Quasi-arithmetic means and ratios of an interval induced from weighted aggregation operations

Abstract

In this paper we deal with weighted quasi-arithmetic means of an interval using utility functions in decision making. The mean values are discussed from the viewpoint of weighted aggregation operators, and they are given as weighted aggregated values of each point in the interval. The properties of the weighted quasi-arithmetic mean and its translation invariance are investigated. For the application in economics, we demonstrate the decision maker’s attitude based on his utility by the weighted quasi-arithmetic mean and the aggregated mean ratio. Several examples of the weighted quasi-arithmetic mean and the aggregated mean ratio for various typical utility functions are shown to understand our motivation and for the applications in decision making.

Appendix

In this section, we give the proofs of the theorems, the propositions and the lemmas in Sects. 3–5.

Proof of Lemma 3.1

Let $[a,b] \in {\mathcal{C}}(D)$ satisfying $a< b.$ (M.i) Since $f$ and $f^{-1}$ are strictly increasing, we have

$$ \begin{aligned} M^{f}([a,b]) & = f^{-1} \left(\left. \intop_{a}^{b} f(x) w(x){\rm d}x \right/ \intop\limits_{a}^{b} w(x){\rm d}x \right)\\ & \le f^{-1} \left( \left. \intop\limits_{a}^{b} f(b) w(x){\rm d}x \right/ \intop\limits_{a}^{b} w(x){\rm d}x \right)\\ & = f^{-1} (f(b))=b. \end{aligned} $$

Thus we get $M^{f}([a,b]) \le b.$ In the same way, we also have $M^{f}([a,b]) \ge a.$ Therefore, we obtain $a \le M^{f}([a,b])\le b.$ This inequality also implies that $M^{f}([a,a]) = a$ for $a \in D$ together with the definition (6). (M.ii) Put a function

$$ F(t)=\left. \intop_{a}^{t} f(x) w(x){\rm d}x \right/ \intop_{a}^{t} w(x){\rm d}x $$

for $t > a.$ Then we have

$$ \begin{aligned} F^{\prime}(t)& =\frac{f(t)w(t) \int_{a}^{t} w(x){\rm d}x - w(t) \int_{a}^{t} f(x) w(x){\rm d}x}{\left( \int_{a}^{t} w(x){\rm d}x \right)^2}\\ &=\frac{w(t) \int_{a}^{t} (f(t) - f(x)) w(x){\rm d}x}{\left( \int_{a}^{t} w(x){\rm d}x \right)^2} > 0 \end{aligned} $$

since f is strictly increasing on D and $w>0$ on D. Thus the map $t \in (a, \infty) \cap D \, \mapsto \, F(t)$ is strictly increasing. We also put a function

$$ G(t)=\left. \intop_{t}^{b} f(x) w(x){\rm d}x \right/ \intop_{t}^{b} w(x){\rm d}x$$

for $t < b.$ Then we have

$$ \begin{aligned} G^{\prime}(t)&= \frac{- f(t) w(t) \int_{t}^{b} w(x){\rm d}x + w(t)\int_{t}^{b} f(x) w(x){\rm d}x}{\left( \int_{t}^{b} w(x){\rm d}x \right)^2}\\ &=\frac{w(t) \int_{t}^{b} (f(x) - f(t)) w(x){\rm d}x}{\left( \int_{t}^{b} w(x){\rm d}x \right)^2} >0 \end{aligned} $$

since f is strictly increasing on D and $w>0$ on D. Thus the map $t \in (-\infty, b) \cap D \, \mapsto \, G(t)$ is also strictly increasing. Let $[a,b], [c,d] \in {\mathcal{C}}(D)$ satisfying $[a,b] \preceq [c,d].$ From the above results and the definition (6), we have $f(M^{f}([a,b])) \le f(M^{f}([a,d])) \le f(M^{f}([c,d])).$ Therefore, we get $M^{f}([a,b]) \le M^{f}([c,d])$ since $f^{-1}$ is strictly increasing. (M.iii) The continuity is trivial from the definition (6). Therefore, the proof is completed.$\hfill\square$

Proof of Proposition 4.1

• (i) Let r be a positive number and let $[a,b] \subset [0,\infty).$ Then we have

  $$ \begin{aligned} r \cdot M^{f}([a,b])&=r \cdot \left(\left. \int\limits_{a}^{b} x^{\gamma} w(x){\rm d}x \right/ \int\limits_{a}^{b} w(x){\rm d}x \right)^{1/\gamma}\\ &=\left(\left. \int\limits_{a}^{b} (rx)^{\gamma} w(x){\rm d}x \right/ \int\limits_{a}^{b} w(x){\rm d}x \right)^{1/\gamma}\\ &=\left(\left. \int\limits_{ra}^{rb} x^{\gamma} w(x/r){\rm d}x \right/ \int\limits_{ra}^{rb} w(x/r){\rm d}x \right)^{1/\gamma}\\ &=\left(\left. \int\limits_{ra}^{rb} x^{\gamma} w(x){\rm d}x \right/ \int\limits_{ra}^{rb} w(x){\rm d}x \right)^{1/\gamma}\\ &=M^{f}([ra,rb]). \end{aligned} $$

• (ii) Let r be a positive number and let $[a,b] \subset (0,\infty).$ Then we have

  $$ \begin{aligned} r \cdot M^f([a,b]) & =\exp \left(\frac{1}{\gamma} \cdot \left. \int\limits_{a}^{b} \gamma \log x \cdot w(x){\rm d}x \right/ \int\limits_{a}^{b} w(x){\rm d}x + \log r \right)\\ &= \exp \left(\frac{1}{\gamma} \cdot \left. \int\limits_{a}^{b} \gamma \log (rx) \cdot w(x){\rm d}x \right/ \int\limits_{a}^{b} w(x){\rm d}x \right)\\ &= \exp \left(\frac{1}{\gamma} \cdot \left. \int\limits_{ra}^{rb} \gamma \log x \cdot w(x/r){\rm d}x \right/ \int\limits_{ra}^{rb} w(x/r){\rm d}x \right)\\ & =\exp \left(\frac{1}{\gamma} \cdot \left. \int\limits_{ra}^{rb} \gamma \log x \cdot w(x){\rm d}x \right/ \int\limits_{ra}^{rb} w(x){\rm d}x \right)\\ & = M^f([ra,rb]). \end{aligned} $$

• (iii) Let s be a real number and let $[a,b] \subset (-\infty,\infty).$ Then we have

  $$ \begin{aligned} M^f([a,b]) + s &= \frac{1}{\gamma} \log \left( \left.\int\limits_{a}^{b} {\rm e}^{\gamma x} w(x){\rm d}x \right/ \int\limits_{a}^{b} w(x){\rm d}x \right) +s\\ &=\frac{1}{\gamma} \log \left(\left.\int\limits_{a}^{b} {\rm e}^{\gamma (x+s)} w(x){\rm d}x \right/ \int\limits_{a}^{b} w(x){\rm d}x \right)\\ &=\frac{1}{\gamma} \log \left(\left.\int\limits_{a+s}^{b+s} {\rm e}^{\gamma x} w(x-s){\rm d}x \right/ \int\limits_{a+s}^{b+s} w(x-s){\rm d}x \right)\\ &=\frac{1}{\gamma} \log \left( \left.\int\limits_{a+s}^{b+s} {\rm e}^{\gamma x} w(x){\rm d}x \right/ \int\limits_{a+s}^{b+s}w(x){\rm d}x \right)\\ &= M^f([a+s,b+s]). \end{aligned} $$

  Thus we obtain this lemma.$\hfill\square$

Proof of Theorem 5.3

• (i) Let f and g satisfy $f^{\prime\prime} / f^{\prime}< g^{\prime\prime} / g^{\prime}$ on $(a,b).$ Define

  $$ \begin{aligned} H(y) & =\int\limits_{a}^{y} f(x) w(x){\rm d}x - f \left( M^{g}([a,y]) \right) \int\limits_{a}^{y} w(x){\rm d}x\\ & = \int\limits_{a}^{y} f(x) w(x){\rm d}x\\ & \quad - f \left( g^{-1} \left( \left. \int\limits_{a}^{y} g(x) w(x){\rm d}x \right/ \int\limits_{a}^{y} w(x){\rm d}x \right) \right)\\ &\quad \times\int\limits_{a}^{y} w(x){\rm d}x \end{aligned} $$

  for $y \in (a,b].$ Then we have

  $$ \begin{aligned} H^{\prime}(y) & = \frac{w(y)}{g^{\prime}(M^{g}([a,y]))}\\ & \quad \times \left( g^{\prime}(M^{g}([a,y])) ( f(y) - f(M^{g}([a,y])) ) \right.\\ & \quad \left. - f^{\prime}(M^{g}([a,y])) ( g(y) - g(M^{g}([a,y])) ) \right). \end{aligned} $$

  On the other hand, the map $x \, \mapsto \, g^{\prime}(x) / f^{\prime}(x)$ is increasing on $(a,b)$ since

  $$ \left(\frac{g^{\prime}(x)}{f^{\prime}(x)} \right)^{\prime}=\frac{g^{\prime\prime}(x) f^{\prime}(x) - g^{\prime}(x) f^{\prime\prime}(x)}{(f^{\prime}(x))^2} > 0 $$

  for $x \in (a,b)$ from $f^{\prime}>0,$ $g^{\prime}>0$ and $f^{\prime\prime} / f^{\prime} < g^{\prime\prime} / g^{\prime}$ on $(a,b).$ Therefore, we get

  $$ \frac{g^{\prime}(M^{g}([a,y]))}{f^{\prime}(M^{g}([a,y]))} <\frac{g^{\prime}(\xi)}{f^{\prime}(\xi)}, $$

  where a constant $\xi \in (M^{g}([a,y], y)$ is given by

  $$ \frac{g^{\prime}(\xi)}{f^{\prime}(\xi)}=\frac{g(y) - g(M^{g}([a,y]))}{f(y)-f(M^{g}([a,y]))} $$

  from Cauchy’s mean value theorem. Thus we get

  $$ \frac{g^{\prime}(M^{g}([a,y]))}{f^{\prime}(M^{g}([a,y]))} < \frac{g(y) - g(M^{g}([a,y]))}{f(y) - f(M^{g}([a,y]))}. $$

  From $f^{\prime}>0,$ $g^{\prime}>0$ and the above equality regarding $H^{\prime}(y),$ this implies $H^{\prime}(y) < 0$ for $y \in (a,b),$ Together with $H(a) = 0,$ we obtain $H(b) < 0$ for $b > a.$ Thus we have

  $$ \intop_{a}^{b} f(x) w(x){\rm d}x \,<\, f \left( M^{g}([a,b]) \right) \intop_{a}^{b} w(x){\rm d}x. $$

  Since, $f^{-1}$ is increasing, we obtain

  $$ \begin{aligned} M^{f}([a,b]) & = f^{-1} \left( \left. \intop_{a}^{b} f(x) w(x){\rm d}x \right/ \intop_{a}^{b} w(x){\rm d}x \right)\\ & < M^{g}([a,b]). \end{aligned} $$

  This also implies

  $$ \begin{aligned} \theta^{f}(a,b)&=\frac{M^{f}([a,b])-a}{b- a}\\ &< \frac{M^{g}([a,b])-a}{b-a}\\ &=\theta^{g}(a,b). \end{aligned} $$

  Thus (i) holds. We also obtain (ii) in the same way as (i). The proof of (iii) is straightforward. Therefore, the proof of this theorem is completed.$\hfill\square$

Proof of Theorem 5.6

$(a) \Rightarrow (b)$: Let $[c,d]$ satisfy $[c,d] \subset [a,b]$ and $c< d.$ Then, we obtain (b) from (a) applying Theorem 5.3(ii) to $f^{\prime\prime} / f^{\prime} \le g^{\prime\prime} / g^{\prime}$ on $(c,d).$$(b) \Rightarrow (a)$: Let $M^{f}$ and $M^{g}$ satisfy $M^{f}(I) \le M^{g}(I)$ for all closed intervals $I (\subset [a,b]).$ Suppose that $f^{\prime\prime} / f^{\prime} \le g^{\prime\prime} / g^{\prime}$ does not hold on $(a,b).$ Then there exits an closed interval $[c,d]$ such that $[c,d] \subset [a,b],$$c< d$ and $f^{\prime\prime} / f^{\prime} > g^{\prime\prime} / g^{\prime}$ on $(c,d).$ By Theorem 5.3(i) we have $M^{f}([c,d]) > M^{g}([c,d]),$ and this contradicts $M^{f}(I) \le M^{g}(I)$ with $I=[c,d].$ Therefore, we obtain $f^{\prime\prime} / f^{\prime} \le g^{\prime\prime} / g^{\prime}$ on $(a,b).$ Thus, the equivalence between (a) and (b) hold. This theorem holds since (b) and (c) are also equivalent clearly.$\hfill\square$

Proof of Proposition 5.8

• (i) Let f and g satisfy $f^{\prime\prime} / f^{\prime} < g^{\prime\prime} / g^{\prime}$ on $(a,b).$ For $h=(f+g)/2,$ we have

  $$ \begin{aligned} \frac{h^{\prime\prime}}{h^{\prime}} &=\frac{f^{\prime\prime} + g^{\prime\prime}}{f^{\prime}+g^{\prime}}\\ &=\frac{f^{\prime\prime}}{f^{\prime}}\cdot\frac{f^{\prime}}{f^{\prime} + g^{\prime}}+\frac{g^{\prime\prime}}{g^{\prime}} \cdot \frac{g^{\prime}}{f^{\prime} + g^{\prime}}\\ & < \frac{g^{\prime\prime}}{g^{\prime}} \cdot \frac{f^{\prime}}{f^{\prime} + g^{\prime}}+\frac{g^{\prime\prime}}{g^{\prime}} \cdot \frac{g^{\prime}}{f^{\prime}+ g^{\prime}}\\ & =\frac{g^{\prime\prime}}{g^{\prime}}. \end{aligned} $$

  Therefore, we get $h^{\prime\prime}/h^{\prime} < g^{\prime\prime}/g^{\prime}.$ In the same way, we also have $f^{\prime\prime}/f^{\prime} < h^{\prime\prime}/h^{\prime}.$ From Theorem 5.3(i), we obtain $M^{f}([a,b]) < M^{h}([a,b]) < M^{g}([a,b])$ and $\theta^{f}(a,b) < \theta^{h}(a,b) <\theta^{g}(a,b).$ We also obtain (ii) in the same way as (i). Therefore the proof is completed.$\hfill\square$

Proof of Theorem 5.9

Fix any $a \in D.$ First, we have

$$ \begin{aligned} \lim_{b \downarrow a} \nu(a,b) & = \lim_{b \downarrow a}\frac{1}{(b - a) \int\nolimits_{a}^{b} w(x){\rm d}x} \int\limits_{a}^{b} (x-a) w(x){\rm d}x\\ & =\lim_{b \downarrow a} \frac{(b-a) w(b)}{\int\nolimits_{a}^{b} w(x){\rm d}x + (b-a) w(b)}\\ & =\lim_{b \downarrow a}\frac{w(b)}{\left. \int\nolimits_{a}^{b} w(x){\rm d}x \right/ (b-a) + w(b)}\\ & =\frac{w(a)}{w(a) + w(a)}=\frac{1}{2}. \end{aligned} $$

Therefore, it holds that $\lim_{b \downarrow a} \nu(a,b) = 1/2.$ Next, by Taylor expansion, we have

$$ f(x) = f(a) + f^{\prime}(a) (x-a) + {{1}\over {2}} f^{\prime\prime}(c(x)) (x-a)^2 $$

for $x \in (a,\infty) \cap D,$ where $c(x)$ satisfies $a < c(x)< x.$ For $b \in D$ such that $a< b,$ it implies that

$$ \begin{aligned} &\frac{\left. \int_{a}^{b} f(x) w(x){\rm d}x \right/ \int_{a}^{b} w(x){\rm d}x- f(a)}{b- a}\\ & \quad =\frac{1}{(b - a) \int_{a}^{b} w(x){\rm d}x} \intop_{a}^{b} (f(x) -f(a)) w(x){\rm d}x\\ & \quad =\frac{1}{(b - a) \int_{a}^{b} w(x){\rm d}x}\\ & \quad \quad\times \intop_{a}^{b} \left( f^{\prime}(a) (x-a) + \frac{1}{2} f^{\prime\prime}(c(x)) (x-a)^2 \right) w(x){\rm d}x. \end{aligned} $$

Then we have

$$ \begin{aligned} &\frac{1}{(b-a) \int_{a}^{b} w(x){\rm d}x} \intop_{a}^{b} f^{\prime}(a) (x-a) w(x) {\rm d}x\\ &=f^{\prime}(a) \nu(a,b) \end{aligned} $$

and

$$ \begin{aligned} &\left|\frac{1}{(b - a) \int_{a}^{b} w(x){\rm d}x} \intop_{a}^{b} \frac{1} {2} f^{\prime\prime}(c(x)) (x-a)^2 w(x){\rm d}x \right|\\ & \quad \le \frac{1}{(b - a) \int_{a}^{b} w(x){\rm d}x} \intop_{a}^{b} K (b-a)^2 w(x){\rm d}x\\ & \quad \le K (b-a) \end{aligned} $$

with a positive constant $K = \max_{y \in [a,b]} | f^{\prime\prime}(y)|/2.$ Thus we get

$$ \begin{aligned} &\lim_{b \downarrow a} \frac{\left. \int_{a}^{b} f(x) w(x) {\rm d}x \right/ \int_{a}^{b} w(x){\rm d}x- f(a)}{b- a}\\& \qquad\qquad = f^{\prime}(a) \lim_{b \downarrow a} \nu(a,b). \end{aligned} $$

(16)

Next, we put a function

$$ F(t)=\left. \intop_{a}^{t} f(x) w(x){\rm d}x \right/ \intop_{a}^{t} w(x){\rm d}x, $$

for $t \in (a,\infty) \cap D.$ Then we have

$$ \begin{aligned} &\frac{f^{-1} \left( \left. \int_{a}^{b} f(x) w(x){\rm d}x \right/ \int_{a}^{b} w(x){\rm d}x \right)-a}{\left. \int_{a}^{b} f(x) w(x){\rm d}x \right/ \int_{a}^{b} w(x){\rm d}x - f(a)}\\ &\qquad\qquad =\frac{f^{-1} ( F(b) ) - a}{F(b) - f(a)}. \end{aligned} $$

(17)

Since $\lim_{b \to a} F(b) = f(a)$ and $\lim_{t \to f(a)} f^{-1} (t) = a,$ we get

$$ \begin{aligned} \lim_{b \to a} \frac{f^{-1} ( F(b) ) - a}{F(b) - f(a)} & =\lim_{t \to f(a)} \frac{f^{-1} (t) - a}{t - f(a)}\\ &= \left. \frac{1}{f^{\prime}(f^{-1}(t))} \right|_{t=f(a)}\\ & =\frac{1}{f^{\prime}(a)}. \end{aligned} $$

(18)

These equalities (16), (17) and (18) imply

$$ \begin{aligned} \lim_{b \downarrow a} \theta^{f}(a,b)& =\lim_{b \downarrow a} \frac{f^{-1} \left( \left. \int_{a}^{b} f(x) w(x){\rm d}x \right/ \int_{a}^{b} w(x){\rm d}x \right) - a}{b- a}\\& = f^{\prime}(a) \lim_{b \downarrow a} \nu(a,b) \frac{1}{f^{\prime}(a)}\\& = \lim_{b \downarrow a} \nu(a,b)\\& =\frac{1}{2}. \end{aligned} $$

Thus we obtain (10). We can easily check (11) in a similar way. Therefore, we obtain this theorem.$\hfill\square$

Proof of Lemma 5.10

• (i) Fix any $b>0.$ (12) is trivial from (M.iii). (ii) If $M^{f}$ satisfies (13), then

  $$ \begin{aligned} \lim_{a \downarrow 0} \theta^{f}(a,b) & = \frac{1}{b} M^{f}([0,b])\\ & =M^{f}([0,1])\\ &=f^{-1} \left( \left.\intop_{0}^{1} f(x) w(x){\rm d}x \right/ \intop_{0}^{1} w(x){\rm d}x \right) \end{aligned} $$

  for $b>0.$ Thus we have (14). Finally fix any $a \in D.$ From (13) and (M.iii),

  $$ \begin{aligned} \lim_{b \to \infty} \theta^{f}(a,b)& = \lim_{b \to \infty} \frac{M^{f}([a,b])-a}{b-a}\\& =\lim_{b \to \infty} \frac{1}{b} M^{f}([a,b])\\& = \lim_{b \to \infty} M^{f}([a/b,1])\\ & = M^{f}([0,1]). \end{aligned} $$

  Thus we also obtain (15). Therefore, we get this lemma.$\hfill\square$