1. Shapley value
The Shapley value is a value attribution method that originated from economic game theory [Reference Shapley, Kuhn and Tucker12]. The intuition is that, considering a team of players together producing a value, the Shapley value attributes this value to individual members of the team. The Shapley value has been successfully applied in many probabilistic and statistical problems, such as collinear regression [Reference Lipovetsky and Conklin6], reliability [Reference Marichal and Mathonet8], queuing theory [Reference Anily and Haviv2], uncertainty quantification [Reference Owen11], inventory [Reference MÜller, Scarsini and Shaked10], multivariate risk analysis [Reference Abbasi and Hosseinifard1] and machine learning [Reference Lundberg7]. See [Reference Moretti and Patrone9] for a survey of applications. The intuition is to regard a set of random variables as the team of players and the statistical/probabilistic index of interest as the produced value. Formally, consider a set of n players (random variables)
$N=\{1,2, \ldots, n\}$
and any possible subset
$J\subseteq N$
, called a coalition. The function
$\nu:2^N \mapsto \mathbb{R}$
, with the condition that
$\nu(\emptyset)=0$
, is called the characteristic function or the value function of the game. The characteristic function of a coalition J,
$\nu(J)$
, is the value produced by all players in the coalition, for any possible coalition. Then, the Shapley value
$\phi_i(\nu)$
is a measure of the value of player i and is given by
\begin{equation}\phi_{i}^{\nu}=\sum_{J\subseteq N\setminus\left\{ i\right\} }\frac{(n-|J|-1)!\,|J|!}{n!}\left[\nu(J\cup\{i\})-\nu(J)\right],\end{equation}
where
$|J|$
is the cardinality of J. It can be recognized that the Shapley value for the ith player is based on the marginal increase in the value function
$\nu(J\cup\{i\})-\nu(J)$
when player i joins coalition J, averaged over all possible coalitions.
The Shapley value can be characterized by several interesting properties:
-
Efficiency:
$\sum_{i=1}^n \phi_i(\nu)=\nu(N)$
. -
Symmetry: If
$ \nu(J\cup \{i\})= \nu(J\cup \{j\})$
for all
$J\subseteq N \setminus \{i,j\}$
, then
$\phi_i(\nu)=\phi_j(\nu)$
. -
Dummy player: If
$\nu(J\cup \{i\})=\nu(J)$
for all
$J\subseteq N$
, then
$\phi_i(\nu)=0$
. -
Linearity: If two value functions
$\nu$
and
$\mu$
have respective Shapley values
$\phi(\nu)$
and
$\phi(\mu)$
, then the game with value
$\alpha \nu+\beta\mu$
has Shapley value
$\alpha \phi(\nu)+\beta \phi(\mu)$
for all
$\alpha,\beta \in \mathbb{R}$
.
Reference [Reference Shapley, Kuhn and Tucker12] proved that the Shapley value is the unique attribution method satisfying these four properties. Equivalently, the Shapley value (1) can be expressed as
\begin{equation}\phi _{i}\left( \nu \right) =\frac{1}{n!}\sum_{\psi \in \mathcal{P}(N)} \left( \nu \left( P^{\psi}( i) \cup \left\{ i\right\} \right) -\nu \left( P^{\psi }(i) \right) \right),\end{equation}
where
$\mathcal{P}(N)$
is the set of all permutations of N and
$ P^{\psi}( i)$
is the set of players who precede i in the order determined by the permutation
$\psi$
. The representation of Shapley value in (2) based on permutations has been adopted for developing algorithms for the calculation of Shapley values [Reference Castro, GÓmez and Tejada3,Reference Song, Nelson and Staum13].
2. Variance and standard deviation games
Reference [Reference Colini-Baldeschi, Scarsini and Vaccari5] recently used Shapley values as the allocation criterion for a portfolio problem.
Consider the random vector
$X=(X_1,X_2,\ldots,X_n)$
with finite second moment, and the sum of the random variables
$S=\sum_{i=1}^n X_i$
. Then, [Reference Colini-Baldeschi, Scarsini and Vaccari5] considers the Shapley values using the characteristic functions
\begin{equation*} \nu(J)=\mathrm{Var}\left( S_J \right), \qquad \lambda(J)=\sqrt{\mathrm{Var}\left( S_J \right)},\end{equation*}
where the partial sums
$S_J=\sum_{i\in J}X_i$
are defined for every
$J\subseteq N$
. Reference [Reference Colini-Baldeschi, Scarsini and Vaccari5] calls
$\nu$
a variance game and
$\lambda$
a standard deviation game, and characterizes the Shapley values for a variance game.
Theorem 1. (Reference [Reference Colini-Baldeschi, Scarsini and Vaccari5].) For a variance game
$\nu$
,
$\phi_i(\nu)=\mathrm{Cov}\left(X_i,S \right)$
. However, an analogous closed-form representation of the Shapley value is not available for a standard deviation game. Thus, comparing Shapley values for variance and for standard deviation games is very difficult. This led the authors of [Reference Colini-Baldeschi, Scarsini and Vaccari5] to formulate a conjecture about their comparison, which we describe in the next section.
3. Conjecture and results
Given two vectors
$\textbf{x},\textbf{y}\in \mathbb{R}^n$
, the vector
$\textbf{x}$
is said to be majorized by
$\textbf{y}$
(
$\textbf{x}\leq \textbf{y}$
) if
$\sum_{i=k}^n x_{(i)}\leq \sum_{i=k}^n y_{(i)}$
for all
$k\in\{ 2,\ldots,n\}$
and
$\sum_{i=1}^n x_i=\sum_{i=1}^n y_i$
, where
$x_{(1)}\leq x_{(2)}\leq \cdots \leq x_{(n)}$
is the increasing rearrangement of
$\textbf{x}$
. Using this notion, the conjecture can be formulated as follows.
Conjecture 1. (Reference [5].) For any
$n\times n$
covariance matrix
$\Sigma$
, if
$\nu$
is the corresponding variance game and
$\lambda$
the corresponding standard deviation game, then
\begin{equation*} \frac{1}{\lambda(N)}\Phi(\lambda)\leq \frac{1}{\nu(N)}\Phi(\nu),\end{equation*}
where
$\Phi$
denotes the vector of the Shapley values.
Reference [Reference Colini-Baldeschi, Scarsini and Vaccari5] showed that the conjecture holds for
$n=2$
dependent random variables and verified it numerically for
$n=3,4,5$
independent random variables. We now prove in Section 3.1 that the conjecture holds true for an arbitrary number of independent random variables. However, we show, using two counterexamples in Section 3.2, that the conjecture is not valid for
$n=3$
dependent random variables. From a reviewer’s report we learnt about [Reference Chen, Hu and Tang4], where it is shown how the conjecture in the case of independent random variables can be seen as an application of a far more general theorem, whose proof includes several lemmas and definitions. Although we expect those results to be sound, we think that our alternative proof of the conjecture for independent random variables, being direct and self-contained, can be interesting and useful.
3.1. Proof of the conjecture for independent random variables
Theorem 2. Assume that the random variables are independent. Then, the conjecture holds for any n.
Proof. Let
$\nu $
be the variance game and
$\lambda $
the standard deviation game. We assume the random variables to be independent with variances
$\sigma_{1}^{2}\leq \sigma _{2}^{2}\leq \cdots \leq \sigma _{n}^{2}$
, so that it is easily checked that
$\phi _{i}( \nu ) =\sigma _{i}^{2}$
. Moreover, straightforward computations show that, when
$i\leq j$
,
$\phi_{i}( \lambda ) \leq \phi _{j}( \lambda ) $
as well.
First of all, we observe that the conjecture holds for
$n=2$
and 3. In fact, the case
$n=2$
is trivial, so that we have to prove the conjecture when
$n=3$
. Since the ordering of the vectors
$\boldsymbol{\Phi }( \boldsymbol{\nu })$
and
$\boldsymbol{\Phi }( \boldsymbol{\lambda })$
by increasing components is the same, and, moreover, we can assume, without loss of generality, that
$\mathrm{Var}\left( X_{1}+X_{2}+X_{3}\right) = \mathrm{SD}\left(X_{1}+X_{2}+X_{3}\right) =1$
, the proof amounts to showing that
\begin{equation}\boldsymbol{\phi }_{1}\left( \lambda \right) \boldsymbol{\geq \phi }_{1}( \nu) , \qquad \phi _{3}\left( \lambda \right) \leq \phi _{3}( \nu ).\end{equation}
We start with the first inequality in (3). In fact, after straightforward computations, that inequality is seen to be equivalent to
\begin{equation*}F\left( a,b\right) =2-2\sqrt{1-a}+2\sqrt{a}+\sqrt{1-b}-\sqrt{1-a-b}+\sqrt{a+b}-\sqrt{b}-6a\geq 0 \end{equation*}
in a suitable set of the
$\left( a,b\right) $
plane, where
$\sigma_{1}^{2}=a$
,
$\sigma _{2}^{2}=b$
, and
$\sigma _{3}^{2}=c$
satisfy
$c=1-a-b\geq b\geq a\geq 0$
. Namely, our first inequality must be verified in the triangle T with vertices in (0,0),
$\left( 0,\frac{1}{2}\right)$
,
$\left( \frac{1}{3},\frac{1}{3}\right) $
. In fact, it is easily computed that
$F\left( 0,b\right) =F\left( \frac{1}{3},\frac{1}{3}\right) =0$
. Moreover, straightforward computations show that
$\underset{a\rightarrow 0+}{\lim }\frac{\partial F}{\partial a}\left(0,b\right) =+\infty $
when
$b>0$
, while
$\frac{\partial^{2}F}{\partial a^{2}}\leq -\frac{1}{2}\left[ \left( \frac{1}{3}\right) ^{-\frac{3}{2}}-\left( \frac{2}{3}\right) ^{-\frac{3}{2}}\right] <0$
throughout T. Hence, for any fixed
$\overline{b}\in \left( 0,\frac{1}{2}\right) $
,
$F\left( a,\overline{b}\right) $
has a concave graph when
$\left( a,\overline{b}\right) \in T$
and is positive when a lies in some right neighborhood of 0. Then, consider the two oblique sides of T. Replacing b by linear functions of a, the function F is given along these sides by the two functions
\begin{equation*}f_{1}( a) =F\left( a,\frac{1}{3}-\frac{1}{2}\left( a-\frac{1}{3}\right) \right) , \qquad f_{2}( a) =F\left( a,a\right) .\end{equation*}
It is again a matter of standard computations to check that
$f_{1}\left(0\right) =f_{1}\left( \frac{1}{3}\right) =f_{2}\left( 0\right) =f_{2}\left(\frac{1}{3}\right) =0$
, that both
$f_{1}^{\prime }$
and
$f_{2}^{\prime }$
are positive in a right neighborhood of 0, and that
$f_{1}^{\prime \prime }$
is negative when
$0<a\leq \frac{1}{3}$
. As to
$f_{2}^{\prime\prime }$
, instead, straightforward computations show that it changes sign, from negative to positive, only once when
$0<a\leq \frac{1}{3}$
, whereas
$f_{2}^{\prime }\left( \frac{1}{3}\right) <0$
. Therefore, both
$f_{1}(a) $
and
$f_{2}( a) $
are positive when
$0<a<\frac{1}{3}$
, which leads us to conclude that
$F( a,\overline{b}) \geq 0$
when
$\left( a,\overline{b}\right) \in T$
and thus
$F(a,b) \geq 0$
all over T.
Analogously, using the same symbols, the second inequality in (3) is easily checked to be equivalent to
\begin{equation*} {G}\left( a,c\right) =2-2\sqrt{1-c} +2\sqrt{c} +\sqrt{1-a} -\sqrt{1-a-c} +\sqrt{a+c} -\sqrt{a}-6c\leq 0 \end{equation*}
in the triangle S of the (a,c) plane with vertices
$\left( 0,\frac{1}{2}\right)$
, (0,1),
$\left( \frac{1}{3},\frac{1}{3}\right) $
. Then, straightforward computations show that the inequality holds on the sides of S, while
$\frac{\partial^{2}G}{\partial a^{2}}>0$
throughout S, which implies the above inequality. Moreover, we can prove (see Appendix A) that, for any set of n independent random variables,
$\phi _{1}( \lambda ) \geq \sigma _{1}^{2}$
, which implies
\begin{equation*}\underset{i=2}{\overset{n}{\sum }}\phi _{i}( \lambda ) \leq\underset{i=2}{\overset{n}{\sum }}\sigma _{i}^{2}.\end{equation*}
Now, assume
$n\geq 4$
. Consider the set of
$n-1$
random variables
$\widetilde{N}=\left(Y,X_{3},\ldots,X_{n}\right) $
, where
$Y=X_{1}+X_{2}$
. Hence,
$\widetilde{N}$
is a set of
$n-1\geq 3$
independent random variables and we can apply the induction hypothesis. Assume, without loss of generality, that
$\nu (N) =\lambda ( N) =\nu ( \widetilde{N}) =\lambda( \widetilde{N}) =1$
. Denoting by
$\widetilde{\phi }_{k}$
the Shapley values relative to the set
$\widetilde{N}$
, we observe that, for any
$k\geq 3$
,
$\phi _{k}( \nu ) =\widetilde{\phi }_{k}( \nu ) =\sigma _{k}^{2}$
.
The next step is to prove that, for any
$k\geq 3$
,
$\phi_{k}( \lambda ) \leq \widetilde{\phi }_{k}( \lambda ) $
. Given a permutation
$\psi $
of the elements of
$\widetilde{N}$
, denote by Z the sum of the random variables different from Y preceding
$X_{k}$
in
$\psi $
(where possibly
$Z=0$
) and by
$a^{2}\geq 0$
the variance of Z. When we pass from the computation of
$\widetilde{\phi }_{k}( \lambda ) $
to that of
$\phi _{k}( \lambda) $
, to each previous permutation
$\psi $
correspond n new permutations. More precisely, consider the function
\begin{equation*}\Phi :\left\{ \text{permutations of }\left(X_{1},X_{2},X_{3},\ldots,X_{n}\right) \right\} \rightarrow \left\{ \text{permutations of }\left( Y,X_{3},\ldots,X_{n}\right) \right\}\end{equation*}
defined by replacing, in each permutation,
$X_{1}$
with
$Y $
and removing
$X_{2}$
. So, taking Z as above and letting
$0\leq h\leq n-3$
be the number of addends in
$Z$
, it follows that there exist
$\left( h+1\right) !\left(n-h-2\right) !$
permutations of
$\left( Y,X_{3},\ldots,X_{n}\right) $
corresponding to the quantity
$\sqrt{a^{2}+\sigma _{Y}^{2}+\sigma _{k}^{2}}-\sqrt{a^{2}+\sigma _{Y}^{2}}$
in the computation of
$\widetilde{\phi }_{k}( \lambda ) $
. Then, for each such permutation, say
$\psi ^{\prime }$
, there are
$h+1$
permutations in
$\Phi ^{-1}\left( \psi ^{\prime}\right) $
corresponding to the above quantity in the computation of
$\phi _{k}( \lambda ) $
and
$n-h-1$
permutations corresponding to
$\sqrt{a^{2}+\sigma _{1}^{2}+\sigma _{k}^{2}}\textbf{-}\sqrt{a^{2}+\sigma_{1}^{2}} $
. Similarly, there are
$h!\left( n-h-1\right) !$
permutations of
$\left( Y,X_{3},\ldots,X_{n}\right) $
corresponding to the quantity
$\sqrt{a^{2}+\sigma _{k}^{2}}-\sqrt{a^{2}}$
in the computation of
$\widetilde{\phi }_{k}( \lambda ) $
. For each such permutation, say
$\psi ^{\prime \prime }$
, there are
$n-h-1$
permutations in
$\Phi ^{-1}\left( \psi ^{\prime\prime }\right) $
corresponding to the above quantity in the computation of
$\phi _{k}( \lambda ) $
and
$h+1$
permutations corresponding to
$\sqrt{a^{2}+\sigma _{2}^{2}+\sigma _{k}^{2}}-\sqrt{a^{2}+\sigma _{2}^{2}} $
. Therefore, having multiplied by
$\frac{n}{n}$
the expression of
$\widetilde{\phi }_{k}( \lambda ) $
, the difference with
$\phi _{k}( \lambda ) $
is constituted by replacing
$\left( h+1\right) !\left( n-h-1\right) !$
times the quantity
\begin{equation*}F\left( a^{2},\sigma _{Y}^{2},\sigma _{k}^{2}\right) =\sqrt{a^{2}+\sigma_{Y}^{2}+\sigma _{k}^{2}}-\sqrt{a^{2}+\sigma _{Y}^{2}}+\sqrt{a^{2}+\sigma_{k}^{2}}-\sqrt{a^{2}} ,\end{equation*}
where
$\sigma _{Y}^{2}=\sigma _{1}^{2}+\sigma _{2}^{2}$
, by the quantity
\begin{equation*}G\left( a^{2},\sigma _{1}^{2},\sigma _{2}^{2},\sigma _{k}^{2}\right) =\sqrt{a^{2}+\left( \sigma _{1}\right) ^{2}+\sigma _{k}^{2}}-\sqrt{a^{2}+\left(\sigma _{1}\right) ^{2}}+\sqrt{a^{2}+\left( \sigma _{2}\right) ^{2}+\sigma_{k}^{2}}-\sqrt{a^{2}+\left( \sigma _{2}\right) ^{2}}.\end{equation*}
Finally, it follows from straightforward computations, which we omit for the sake of synthesis, that
$F\geq G$
whenever
$a^{2}+\sigma _{1}^{2}+\sigma _{2}^{2}+\sigma _{k}^{2}\leq 1,$
where either
$a=0$
or
$\sigma _{1}^{2}\leq \sigma _{2}^{2}\leq \min \left(a^{2},\sigma _{k}^{2}\right) $
. This proves that
\begin{equation}\phi _{k}( \lambda ) \leq \widetilde{\phi }_{k}( \lambda) \qquad \text{when }k\geq 3.\end{equation}
Now, our first step will be to prove that
\begin{equation}\underset{i=3}{\overset{n}{\sum }}\phi _{i}( \lambda ) \leq\underset{i=3}{\overset{n}{\sum }}\sigma _{i}^{2}.\end{equation}
Denoting in the following each
$\sigma _{i}^{2}$
as
$a_{i}\geq 0$
, we observe that the above inequality holds both when
$a_{1}+a_{2}\leq a_{3}$
, by the induction hypothesis, as a consequence of (4), and when
$a_{1}=a_{2}$
, since in such a case
$\phi _{2}(\lambda ) =\phi _{1}( \lambda ) \geq a_{1}=a_{2}$
, as proved in Appendix A. Then, either
$a_{1}=a_{2}$
or
$a_{1}+a_{2}\leq a_{3}$
hold, implying (5), or else we can determine
$\alpha >0$
,
$\beta \geq 0$
such that
$a_{1}+\alpha =a_{2}-\beta$
,
$\alpha =\left( n-1\right) \beta $
. Therefore, we consider a path
$\textbf{a}( x) =\left( a_{1}(x) ,a_{2}( x) ,a_{3}( x) ,\ldots,a_{n}(x) \right) $
,
$0\leq x\leq \overline{x}$
,
$\overline{x}>1$
, such that
$a_{1}( 0) =a_{1}+\alpha$
and
$a_{i}( 0) =a_{i}-\beta$
,
$i=2,3,\ldots,n$
, while
$a_{1}( x) =a_{1}( 0) -\alpha x$
and
$a_{i}( x) =a_{i}( 0) +\beta x$
,
$i=2,3,\ldots,n$
, until
$a_{1}\left( \overline{x}\right) +a_{2}\left( \overline{x}\right)=a_{3}\left( \overline{x}\right)$
for some
$\overline{x}>1$
. Hence, denoting by
$\phi _{i}( \lambda ) (x) $
the Shapley values for the standard deviation game corresponding to
$\mathrm{Var}(X_{i})=a_{i}( x) $
,
$i=1,\ldots,n$
,
\begin{equation*}\underset{i=3}{\overset{n}{\sum }}\phi _{i}( \lambda ) (0) \leq \underset{i=3}{\overset{n}{\sum }}a_{i}( 0) , \qquad \underset{i=3}{\overset{n}{\sum }}\phi _{i}( \lambda ) (\overline{x}) \leq \underset{i=3}{\overset{n}{\sum }}a_{i}(\overline{x}) .\end{equation*}
Let
$\phi _{i}( \lambda ) ( x) =f_{i}(x) $
,
$i=3,\ldots,n$
. What we want to prove is that
\begin{equation*}\underset{i=3}{\overset{n}{\sum }}f_{i}^{\prime \prime }( x) =\underset{i=3}{\overset{n}{\sum }}\left( f_{i}( x) -a_{i}(x) \right) ^{\prime \prime }\geq 0 , \qquad 0\leq x\leq \overline{x} ,\end{equation*}
which clearly implies
\begin{equation*}\underset{i=3}{\overset{n}{\sum }}\phi _{i}( \lambda ) (x) \leq \underset{i=3}{\overset{n}{\sum }}a_{i}( x) , \qquad 0\leq x\leq \overline{x}.\end{equation*}
In order to do this, recall the definition of the Shapley value, i.e.
\begin{equation*}\phi _{i}( \lambda ) ( x) =f_{i}( x) =\frac{1}{n!}\sum \left( \lambda \left( P^{\lambda }( i) \cup \left\{i\right\} \right) -\lambda \left( P^{\lambda }( i) \right) \right) ,\end{equation*}
and call
$f_{ij}( x) $
the addend of
$f_{i}( x) $
corresponding to j terms preceding
$a_{i}( x) $
in a permutation
$P^{\lambda}( i) $
,
$j=0,1,\ldots,n-1$
. Consider the functions
\begin{equation*}g_{1}( x) =\underset{i=3}{\overset{n}{\sum }}\left( f_{i0}(x) +f_{i1}( x) \right) , \qquad g_{j}( x) =\underset{i=3}{\overset{n}{\sum }}f_{ij}(x) ,\text{ }\quad j=2,\ldots,n-1 .\end{equation*}
Then, it can be shown that
$g_{j}^{\prime \prime }(x) >0$
as
$0\leq x\leq \overline{x}$
(see Appendix B). In this way, (5) is proved.
We have to consider, now,
$k\geq 4$
; i.e. we have to prove
\begin{equation}\underset{i=k}{\overset{n}{\sum }}\phi _{i}( \lambda ) \leq\underset{i=k}{\overset{n}{\sum }}\sigma _{i}^{2} , \qquad k=4,\ldots,n . \end{equation}
Let us start from
$k=4$
. Again setting
$\sigma _{i}^{2}=a_{i}$
, it follows from the same previous arguments that the inequality holds if
$a_{1}=a_{2}=a_{3}$
or
$a_{1}+a_{2}\leq a_{4}$
. Suppose neither one is the case. Then we can find
$\alpha ,\beta ,\gamma $
such that
\begin{align*} a_{1}+\alpha & = a_{2}+\beta =a_{3}-\gamma , \qquad \alpha >\gamma >0 , \\[3pt] \alpha & = s\gamma , \qquad s>0 , \\[3pt] \beta & = h\gamma , \\[3pt] s+h & = n-2 ,\end{align*}
the extreme cases being given by
$h=s=\frac{n-2}{2}$
, when
$a_{2}=a_{3}$
, and
$h=-1$
, when
$a_{2}=a_{1}$
. We consider, as above, a path joining the two n-tuples where (6) holds, but inverting, for our convenience, the direction. Thus, a(0) corresponds to
$a_{1}( 0) +a_{2}( 0) =a_{4}( 0) $
, while
$a(\overline{x})$
will correspond to
$a_{1}\left(\overline{ {x}}\right) =a_{2}\left( \overline{ {x}}\right)=a_{3}\left( \overline{ {x}}\right) $
, so that
${a}( {x}) $
is defined by
\begin{equation*}a_{1}( x) =a_{1}( 0) +\alpha x , \qquad a_{2}( x) =a_{2}( 0) +\beta x , \qquad a_{i}( x) =a_{i}( 0) -\gamma x, \quad i=3,\ldots,n.\end{equation*}
Using the previous notation, we denote the functions
$\phi _{i}(\lambda ) ( x) $
as
$f_{i}( x) $
,
$i=1,2,\ldots,n$
. Hence, from what we have seen,
$f_{1}( x) +f_{2}( x) \geq a_{1}( x)+a_{2}( x)$
along the whole path. First of all, we can show that, for
$x\in \left[ 0,\overline{x}\right] $
,
\begin{equation}f_{3}^{\prime }( x) \leq f_{4}^{\prime }( x) \leq\cdots \leq f_{n}^{\prime }( x) ,\end{equation}
while we recall that
\begin{equation*}\underset{i=3}{\overset{n}{\sum }}a_{i}( x) \underset{i=3}{\overset{n}{=\sum }}a_{i}( 0) -\left( n-2\right) \gamma x.\end{equation*}
In fact, compare, for example,
$f_{3}^{\prime }( x) $
and
$f_{4}^{\prime }( x) $
: clearly they are equal, as are
$f_{3}( x) $
and
$f_{4}( x) $
, when
$a_{3}=a_{4}$
. So, change
$a_{3}$
into
$a_{3}-z$
and
$a_{4}$
into
$a_{4}+z$
, for any small
$z>0$
. Assume, by contradiction, that, for some
$\widehat{x}\in \left( 0,\overline{x}\right) $
,
\begin{equation*}\frac{\partial f_{3}^{\prime }\left( \widehat{x},0\right) }{\partial z}-\frac{\partial f_{4}^{\prime }\left( \widehat{x},0\right) }{\partial z}>0.\end{equation*}
Then, for a sufficiently small
$h>0$
, letting
$x\in \left( \widehat{x},\widehat{x}+h\right) $
and setting
$z=x$
as well,
$f_{3}^{\prime }( x) >f_{4}^{\prime }(x) $
. But, since
$f_{3}\left( \widehat{x}\right)=f_{4}\left( \widehat{x}\right) $
, this implies that
$f_{3}\left( \widehat{x}+h\right) >f_{4}\left( \widehat{x}+h\right)$
, which leads to a contradiction, since we have observed that
$\sigma_{i}^{2}<\sigma _{j}^{2}$
,
$i<j$
, implies
$\phi _{i}(\lambda ) \leq \phi _{j}( \lambda ) $
. Hence, moving, say, from
$\overline{a}=\frac12({a_{3}( x)+a_{4}( x) })$
and setting
$a_{3}( x) =\overline{a}-\overline{z}$
,
$a_{4}( x) =\overline{a}+\overline{z}$
at each
$z\in \left( 0,\overline{z}\right) $
the above steps can be repeated, observing that
$\phi _{j}( \lambda) -\phi _{j}( \lambda ) $
increases with
$\sigma_{j}^{2}-\sigma _{i}^{2}$
. Thus, the inequality
$f_{3}^{\prime}( x) \leq f_{4}^{\prime }( x) $
follows and so, by the same arguments, do the others. Moreover, we observe that
$\sum_{i=4}^nf_{i}^{\prime \prime }( x) $
can change sign at most once in
$\left( 0,\overline{x}\right) $
, from positive to negative, since it can be computed that
$\sum_{i=4}^nf_{i}^{\prime\prime }( x) =0$
implies
$\sum_{i=4}^nf_{i}^{\prime \prime \prime }( x) <0$
. Finally, we recall that, at
$x=\overline{x}$
,
$f_{3}( \overline{x}) \geq a_{3}( \overline{x}) $
. Hence, assume, by contradiction, that there exists a subinterval
$\left[ p,q\right] $
of
$\left[ 0,\overline{x}\right] $
such that
\begin{equation*}\underset{i=4}{\overset{n}{\sum }}f_{i}( x) =\underset{i=4}{\overset{n}{\sum }}a_{i}( x) \text{ when }x=p,q , \qquad \underset{i=4}{\overset{n}{\sum }}f_{i}( x) >\underset{i=4}{\overset{n}{\sum }}a_{i}( x) \text{ when }p<x<q .\end{equation*}
Then, when
$p<x<q$
,
$f_{3}( x) <a_{3}( x) $
; otherwise,
$f_{1}( x) +f_{2}( x) + f_{3}(x) \geq a_{1}( x) +a_{2}( x) +a_{3}(x) $
would imply
$\sum_{i=4}^nf_{i}(x) \leq \underset{i=4}{\overset{n}{\sum }}a_{i}( x) $
. In particular, we will have
$f_{3}( x) <a_{3}(x) $
when
$x\in \left[ r,q\right) $
, where r is such that
$\sum_{i=4}^nf_{i}^{\prime}( r) ={-}( n-3) \gamma $
and
$\sum_{i=4}^nf_{i}^{\prime }( x) <{-}( n-3)\gamma $
when
$x\in \left( r,q\right) $
. Hence, because of (7),
$f_{3}^{\prime }( x) <-\gamma $
for
$x\in \left( r,q\right) $
. Also, since
$\sum_{i=4}^nf_{i}^{\prime \prime }( x) \leq 0$
in
$\left[ r,\overline{x}\right] $
, it follows that
$\sum_{i=4}^nf_{i}^{\prime }( x) <{-}( n-3) \gamma $
in
$\left[ r,\overline{x}\right] $
, so that
$f_{3}^{\prime }( x) <-\gamma $
in
$\left[ r,\overline{x}\right] $
as well, implying
$f_{3}( x)<a_{3}( x) $
in
$\left[ r,\overline{x}\right] $
and thus leading to a contradiction, as
$f_{3}( \overline{x}) \geq a_{3}(\overline{x}) $
.
In fact, by recurrent arguments, the other cases
$k>4$
are proved analogously. This concludes the proof of the conjecture in the case of independent random variables.
3.2. The conjecture does not hold for
$n>2$
dependent random variables
We will show that the conjecture does not hold in the case of three dependent random variables. Before providing numerical counterexamples, we consider a more general setting. Namely, let
$X_{1}$
,
$X_{2}$
,
$X_{3}$
be random variables such that
$\mathrm{Var}( X_{1}+X_{2}+X_{3}) =1$
with
$\mathrm{Cov}(X_{i},X_{1}+X_{2}+X_{3}) =\frac{1}{3}$
(i.e.
$\phi _{i}( \nu) =\frac{1}{3}$
) for
$i=1,2,3$
and, moreover,
$\sigma _{1}<\sigma_{2}<\sigma _{3}$
. Then, we will prove that
$\phi _{1}( \lambda ) <\phi _{2}( \lambda ) <\phi_{3}( \lambda )$
.
First of all, recalling
$\mathrm{Var}( X_{1}+X_{2}+X_{3}) =1$
, it is easily calculated that
\begin{equation*}6\phi _{i}( \lambda ) =2-2\sqrt{\mathrm{Var}(X_{j}+X_{k})}+2\sigma _{i}+\sqrt{\mathrm{Var}(X_{i}+X_{j})}-\sigma _{j}+\sqrt{\mathrm{Var}(X_{i}+X_{k})}-\sigma _{k}\end{equation*}
where
$i\neq j\neq k$
. Hence, let us compare
$\phi _{1}( \lambda) $
and
$\phi _{2}( \lambda ) $
(the other case is analogous). Again, by straightforward computations it can be checked that
$\phi_{1}( \lambda ) <\phi _{2}( \lambda ) $
is equivalent to
$\sigma _{1}+\sqrt{\mathrm{Var}(X_{1}+X_{3})}<\sigma _{2}+\sqrt{\mathrm{Var}(X_{2}+X_{3})}$
. Squaring and observing that
$\phi _{1}( \nu ) =\phi_{2}( \nu ) $
is equivalent to
$\sigma _{1}^{2}+\rho _{13}\sigma_{1}\sigma _{3}=\sigma _{2}^{2}+\rho _{23}\sigma _{2}\sigma _{3}$
, we are eventually led to show that
$\sigma _{1}\sqrt{\mathrm{Var}( X_{1}+X_{3}) }<\sigma _{2}\sqrt{\mathrm{Var}(X_{2}+X_{3}) }$
, i.e.
\begin{equation*}\frac{\mathrm{Var}\left( X_{1}+X_{3}\right) }{\mathrm{Var}\left( X_{2}+X_{3}\right) }<\frac{\sigma _{2}^{2}}{\sigma _{1}^{2}}.\end{equation*}
Since
$\sigma _{1}^{2}+\rho _{13}\sigma _{1}\sigma _{3}= \sigma_{2}^{2}+\rho _{23}\sigma _{2}\sigma _{3}$
implies
$\rho _{13}\sigma_{1}\sigma _{3}= \sigma _{2}^{2}-\sigma _{1}^{2}+\rho _{23}\sigma_{2}\sigma _{3}$
, the above inequality is easily seen to be equivalent to
\begin{equation*}\frac{\sigma _{2}^{2}-\sigma _{1}^{2}}{\mathrm{Var}\left( X_{2}+X_{3}\right) }<\frac{\sigma _{2}^{2}-\sigma _{1}^{2}}{\sigma _{1}^{2}} ,\end{equation*}
implying
$\sigma _{1}^{2}<\mathrm{Var}( X_{2}+X_{3}) $
. Then, assume, by contradiction, that
$\sigma _{1}^{2}\geq \mathrm{Var}( X_{2}+X_{3}) $
, i.e.
$\sigma _{1}^{2}\geq \sigma _{2}^{2}+\rho _{23}\sigma _{2}\sigma _{3}+\rho_{23}\sigma _{2}\sigma _{3}+\sigma _{3}^{2}=\sigma _{1}^{2}+\rho _{13}\sigma_{1}\sigma _{3}+\rho _{23}\sigma _{2}\sigma _{3}+\sigma _{3}^{2}$
. Hence, we should have
$0\geq \rho _{13}\sigma _{1}\sigma _{3}+\rho _{23}\sigma _{2}\sigma_{3}+\sigma _{3}^{2}=\mathrm{Cov}\left( X_{3},X_{1}+X_{2}+X_{3}\right)$
. But we have assumed that
$\mathrm{Cov}\left( X_{3},X_{1}+X_{2}+X_{3}\right) =\phi_{3}( \nu ) =\frac{1}{3}$
, so we are led to a contradiction. As we have observed,
$\phi _{2}( \lambda )<\phi _{3}( \lambda ) $
is proved in the same way. In fact,
$\phi _{2}( \lambda ) <\phi _{3}( \lambda ) $
is easily checked to be equivalent to
\begin{equation} {\sigma }_{1} +\sqrt{\mathrm{Var}(X_{1}+X_{3})} <\sigma_{2} +\sqrt{\mathrm{Var}(X_{2}+X_{3})}.\end{equation}
Repeating the above steps relative to the comparison of
$\phi_{1}( \lambda ) $
and
$\phi _{2}( \lambda ) $
, since
$\phi _{2}( \nu ) =\phi _{3}( \nu ) $
implies
$\sigma _{2}^{2}+\rho _{12}\sigma _{1}\sigma _{2}=\sigma_{3}^{2}+\rho _{13}\sigma _{1}\sigma _{3}$
, it follows that (8) is equivalent to
\begin{equation*}\frac{\mathrm{Var}\left( X_{1}+X_{2}\right) }{\mathrm{Var}\left( X_{1}+X_{3}\right) }<\frac{\sigma _{3}^{2}}{\sigma _{2}^{2}} ,\end{equation*}
i.e., recalling again that
$\sigma _{2}^{2}+\rho _{12}\sigma _{1}\sigma_{2}=\sigma _{3}^{2}+\rho _{13}\sigma _{1}\sigma _{3}$
, after simple computations,
\begin{equation*}\frac{\sigma _{3}^{2}-\sigma _{2}^{2}}{\mathrm{Var}\left( X_{1}+X_{3}\right) }<\frac{\sigma _{3}^{2}-\sigma _{2}^{2}}{\sigma _{2}^{2}}.\end{equation*}
Then, assume, by contradiction, that
$\sigma _{2}^{2}\geq \mathrm{Var}\left( X_{1}+X_{3}\right) $
. Hence,
$\sigma _{2}^{2}\geq \sigma _{1}^{2}+\sigma _{3}^{2}+2\rho _{13}\sigma_{1}\sigma _{3}=\sigma _{1}^{2}+\sigma _{2}^{2}+\rho _{12}\sigma _{1}\sigma_{2}+\rho _{13}\sigma _{1}\sigma _{3}$
, i.e.
\begin{equation*} 0\geq \sigma _{1}^{2}+\rho _{12}\sigma _{1}\sigma _{2}+\rho _{13}\sigma_{1}\sigma _{3}=\mathrm{Cov}\left( X_{1},X_{1}+X_{2}+X_{3}\right) =\phi _{1}(\nu ) ,\end{equation*}
and we are led to a contradiction, having hypothesised
$\phi_{1}( \nu ) =\frac{1}{3}$
. Thus,
$\phi _{1}( \lambda ) <\phi _{2}( \lambda ) <\phi_{3}( \lambda ) $
, which, since
$\phi _{1}( \lambda )+\phi _{2}( \lambda ) +\phi _{3}( \lambda ) =1$
, implies
$\phi _{2}( \lambda ) +\phi _{3}( \lambda ) >\frac{2}{3}=\phi _{2}( \nu ) +\phi _{3}( \nu ) $
, disproving the conjecture.
Example 1. We provide a numerical example of the above description:
\begin{equation*} \sigma _{1}^{2}=\frac{1}{6}, \quad \sigma _{2}^{2}=\frac{2}{3}, \quad \sigma _{3}^{2}=\frac{5}{6}, \quad \rho _{12}=\frac{1}{2} \quad \rho _{13}=0, \quad \rho _{23}=-\frac{3}{2\sqrt{5}\,}. \end{equation*}
In particular, in such a case,
$\mathrm{Var}\left( X_{2}+X_{3}\right) =\frac{1}{2}$
and
$\mathrm{Var}\left( X_{1}+X_{3}\right) =1$
, while the covariance matrix is given by
\begin{equation*}\sum =\left(\begin{array}{c@{\quad}c@{\quad}c}\frac{1}{6} & \frac{1}{6} & 0 \\ \\[-7pt]\frac{1}{6} & \frac{2}{3} & -\frac{1}{2} \\ \\[-7pt]0 & -\frac{1}{2} & \frac{5}{6}\end{array}\right).\end{equation*}
Example 2. This is another example with all non-negative covariances:
\begin{equation*} \sigma _{1}^{2}=\frac{1}{6}, \quad \sigma _{2}^{2}=\frac{1}{5}, \quad \sigma _{3}^{2}=\frac{3}{10}, \quad \rho _{12}=\frac{0.8}{\sqrt{1.2}\,}, \quad \rho _{13}=\frac{0.2}{\sqrt{1.8}\,}, \quad \rho _{23}=0 . \end{equation*}
The covariance matrix is
\begin{equation*}\sum =\left(\begin{array}{c@{\quad}c@{\quad}c}\frac{1}{6} & \frac{2}{15} & \frac{1}{30} \\ \\[-7pt]\frac{2}{15} & \frac{1}{5} & 0 \\ \\[-7pt]\frac{1}{30} & 0 & \frac{3}{10}\end{array}\right).\end{equation*}
4. Future research
In Section 3.1 we have shown that the conjecture holds true for any number of random variables in the particular case of a diagonal covariance matrix. We present another case in which the conjecture is valid for any number of random variables.
Theorem 3. Assume that all the correlation coefficients of the covariance matrix are unitary, i.e.
$\rho_{ij}\equiv1$
for all
$i\neq j$
. Then, the normalized Shapley values for the variance and the standard deviation games coincide. In particular, the conjecture holds true.
Proof. Under the theorem assumption, the standard deviation value function can be written as
\begin{equation*} \begin{aligned} \lambda(J) & = \left\{\mathrm{Var}\Bigg[\sum_{j\in J}X_{j}\Bigg]\right\}^{1/2} = \Bigg\{\sum_{j\in T}\sigma_{j}^{2}+2\sum_{k,j\in T}\sigma_{k}\sigma_{j}\Bigg\}^{1/2} \\ & = \left\{\Bigg(\sum_{j\in T}\sigma_{j}\Bigg)^{2}\right\}^{1/2} =\sum_{j\in T}\sigma_{j} . \end{aligned}\end{equation*}
Hence, it easily follows that
$\lambda(N)=\sum_{j\in N}\sigma_{j}$
is the sum of all the standard deviations and that the marginal value increase simplifies to
$\lambda(J\cup i)-\lambda(J)=\sigma_{i}$
. This implies that the normalized Shapley values for the standard deviation games becomes
\begin{equation}\frac{1}{\lambda(N)}\phi_{i}(\lambda)=\frac{1}{\lambda(N)}\sum_{T\subseteq N\setminus\left\{ i\right\} }\frac{|J|!\,(N-|J|-1)!}{N!}\left(\lambda(J\cup i)-\lambda(J)\right)=\frac{\sigma_{i}}{\sum_{j\in N}\sigma_{j}}\end{equation}
for all
$i=1,2,\ldots,n$
. If we multiply the numerator and the denominator in (9) by
$\sum_{j\in N}\sigma_{j}$
, we find
\begin{equation*}\frac{1}{\lambda(N)}\phi_{i}(\lambda)=\frac{\sigma_{i}\left(\sum_{j\in N}\sigma_{j}\right)}{\left(\sum_{j\in N}\sigma_{j}\right)^{2}}=\frac{\sigma_{i}^{2}+\sum_{j\neq i}\sigma_{j}}{\left(\sum_{j\in N}\sigma_{j}\right)^{2}}=\frac{\mathrm{Cov}(X_{i},S)}{\mathrm{Var}(S)}=\frac{1}{\nu(N)}\phi_{i}(\nu) ,\end{equation*}
where
$\nu$
is the variance value function.
In conclusion, we believe that the approach we have followed to prove the conjecture in the independent case for general n can be used to check whether the conjecture of [Reference Colini-Baldeschi, Scarsini and Vaccari5] holds for some specific dependence structures.
A Appendix
In order to show that
$\sum_{i=2}^n \phi_{i}( \lambda ) \leq \sum_{i=2}^n \sigma_{i}^{2}$
, we prove that
$\phi _{1}( \lambda ) \geq \sigma _{1}^{2}$
. First of all, set
$\sigma _{i}^{2}=a_{i}$
. Then define
\begin{align*}f( a_{1},\ldots,a_{n}) & = \left( n-1\right) !\sqrt{a_{1}} \\ & \quad + \underset{}{\overset{n}{\underset{k=2}{\sum }}\left( k-1\right) !\left( n-k\right) !}\left( \Bigg\{a_{1}+\underset{j_{1}< \cdots <j_{k-1}}{\sum }\!\!\!a_{j_{r}}\Bigg\}^{1/2}-\Bigg\{\underset{j_{1}< \cdots <j_{k-1}}{\sum }\!\!\!a_{j_{r}}\Bigg\}^{1/2}\right) ,\end{align*}
where
$2\leq j_{1}< \cdots <j_{k-1}\leq n$
, and
$g( a_{1}) =n!\,a_{1}$
. Thus, it is clear that
$\phi _{1}( \lambda ) \geq \sigma _{1}^{2}$
is equivalent to
$f( a_{1},\ldots,a_{n}) \geq g( a_{1})$
whenever
$0\leq a_{1}\leq a_{2}\leq \cdots \leq a_{n}$
and
$a_{1}+a_{2}+ \cdots +a_{n}=1$
.
In fact, it is easily computed that, whatever the
$a_{2},\ldots,a_{n}$
satisfying the previous conditions,
$f( 0,a_{2},\ldots,a_{n})=g( 0) $
,
$f\left( \frac{1}{n},\frac{1}{n},\ldots,\frac{1}{n}\right)=g\left( \frac{1}{n}\right) $
, and finally
$f( a_{1},\ldots,a_{n})>g( a_{1}) $
when
$0<a_{1}\leq \frac{1}{n^{2}}$
.
So, assume, by contradiction, that
$f( a_{1}^{\prime},\ldots,a_{n}^{\prime }) =g( a_{1}^{\prime }) $
for some vector
$\textbf{a}^{\boldsymbol{\prime }}=\left( a_{1}^{\prime},\ldots,a_{n}^{\prime }\right) $
satisfying the above conditions with
$\frac{1}{n^{2}}<a_{1}^{\prime }<\frac{1}{n}$
, and, moreover, that there exist vectors
$\textbf{a}$
, still satisfying the previous conditions, with
$\|\textbf{a-a}^{\prime }\| <\varepsilon $
for a small enough
$\varepsilon >0$
and
$a_{1}^{\prime}<a_{1}$
, such that
$f( \textbf{a}) <g( a_{1}) $
.
Hence, since
$f\left( \frac{1}{n},\frac{1}{n},\ldots,\frac{1}{n}\right) =g\left( \frac{1}{n}\right) $
, there must also exist a vector
$\textbf{a}^{\prime \prime }$
, with
$a_{1}^{\prime}<a_{1}^{\prime \prime }\leq \frac{1}{n}$
, such that
$f(\textbf{a}^{\prime \prime }) =g( a_{1}^{\prime \prime }) $
, while, for suitable values of
$\textbf{a}$
close to
$\textbf{a}^{\prime \prime }$
, satisfying the above conditions with
$a_{1}<a_{1}^{\prime \prime }$
, it again holds that
$f( \textbf{a}) <g( a_{1}) $
.
Let us start, precisely, from such an
$\textbf{a}^{\prime \prime }$
. It is easily computed that, for
$a_{1}>0$
,
$\frac{\partial f}{\partial a_{1}}>0$
, while, for
$k=2,\ldots,n$
,
$\frac{\partial f}{\partial a_{k}}<0$
. Hence, for a suitably small
$x>0$
and suitable non-negative values
$b_{2},\ldots,b_{n}$
such that
$b_{2}+ \cdots +b_{n}=1$
, the vector
$\textbf{a}=\left( a_{1}^{\prime \prime }-x,a_{2}^{\prime \prime}+b_{2}x,\ldots,a_{n}^{\prime \prime }+b_{n}x\right) $
satisfies
$f(\textbf{a}) <g( a_{1}) $
. Therefore, it is possible to construct a path a(x),
$0\leq x\leq\overline{x}$
, joining
$a^{\prime \prime }$
to some
$a^{\prime }$
defined as above, such that
$f( \textbf{a}( x) ) <g( a_{1}( x))$
when
$0<x<\overline{x}$
,
$f( \textbf{a}(0) ) =g( a_{1}( 0) )$
,
$f(\textbf{a}( \overline{x}) ) =g( a_{1}( \overline{x}) )$
. In fact, we can expect, at most, such a path to be piecewise linear, since, moving, say, linearly from
$\textbf{a}^{\prime \prime }$
, some
$a_{i}( x) $
,
$2\leq i<n$
, might reach an initially higher
$a_{i+1}$
before the path reaches
$\textbf{a}^{\prime }$
. However, a piecewise linear path can be approximated, as well as we want, by a smooth one. Hence, after inverting, for our convenience, the direction of the path, i.e. letting it go from
$\textbf{a}^{\prime }$
to
$\textbf{a}^{\prime \prime }$
, we can write
$\textbf{a}( x) =\left( a_{1}^{\prime }+x,a_{2}^{\prime }-\beta_{2}( x) ,\ldots,a_{n}^{\prime }-\beta _{n}( x) \right)$
,
$0\leq x\leq \overline{x}$
, with
$0\leq \beta _{2}( x) ,\ldots,\beta _{n}(x) $
,
$\beta _{2}( x) + \cdots +\beta _{n}(x) =x$
,
$\beta _{2}^{\prime }( x) ,\ldots,\beta_{n}^{\prime }( x) \geq 0$
,
$\beta _{2}^{\prime \prime }(x) ,\ldots,\beta _{n-1}^{\prime \prime }( x) \leq 0$
, so that, setting
$\overline{f}( x) =f( \textbf{a}(x) ) ,\widetilde{g}( x) =g( a_{1}( x)) =n!\left( a_{1}^{\prime }+x\right) $
,
$\widetilde{f}( 0) = \widetilde{g}( 0)$
,
$\widetilde{f}( \overline{x}) = \widetilde{g}(\overline{x})$
,
$\widetilde{f}( x) <\widetilde{g}( x)$
when
$0<x<\overline{x}$
. Moreover, as
$\widetilde{g}^{\prime }( x) =n!$
, it follows that
$\widetilde{f}^{\prime }( x) <n!$
when
$0<x<\varepsilon $
and
$\widetilde{f}^{\prime }( x) >n!$
when
$x^{\prime }-\varepsilon <x<\overline{x}$
for a suitably small
$\varepsilon >0$
. However, straightforward computations show that
$\widetilde{f}^{\prime \prime }( x) <0$
when
$0<x<\overline{x}$
, leading to a contradiction. In fact, since
$\beta_{n}( x) =x-\beta _{2}( x) - \cdots -\beta _{n-1}(x) $
,
$\beta _{n}^{\prime \prime }( x) =-\beta _{2}^{\prime\prime }( x) - \cdots -\beta _{n-1}^{\prime \prime }( x) $
. Hence, in particular, in the expression of
$\widetilde{f}^{\prime \prime}( x) $
, to each term of the type
\begin{equation*}\frac{1}{2}\beta _{n}^{\prime \prime }( x) \left( \frac{1}{\sqrt{\underset{r}{\sum }a_{j_{r}}+a_{h}}\,}-\frac{1}{\sqrt{\underset{r}{\sum }a_{j_{r}}+a_{h}+a_{1}}\,}\right)\end{equation*}
corresponds a sum
\begin{align*}&\frac{1}{2}\left[ \underset{r}{\sum }\beta _{j_{r}}^{\prime \prime }(x) \left( \frac{1}{\sqrt{\underset{r}{\sum }a_{j_{r}}+a_{n}}\,}-\frac{1}{\sqrt{\underset{r}{\sum }a_{j_{r}}+a_{n}+a_{1}}\,}\right) \right. \\ &\quad \left. + \underset{2\leq h\neq j_{r},n}{\sum }\beta _{h}^{\prime \prime }( x) \left( \frac{1}{\sqrt{\underset{r}{\sum }a_{j_{r}}+a_{h}}\,}-\frac{1}{\sqrt{\underset{r}{\sum }a_{j_{r}}+a_{h}+a_{1}}\,}\right) \right]\!.\end{align*}
Since, when
$h\neq n$
,
\begin{equation*}\frac{1}{\sqrt{\underset{r}{\sum }a_{j_{r}}+a_{h}}\,}-\frac{1}{\sqrt{\underset{r}{\sum }a_{j_{r}}+a_{h}+a_{1}}\,}\geq \frac{1}{\sqrt{\underset{r}{\sum }a_{j_{r}}+a_{h}}\,}-\frac{1}{\sqrt{\underset{r}{\sum }a_{j_{r}}+a_{h}+a_{1}}\,},\end{equation*}
the sum of the above two quantities is
$\leq 0$
.
B Appendix
Let us consider, first,
$g_{1}( x) $
. For any fixed
$x\in \left[ 0,\overline{x}\right] $
, set
$a_{i}( x) =a_{i}$
. Observe that, whenever
$3\leq l<m\leq n$
,
\begin{equation*}\frac{1}{2}\left[ 2\frac{\beta ^{2}}{\left( a_{l}+a_{m}\right) ^{\frac{3}{2}}}-\frac{\beta ^{2}}{4\left( a_{m}\right) ^{\frac{3}{2}}}-\frac{\beta ^{2}}{4\left( a_{l}\right) ^{\frac{3}{2}}}\right] =\frac{\beta ^{2}}{4}\left(\widehat{a}\right) ^{-\frac{3}{2}}\left( \sqrt{2}-1\right)\end{equation*}
for some
$a_{l}\leq \widehat{a}\leq a_{m}$
, so that
$\widehat{a}\geq a_{2}\geq a_{1}$
. Hence, straightforward computations lead us to conclude that
$g_{1}^{\prime \prime }( x) \geq 0$
certainly holds if
\begin{equation*}\left( n-1\right) \left( 1+\frac{1}{2\sqrt{2}\,}\right) <\left( \left(n-1\right) ^{2}+n-2\right) \left( 1-\frac{1}{2\sqrt{2}\,}\right)\end{equation*}
when
$n\geq 4$
. In fact, the worst case is precisely
$n=4$
, when it is easily checked that
\begin{equation*}3\left( 1+\frac{1}{2\sqrt{2}\,}\right) <11\left( 1-\frac{1}{2\sqrt{2}\,}\right) .\end{equation*}
As to
$g_{j}( x) $
, when
$2\leq j\leq n-2$
, it can be shown that, in order to prove
$g_{j}^{\prime \prime }( x) \leq 0$
, it is sufficient to check the inequality
\begin{align*} & \left( j+1\right) ^{-\frac{3}{2}}\left[ \binom{n-2}{j-1}\left( n-j-1\right)^{2}+\binom{n-2}{j}\left( j+1\right) ^{2}\right] \\ &\quad \leq \left( j\right) ^{-\frac{3}{2}}\left[ \binom{n-2}{j-1}\left( n-j\right)^{2}+\binom{n-2}{j}\left( j\right) ^{2}\right] ,\end{align*}
which follows through straightforward steps. Finally,
$f_{i,n-1}( x) =\frac{1}{n}( 1-\sqrt{1-a_{i}\left( x\right) }) $
, so that it is obvious that
$f_{i,n-1}^{\prime \prime }(x) \geq 0$
.
Acknowledgement
Marcello Galeotti is a member of INDAM (National Institute of High Mathematics) affiliated to the group of Mathematical Analysis and Probability (GNAMPA).
