Closed-form maximum likelihood estimator for generalized linear models in the case of categorical explanatory variables: application to insurance loss modeling

Abstract

Generalized linear models with categorical explanatory variables are considered and parameters of the model are estimated by an exact maximum likelihood method. The existence of a sequence of maximum likelihood estimators is discussed and considerations on possible link functions are proposed. A focus is then given on two particular positive distributions: the Pareto 1 distribution and the shifted log-normal distributions. Finally, the approach is illustrated on an actuarial dataset to model insurance losses.

Appendices

Proofs of Sect. 3

1.1 Proof for the one-variable case

Proof of Theorem 3.1

We have to solve the system

$$\begin{aligned} \left\{ \begin{array}{ll}S(\varvec{\vartheta }) = 0\\ \varvec{R}\varvec{\vartheta }=0. \end{array}\right. \end{aligned}$$

(23)

The system $S(\varvec{\vartheta })=0$ is

$$\begin{aligned} \left\{ \begin{array}{ll}\displaystyle \sum _{i=1}^n\ell '(\eta _i)\left( y_i - b'\circ \ell (\eta _i)\right) = 0\\ \displaystyle \sum _{i=1}^nx_i^{(2),j}\ell '(\eta _i)\left( y_i - b'\circ \ell (\eta _i)\right) = 0,\quad \forall j\in J. \end{array}\right. \end{aligned}$$

that is

$$\begin{aligned} \left\{ \begin{array}{ll}\displaystyle \sum _{j\in J}\ell '(\vartheta _{(1)}+\vartheta _{(2),j})\left( \sum _{i=1}^nx_i^{(2),j}y_i - m_jb'\circ \ell (\vartheta _{(1)} + \vartheta _{(2),j})\right) = 0\\ \ell '(\vartheta _{(1)}+\vartheta _{(2),j})\left( \displaystyle \sum _{i=1}^nx_i^{(2),j}y_i - m_jb'\circ \ell (\vartheta _{(1)} + \vartheta _{(2),j})\right) = 0,\quad \forall j\in J. \end{array}\right. \end{aligned}$$

The first equation in the previous system is redundancy, and

$$\begin{aligned} S(\varvec{\vartheta }) = 0 \Leftrightarrow \ell '(\vartheta _{(1)}+\vartheta _{(2),j})\left( \sum _{i=1}^nx_i^{(2),j}y_i - m_jb'\circ \ell (\vartheta _{(1)} + \vartheta _{(2),j})\right) = 0,\quad \forall j\in J. \end{aligned}$$

Hence if $Y_i$ takes values in $\mathbb {Y}\subset b'(\varLambda )$, and $\ell$ injective, we have

$$\begin{aligned} \vartheta _{(1)}+\vartheta _{(j)} = g(\overline{Y}_n^{(j)})\quad \forall j\in J. \end{aligned}$$

The system (23) is

$$\begin{aligned} \left\{ \begin{array}{ll}\varvec{Q}\varvec{\vartheta }= \varvec{g({\bar{Y}})}\\ \varvec{R}\varvec{\vartheta }=0. \end{array}\right. \Leftrightarrow \left( \begin{array}{c} \varvec{Q} \\ \varvec{R}\end{array}\right) \varvec{\vartheta }=\left( \begin{array}{c}\varvec{g({\bar{Y}})}\\ 0\end{array}\right) . \end{aligned}$$

(24)

Let us compute the determinant of the matrix $M_d = \left( \begin{array}{c} \varvec{Q} \\ \varvec{R}\end{array}\right)$. Consider $\varvec{R} = (r_0,r_1,\ldots ,r_d)$. We have

$$\begin{aligned} M_d = \left( \begin{array}{c@{\quad }c} \varvec{1}_d &{} I_d \\ r_0 &{} \varvec{r} \end{array}\right) = \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} 1 &{} 1 &{} 0 &{} \dots \\ 1 &{} 0 &{} 1 &{} 0 \\ \vdots &{} \vdots &{} \ddots &{} \ddots &{} \ddots \\ 1 &{} 0 &{} \dots &{} 0 &{} 1 \\ r_0 &{} r_1 &{} &{} \dots &{} r_d \\ \end{array}\right) , \text { with } \varvec{r}= \left( \begin{array}{c@{\quad }cc} r_1 &{} \dots &{} r_d \\ \end{array}\right) , \varvec{1}_d = \left( \begin{array}{c} 1 \\ \vdots \\ 1 \end{array}\right) . \end{aligned}$$

The determinant can be computed recursively

$$\begin{aligned} \det (M_d) = r_d \left| \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} 1 &{} 1 &{} 0 &{} \dots \\ 1 &{} 0 &{} \ddots &{} 0 \\ \vdots &{} \vdots &{} \ddots &{} 1 \\ 1 &{} 0 &{} \dots &{} 0 \\ \end{array}\right| - \left| \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} 1 &{} 1 &{} 0 &{} \dots \\ 1 &{} 0 &{} \ddots &{} 0 \\ \vdots &{} \vdots &{} \ddots &{} 1 \\ r_0 &{} r_1 &{} \dots &{} r_{d-1} \\ \end{array}\right| = (-1)^{d+1} r_d - \det (M_{d-1}). \end{aligned}$$

Since $\det (M_1) = -r_0+ r_1$ and $\det (M_2) = -r_2 -(-r_0 +r_1) = r_0 - r_1 -r _2,$ we get $\det (M_d) = (-1)^d r_0+ (-1)^{d+1}(r_1+\dots + r_d) =(-1)^d( r_0 - r_1-\dots -r_d)$. This determinant is non zero as long as $r_0 \ne \sum _{j=1}^d r_j$.

Now we compute the inverse of matrix $M_d$ by a direct inversion.

$$\begin{aligned} \left( \begin{array}{c@{\quad }c} \varvec{1}_d &{} I_d \\ r_0 &{} \varvec{r} \end{array}\right) \left( \begin{array}{c@{\quad }c} \varvec{a}' &{} b \\ C &{} \varvec{d} \end{array}\right) = \left( \begin{array}{c@{\quad }c} I_d &{} \varvec{0} \\ \varvec{0}' &{} 1 \end{array}\right) \Leftrightarrow \left\{ \begin{array}{ll}\varvec{1}_d \varvec{a}' + I_d C = I_d \\ b \varvec{1}_d + I_d \varvec{d} = \varvec{0} \\ r_0 \varvec{a}' + \varvec{r} C = \varvec{0}' \\ b r_0 + \varvec{r} \varvec{d} = 1 \end{array}\right. \Leftrightarrow \left\{ \begin{array}{ll}C = I_d - \frac{1}{-r_0 + \varvec{r} \varvec{1}_d}\varvec{1}_d\varvec{r} \\ \varvec{d}= \frac{1}{-r_0+\varvec{r} \varvec{1}_d} \varvec{1}_d \\ \varvec{a}' = \frac{\varvec{r}}{-r_0 + \varvec{r} \varvec{1}_d} \\ b = \frac{-1}{-r_0+\varvec{r} \varvec{1}_d} \\ \end{array}\right. \end{aligned}$$

Let us check the inverse of $M_d$

$$\begin{aligned}&\left( \begin{array}{c@{\quad }c} \varvec{1}_d &{} I_d \\ r_0 &{} \varvec{r} \end{array}\right) \left( \begin{array}{c@{\quad }c} \frac{\varvec{r}}{-r_0 + \varvec{r} \varvec{1}_d} &{} \frac{-1}{-r_0+\varvec{r} \varvec{1}_d} \\ I_d - \frac{\varvec{1}_d\varvec{r}}{-r_0 + \varvec{r} \varvec{1}_d} &{} \frac{\varvec{1}_d}{-r_0+\varvec{r} \varvec{1}_d} \end{array}\right) \\&\qquad = \left( \begin{array}{c@{\quad }c} \frac{\varvec{1}_d \varvec{r}}{-r_0 + \varvec{r} \varvec{1}_d} + I_d - \frac{\varvec{1}_d\varvec{r} }{-r_0 + \varvec{r} \varvec{1}_d} &{} \frac{-\varvec{1}_d}{-r_0+\varvec{r} \varvec{1}_d} +\frac{\varvec{1}_d}{-r_0+\varvec{r} \varvec{1}_d}\\ r_0 \frac{\varvec{r}}{-r_0 + \varvec{r} \varvec{1}_d}+ \varvec{r} - \frac{\varvec{r}\varvec{1}_d\varvec{r}}{-r_0 + \varvec{r} \varvec{1}_d} &{} \frac{-r_0}{-r_0+\varvec{r} \varvec{1}_d}+ \frac{\varvec{r}\varvec{1}_d}{-r_0+\varvec{r} \varvec{1}_d} \end{array}\right) \\&\qquad = \left( \begin{array}{c@{\quad }c} I_d &{} 0 \\ 0 &{} 1 \end{array}\right) . \end{aligned}$$

So as long as $r_0 \ne \sum _{j=1}^d r_j$

$$\begin{aligned} \widehat{\varvec{\vartheta }}_n = \left( \begin{array}{c@{\quad }c} \frac{\varvec{r}}{-r_0 + \varvec{r} \varvec{1}_d} &{} \frac{-1}{-r_0+\varvec{r} \varvec{1}_d} \\ I_d - \frac{\varvec{1}_d\varvec{r}}{-r_0 + \varvec{r} \varvec{1}_d} &{} \frac{\varvec{1}_d}{-r_0+\varvec{r} \varvec{1}_d} \end{array}\right) \left( \begin{array}{c} \varvec{g}({\varvec{{\bar{Y}}}})\\ 0 \end{array}\right) = \left( \begin{array}{c} \frac{\varvec{r} \varvec{g({\bar{Y}})}}{-r_0 + \varvec{r} \varvec{1}_d} \\ {\varvec{g({\bar{Y}})}} - \varvec{1}_d\frac{\varvec{r} \varvec{g({\bar{Y}})}}{-r_0 + \varvec{r} \varvec{1}_d} \end{array}\right) . \end{aligned}$$

In an other way, the system (24) is equivalent to

$$\begin{aligned} (\varvec{Q}',\varvec{R}')\left( \begin{array}{c} \varvec{Q} \\ \varvec{R}\end{array}\right) \varvec{\vartheta }= \varvec{Q}' \varvec{g({\bar{Y}})}, \end{aligned}$$

and for $(\varvec{Q}\, \varvec{R})$ of full rank, the matrix $(\varvec{Q}'\varvec{Q} + \varvec{R}'\varvec{R})$ is invertible and ${\varvec{\vartheta }} = (\varvec{Q}'\varvec{Q} + \varvec{R}'\varvec{R})^{-1}\varvec{Q}'\varvec{g({\bar{Y}})}.$$\square$

Examples—Choice of the contrast vector $\varvec{R}$

• 1. Taking $r_0=1, \varvec{r}=\varvec{0}$ leads to $-r_0 + \varvec{r} \varvec{1}_d=-1 \Rightarrow \widehat{\varvec{\vartheta }}_n = \left( \begin{array}{c} 0 \\ \varvec{g({\bar{Y}})} \end{array}\right) .$

• 2. Taking $r_0=0, \varvec{r}=(1,\varvec{0})$ leads to

  $$\begin{aligned} -r_0 + \varvec{r} \varvec{1}_d=1 \Rightarrow \widehat{\varvec{\vartheta }}_n = \left( \begin{array}{c} g({\bar{Y}}_n^{(1)})\\ 0\\ g({\bar{Y}}_n^{(2)}) - g({\bar{Y}}_n^{(1)})\\ \vdots \\ g({\bar{Y}}_n^{(d)}) - g({\bar{Y}}_n^{(1)})) \end{array}\right) . \end{aligned}$$

• 3. Taking $r_0=0, \varvec{r}=\varvec{1}$ leads to

  $$\begin{aligned} -r_0 + \varvec{r} \varvec{1}_d=d \Rightarrow \widehat{\varvec{\vartheta }}_n = \left( \begin{array}{c} \overline{\varvec{g({\bar{Y}})}}\\ g({\bar{Y}}_n^{(1)}) - \overline{\varvec{g({\bar{Y}})}}\\ \dots \\ g({\bar{Y}}_n^{(d)}) - \overline{\varvec{g({\bar{Y}})}} \end{array}\right) , \text { with } \overline{\varvec{g({\bar{Y}})}} = \dfrac{1}{d}\displaystyle \sum _{j=1}^dg(\overline{Y}_n^{(j)}). \end{aligned}$$

Proof of Remark 3.4

We have to solve the system

$$\begin{aligned} S(\vartheta ) = 0 \Leftrightarrow \displaystyle \sum _{i=1}^n\ell '(\eta )\left( y_i - b'\circ \ell (\eta )\right) = 0. \end{aligned}$$

If $\ell$ is injective, the system simplifies to

$$\begin{aligned} \displaystyle \sum _{i=1}^n y_i - n b'\circ (b^\prime )^{-1}\circ g^{-1}(\eta ) = 0 \Leftrightarrow \eta = g\left( \begin{array}{c}{\overline{y}}_n\end{array}\right) \Leftrightarrow \theta = g\left( \begin{array}{c}{\overline{y}}_n\end{array}\right) . \end{aligned}$$

$\square$

Proof of Remark 3.5

Let $Y_i$ from the exponential family $F_{exp}(a,b,c,\lambda ,\phi )$. It is well known, that the moment generating function of $Y_i$ is

$$\begin{aligned} \mathbf {E}e^{t Y_i} =\exp \left( \frac{b(\lambda +ta(\phi )) - b(\lambda )}{a(\phi )}\right) . \end{aligned}$$

Hence, the moment generating function of the average ${\overline{Y}}_m$ is

$$\begin{aligned} M_{{\overline{Y}}_m}(t) = \left( \exp \left( \frac{b(\lambda +\frac{t}{m} a(\phi )) - b(\lambda )}{a(\phi )}\right) \right) ^m = \exp \left( \frac{b(\lambda +t a(\phi )/m) - b(\lambda )}{a(\phi )/m}\right) . \end{aligned}$$

So we get back to a known result that ${\overline{Y}}_m$ belongs to the exponential family $F_{exp}(x\mapsto a(x)/m,b,c,\lambda ,\phi )$ (e.g. McCullagh and Nelder 1989).

In our setting, random variables in the average $\overline{Y}_n^{(j)}$ are i.i.d. with functions a, b, c and parameters $\lambda =\ell (\vartheta _{(1)}+\vartheta _{(j)})$ and $\phi$. And ${\overline{Y}}_n^{(j)}$ also belongs to the exponential family with the same parameter but with the function ${\bar{a}}:x\mapsto a(x)/m_j$. In particular,

$$\begin{aligned} \mathbf {E}{\overline{Y}}_n^{(j)} = b'(\ell (\vartheta _{(1)}+\vartheta _{(j)})) = g^{-1}(\vartheta _{(1)}+\vartheta _{(j)}),~ \text{ Var }{\overline{Y}}_n^{(j)} = \frac{a(\phi )}{m_j} b''(\ell (\vartheta _{(1)}+\vartheta _{(j)})). \end{aligned}$$

But the computation of $\mathbf {E}g({\overline{Y}}_n^{(j)})$ remains difficult unless g is a linear function. By the strong law of large numbers, as $m_j\rightarrow +\,\infty$, the estimator is consistent since

$$\begin{aligned} {\overline{Y}}_n^{(j)}{\mathop {\underset{n\rightarrow +\infty }{\longrightarrow }}\limits ^{\text {a.s.}}} g^{-1}(\vartheta _{(1)}+\vartheta _{(j)}) \Rightarrow g({\overline{Y}}_n^{(j)}){\mathop {\underset{n\rightarrow +\infty }{\longrightarrow }}\limits ^{\text {a.s.}}} g(g^{-1}(\vartheta _{(1)}+\vartheta _{(j)}))=\vartheta _{(1)}+\vartheta _{(j)}. \end{aligned}$$

By the Central Limit Theorem (i.e. ${\overline{Y}}_n^{(j)}$ converges in distribution to a normal distribution) and using the Delta Method, we obtain that the following

$$\begin{aligned}&\sqrt{m_j}\left( g({\overline{Y}}_n^{(j)}) - \vartheta _{(1)}+\vartheta _{(j)}\right) {\mathop {\underset{n\rightarrow +\infty }{\longrightarrow }}\limits ^{\mathcal {L}}} \\&\quad {\mathcal {N}}\left( 0, a(\phi )b''(\ell (\vartheta _{(1)}+\vartheta _{(j)})) g'(g^{-1}(\vartheta _{(1)}+\vartheta _{(j)}))^2 \right) . \end{aligned}$$

$\square$

Proof of Corollaries 3.1

The log likelihood of $\widehat{\varvec{\vartheta }}_n$ is defined by

$$\begin{aligned} \log L(\widehat{\varvec{\vartheta }}_n\,|\,\underline{\varvec{y}}) = \frac{1}{a(\phi )}\sum _{i=1}^n \left( y_i \ell ({\widehat{\eta }}_i) - b(\ell ({\widehat{\eta }}_i))\right) + \sum _{i=1}^nc(y_i,\phi ). \end{aligned}$$

In fact, we must be verified than $\ell ({\widehat{\eta }}_i)$ does not depend on g function. If we consider $\widehat{\varvec{\vartheta }}_n$ defined by (8), we have $\varvec{Q}\widehat{\varvec{\vartheta }}_n = \varvec{g({\bar{y}})}$ , since $\widehat{\varvec{\vartheta }}_n$ is solution of the system (23), i.e. $\varvec{Q}(\varvec{Q}'\varvec{Q} + \varvec{R}'\varvec{R})^{-1}\varvec{Q}'=I$ Using ${\widehat{\eta }}_i= (\varvec{Q}\widehat{\varvec{\vartheta }}_n)_j$ for i such that $x_i^{(2),j}=1$ we obtain

$$\begin{aligned} \ell ({\widehat{\eta }}_i)= \displaystyle \sum _{j=1}^d\ell \circ g(\bar{y}_n^{(j)})x_i^{(2),j} = \displaystyle \sum _{j=1}^d\ell \circ \ell ^{-1}\circ (b')^{-1}({\bar{y}}_n^{(j)})x_i^{(2),j} = \displaystyle \sum _{j=1}^d (b')^{-1}({\bar{y}}_n^{(j)})x_i^{(2),j}, \end{aligned}$$

and

$$\begin{aligned} \log L(\widehat{\varvec{\vartheta }}_n\,|\,\underline{\varvec{y}}) = \frac{1}{a(\phi )}\sum _{j=1}^d\sum _{i, x_i^{(2)}=v_j} \left( y_i (b')^{-1}\left( {\overline{y}}_n^{(j)}\right) - b\left( \left( b'\right) ^{-1}\left( {\overline{y}}_n^{(j)}\right) \right) \right) + \sum _{i=1}^nc(y_i,\phi ). \end{aligned}$$

In the same way,

$$\begin{aligned} \widehat{\mathbf {E}Y_i}= & {} b'(\ell ({\widehat{\eta }}_i)) = \sum _{j=1}^d \bar{y}_n^{(j)}x_i^{(2),j}, \quad \widehat{\text{ Var }Y_i} = a(\phi )b''(\ell ({\widehat{\eta }}_i)) \\= & {} a(\phi )\sum _{j=1}^d b''\circ (b')^{-1}({\bar{y}}_n^{(j)})x_i^{(2),j}. \end{aligned}$$

$\square$

1.2 Proof for the two-variable case

Proof of Theorem 3.2

The system $S(\varvec{\vartheta })=0$ is

$$\begin{aligned} \left\{ \begin{array}{ll}\displaystyle \sum _{i=1}^n\ell '(\eta _i)\left( y_i - b'\circ \ell (\eta _i)\right) = 0\\ \displaystyle \sum _{i=1}^nx_i^{(3),l}\ell '(\eta _i)\left( y_i - b'\circ \ell (\eta _i)\right) = 0,\quad \forall l\in L\\ \displaystyle \sum _{i=1}^nx_i^{(2),k}\ell '(\eta _i)\left( y_i - b'\circ \ell (\eta _i)\right) = 0,\quad \forall k\in K\\ \displaystyle \sum _{i=1}^nx_i^{kl}\ell '(\eta _i)\left( y_i - b'\circ \ell (\eta _i)\right) = 0,\quad \forall (k,l)\in KL^\star . \end{array}\right. \end{aligned}$$

that is

$$\begin{aligned} \left\{ \begin{array}{ll}\displaystyle \sum _{(k,l)\in KL^\star }\ell '(\vartheta _{(1)}+\vartheta _{(2),k} + \vartheta _{(3),l} + \vartheta _{kl})\left( \sum _{i=1}^nx_i^{(k,l)}y_i - m_{k,l}b'\circ \ell (\vartheta _{(1)}+\vartheta _{(2),k} + \vartheta _{(3),l} + \vartheta _{kl})\right) = 0\\ \displaystyle \sum _{k\in K_l^\star }\ell '(\vartheta _{(1)}+\vartheta _{(2),k} + \vartheta _{(3),l} + \vartheta _{kl})\left( \sum _{i=1}^nx_i^{(k,l)}y_i - m_{k,l}b'\circ \ell (\vartheta _{(1)}+\vartheta _{(2),k} + \vartheta _{(3),l} + \vartheta _{kl})\right) = 0\quad \forall l\in L\\ \displaystyle \sum _{l\in L_k^\star }\ell '(\vartheta _{(1)}+\vartheta _{(2),k} + \vartheta _{(3),l} + \vartheta _{kl})\left( \sum _{i=1}^nx_i^{(k,l)}y_i - m_{k,l}b'\circ \ell (\vartheta _{(1)}+\vartheta _{(2),k} + \vartheta _{(3),l} + \vartheta _{kl})\right) = 0\quad \forall k\in K\\ \ell '(\vartheta _{(1)}+\vartheta _{(2),k} + \vartheta _{(3),l} + \vartheta _{kl})\left( \displaystyle \sum _{i=1}^nx_i^{(k,l)}y_i - m_{k,l}b'\circ \ell (\vartheta _{(1)}+\vartheta _{(2),k} + \vartheta _{(3),l} + \vartheta _{kl})\right) = 0\quad \forall (k,l)\in KL^\star . \end{array}\right. \end{aligned}$$

The system have exactly $1+d_2+d_3$ redundancies, and $S(\varvec{\vartheta })=0$ reduces to

$$\begin{aligned}&\ell '(\vartheta _{(1)}+\vartheta _{(2),k} + \vartheta _{(3),l} + \vartheta _{kl})\left( \displaystyle \sum _{i=1}^nx_i^{(k,l)}y_i - m_{k,l}b'\circ \ell (\vartheta _{(1)}+\vartheta _{(2),k} \right. \nonumber \\&\quad \left. + \vartheta _{(3),l} + \vartheta _{kl}) \right) = 0\quad \forall (k,l)\in { KL}^\star . \end{aligned}$$

Hence the system has rank ${ KL}^\star$ and if $Y_i$ takes values in $\mathbb {Y}\subset b'(\varLambda )$, and $\ell$ injective, we have

$$\begin{aligned} \vartheta _{(1)}+\vartheta _{(2),k} + \vartheta _{(3),l} + \vartheta _{kl} = g({\bar{Y}}_n^{(k,l)})\quad \forall (k,l)\in KL^\star . \end{aligned}$$

In the same way of proof of Theorem 3.1, we have to solve

$$\begin{aligned} \left\{ \begin{array}{ll}\varvec{Q}\varvec{\vartheta }= \varvec{g({\bar{Y}})}\\ \varvec{R}\varvec{\vartheta }=\varvec{0}. \end{array}\right. \end{aligned}$$

(25)

that is, because $\varvec{Q}\varvec{Q}'+\varvec{R}\varvec{R}'$ is full rank, in the same way of proof of Theorem 3.1

$$\begin{aligned} {\varvec{\vartheta }} = (\varvec{Q}'\varvec{Q} + \varvec{R}'\varvec{R})^{-1}\varvec{Q}'\varvec{g({\bar{Y}})}. \end{aligned}$$

In that case, the MLE solves a least square problem with response variable $\varvec{g({\bar{Y}})}$, explanatory variable $\varvec{Q}$ under a linear constraint $\varvec{R}$.

• 1. Under linear contrasts (${\tilde{C}}_0$), the model (10) is equivalent to model (6) with $J=KL^\star$ modalities. Hence the solution is evident.

• 2. Under linear contrasts (${\tilde{C}}_\varSigma$ ), the system

  $$\begin{aligned} \vartheta _{(1)}+\vartheta _{(2),k} + \vartheta _{(3),l} + \vartheta _{kl} = g({\bar{Y}}_n^{(k,l)})\quad \forall (k,l)\in KL^\star \end{aligned}$$

  implies that

  $$\begin{aligned} \sum _{(k,l)\in KL^\star }m_{k,l}(\vartheta _{(1)}+\vartheta _{(2),k} + \vartheta _{(3),l} + \vartheta _{kl}) = \sum _{(k,l)\in KL^\star }m_{k,l} g({\bar{Y}}_n^{(k,l)}). \end{aligned}$$

  Using

  $$\begin{aligned} \sum _{(k,l)\in KL^\star }m_{k,l}= & {} n,\quad \sum _{(k,l)\in KL^\star }m_{k,l}\vartheta _{(2),k} = \sum _{k\in K}\sum _{l\in L^\star _k}m_{k,l}\vartheta _{(2),k}\nonumber \\= & {} \sum _{k\in K}m^{(2)}_k\vartheta _{(2),k}= 0,\\ \sum _{(k,l)\in KL^\star }m_{k,l}\vartheta _{(3),l}= & {} \sum _{l\in L}\sum _{k\in K^\star _l}m_{k,l}\vartheta _{(3),l}= \sum _{l\in L}m^{(3)}_l\vartheta _{(3),l}= 0,\nonumber \\&\quad \sum _{(k,l)\in KL^\star }m_{k,l}\vartheta _{kl} =0, \end{aligned}$$

  we get $\vartheta _{(1)} = \dfrac{1}{n}\displaystyle \sum \nolimits _{(k,l)\in KL^\star }m_{k,l} g({\bar{Y}}_n^{(k,l)}).$ In the same way, taking summation over $K^\star _l$ for $l\in L$ and over $L^\star _k$ for $k\in K$, we found $\vartheta _{(2),k}$ and $\vartheta _{(3),l}$, and then $\vartheta _{kl}$.

With main effect only, the system $S(\varvec{\vartheta })=0$ is

$$\begin{aligned} \left\{ \begin{array}{ll}\displaystyle \sum _{i=1}^n\ell '(\eta _i)y_i = \sum _{i=1}^n g^{-1}(\eta _i)\ell '(\eta _i) \\ \displaystyle \sum _{i=1}^nx_i^{(3),l}\ell '(\eta _i) y_i = \sum _{i=1}^nx_i^{(3),l} g^{-1}(\eta _i)\ell '(\eta _i) \quad \forall l\in L\\ \displaystyle \sum _{i=1}^nx_i^{(2),k}\ell '(\eta _i)y_i = \sum _{i=1}^nx_i^{(2),k} g^{-1}(\eta _i)\ell '(\eta _i),\quad \forall k\in K \end{array}\right. \end{aligned}$$

There are $1+d_2+d_3$ equations for $1+d_2+d_3$ parameters, but each explanatory variable are colinear. So, the two additional constraints $\varvec{R}\varvec{\vartheta }=0$ ensures that a solution exist for the remaining $d_2+d_3-1$ parameters. Using $\sum _k x_i^{(2),k}=1$, the second set of equations becomes $\forall l\in L$

$$\begin{aligned}&\displaystyle \sum _{k\in K}\ell '(\vartheta _{(1)}+\vartheta _{(2),k} + \vartheta _{(3),l}) {\bar{y}}_n^{(k,l)} m_{k,l} \\&\quad = \sum _{k\in K} g^{-1}(\vartheta _{(1)}+\vartheta _{(2),k} + \vartheta _{(3),l}) \ell '(\vartheta _{(1)}+\vartheta _{(2),k} + \vartheta _{(3),l}) m_{k,l} \end{aligned}$$

Similarly, the third set of equations becomes $\forall k\in K$

$$\begin{aligned}&\displaystyle \sum _{l\in L}\ell '(\vartheta _{(1)}+\vartheta _{(2),k} + \vartheta _{(3),l}) {\bar{y}}_n^{(k,l)} m_{k,l} \\&\quad = \sum _{l\in L} g^{-1}(\vartheta _{(1)}+\vartheta _{(2),k} + \vartheta _{(3),l}) \ell '(\vartheta _{(1)}+\vartheta _{(2),k} + \vartheta _{(3),l}) m_{k,l} \end{aligned}$$

Even with a canonical link $\ell (x)=x$ so that $\ell '(x)=1$, this system is not a least-square problem for a nonlinear g function. $\square$

Calculus of the Log-likelihoods appearing in Sects. 4 and 5

Consider the Pareto GLM described on (13) and (15). The b function is $b(\lambda ) = -\log (\lambda )$, using corollary 3.1 we have $\ell (\hat{\eta }_i) = (b')^{-1}(\overline{z}_n^{(j)})=-(\overline{z}_n^{(j)})^{-1}$ for j such that $x_i^{(2),j}=1$ and

$$\begin{aligned} \log L(\widehat{\varvec{\vartheta }}_n\,|\,\underline{\varvec{z}}) = \sum _{j=1}^d\sum _{i, x_i^{(2),j}=1} \left( z_i/{\overline{z}}_n^{(j)} - \log \left( -{\overline{z}}_n^{(j)} \right) \right) = n -\sum _{j=1}^d m_j \log \left( -{\overline{z}}_n^{(j)} \right) . \end{aligned}$$

Compute the original log likelihood of Pareto 1 distribution:

$$\begin{aligned} \log L(\varvec{\vartheta }\,|\,\underline{\varvec{y}}) = \sum _{i=1}^n \big (\log \ell (\eta _i) + \ell (\eta _i)\log \mu - (\ell (\eta _i) +1)\log y_i \big ). \end{aligned}$$

Hence with $z_i=-\log (y_i/\mu )$,

$$\begin{aligned} \log L(\widehat{\varvec{\vartheta }}_n\,|\,\underline{\varvec{y}})= & {} \sum _{j=1}^d\sum _{i, x_i^{(2),j}=1}\left( -\log (-\overline{z}_n^{(j)}) -\frac{\log \mu }{{\overline{z}}_n^{(j)}} + \frac{\log (y_i)}{ {\overline{z}}_n^{(j)}} - \log y_i \right) \\= & {} n - \sum _{j=1}^d m_j\log (-{\overline{z}}_n^{(j)}) - \sum _{i=1}^n\log y_i =\log L(\widehat{\varvec{\vartheta }}_n\,|\,\underline{\varvec{z}})- \sum _{i=1}^n\log y_i. \end{aligned}$$

Now consider the shifted log-normal GLM described on (18) and (19). Here, the b function is $b(\lambda )=\lambda ^2/2$, hence using Corollary 3.1, we have $\ell (\hat{\eta }_i) = (b')^{-1}(\overline{z}_n^{(j)})=\overline{z}_n^{(j)}$ for j such that $x_i^{(2),j}=1$ and Eq. (21) holds.

Let us compute the original log likelihood of the shifted log normal distribution:

$$\begin{aligned} \log L(\varvec{\vartheta }\,|\,\underline{\varvec{y}})= & {} \sum _{i=1}^n\left( - \log (x_i-\mu ) - \log (\sqrt{2\pi \phi }) -\dfrac{(\log (x_i-\mu ) - \ell (\eta _i))^2}{2\phi }\right) \\= & {} - \sum _{i=1}^n z_i - n\log (\sqrt{2\pi \phi }) - \sum _{i=1}^n \dfrac{(z_i - \ell (\eta _i))^2}{2\phi }, \end{aligned}$$

with $z_i=\log (y_i-\mu )$. Hence

$$\begin{aligned} \log L(\widehat{\varvec{\vartheta }}\,|\,\underline{\varvec{y}})= & {} - \sum _{i=1}^n z_i - n\log (\sqrt{2\pi \phi }) - \frac{1}{2\phi }\sum _{j=1}^d\sum _{i, x_i^{(2),j}=1} (z_i - \overline{z}_n^{(j)})^2. \end{aligned}$$

Using ${\widehat{\phi }} = \frac{1}{n}\sum _{j\in J}\sum _{i, x_i^{(2),j}=1}\left( z_i - {\bar{z}}_n^{(j)}\right) ^2$ leads to the desired result.

Link functions and descriptive statistics

See Fig. 4 and Table 10.

Fig. 4

Graphs of link functions

Table 10 Empirical quantiles and moments (in euros)