A generalization of the Thurstone method for multiple choice and incomplete paired comparisons

Abstract

A ranking method based on paired comparisons is proposed. The object’s characteristics are considered as random variables and the observers judge about their differences. The differences are classified. More than two classes are allowed. Assuming Gauss distributed latent random variables we set up the likelihood function and estimate the parameters by the maximum likelihood method. The rank of the objects is the order of the expectations. We analyse the log-likelihood function and provide reasonable conditions for the existence of the maximum value and the uniqueness of the maximizer. Some illustrative examples are also presented. The method can be applied in case of incomplete comparisons as well. It allows constructing confidence intervals for the probabilities and testing the hypothesis that there are no significant differences between the expectations.

Appendices

Appendix A: Preparation

We introduce the new variables $m_{i,j}=m_{i}-m_{j},$$i=1,{\ldots },n-1,j=i+1,{\ldots },n$ and we investigate the likelihood and the log-likelihood function in the function of these variables. Let

$$\begin{aligned}&L\left( A|m_{1,2},{\ldots },m_{n-1,n},d\right) \nonumber \\&\quad = \prod _{i=1}^{n-1}\prod _{j=i+1}^{n}\varPhi \left( -(s-2)d-m_{i,j}\right) ^{A_{i,j,1}} \nonumber \\&\qquad \cdot \prod _{i=1}^{n-1}\prod _{j=i+1}^{n}\prod _{k=2}^{s-1}\left( \varPhi \left( (-s+2k)d-m_{i,j}\right) -\varPhi \left( \left( -s+2\left( k-1\right) \right) d-m_{i,j}\right) \right) ^{A_{i,j,k}}\nonumber \\&\qquad \cdot \prod _{i=1}^{n-1}\prod _{j=i+1}^{n} \left( 1-\varPhi \left( \left( s-2\right) d-m_{i,j}\right) \right) ^{A_{i,j,s}}. \end{aligned}$$

(51)

and its logarithm

$$\begin{aligned} \log L(A|m_{1,2},m_{1,3},{\ldots },m_{n-1,n},d) \end{aligned}$$

(52)

subject to

$$\begin{aligned} m_{i,j}+m_{j,k}=m_{i,k\text { }},1\le i<j<k\le n. \end{aligned}$$

(53)

This way we increase the number of variables, but there are linear relations among the new variables. The set defined by (53) is convex and closed. $m_{i,j}\in {\mathbb {R}},$$0<d$ and we optimize (51) or (52) restricted to the set defined by (53) for all triplets $1\le i<j<k\le n$.

The existence of the maximum is proved by Weierstrass’ theorem. We remark that all the factors in the product are between 0 and 1, therefore 1 is an upper bound, 0 is a lower bound for the likelihood function. It is obvious that all factors are continuous in all variables. We prove that under certain assumptions the values of the factors of the likelihood function are close to zero outside of a bounded closed set, therefore the maximization can be executed on a bounded closed set where, by Weierstrass’ theorem, a continuous function takes its maximal value.

The proof of the uniqueness of the maximizer relies on the log-concave property of all factors and the strictly log-concave property of some of them. We use the following propositions: the optimal set of a concave function maximized on a convex set is convex, moreover, if the function is strictly concave, then its maximizer is unique supposing that the maximum exists (Boyd and Vandenberghe 2004). To apply these propositions first we investigate the strictly concave property of (52) in d, in all the variables $m_{i,j}$ separately, then all together. We provide conditions on $A_{i,j,k}$ for the strictly log-concavity of (52). These guarantee the uniqueness of the maximizer with (53). Finally we return to the original variables $(m_{1},m_{2},{\ldots }$,$m_{n},d)$

Appendix B: The existence of the maximal value

First we prove that the likelihood function has a maximal value. We consider the likelihood function $L(A|m_{1,2},{\ldots },m_{n-1,n},d)$ as a function mapping from ${\mathbb {R}}^{n(n-1)/2}\hbox {x}{\mathbb {R}}_{0}^{+}$ to ${\mathbb {R}}$ and we provide conditions under which it can be maximized in a convex closed bounded set. We separate the variable d from zero and we find reasonable upper bounds for the variable d and for the absolute value of $m_{i,j}$ when searching for the maximum value. For the sake of simplicity, while investigating the likelihood function we allow $d=0$ as well, in this case the value of the likelihood function is 0. We note that all factors are nonnegative and less than or equal to 1, and $L(A|m_{1,2},{\ldots },m_{n-1,n},d)$ is less than or equal to any of its factors. Moreover, $L(A|0,0,{\ldots },0,1)$ is positive and its argument satisfies (53). Let $0<\varepsilon <L(A|0,0,{\ldots },0,1)$.

The purpose of the first lemma is to separate the value of d from zero.

Lemma 1

Suppose that for a fixed pair $i<j$ and $1<k<s$ the inequality $0<A_{i,j,k}$ is satisfied. Then, there exists a value $0<K_{\varepsilon }^{a}$ for which

$$\begin{aligned} 0\le L(A|m_{1,2},{\ldots },m_{n-1,n},d)<\varepsilon , \end{aligned}$$

(54)

if $0\le d<K_{\varepsilon }^{a}$.

Proof

The proof is elementary, it is left to the reader.

Lemma 1 assures that the maximal value can be attained only such values of d for which $K_{\varepsilon }^{a}\le d$ is satisfied. A lower bound has been built for the variable d.

Now, let us see an upper bound for it, and together an upper bound for the absolute value for one of the differences.

Fix the indices $i<j$ for which the assumption $0<A_{i,j,l}$, $0<A_{i,j,k} ,1<l<k-1<s-1$ is satisfied. Now let us deal with large values of $\left| m_{i,j}\right|$ and d.

Lemma 2

Fix the pair $1\le$$i<j\le n$. If there exist two indices l and k, $1<l<k-1<s-1$ for which $0<A_{i,j,l}$, $0<A_{i,j,k}$ hold, then, for a sufficiently large value of $K_{\varepsilon }$

$$\begin{aligned} \left( \varPhi (a_{l}-m_{i,j})-\varPhi (a_{l-1}-m_{i,j})\right) \left( \varPhi (a_{k}-m_{i,j})-\varPhi (a_{k-1}-m_{i,j})\right) <\varepsilon \end{aligned}$$

(55)

holds if

$$\begin{aligned} K_{\varepsilon }<d \end{aligned}$$

(56)

or

$$\begin{aligned} K_{\varepsilon }<\left| m_{i,j}\right| . \end{aligned}$$

(57)

Proof

Let $x=a_{l-1}-m_{i,j}$, $0<y=2d$. Then $a_{l}-m_{i,j}=x+y$, $a_{k-1} -m_{i,j}=x+(k-l)y$, $a_{k}-m_{i,j}=x+(k-l+1)y$,

$$\begin{aligned}&\left( \varPhi (a_{l}-m_{i,j})-\varPhi (a_{l-1}-m_{i,j})\right) \left( \varPhi (a_{k}-m_{i,j})-\varPhi (a_{k-1}-m_{i,j})\right) \end{aligned}$$

(58)

$$\begin{aligned}&\quad =\left( \varPhi (x+y)-\varPhi (x)\right) \left( \varPhi (x+hy)-\varPhi (x+(h-1)y)\right) \end{aligned}$$

(59)

where $h=k-l+1$. Now, 3$\le h$. We introduce polar coordinates. Let $x=R\cos \alpha ,$$y=R\sin \alpha ,0\le R,$$0\le \alpha \le \pi$. With these the investigated function is

$$\begin{aligned}&\left( \varPhi (R\cos \alpha +R\sin \alpha )-\varPhi (R\cos \alpha )\right) \nonumber \\&\quad \cdot \left( \varPhi (R\cos \alpha +hR\sin \alpha )-\varPhi (R\cos \alpha +(h-1)R\sin \alpha )\right) . \end{aligned}$$

(60)

Notice that $\cos \alpha =0$ holds if $\alpha =\frac{\pi }{2}=\alpha _{0},$$\cos \alpha +\sin \alpha =0$ is satisfied if $\alpha =\frac{3\pi }{4}=\alpha _{1}$. Moreover, $\cos \alpha +(h-1)\sin \alpha =0$ holds for a unique value $\alpha =\alpha _{h-1}$ in [0,$\pi ]$ and $\cos \alpha +h\sin \alpha =0$ is satisfied for a unique value $\alpha =\alpha _{h}$ in $[0,\pi ]$. Taking into account the inequality $2\le h-1$ it can be easily seen that 0$<\alpha _{0}<\alpha _{1}<\alpha _{h-1}<\alpha _{h}<\pi$. Now we cover the interval $\left[ 0,\pi \right]$ with two closed subintervals in which one of the factors in (60) is close to zero if R takes large values. These subintervals are $\left[ 0,\alpha _{h-1}-\delta \right]$ and $\left[ \alpha _{1}+\delta ,\pi \right] ,$ if $0<\delta <\frac{\alpha _{h-1}-\alpha _{1}}{2}$. These intervals are bounded and closed, $\varPhi (R\cos \alpha +hR\sin \alpha ),$$\varPhi (R\cos \alpha +(h-1)R\sin \alpha ),$$\varPhi (R\cos \alpha +R\sin \alpha )$ and $\varPhi (R\cos \alpha )$ are continuous, consequently we can deduce the following inequalities: if $\alpha \in \left[ 0,\alpha _{h-1}-\delta \right] ,$ then

$$\begin{aligned} 0<\nu _{h}^{(a)}\le \cos \alpha +h\sin \alpha \le \nu _{h}^{(f)}, \end{aligned}$$

(61)

and

$$\begin{aligned} 0<\nu _{h-1}^{(a)}\le \cos \alpha +\left( h-1\right) \sin \alpha \le \nu _{h-1}^{(f)}. \end{aligned}$$

(62)

These imply

$$\begin{aligned}&\left( \varPhi \left( R\cos \alpha +hR\sin \alpha \right) -\varPhi \left( R\cos \alpha +(h-1)R\sin \alpha \right) \right) \nonumber \\&\quad \le \varPhi \left( R\nu _{h}^{(f)}\right) -\varPhi \left( R\nu _{h-1}^{(a)}\right) \end{aligned}$$

(63)

As $R\rightarrow \infty$ , $R\nu _{h}^{(f)}\rightarrow \infty$ and $R\nu _{h-1}^{(a)}\rightarrow \infty$, and $\underset{z\rightarrow \infty }{lim} \varPhi (z)=1$, hence

$$\begin{aligned} 0\le \varPhi \left( R\nu _{h}^{(f)}\right) - \varPhi \left( R\nu _{h-1}^{(a)}\right) <\varepsilon \end{aligned}$$

(64)

if $K_{1}<R$ with an appropriate value of $K_{1}$.

Similarly, if $\alpha \in \left[ \alpha _{1}+\delta ,\pi \right]$, $\cos \alpha <0$ , $\cos \alpha +\sin \alpha <0$,

$$\begin{aligned} \nu _{0}^{(a)}\le \cos \alpha \le \nu _{0}^{(f)}<0 \end{aligned}$$

(65)

and

$$\begin{aligned} \nu _{1}^{(a)}\le \cos \alpha +\sin \alpha \le \nu _{1}^{(f)}<0 \end{aligned}$$

(66)

which imply

$$\begin{aligned} 0\le \left( \varPhi \left( R\cos \alpha +R\sin \alpha \right) -\varPhi \left( R\cos \alpha \right) \right) \le \varPhi \left( R\nu _{1}^{(f)}\right) -\varPhi \left( R\nu _{0}^{(a)}\right) \end{aligned}$$

(67)

As $R\rightarrow \infty ,$$R\nu _{1}^{(f)}\rightarrow -\infty$ and $R\nu _{0}^{(a)}\rightarrow -\infty$, and $\underset{z\rightarrow -\infty }{lim} \varPhi (z)=0$, hence

$$\begin{aligned} 0\le \varPhi \left( R\nu _{1}^{(f)}\right) -\varPhi \left( R\nu _{0}^{(a)}\right) <\varepsilon \end{aligned}$$

(68)

if $K_{2}<R$ with an appropriate value of $K_{2}$. Consequently, with $K=max(K_{1},K_{2}),$ if $K<\left| x\right|$ or $K<y$ then (59) is less than $\varepsilon$. Returning to $m_{i,j}$ and d,  the inequality

$$\begin{aligned} K<\left| -s+2(l-1)d-m_{i,j}\right| \end{aligned}$$

(69)

holds if $(s+1)K<\left| m_{i,j}\right| ,$ consequently with $K_{\varepsilon }=(s+1)K,$ the inequality

$$\begin{aligned} \left( \varPhi (a_{l}-m_{i,j})-\varPhi (a_{l-1}-m_{i,j})\right) \left( \varPhi (a_{k}-m_{i,j})-\varPhi (a_{k-1}-m_{i,j})\right) <\varepsilon \end{aligned}$$

(70)

holds, therefore $L(A|m_{1,2},..,m_{n-1,n},d)<\varepsilon$.

Remark 1

It is easy to see that the proof remains true if $l=1$ or $k=s$.

If we know that the likelihood function is small if d takes large values, and if $\left| m_{i,j}\right|$ is large, we can focus on the values of $m_{e,f}$, for the other fixed values of $1\le e<f\le n$.

First for the pairs which have opinions in a middle intervals we state the following:

Lemma 3

Suppose that for the pair $1\le e<f\le n$$\ 0<A_{e,f,k}$ holds for some value of $k=2,3,{\ldots },s-1$. Then, in case of $d\le K_{\varepsilon },$

$$\begin{aligned} \varPhi (\left( -s+2k\right) d-m_{e,f})-\varPhi (\left( -s+2(k-1)\right) d-m_{e,f})<\varepsilon \end{aligned}$$

(71)

if $K_{\varepsilon }^{e,f}<\left| m_{e,f}\right|$.

Proof

The proof is left to the reader.

Secondly for the pairs with opinion in both terminal intervals, we state the following:

Lemma 4

Suppose that for the pair $1\le e<f\le n$ the inequalities $\ 0<A_{e,f,1}$ and $0<A_{e,f,s}$ hold. Then, assuming$\ d\le K_{\varepsilon },$

$$\begin{aligned} \varPhi (-(s-1)d-m_{e,f})(\left( 1-\varPhi ((s-1)d-m_{e,f})\right) <\varepsilon \end{aligned}$$

(72)

if $K_{\varepsilon }^{e,f*}<\left| m_{e,f}\right|$ .

Proof

The proof is left to the reader.

Now, we can state that we could build upper and lower bound for the variable d, and also for all the pairs which have opinion either in a middle interval or both terminal intervals. If all pairs satisfy at least one of these requirements we can state that the maximal value of the likelihood function is attained.

Theorem 2

Suppose that there is at least one pair of indices $i_{1}<j_{1}$ for which 0 < $\hbox {A}_{i_{1},j_{1},k}$ holds for some $1<k<s$. Moreover there is at least one pair of indices $i_{2}<j_{2}$ for which $0<A_{i_{2},j_{2},l,}$ and $0<A_{i_{2},j_{2},k}$ are satisfied with some $1\le l<k-1\le s-1$. If for all pairs $1\le e<f\le n$

$$\begin{aligned} 0<A_{e,f,k} \end{aligned}$$

(73)

holds with some $1<k<s$, or

$$\begin{aligned} 0<A_{e,f,1}\text { and 0 }< A_{e,f,s} \end{aligned}$$

(74)

is satisfied then the maximal value of the likelihood function (51) is attained.

Proof

The proof is the straightforward consequence of the above Lemmas as they guarantee that the inequality

$$\begin{aligned} L(A|m_{1,2},{\ldots },m_{n-1,n},d)<\varepsilon \end{aligned}$$

(75)

holds outside of a bounded closed set for all variables, consequently the optimization can be executed on a closed bounded non-empty set. Therefore the maximal value is reached.

Now due to (53), weaker conditions than the above ones can also guarantee the existence of the maximal value. We do not require (73) or (74) for all pairs 1$\le e<f\le 1,$ but only for some of them. For this purpose we first state the following

Lemma 5

If (75) holds supposing $K_{\varepsilon }^{e,f} <\left| m_{e,f}\right|$ for the pair of indices 1$\le e<f<n$, and (75) also holds if $K_{\varepsilon }^{f,g}<\left| m_{f,g} \right|$ for the pair of indices $1\le f<g\le n,$ then (75) holds for $K_{\varepsilon }^{e,f}+K_{\varepsilon }^{f,g}<\left| m_{e,g}\right|$.

Proof

The proof is simple, it is left to the reader.

Lemma 5 shows that the bounded property of the variables is descending by the linear relationships. Therefore if we have a “base collection” of bounded variables, then we can build up bounds for the other variables as well. If all variables can be restricted to a closed bounded set, then the Weierstrass’ theorem can be applied. The “base collection” is defined by the connected graph.

Theorem 3

Suppose that there is at least one pair of indices $i_{1}<j_{1}$ for which $0<A_{i_{1},j_{1},k}$ holds for some $k=2,3,{\ldots },s-1$. Moreover, there is at least one pair of indices $i<j$ for which $0<A_{i,j,l}$ and $0<A_{i,j,k}$ hold for some $1\le l<k-1\le s-1$. If the graph GR is connected, then the likelihood function reaches its maximal value.

Proof

The maximization can be executed in a closed bounded set in all variables. The set $d<K_{\varepsilon }^{a}$ is excluded by the assumption $0<A_{i_{1},j_{1} ,k}$ for some $k=2,3,{\ldots },s-1$, the set $K_{\varepsilon }<d$ by the assumption $0<A_{i,j,l}$ and $0<A_{i,j,k}$ for some $1\le l<k-1\le s-1$. If for a pair $1\le e<f\le n$ (7) or (8) holds, then the maximization in this variable can be restricted to a closed bounded set. If neither conditions (7) nor (8) hold for a pair $i<j$, then, take the sequence of edges from i to j: ($i=v_{1} \rightarrow v_{2}\rightarrow {\cdots }\rightarrow v_{h}=j)$. This sequence exists as the graph is connected. Now applying the previous lemma step by step, we get that (75) holds outside of a bounded closed set of $m_{i,j}$. Consequently the maximization of the likelihood function can be executed on a closed bounded non-empty set in all variables, consequently the maximal value is attained.

Appendix C: The uniqueness of the maximizer

The main idea of the proof is the log-concave property of the likelihood function, which is the concave property of the log-likelihood function. More precisely, we prove the strictly concave property, because it guarantees the uniqueness of the maximizer, if the maximum is attained. We use the well known fact that the standard normal distribution function is strictly log-concave and so is its difference from 1 (Prékopa 1973). We point out that the likelihood function itself is not concave, therefore it is important to deal with the logarithm.

We deal with (52) as a function of the transformed variables, then, in the main theorem, we return to $m_{i},$$i=1,2,{\ldots },n$. As the factors of (52) are functions of two variables containing the differences of Gauss cumulative distribution functions, first we prove the

Theorem 4

The $\hbox {function}\quad H(x,y)=\log \left( \varPhi (x)-\varPhi (y)\right)$ is strictly concave in the set $y<x\in {\mathbb {R}}$.

Proof

Let $\varphi$ denote the standard normal probability density function. The set

$$\begin{aligned} C=\left\{ (x,y)\mid y<x\in {\mathbb {R}}\right\} \subset {\mathbb {R}}^{2} \end{aligned}$$

(76)

is convex, $H\in C^{\infty }$, therefore it is enough to prove that the matrix containing the second order partial derivatives

$$\begin{aligned} D=\left[ \begin{array}{c@{\quad }c} -\varphi (x)\dfrac{x\left( \varPhi (x)-\varPhi (y)\right) +\varphi (x)}{\left( \varPhi (x)-\varPhi (y)\right) ^{2}} &{} \dfrac{\varphi (x)\varphi (y)}{\left( \varPhi (x)-\varPhi (y)\right) ^{2}}\\ \dfrac{\varphi (x)\varphi (y)}{\left( \varPhi (x)-\varPhi (y)\right) ^{2}} &{} \varphi (y)\dfrac{y\left( \varPhi (x)-\varPhi (y)\right) -\varphi (y)}{\left( \varPhi (x)-\varPhi (y)\right) ^{2}} \end{array} \right] \end{aligned}$$

(77)

is negative definite on C.

To prove

$$\begin{aligned} -\varphi (x)\dfrac{x\left( \varPhi (x)-\varPhi (y)\right) +\varphi (x)}{\left( \varPhi (x)-\varPhi (y)\right) ^{2}}<0 \end{aligned}$$

(78)

on C, consider the function

$$\begin{aligned} f_{y}(x)=x\left( \varPhi (x)-\varPhi (y)\right) +\varphi (x),\quad y\le x\in {\mathbb {R}} \end{aligned}$$

(79)

As $f_{y}(y)=\varphi (y)$ and $f_{y}^{\prime }(x)=\varPhi (x)-\varPhi (y)\ge 0,\quad y\le x\in {\mathbb {R}}$, therefore

$$\begin{aligned} f_{y}(x)\ge f_{y}(y)=\varphi (y)>0,\quad y<x\in {\mathbb {R}} \end{aligned}$$

(80)

which implies (78). Now we turn to the inequality

$$\begin{aligned} \det (D)= & {} \dfrac{\varphi (x)\varphi (y)}{\left( \varPhi (x)-\varPhi (y)\right) ^{4} }\\&\cdot \Big ( \big (x\left( \varPhi (x)-\varPhi (y)\right) +\varphi (x) \big ) \big (\varphi (y)-y\left( \varPhi (x)-\varPhi (y)\right) \big )-\varphi (x)\varphi (y) \Big ) >0,\\&y<x\in {\mathbb {R}}. \end{aligned}$$

Let $x\in {\mathbb {R}}$, and consider $g_{x}(y)=\varphi (y)-y\left( \varPhi (x)-\varPhi (y)\right) ,\quad x\ge y\in {\mathbb {R}}$.

As $g_{x}(x)=\varphi (x)$ and

$$\begin{aligned} g_{x}^{\prime }(y)=-\left( \varPhi (x)-\varPhi (y)\right) \le 0,\quad y\le x\in {\mathbb {R}} \end{aligned}$$

(81)

therefore

$$\begin{aligned} g_{x}(y)\ge g_{x}(x)=\varphi (x)>0,\quad y<x\in {\mathbb {R}} \end{aligned}$$

(82)

which implies

$$\begin{aligned} \det (D)=\dfrac{\varphi (x)\varphi (y)}{\left( \varPhi (x)-\varPhi (y)\right) ^{4} }\cdot \left( f_{y}(x)g_{x}(y)-\varphi (x)\varphi (y)\right) >0\text {.} \end{aligned}$$

(83)

Now we turn to the terminal intervals.

Lemma 6

If $0<c_{1}$ and $0<c_{2}$, then $G:{\mathbb {R}}^{2}\rightarrow {\mathbb {R}}^{2},$

$$\begin{aligned} G(x,y)=c_{1}\log \varPhi \left( x\right) +c_{2}\log (1-\varPhi (y)) \end{aligned}$$

(84)

is strictly concave.

Proof

Consider

$$\begin{aligned} D=\left[ \begin{array}{c@{\quad }c} c_{1}\dfrac{\partial ^{2}\log \varPhi (x)}{\partial x^{2}} &{} 0\\ 0 &{} c_{2}\dfrac{\partial ^{2}\log (1-\varPhi (y))}{\partial y^{2}} \end{array} \right] ,\quad y,x\in {\mathbb {R}} \end{aligned}$$

(85)

Since

$$\begin{aligned} \dfrac{\partial ^{2}\log \varPhi (x)}{\partial x^{2}}<0 \end{aligned}$$

(86)

is the consequence of the strictly log-concave property of $\varPhi$ and

$$\begin{aligned} \frac{\partial ^{2}\log (1-\varPhi (y))}{\partial y^{2}}<0 \end{aligned}$$

(87)

is the consequence of strictly log-concave property of $1-\varPhi ,$ therefore the matrix D is negative definite.

Now we turn to the strictly concave property of the log-likelihood function of the factors of (52). As we use transformations in variables, we have to guarantee that the strictly log-concave property is preserved after the transformation, that is the function is a strictly log-concave function of the new variables. For that, we state Lemma 7, Corollary 1 and Lemma 8.

Lemma 7

If B : ${\mathbb {R}}^{2}\rightarrow {\mathbb {R}}^{2}$ is an invertible linear transformation, moreover $G:{\mathbb {R}}^{2}\rightarrow {\mathbb {R}}$ is strictly concave, then $G\circ B:{\mathbb {R}}^{2}\rightarrow {\mathbb {R}}$ is strictly concave as well.

Proof

The proof is technical, it is left to the reader.

Corollary 1

$T_{i,j}:{\mathbb {R}}\hbox {x}{\mathbb {R}}^{+}\rightarrow {\mathbb {R}},T_{i,j}\left( m_{i,j},d\right) =\log (\varPhi ((-s+2k)d-m_{i,j} )-\varPhi ((-s+2\left( k-1\right) )d-m_{i,j}))$ is a strictly concave function. $T_{i,j}^{*}={\mathbb {R}}\hbox {x}{\mathbb {R}}^{+}\rightarrow {\mathbb {R}},T_{i,j} ^{*}\left( m_{i,j},d\right) =c_{1}\log \varPhi ((-(s-1)d-m_{i,j})+c_{2}) \log (1-\varPhi ((s-1)d-m_{i,j}))$ is a strictly concave function.

We note that as (52) is the sum of concave functions it is concave as well. But we need the strictly concave property. Step by step we return to the multivariate function by preserving strictly concave property. For that we state

Lemma 8

If $G_{i,j}:{\mathbb {R}}\hbox {x}{\mathbb {R}}^{+}\rightarrow {\mathbb {R}},$$G_{i,j}(m_{i,j},d)$ is strictly concave and so is $G_{e,f}(m_{e,f},d),$ then $G:{\mathbb {R}}^{2}\hbox {x}{\mathbb {R}}^{+}\rightarrow {\mathbb {R}},G(m_{i,j} ,m_{e,f},d)=G_{i,j}(m_{i,j},d)+G_{e,f}(m_{e,f},d)$ is a strictly concave function.

Proof

The proof is technical, it is left to the reader.

Now we prove the uniqueness if all variables are contained in (52).

Theorem 5

If for all pairs $i<j$ (7) or (8) holds, and the likelihood function reaches its maximal value, then the maximizer is unique.

Proof

(7) or (8) guarantee the strictly concave property of the log-likelihood function for all variables $m_{i,j}$ and d. The maximizer of a strictly concave function in a convex set is unique if the maximal value is attained.

The condition concerning all pairs can be weakened. We do not require the strictly concave property for all pairs of indices. If some of the factors are not contains in (52) because the multipliers are zero, then the uniqueness in these variables has to be guaranteed by another way. For this purpose first we prove the following

Lemma 9

Let $G(x,y,z)=G_{1}(x,y)+G_{2}(y,z),$$G_{1}$ is strictly concave, $G_{2}$ is concave and assume that the maximal value of G is reached. Then there exists a unique pair $(x_{1},y_{1})$ for which G is maximal.

Proof

The proof is technical based on definitions, we do not detail it.

Lemma 9 states that strictly log-concave factors produce a unique maximizer in their variables, if the maximum exists. It does not mean the uniqueness of the maximizer yet.

Now we can turn to our final statement in connection with the uniqueness of the maximizer. The main idea is that, if the maximizer is unique in a set of the transformed variables, then the linear conditions may guarantee the uniqueness in the other variables as well.

Theorem 6

Suppose that the maximal value of the log-likelihood function exists and the graph GR defined by (1) is connected. Then the maximizer of (52) is unique.

Proof

The strictly log-concave property in d is guaranteed. In the case when the pair $i<j$ is connected, i.e. (7) or (8) holds, then for the pair $i<j$ the log-likelihood function is strictly concave in $m_{i,j}$. It implies that the uniqueness of the maximizer holds for this variable. Notice that, if the maximal value of $m_{i,j}$ is unique, and so is that of $m_{j,e},$ then by (53) $m_{i,e}$ is unique at the maximum as well. In the case when if the vertices i and j$(i<j)$ do not have an edge between them, then consider the path from i to j ($i=v_{1}\rightarrow v_{2}\rightarrow {\ldots }\rightarrow v_{h}=j)$. Now, according to the previous case $m_{v_{1},v_{2}}$ is unique at the maximum, so is $m_{v_{2},v_{3}},$ therefore so is $m_{v_{1},v_{3}}$. Repeating this step by step we get that $m_{v_{1},v_{h}}=m_{i,j}$ is unique at the maximal value as well.

Appendix D: The proof of the main theorem

Proof

of Theorem 1 The assumptions guarantee both the existence and the uniqueness of the maximizer if we consider the log-likelihood function as a function of $m_{i,j}$ and d. The details are in Appendices B and C. One can see that the mapping

$$\begin{aligned} (0,m_{2},{\ldots },m_{n})\rightarrow \left( m_{1,2},m_{1,3},{\ldots },m_{n-1,n}\right) \end{aligned}$$

(88)

is invertible, therefore if the maximum exists and $m_{i,j}$ are unique at the maximum, then $m_{i}=-m_{1,i},$$i=2,3,{\ldots },n$ are unique at the maximum as well.