{split} R\_{\gamma}(\mathcal{N}(\mu\_{1},\sigma\_{1}),\mathcal{N}(\mu\_{2},\sigma\_{2})) &= \quad \frac{1}{\gamma+1} \log \int \mathcal{N}(\mu\_{1},\sigma\_{1})^{\gamma+1} d\mathbf{x} \\ &\quad + \frac{1}{\gamma(\gamma+1)} \log \int \mathcal{N}(\mu\_{2},\sigma\_{2})^{\gamma+1} d\mathbf{x} - \frac{1}{\gamma} \log \int \mathcal{N}(\mu\_{1},\sigma\_{1})^{\gamma} \mathcal{N}(\mu\_{2},\sigma\_{2}) d\mathbf{x} \\ &= \quad \frac{1}{\gamma+1} \log L\_{1} + \frac{1}{\gamma(\gamma+1)} \log L\_{2} - \frac{1}{\gamma} \log L\_{3}. \end{split}$$

We first compute

$$\int \mathcal{N}(\mu, \sigma)^{\beta} dx$$

for the seek of simplicity in later calculations.

$$\begin{split} \int \mathcal{N}(\mu, \sigma)^{\beta} d\mathbf{x} &= \quad \int \left( \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^{2}} \right)^{\beta} d\mathbf{x} \\ &= \quad \frac{1}{\sigma^{\beta-1} \left( \sqrt{2\pi} \right)^{\beta-1}} \frac{1}{\sqrt{\beta}} \int \frac{1}{\frac{\sigma}{\sqrt{\beta}} \sqrt{2\pi}} e^{-\frac{1}{2} \left( \frac{x-\mu}{\sqrt{\beta}} \right)^{2}} d\mathbf{x} \\ &= \quad \frac{1}{\sigma^{\beta-1} \left( \sqrt{2\pi} \right)^{\beta-1}} \frac{1}{\sqrt{\beta}}. \end{split}$$

Therefore,

$$L\_1 = \frac{1}{\sigma\_1^{\gamma} \left(\sqrt{2\pi}\right)^{\gamma}} \frac{1}{\sqrt{\gamma+1}} \text{ and } L\_2 = \frac{1}{\sigma\_2^{\gamma} \left(\sqrt{2\pi}\right)^{\gamma}} \frac{1}{\sqrt{\gamma+1}}$$

.

In relation with *L*<sup>3</sup> we have,

*L*<sup>3</sup> = <sup>N</sup> (*μ*1, *<sup>σ</sup>*1)*γ*<sup>N</sup> (*μ*2, *<sup>σ</sup>*2)*dx* = 1 *σγ* 1 -<sup>√</sup>2*<sup>π</sup> <sup>γ</sup> <sup>e</sup>* − 1 2 ⎛ <sup>⎝</sup> *<sup>x</sup>*−*μ*<sup>1</sup> *<sup>σ</sup>* √1 *γ* ⎞ ⎠ 2 1 *σ*2 <sup>√</sup>2*<sup>π</sup> <sup>e</sup>* − 1 2 *<sup>x</sup>*−*μ*<sup>2</sup> *<sup>σ</sup>*<sup>2</sup> 2 <sup>=</sup> <sup>1</sup> *σγ* 1 -<sup>√</sup>2*<sup>π</sup> γ* 1 *σ*2 <sup>√</sup>2*<sup>π</sup>* <sup>×</sup> × exp ⎧ ⎪⎨ ⎪⎩ −1 2 ⎡ ⎢ ⎣*x*2 ⎛ ⎜⎝ 1 - √ *σ*1 *γ* <sup>2</sup> <sup>+</sup> 1 *σ*2 2 ⎞ ⎟⎠ <sup>−</sup> <sup>2</sup>*<sup>x</sup>* ⎛ ⎜⎝ *μ*1 - √ *σ*1 *γ* <sup>2</sup> <sup>+</sup> *<sup>μ</sup>*<sup>2</sup> *σ*2 2 ⎞ ⎟⎠ <sup>+</sup> *<sup>μ</sup>*<sup>2</sup> 1 - √ *σ*1 *γ* <sup>2</sup> <sup>+</sup> *<sup>μ</sup>*<sup>2</sup> 2 *σ*2 2 ⎤ ⎥ ⎦ ⎫ ⎪⎬ ⎪⎭ *dx* <sup>=</sup> <sup>1</sup> *σγ* 1 -<sup>√</sup>2*<sup>π</sup> γ* 1 *σ*2 <sup>√</sup>2*<sup>π</sup>* exp ⎧ ⎪⎨ ⎪⎩ −1 2 ⎡ ⎢ ⎣ *μ*2 1 - √ *σ*1 *γ* <sup>2</sup> <sup>+</sup> *<sup>μ</sup>*<sup>2</sup> 2 *σ*2 2 ⎤ ⎥ ⎦ ⎫ ⎪⎬ ⎪⎭ × × exp ⎧ ⎪⎨ ⎪⎩ −1 2 ⎡ ⎢ ⎣*x*2 ⎛ ⎜⎝ 1 - √ *σ*1 *γ* <sup>2</sup> <sup>+</sup> 1 *σ*2 2 ⎞ ⎟⎠ <sup>−</sup> <sup>2</sup>*<sup>x</sup>* ⎛ ⎜⎝ *μ*1 - √ *σ*1 *γ* <sup>2</sup> <sup>+</sup> *<sup>μ</sup>*<sup>2</sup> *σ*2 2 ⎞ ⎟⎠ ⎤ ⎥ ⎦ ⎫ ⎪⎬ ⎪⎭ *dx* <sup>=</sup> <sup>1</sup> *σγ* 1 -<sup>√</sup>2*<sup>π</sup> γ* 1 *σ*2 <sup>√</sup>2*<sup>π</sup>* exp ⎧ ⎪⎨ ⎪⎩ −1 2 ⎡ ⎢ ⎣ *μ*2 1 - √ *σ*1 *γ* <sup>2</sup> <sup>+</sup> *<sup>μ</sup>*<sup>2</sup> 2 *σ*2 2 ⎤ ⎥ ⎦ ⎫ ⎪⎬ ⎪⎭ exp<sup>1</sup> 2 *A*2 *B*2 *B* √ 2*π* × × 1 <sup>√</sup>2*π<sup>B</sup>* exp& −1 2 *<sup>x</sup>* <sup>−</sup> *<sup>A</sup> B* 2 ' *dx* <sup>=</sup> <sup>1</sup> *σγ* 1 -<sup>√</sup>2*<sup>π</sup> γ* 1 *σ*2 <sup>√</sup>2*<sup>π</sup>* exp ⎧ ⎪⎨ ⎪⎩ −1 2 ⎡ ⎢ ⎣ *μ*2 1 - √ *σ*1 *γ* <sup>2</sup> <sup>+</sup> *<sup>μ</sup>*<sup>2</sup> 2 *σ*2 2 ⎤ ⎥ ⎦ ⎫ ⎪⎬ ⎪⎭ exp<sup>1</sup> 2 *A*2 *B*2 *B* √ 2*π*.

Now it is necessary to obtain *A* and *B*. However, for this, we have,

$$\begin{cases} \begin{array}{c} \frac{1}{B^2} = \frac{1}{\left(\frac{\sigma\_1}{\sqrt{\gamma}}\right)^2} + \frac{1}{\sigma\_2^2} \\\\ \frac{A}{B^2} = \left(\frac{\mu\_1}{\left(\frac{\sigma\_1}{\sqrt{\gamma}}\right)^2} + \frac{\mu\_2}{\sigma\_2^2}\right) \end{array} . \end{cases}$$

Then,

$$A\left(\frac{1}{\left(\frac{\sigma\_1}{\sqrt{\gamma}}\right)^2} + \frac{1}{\sigma\_2^2}\right) = \frac{\mu\_1}{\left(\frac{\sigma\_1}{\sqrt{\gamma}}\right)^2} + \frac{\mu\_2}{\sigma\_2^2}$$

and

$$A = \frac{\frac{\mu\_1}{\left(\frac{\sigma\_1}{\sqrt{\gamma}}\right)^2 + \frac{\mu\_2}{\sigma\_2^2}}}{\frac{1}{\left(\frac{\sigma\_1}{\sqrt{\gamma}}\right)^2 + \frac{1}{\sigma\_2^2}} + \frac{1}{\sigma\_2^2}} = \frac{\frac{\sigma\_2^2 \mu\_1 + \mu\_2 \left(\frac{\sigma\_1}{\sqrt{\gamma}}\right)^2}{\sigma\_2^2 \left(\frac{\sigma\_1}{\sqrt{\gamma}}\right)^2}}{\frac{\sigma\_2^2 + \left(\frac{\sigma\_1}{\sqrt{\gamma}}\right)^2}{\sigma\_2^2 \left(\frac{\sigma\_1}{\sqrt{\gamma}}\right)^2}} = \frac{\sigma\_2^2 \mu\_1 + \mu\_2 \frac{\sigma\_1^2}{\gamma}}{\sigma\_2^2 + \frac{\sigma\_1^2}{\gamma}} = \frac{\gamma \sigma\_2^2 \mu\_1 + \mu\_2 \sigma\_1^2}{\gamma \sigma\_2^2 + \sigma\_1^2}.$$

We have,

$$\frac{1}{B^2} = \frac{1}{\left(\frac{\sigma\_1}{\sqrt{\gamma}}\right)^2} + \frac{1}{\sigma\_2^2} = \frac{\gamma}{\sigma\_1^2} + \frac{1}{\sigma\_2^2} = \frac{\sigma\_2^2 \gamma + \sigma\_1^2}{\sigma\_1^2 \sigma\_2^2}$$

Therefore,

$$B = \frac{\sigma\_1 \sigma\_2}{\sqrt{\sigma\_2^2 \gamma + \sigma\_1^2}}$$

.

On the other hand,

$$\frac{A^2}{B^2} = \left(\frac{\gamma \sigma\_2^2 \mu\_1 + \mu\_2 \sigma\_1^2}{\gamma \sigma\_2^2 + \sigma\_1^2}\right)^2 \frac{\sigma\_2^2 \gamma + \sigma\_1^2}{\sigma\_1^2 \sigma\_2^2} = \frac{\left(\gamma \sigma\_2^2 \mu\_1 + \mu\_2 \sigma\_1^2\right)^2}{\left(\gamma \sigma\_2^2 + \sigma\_1^2\right) \sigma\_1^2 \sigma\_2^2}$$

and

*<sup>L</sup>*<sup>3</sup> <sup>=</sup> <sup>1</sup> *σγ* 1 -<sup>√</sup>2*<sup>π</sup> γ* 1 *σ*2 <sup>√</sup>2*<sup>π</sup>* exp ⎧ ⎪⎨ ⎪⎩ −1 2 ⎡ ⎢ ⎣ *μ*2 1 - √ *σ*1 *γ* <sup>2</sup> <sup>+</sup> *<sup>μ</sup>*<sup>2</sup> 2 *σ*2 2 ⎤ ⎥ ⎦ ⎫ ⎪⎬ ⎪⎭ exp<sup>1</sup> 2 *A*2 *B*2 *B* √ 2*π* <sup>=</sup> <sup>1</sup> *σγ* 1 -<sup>√</sup>2*<sup>π</sup> γ* 1 *σ*2 exp ⎧ ⎪⎨ ⎪⎩ −1 2 ⎡ ⎢ ⎣ *μ*2 1 - √ *σ*1 *γ* <sup>2</sup> <sup>+</sup> *<sup>μ</sup>*<sup>2</sup> 2 *σ*2 2 ⎤ ⎥ ⎦ ⎫ ⎪⎬ ⎪⎭ exp& 1 2 *γσ*<sup>2</sup> <sup>2</sup>*μ*<sup>1</sup> + *<sup>μ</sup>*2*σ*<sup>2</sup> 1 2 *γσ*<sup>2</sup> <sup>2</sup> + *<sup>σ</sup>*<sup>2</sup> 1 *σ*2 <sup>1</sup> *<sup>σ</sup>*<sup>2</sup> 2 ' *<sup>σ</sup>*1*σ*<sup>2</sup> *σ*2 <sup>2</sup>*<sup>γ</sup>* + *<sup>σ</sup>*<sup>2</sup> 1 <sup>=</sup> *<sup>σ</sup>*1*σ*<sup>2</sup> *σ*2 <sup>2</sup>*<sup>γ</sup>* + *<sup>σ</sup>*<sup>2</sup> 1 1 *σγ* 1 -<sup>√</sup>2*<sup>π</sup> γ* 1 *σ*2 exp& 1 2 ( *γσ*<sup>2</sup> <sup>2</sup>*μ*<sup>1</sup> + *<sup>μ</sup>*2*σ*<sup>2</sup> 1 2 *γσ*<sup>2</sup> <sup>2</sup> + *<sup>σ</sup>*<sup>2</sup> 1 *σ*2 <sup>1</sup> *<sup>σ</sup>*<sup>2</sup> 2 <sup>−</sup> *γμ*<sup>2</sup> 1*σ*2 <sup>2</sup> + *<sup>σ</sup>*<sup>2</sup> 1*μ*<sup>2</sup> 2 *σ*2 <sup>2</sup> *<sup>σ</sup>*<sup>2</sup> 1 )'.

However,

 *γσ*<sup>2</sup> <sup>2</sup>*μ*<sup>1</sup> + *<sup>μ</sup>*2*σ*<sup>2</sup> 1 2 *γσ*<sup>2</sup> <sup>2</sup> + *<sup>σ</sup>*<sup>2</sup> 1 *σ*2 <sup>1</sup> *<sup>σ</sup>*<sup>2</sup> 2 <sup>−</sup> *γμ*<sup>2</sup> 1*σ*2 <sup>2</sup> + *<sup>σ</sup>*<sup>2</sup> 1*μ*<sup>2</sup> 2 *σ*2 <sup>2</sup> *<sup>σ</sup>*<sup>2</sup> 1 = *γσ*<sup>2</sup> <sup>2</sup>*μ*<sup>1</sup> + *<sup>μ</sup>*2*σ*<sup>2</sup> 1 <sup>2</sup> <sup>−</sup> *γμ*<sup>2</sup> 1*σ*2 <sup>2</sup> + *<sup>σ</sup>*<sup>2</sup> 1*μ*<sup>2</sup> 2 *γσ*<sup>2</sup> <sup>2</sup> + *<sup>σ</sup>*<sup>2</sup> 1 *γσ*<sup>2</sup> <sup>2</sup> + *<sup>σ</sup>*<sup>2</sup> 1 *σ*2 <sup>1</sup> *<sup>σ</sup>*<sup>2</sup> 2 <sup>=</sup> *<sup>γ</sup>*2*σ*<sup>4</sup> 2*μ*<sup>2</sup> <sup>1</sup> + *<sup>μ</sup>*<sup>2</sup> 2*σ*4 <sup>1</sup> + <sup>2</sup>*γσ*<sup>2</sup> 2*μ*1*μ*2*σ*<sup>2</sup> <sup>1</sup> *γσ*<sup>2</sup> <sup>2</sup> + *<sup>σ</sup>*<sup>2</sup> 1 *σ*2 <sup>1</sup> *<sup>σ</sup>*<sup>2</sup> 2 −*γ*2*μ*<sup>2</sup> 1*σ*4 <sup>2</sup> + *γμ*<sup>2</sup> 1*σ*2 <sup>2</sup> *<sup>σ</sup>*<sup>2</sup> <sup>1</sup> + *<sup>μ</sup>*<sup>2</sup> 2*γσ*<sup>2</sup> <sup>2</sup> *<sup>σ</sup>*<sup>2</sup> <sup>1</sup> + *<sup>μ</sup>*<sup>2</sup> 2*σ*4 <sup>1</sup> *γσ*<sup>2</sup> <sup>2</sup> + *<sup>σ</sup>*<sup>2</sup> 1 *σ*2 <sup>1</sup> *<sup>σ</sup>*<sup>2</sup> 2 <sup>=</sup> <sup>2</sup>*γσ*<sup>2</sup> 2*μ*1*μ*2*σ*<sup>2</sup> <sup>1</sup> <sup>−</sup> *γμ*<sup>2</sup> 1*σ*2 <sup>2</sup> *<sup>σ</sup>*<sup>2</sup> <sup>1</sup> <sup>−</sup> *<sup>μ</sup>*<sup>2</sup> 2*γσ*<sup>2</sup> <sup>2</sup> *<sup>σ</sup>*<sup>2</sup> <sup>1</sup> *γσ*<sup>2</sup> <sup>2</sup> + *<sup>σ</sup>*<sup>2</sup> 1 *σ*2 <sup>1</sup> *<sup>σ</sup>*<sup>2</sup> 2 <sup>=</sup> *<sup>σ</sup>*<sup>2</sup> <sup>2</sup> *<sup>σ</sup>*<sup>2</sup> 1*γ* <sup>2</sup>*μ*1*μ*<sup>2</sup> <sup>−</sup> *<sup>μ</sup>*<sup>2</sup> <sup>1</sup> <sup>−</sup> *<sup>μ</sup>*<sup>2</sup> 2 *γσ*<sup>2</sup> <sup>2</sup> + *<sup>σ</sup>*<sup>2</sup> 1 *σ*2 <sup>1</sup> *<sup>σ</sup>*<sup>2</sup> 2 <sup>=</sup> <sup>−</sup>*γ*(*μ*<sup>1</sup> <sup>−</sup> *<sup>μ</sup>*2) 2 *γσ*<sup>2</sup> <sup>2</sup> + *<sup>σ</sup>*<sup>2</sup> 1 

Therefore,

$$L3 = \frac{1}{\sigma\_1^{\gamma - 1} \sqrt{\sigma\_2^2 \gamma + \sigma\_1^2}} \frac{1}{\left(\sqrt{2\pi}\right)^{\gamma}} \exp\left\{-\frac{1}{2} \frac{\gamma \left(\mu\_1 - \mu\_2\right)^2}{\left(\gamma \sigma\_2^2 + \sigma\_1^2\right)}\right\}$$

.

Then,

*<sup>R</sup>γ*(<sup>N</sup> (*μ*1, *<sup>σ</sup>*1), <sup>N</sup> (*μ*2, *<sup>σ</sup>*2)) = <sup>1</sup> *<sup>γ</sup>* <sup>+</sup> <sup>1</sup> ln *<sup>L</sup>*<sup>1</sup> <sup>+</sup> 1 *<sup>γ</sup>*(*<sup>γ</sup>* <sup>+</sup> <sup>1</sup>) ln *<sup>L</sup>*<sup>2</sup> <sup>−</sup> <sup>1</sup> *γ* ln *L*<sup>3</sup> <sup>=</sup> <sup>1</sup> *<sup>γ</sup>* <sup>+</sup> <sup>1</sup> ln <sup>1</sup> *σγ* 1 -<sup>√</sup>2*<sup>π</sup> γ* 1 <sup>√</sup>*<sup>γ</sup>* <sup>+</sup> <sup>1</sup> <sup>+</sup> 1 *<sup>γ</sup>*(*<sup>γ</sup>* <sup>+</sup> <sup>1</sup>) ln <sup>1</sup> *σγ* 2 -<sup>√</sup>2*<sup>π</sup> γ* 1 <sup>√</sup>*<sup>γ</sup>* <sup>+</sup> <sup>1</sup> − 1 *γ* ln <sup>1</sup> *σγ*−<sup>1</sup> 1 *σ*2 <sup>2</sup>*<sup>γ</sup>* + *<sup>σ</sup>*<sup>2</sup> 1 1 -<sup>√</sup>2*<sup>π</sup> <sup>γ</sup>* <sup>+</sup> 1 2 *γ*(*μ*<sup>1</sup> − *μ*2) 2 *γ γσ*<sup>2</sup> <sup>2</sup> + *<sup>σ</sup>*<sup>2</sup> 1 <sup>=</sup> <sup>1</sup> *γ*(*γ* + 1) ln *<sup>σ</sup>γ*−<sup>1</sup> 1 *σγ* 2 <sup>√</sup>*<sup>γ</sup>* <sup>+</sup> <sup>1</sup> *σ*2 <sup>1</sup> + *γσ*<sup>2</sup> <sup>2</sup> <sup>+</sup> *<sup>γ</sup>* ln <sup>1</sup> *σ*1 <sup>√</sup>*<sup>γ</sup>* <sup>+</sup> <sup>1</sup> *σ*2 <sup>1</sup> + *γσ*<sup>2</sup> 2 + 1 2 *γ*(*μ*<sup>1</sup> − *μ*2) 2 *γ γσ*<sup>2</sup> <sup>2</sup> + *<sup>σ</sup>*<sup>2</sup> 1 <sup>=</sup> <sup>1</sup> *<sup>γ</sup>*(*<sup>γ</sup>* <sup>+</sup> <sup>1</sup>) ln <sup>1</sup> *<sup>σ</sup>*1*σ<sup>γ</sup>* 2 ⎛ ⎝ *σ*2 <sup>1</sup> + *γσ*<sup>2</sup> <sup>2</sup> <sup>√</sup>*<sup>γ</sup>* <sup>+</sup> <sup>1</sup> ⎞ ⎠ *γ*+1 + 1 2 (*μ*<sup>1</sup> − *μ*2) 2 *γσ*<sup>2</sup> <sup>2</sup> + *<sup>σ</sup>*<sup>2</sup> 1 

For *γ* → 0 we have,

$$\lim\_{\gamma\_{\parallel}\to 0} R\_{\gamma}(\mathcal{N}(\mu\_1, \sigma\_1), \mathcal{N}(\mu\_2, \sigma\_2)) = \frac{\sigma\_2^2 - \sigma\_1^2}{2\sigma\_1^2} + \ln \frac{\sigma\_1}{\sigma\_2} + \frac{1}{2} \frac{(\mu\_1 - \mu\_2)^2}{\sigma\_1^2}.\tag{A11}$$

*Appendix A.4. Computation of the Nonzero Eigenvalues of Aγ*(*θ*0)*Bτ*(*θ*0)*Kτ*(*θ*0)*Bτ*(*θ*0)

We know that the matrix *ξ*(*θ*) can be expressed as

$$
\xi(\theta) = c\_\tau(\theta)\kappa(\theta).
$$

with

$$\kappa(\theta) = \int f\_{\theta}(\mathbf{x})^{\mathbf{r}+1} d\mathbf{x} = \frac{1}{\sigma^{\tau} \left(\sqrt{2\pi}\right)^{\tau} \sqrt{1+\tau}}.$$

Then,

$$\mathcal{S}(\boldsymbol{\theta}) = \frac{1}{\sigma^{\tau} \left(\sqrt{2\pi}\right)^{\tau} \sqrt{1+\tau}} \left(0, -\frac{\tau}{\left(\tau+1\right)} \frac{1}{\sigma}\right)^{\tau}.$$

Therefore,

$$c\_{\pi}(\theta) = \frac{\xi(\theta)}{\kappa(\theta)} = \left(0, -\frac{\pi}{(\pi+1)}\frac{1}{\sigma}\right).$$

On the other hand

$$\frac{\partial \log f\_{\mu \mathcal{F}}(X\_i)}{\partial \mu} = \frac{X\_i - \mu}{\sigma^2} \text{ and } \frac{\partial \log f\_{\mu \mathcal{F}}(X\_i)}{\partial \sigma} = -\frac{1}{\sigma} + \frac{1}{\sigma^3}(X\_i - \mu)^2$$

and

$$
\mu\_{\varnothing}(X\_i) = \left(\frac{X\_i - \mu}{\sigma^2}, -\frac{1}{\sigma} + \frac{1}{\sigma^3}(X\_i - \mu)^2\right).
$$

Then,

$$\Psi\_{\pi}(X;\theta) = \left(\Psi\_{\pi}^{1}(X;\theta), \Psi\_{\pi}^{2}(X;\theta)\right).$$

is given by

$$\Psi\_{\mathsf{T}}(\boldsymbol{X};\boldsymbol{\theta}) = \left(\frac{\boldsymbol{X}-\boldsymbol{\mu}}{\sigma^{2}} \frac{1}{\left(\sigma\sqrt{2\pi}\right)^{\mathsf{T}}} e^{-\frac{\mathsf{T}}{2}\left(\frac{\boldsymbol{X}-\boldsymbol{\mu}}{\sigma}\right)^{2}}, \left(\left(\frac{\boldsymbol{X}-\boldsymbol{\mu}}{\sigma}\right)^{2} - \frac{1}{1+\mathsf{T}}\right) \frac{1}{\sigma} \frac{1}{\left(\sigma\sqrt{2\pi}\right)^{\mathsf{T}}} e^{-\frac{\mathsf{T}}{2}\left(\frac{\boldsymbol{X}-\boldsymbol{\mu}}{\sigma}\right)^{2}}\right)^{\mathsf{T}}.$$
and 
$$\mathbb{K}\_{\mathsf{T}}(\boldsymbol{\theta}) = \mathbb{E}\left[\boldsymbol{\mathsf{Y}}\_{\mathsf{T}}(\boldsymbol{X};\boldsymbol{\theta}) \boldsymbol{\mathsf{Y}}\_{\mathsf{T}}(\boldsymbol{X};\boldsymbol{\theta})^{\mathsf{T}}\right].$$

Now we obtain the elements of that matrix,

$$\begin{aligned} \begin{array}{rcl} K\_{\tau}^{11}(\boldsymbol{\theta}) &=& \mathbb{E} \left[ \left( \frac{\boldsymbol{X} - \boldsymbol{\mu}}{\sigma^{2}} \right)^{2} \frac{1}{\left( \sigma \sqrt{2\pi} \right)^{2\tau}} e^{-\frac{2\boldsymbol{\mu}}{\sigma} \left( \frac{\boldsymbol{X} - \boldsymbol{\mu}}{\sigma} \right)^{2}} \right] \\ &=& \frac{1}{\left( \sigma \sqrt{2\pi} \right)^{2\tau} (1 + 2\tau)^{3/2}} \frac{1}{\sigma^{2}} \\\\ \end{array} \\\\ \begin{array}{rcl} \frac{1}{\tau}(\boldsymbol{\theta}) &=& K\_{\tau}^{21}(\boldsymbol{\theta}) =& \mathbb{E} \left[ \left( \frac{\boldsymbol{X} - \boldsymbol{\mu}}{\sigma^{2}} \right) \left( \left( \frac{\boldsymbol{X} - \boldsymbol{\mu}}{\sigma} \right)^{2} - \frac{1}{1 + \tau} \right) \frac{1}{\sigma} \frac{1}{\left( \sigma \sqrt{2\pi} \right)^{2\tau}} e^{-\frac{2\boldsymbol{\mu}}{\sigma} \left( \frac{\boldsymbol{X} - \boldsymbol{\mu}}{\sigma} \right)} \end{array} \end{aligned}$$

*σ*

2 ⎤ ⎥ ⎦

and

*K*<sup>12</sup>

= 0

$$\begin{split} K\_{\tau}^{2}(\boldsymbol{\theta}) &= \ & \mathbb{E}\left[ \left( \left( \frac{\boldsymbol{X} - \boldsymbol{\mu}}{\sigma} \right)^{2} - \frac{1}{1 + \tau} \right)^{2} \frac{1}{\sigma^{2}} \frac{1}{\left( \sigma \sqrt{2\pi} \right)^{2\tau}} e^{-\frac{2\tau}{2} \left( \frac{\boldsymbol{X} - \boldsymbol{\mu}}{\sigma^{2}} \right)^{2}} \right] \\ &= & \ & \frac{1}{\sigma^{2}} \frac{3\tau^{2} + 2 + 4\tau}{\left( \sigma \sqrt{2\pi} \right)^{2\tau} \left( 1 + 2\tau \right)^{5/2} \left( 1 + \tau \right)^{2}} \end{split}$$

and

$$\begin{split} \mathbf{K}\_{\tau}(\boldsymbol{\theta}) &= \begin{pmatrix} \frac{1}{\left(\sigma\sqrt{2\pi}\right)^{2\tau}\left(1+2\tau\right)^{3/2}}\frac{1}{\sigma^{2}} & \mathbf{0} \\ \mathbf{0} & \frac{1}{\sigma^{2}}\frac{3\tau^{2}+2+4\tau}{\left(\sigma\sqrt{2\pi}\right)^{2\tau}\left(1+2\tau\right)^{5/2}\left(1+\tau\right)^{2}} \end{pmatrix} \\ &= \frac{1}{\sigma^{2}}\frac{1}{\left(\sigma\sqrt{2\pi}\right)^{2\tau}\left(1+2\tau\right)^{3/2}}\begin{pmatrix} 1 & 0 \\ 0 & \frac{3\tau^{2}+2+4\tau}{\left(1+\tau\right)^{2}\left(1+2\tau\right)} \end{pmatrix}. \end{split}$$

Now we obtain the matrix *Sτ*(*θ*). We have

$$
\xi(\theta) = c\_\tau(\theta)\kappa(\theta).
$$

with

$$\kappa(\theta) = \int f\_{\theta}(\mathbf{x})^{\mathbf{r}+1} d\mathbf{x} = \frac{1}{\sigma^{\tau} \left(\sqrt{2\pi}\right)^{\tau} \sqrt{1+\tau}}.$$

Then,

$$\xi(\theta) = \frac{1}{\sigma^{\pi} \left(\sqrt{2\pi}\right)^{\pi} \sqrt{1+\tau}} \left(0, -\frac{\tau}{\left(\tau+1\right)} \frac{1}{\sigma}\right)^{T}$$

and

$$\frac{1}{\kappa(\theta)}\xi(\theta)\xi(\theta)^T = \frac{1}{\sigma^{\pi+2}\left(\sqrt{2\pi}\right)^{\frac{\pi}{\pi}}\sqrt{1+\pi}}\begin{pmatrix} 0 & 0\\ 0 & \frac{\tau^2}{\left(\tau+1\right)^2} \end{pmatrix}.$$

On the other hand

$$J\_{\tau}(\theta) = \mathrm{E}\begin{bmatrix} \begin{pmatrix} \frac{1}{\sigma^{4}}(X-\mu)^{2} & \frac{1}{\sigma^{2}}\left(\frac{1}{\sigma}-\frac{1}{\sigma^{2}}(X-\mu)^{2}\right)(X-\mu) \\\\ \frac{1}{\sigma^{2}}\left(\frac{1}{\sigma}-\frac{1}{\sigma^{3}}(X-\mu)^{2}\right)(X-\mu) & \left(\frac{1}{\sigma}-\frac{1}{\sigma^{3}}(X-\mu)^{2}\right)^{2} \end{pmatrix} \frac{1}{\left(\sigma\sqrt{2\pi}\right)^{\tau}}e^{-\frac{\pi}{2}\left(\frac{X-\mu}{\sigma^{2}}\right)^{2}} \end{bmatrix}$$

$$\begin{split} \begin{split} J\_{\tau}^{11}(\boldsymbol{\theta}) &=& \mathbb{E} \left[ \frac{1}{\sigma^{4}} (\boldsymbol{\mu} - \boldsymbol{X})^{2} \frac{1}{\left( \sigma \sqrt{2 \pi} \right)} e^{-\frac{\boldsymbol{\tau}}{2} \left( \frac{\boldsymbol{X} - \boldsymbol{\mu}}{\sigma} \right)^{2}} \right] = \frac{1}{\sigma^{\tau + 2}} \frac{1}{\left( \boldsymbol{\tau} + 1 \right)^{3/2}} \frac{1}{\left( \sqrt{2 \pi} \right)^{\tau}} \\ J\_{\tau}^{12}(\boldsymbol{\theta}) &=& J\_{\tau}^{21}(\boldsymbol{\theta}) = 0 \\ J\_{\tau}^{22}(\boldsymbol{\theta}) &=& \mathbb{E} \left[ \left( \frac{1}{\sigma} - \frac{1}{\sigma^{3}} (\boldsymbol{\mu} - \boldsymbol{X})^{2} \right)^{2} \frac{1}{\left( \sigma \sqrt{2 \pi} \right)^{\tau}} e^{-\frac{\boldsymbol{\tau}}{2} \left( \frac{\boldsymbol{X} - \boldsymbol{\mu}}{\sigma} \right)^{2}} \right] = \frac{1}{\sigma^{\tau + 2}} \frac{1}{\left( \sqrt{2 \pi} \right)^{\tau}} \frac{1}{\sqrt{1 + \tau}} \frac{2 + \tau^{2}}{\left( 1 + \tau \right)^{2}} \end{split}$$

Therefore

$$J\_{\tau}(\theta) = \frac{1}{\sigma^{\tau+2}} \frac{1}{\left(\sqrt{2\pi}\right)^{\tau}} \frac{1}{\sqrt{1+\tau}} \begin{pmatrix} \frac{1}{1+\tau} & 0\\ 0 & \frac{2+\tau^2}{\left(1+\tau\right)^2} \end{pmatrix}.$$

$$\begin{split} \mathcal{S}\_{\tau}(\boldsymbol{\theta}) &= \quad J\_{\tau}(\boldsymbol{\theta}) - \frac{1}{\kappa(\boldsymbol{\theta})} \mathfrak{F}(\boldsymbol{\theta}) \mathfrak{F}(\boldsymbol{\theta})^{T} \\ &= \quad \frac{1}{\sigma^{\tau+2}} \frac{1}{\left(\sqrt{2\pi}\right)^{\tau}} \frac{1}{\sqrt{1+\tau}} \left( \begin{pmatrix} \frac{1}{1+\tau} & 0 \\ 0 & \frac{2+\tau^{2}}{\left(1+\tau\right)^{2}} \end{pmatrix} - \begin{pmatrix} 0 & 0 \\ 0 & \frac{\tau^{2}}{\left(\tau+1\right)^{2}} \end{pmatrix} \right) \\ &= \quad \frac{1}{\sigma^{\tau+2}} \frac{1}{\left(\sqrt{2\pi}\right)^{\tau}} \frac{1}{\sqrt{1+\tau}} \begin{pmatrix} \frac{1}{1+\tau} & 0 \\ 0 & \frac{2}{\left(\tau+1\right)^{2}} \end{pmatrix} \end{split}$$

Now we have,

• The matrix , *G*(*θ*0)*TSτ*(*θ*0)−1*G*(*θ*0) -−<sup>1</sup> (*G*(*θ*) = (0, 1) *T*)

$$\begin{aligned} \mathcal{G}(\boldsymbol{\theta}\_{0})^{\mathrm{T}}\mathcal{S}\_{\boldsymbol{\pi}}(\boldsymbol{\theta}\_{0})^{-1}\mathcal{G}(\boldsymbol{\theta}\_{0}) &= \quad \left( \begin{array}{c} 0 & 1 \end{array} \right) \left( \frac{1}{\sigma^{\tau+2}} \frac{1}{\left( \sqrt{2\pi} \right)^{\tau}} \frac{1}{\sqrt{1+\tau}} \begin{pmatrix} \frac{1}{1+\tau} & 0\\ 0 & \frac{2}{\left(\tau+1\right)^{2}} \end{pmatrix} \right)^{-1} \begin{pmatrix} 0\\ 1 \end{pmatrix} \\ &= \quad \frac{1}{2}\sigma^{2}\sigma^{\tau}(\tau+1)^{\frac{5}{2}} \Big( \sqrt{2}\sqrt{\pi} \Big)^{\tau} \end{aligned}$$

• The matrix *Qτ*(*θ*0) = *S*−<sup>1</sup> *<sup>τ</sup>* (*θ*0)*G*(*θ*0) *GT*(*θ*0)*S*−<sup>1</sup> *<sup>τ</sup>* (*θ*0)*G*(*θ*0) −<sup>1</sup>

$$\begin{aligned} \mathbf{Q}\_{\tau}(\boldsymbol{\theta}\_{0}) &= \left( \frac{1}{\sigma^{\tau+2}} \frac{1}{\left(\sqrt{2\pi}\right)^{\tau}} \frac{1}{\sqrt{1+\tau}} \begin{pmatrix} \frac{1}{1+\tau} & 0\\ 0 & \frac{2}{\left(\tau+1\right)^{2}} \end{pmatrix} \right)^{-1} \begin{pmatrix} 0\\ 1 \end{pmatrix} \left( \frac{1}{2} \sigma^{2} \sigma^{\tau} (\tau+1)^{\frac{5}{2}} \left(\sqrt{2}\sqrt{\pi}\right)^{\tau} \right)^{-1} \\ &= \quad \begin{pmatrix} 0\\ 1 \end{pmatrix} \end{aligned}$$

• The matrix *Bτ*(*θ*0) = *Sτ*(*θ*0)−1*G*(*θ*0) , *G*(*θ*0)*TSτ*(*θ*0)−1*G*(*θ*0) -−<sup>1</sup> *G*(*θ*0)*TSτ*(*θ*0)−<sup>1</sup> = *Qτ*(*θ*0)*G*(*θ*0)*TSτ*(*θ*0)−<sup>1</sup>

$$\begin{aligned} \mathcal{B}\_{\tau}(\theta\_0) &= \begin{array}{rcl} \mathcal{Q}\_{\tau}(\theta\_0) \mathcal{G}^{\top}(\theta\_0) \mathcal{S}\_{\tau}^{-1}(\theta\_0) &= \begin{pmatrix} 0 \\ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} \left( \frac{1}{\sigma^{\tau+2}} \frac{1}{\left(\sqrt{2\pi}\right)^{\tau}} \frac{1}{\sqrt{1+\tau}} \begin{pmatrix} \frac{1}{1+\tau} & 0 \\ 0 & \frac{2}{\left(\tau+1\right)^2} \end{pmatrix} \right)^{-1} \\ &= \begin{pmatrix} 0 & 0 \\ 0 & \frac{1}{2} \sigma^2 \sigma^{\tau} (\tau+1)^{\frac{5}{2}} \left(\sqrt{2}\sqrt{\pi}\right)^{\tau} \end{pmatrix} \end{aligned}$$

• The matrix *<sup>M</sup>γ*,*τ*(*θ*0) <sup>=</sup> *<sup>S</sup>γ*(*θ*0) *<sup>κ</sup>γ*(*θ*0) *<sup>B</sup>τ*(*θ*0)*Kτ*(*θ*0)*Bτ*(*θ*0)

$$\begin{split} \mathbf{M}\_{\gamma,\tau}(\boldsymbol{\theta}\_{0}) &= \begin{array}{cc} \sigma^{\gamma} \left(\sqrt{2\pi}\right)^{\gamma} \sqrt{1+\gamma} \\ \hline \sigma^{\gamma+2} \end{array} \frac{1}{\left(\sqrt{2\pi}\right)^{\gamma}\sqrt{1+\gamma}} \begin{pmatrix} 1 & 0 \\ 0 & \frac{2}{\left(\gamma+1\right)^{\tau}} \end{pmatrix} \\ &\times \begin{pmatrix} 0 & 0 \\ 0 & \frac{1}{2}\sigma^{2}\sigma^{\tau} \left(\tau+1\right)^{\frac{5}{2}} \left(\sqrt{2}\sqrt{\pi}\right)^{\tau} \end{pmatrix} \\ &\times \frac{1}{\sigma^{2}} \frac{1}{\left(\sigma\sqrt{2\pi}\right)^{2\tau}\left(1+2\tau\right)^{3/2}} \begin{pmatrix} 1 & 0 \\ 0 & \frac{3\tau^{2}+2+4\tau}{\left(1+\tau\right)^{2}\left(1+2\tau\right)} \end{pmatrix} \\ &\times \begin{pmatrix} 0 & 0 \\ 0 & \frac{1}{2}\sigma^{2}\sigma^{\tau} \left(\tau+1\right)^{\frac{5}{2}} \left(\sqrt{2}\sqrt{\pi}\right)^{\tau} \end{pmatrix} \\ &= \begin{pmatrix} 0 & 0 \\ 0 & \frac{1}{2}\frac{(\tau+1)^{3}}{\left(\gamma+1\right)^{2}\left(2\tau+1\right)^{\frac{3}{2}}} \left(3\tau^{2}+4\tau+2\right) \end{pmatrix}. \end{split}$$
