Decision Making Under Interval Uncertainty (and Beyond)

Abstract

To make a decision, we must find out the user’s preference, and help the user select an alternative which is the best—according to these preferences. Traditional utility-based decision theory is based on a simplifying assumption that for each two alternatives, a user can always meaningfully decide which of them is preferable. In reality, often, when the alternatives are close, the user is often unable to select one of these alternatives. In this chapter, we show how we can extend the utility-based decision theory to such realistic (interval) cases.

A Proof of the Theorem

$1^\circ$. Due to scale-invariance (9), for every $y_1$, ..., $y_n$, $y'_1$, ..., $y'_n$, we can take $\lambda _i=\displaystyle \frac{1}{y_i}$ and conclude that

$$\begin{aligned} (y'_1,\ldots ,y'_n)\sim (y_1,\ldots ,y_n)\Leftrightarrow \left( \frac{y'_1}{y_1},\ldots ,\frac{y'_n}{y_n}\right) \sim (1,\ldots ,1). \end{aligned}$$

(12)

Thus, to describe the equivalence relation $\sim$, it is sufficient to describe the set of all the vectors $z=(z_1,\ldots ,z_n)$ for which $z\sim (1,\ldots ,1)$. Similarly,

$$\begin{aligned} (y'_1,\ldots ,y'_n)\succ (y_1,\ldots ,y_n)\Leftrightarrow \left( \frac{y'_1}{y_1},\ldots ,\frac{y'_n}{y_n}\right) \succ (1,\ldots ,1). \end{aligned}$$

(13)

So, to describe the ordering relation $\succ$, it is sufficient to describe the set of all the vectors $z=(z_1,\ldots ,z_n)$ for which $z\succ (1,\ldots ,1)$.

Alternatively, we can take $\lambda _i=\displaystyle \frac{1}{y'_i}$ and conclude that

$$\begin{aligned} (y'_1,\ldots ,y'_n)\succ (y_1,\ldots ,y_n)\Leftrightarrow (1,\ldots ,1)\succ \left( \frac{y_1}{y'_1},\ldots ,\frac{y_n}{y'_n}\right) . \end{aligned}$$

(14)

Thus, it is also sufficient to describe the set of all the vectors $z=(z_1,\ldots ,z_n)$ for which $(1,\ldots ,1)\succ z$.

$2^\circ$. The above equivalence involves division. To simplify the description, we can take into account that in the logarithmic space, division becomes a simple difference: $\ln \left( \displaystyle \frac{y'_i}{y_i}\right) =\ln (y'_i)-\ln (y_i)$. To use this simplification, let us consider the logarithms $Y_i\mathop {=}\limits ^{\mathrm{{def}}}\ln (y_i)$ of different values. In terms of these logarithms, the original values can be reconstructed as $y_i=\exp (Y_i)$. In terms of these logarithms, we thus need to consider:

• the set $S_\sim$ of all the tuples $Z=(Z_1,\ldots ,Z_n)$ for which

  $$\begin{aligned} z=(\exp (Z_1),\ldots ,\exp (Z_n))\sim (1,\ldots ,1), \end{aligned}$$

  (15)

  and

• the set $S_\succ$ of all the tuples $Z=(Z_1,\ldots ,Z_n)$ for which

  $$\begin{aligned} z=(\exp (Z_1),\ldots ,\exp (Z_n))\succ (1,\ldots ,1). \end{aligned}$$

  (16)

We will also consider the set $S_\prec$ of all the tuples $Z=(Z_1,\ldots ,Z_n)$ for which

$$\begin{aligned} (1,\ldots ,1)\succ z=(\exp (Z_1),\ldots ,\exp (Z_n)). \end{aligned}$$

(17)

Since the pre-ordering relation is total, for every tuple $z$,

• either $z\sim (1,\ldots ,1)$,

• or $z\succ (1,\ldots ,1)$,

• or $(1,\ldots ,1)\succ z$.

In particular, this is true for $z=(\exp (Z_1),\ldots ,\exp (Z_n))$. Thus, for every tuple $Z$, either $Z\in S_\sim$ or $Z\in S_\succ$ or $Z\in S_\prec$.

$3^\circ$. Let us prove that the set $S_\sim$ is closed under addition, i.e., that if the tuples $Z=(Z_1,\ldots ,Z_n)$ and $Z'=(Z'_1,\ldots ,Z'_n)$ belong to the set $S_\sim$, then their component-wise sum

$$\begin{aligned} Z+Z'=(Z_1+Z'_1,\ldots ,Z_n+Z'_n) \end{aligned}$$

(18)

also belongs to the set $S_\sim$.

Indeed, by definition (15) of the set $S_\sim$, the condition $Z\in S_\sim$ means that

$$\begin{aligned} (\exp (Z_1),\ldots ,\exp (Z_n))\sim (1,\ldots ,1). \end{aligned}$$

(19)

Using scale-invariance (9) with $\lambda _i=\exp (Z'_i)$, we conclude that

$$\begin{aligned} (\exp (Z_1)\cdot \exp (Z'_1),\ldots ,\exp (Z_n)\cdot \exp (Z'_n))\sim (\exp (Z'_1),\ldots ,\exp (Z'_n)). \end{aligned}$$

(20)

On the other hand, the condition $Z'\in S_\sim$ means that

$$\begin{aligned} (\exp (Z'_1),\ldots ,\exp (Z'_n))\sim (1,\ldots ,1). \end{aligned}$$

(21)

Thus, due to transitivity of the equivalence relation $\sim$, we conclude that

$$\begin{aligned} (\exp (Z_1)\cdot \exp (Z'_1),\ldots ,\exp (Z_n)\cdot \exp (Z'_n))\sim (1,\ldots ,1). \end{aligned}$$

(22)

Since for every $i$, we have $\exp (Z_i)\cdot \exp (Z'_i)=\exp (Z_i+Z'_i)$, we thus conclude that

$$\begin{aligned} (\exp (Z_1+Z'_1),\ldots ,\exp (Z_n+Z'_n))\sim (1,\ldots ,1). \end{aligned}$$

(23)

By definition (15) of the set $S_\sim$, this means that the tuple $Z+Z'$ belongs to the set $S_\sim$.

$4^\circ$. Similarly, we can prove that the set $S_\succ$ is closed under addition, i.e., that if the tuples $Z=(Z_1,\ldots ,Z_n)$ and $Z'=(Z'_1,\ldots ,Z'_n)$ belong to the set $S_\succ$, then their component-wise sum

$$\begin{aligned} Z+Z'=(Z_1+Z'_1,\ldots ,Z_n+Z'_n) \end{aligned}$$

(24)

also belongs to the set $S_\succ$.

Indeed, by definition (16) of the set $S_\succ$, the condition $Z\in S_\succ$ means that

$$\begin{aligned} (\exp (Z_1),\ldots ,\exp (Z_n))\succ (1,\ldots ,1). \end{aligned}$$

(25)

Using scale-invariance (8) with $\lambda _i=\exp (Z'_i)$, we conclude that

$$\begin{aligned} (\exp (Z_1)\cdot \exp (Z'_1),\ldots ,\exp (Z_n)\cdot \exp (Z'_n))\succ (\exp (Z'_1),\ldots ,\exp (Z'_n)). \end{aligned}$$

(26)

On the other hand, the condition $Z'\in S_\succ$ means that

$$\begin{aligned} (\exp (Z'_1),\ldots ,\exp (Z'_n))\succ (1,\ldots ,1). \end{aligned}$$

(27)

Thus, due to transitivity of the strict preference relation $\succ$, we conclude that

$$\begin{aligned} (\exp (Z_1)\cdot \exp (Z'_1),\ldots ,\exp (Z_n)\cdot \exp (Z'_n))\succ (1,\ldots ,1). \end{aligned}$$

(28)

Since for every $i$, we have $\exp (Z_i)\cdot \exp (Z'_i)=\exp (Z_i+Z'_i)$, we thus conclude that

$$\begin{aligned} (\exp (Z_1+Z'_1),\ldots ,\exp (Z_n+Z'_n))\succ (1,\ldots ,1). \end{aligned}$$

(29)

By definition (16) of the set $S_\succ$, this means that the tuple $Z+Z'$ belongs to the set $S_\succ$.

$5^\circ$. A similar argument shows that the set $S_\prec$ is closed under addition, i.e., that if the tuples $Z=(Z_1,\ldots ,Z_n)$ and $Z'=(Z'_1,\ldots ,Z'_n)$ belong to the set $S_\prec$, then their component-wise sum

$$\begin{aligned} Z+Z'=(Z_1+Z'_1,\ldots ,Z_n+Z'_n) \end{aligned}$$

(30)

also belongs to the set $S_\prec$.

$6^\circ$. Let us now prove that the set $S_\sim$ is closed under the “unary minus” operation, i.e., that if $Z=(Z_1,\ldots ,Z_n)\in S_\sim$, then $-Z\mathop {=}\limits ^{\mathrm{{def}}}(-Z_1,\ldots ,-Z_n)$ also belongs to $S_\sim$.

Indeed, $Z\in S_\sim$ means that

$$\begin{aligned} (\exp (Z_1),\ldots ,\exp (Z_n))\sim (1,\ldots ,1). \end{aligned}$$

(31)

Using scale-invariance (9) with $\lambda _i=\exp (-Z_i)=\displaystyle \frac{1}{\exp (Z_i)}$, we conclude that

$$\begin{aligned} (1,\ldots ,1)\sim (\exp (-Z_1),\ldots ,\exp (-Z_n)), \end{aligned}$$

(32)

i.e., that $-Z\in S_\sim$.

$7^\circ$. Let us prove that if $Z=(Z_1,\ldots ,Z_n)\in S_\succ$, then $-Z\mathop {=}\limits ^{\mathrm{{def}}}(-Z_1,\ldots ,-Z_n)$ belongs to $S_\prec$.

Indeed, $Z\in S_\succ$ means that

$$\begin{aligned} (\exp (Z_1),\ldots ,\exp (Z_n))\succ (1,\ldots ,1). \end{aligned}$$

(33)

Using scale-invariance (8) with $\lambda _i=\exp (-Z_i)=\displaystyle \frac{1}{\exp (Z_i)}$, we conclude that

$$\begin{aligned} (1,\ldots ,1)\succ (\exp (-Z_1),\ldots ,\exp (-Z_n)), \end{aligned}$$

(34)

i.e., that $-Z\in S_\prec$.

Similarly, we can show that if $Z\in S_\prec$, then $-Z\in S_\succ$.

$8^\circ$. From Part 3 of this proof, it now follows that if $Z=(Z_1,\ldots ,Z_n)\in S_\sim$, then $Z+Z\in S_\sim$, then that $Z+(Z+Z)\in S_\sim$, etc., i.e., that for every positive integer $p$, the tuple

$$\begin{aligned} p\cdot Z=(p\cdot Z_1,\ldots , p\cdot Z_n) \end{aligned}$$

(35)

also belongs to the set $S_\sim$.

By using Part 6 of this proof, we can also conclude that this is true for negative integers $p$ as well. Finally, by taking into account that the zero tuple $0\mathop {=}\limits ^{\mathrm{{def}}}(0,\ldots ,0)$ can be represented as $Z+(-Z)$, we conclude that $0\cdot Z=0$ also belongs to the set $S_\sim$.

Thus, if a tuple $Z$ belongs to the set $S_\sim$, then for every integer $p$, the tuple $p\cdot Z$ also belongs to the set $S_\sim$.

$9^\circ$. Similarly, from Parts 4 and 5 of this proof, it follows that

• if $Z=(Z_1,\ldots ,Z_n)\in S_\succ$, then for every positive integer $p$, the tuple $p\cdot Z$ also belongs to the set $S_\succ$, and

• if $Z=(Z_1,\ldots ,Z_n)\in S_\prec$, then for every positive integer $p$, the tuple $p\cdot Z$ also belongs to the set $S_\prec$.

$10^\circ$. Let us prove that for every rational number $r=\displaystyle \frac{p}{q}$, where $p$ is an integer and $q$ is a positive integer, if a tuple $Z$ belongs to the set $S_\sim$, then the tuple $r\cdot Z$ also belongs to the set $S_\sim$.

Indeed, according to Part 8, $Z\in S_\sim$ implies that $p\cdot Z\in S_\sim$.

According to Part 2, for the tuple $r\cdot Z$, we have either $r\cdot Z\in S_\sim$, or $r\cdot Z\in S_\succ$, or $r\cdot Z\in S_\prec$.

• If $r\cdot Z\in S_\succ$, then, by Part 9, we would get $p\cdot Z=q\cdot (r\cdot Z)\in S_\succ$, which contradicts our result that $p\cdot Z\in S_\sim$.

• Similarly, if $r\cdot Z\in S_\prec$, then, by Part 9, we would get $p\cdot Z=q\cdot (r\cdot Z)\in S_\prec$, which contradicts our result that $p\cdot Z\in S_\sim$.

Thus, the only remaining option is $r\cdot Z\in S_\sim$. The statement is proven.

$11^\circ$. Let us now use continuity to prove that for every real number $x$, if a tuple $Z$ belongs to the set $S_\sim$, then the tuple $x\cdot Z$ also belongs to the set $S_\sim$.

Indeed, a real number $x$ can be represented as a limit of rational numbers: $r^{(k)}\rightarrow x$. According to Part 10, for every $k$, we have $r^{(k)}\cdot Z\in S_\sim$, i.e., the tuple

$$\begin{aligned} Z^{(k)}\mathop {=}\limits ^{\mathrm{{def}}}(\exp (r^{(k)}\cdot Z_1),\ldots ,\exp (r^{(k)}\cdot Z_n))\sim (1,\ldots ,1). \end{aligned}$$

(36)

In particular, this means that $Z^{(k)}\succeq (1,\ldots ,1)$. In the limit,

$$\begin{aligned} Z^{(k)}\rightarrow (\exp (x\cdot Z_1),\ldots ,\exp (x\cdot Z_n))\succeq (1,\ldots ,1). \end{aligned}$$

(37)

By definition of the sets $S_\sim$ and $S_\succ$, this means that $x\cdot Z\in S_\sim$ or $x\cdot Z\in S_\succ$.

Similarly, for $-(x\cdot Z)=(-x)\cdot Z$, we conclude that $-x\cdot Z\in S_\sim$ or

$$\begin{aligned} (-x)\cdot Z\in S_\succ . \end{aligned}$$

(38)

If we had $x\cdot Z\in S_\succ$, then by Part 7 we would get $(-x)\cdot Z\in S_\prec$, a contradiction. Thus, the case $x\cdot Z\in S_\succ$ is impossible, and we have $x\cdot Z\in S_\sim$. The statement is proven.

$12^\circ$. According to Parts 3 and 11, the set $S_\sim$ is closed under addition and under multiplication by an arbitrary real number. Thus, if tuples $Z,\ldots ,Z'$ belong to the set $S_\sim$, their arbitrary linear combination $x\cdot Z+\ldots +x'\cdot Z'$ also belongs to the set $S_\sim$. So, the set $S_\sim$ is a linear subspace of the $n$-dimensional space of all the tuples.

$13^\circ$. The subspace $S_\sim$ cannot coincide with the entire $n$-dimensional space, because then the pre-ordering relation would be trivial. Thus, the dimension of this subspace must be less than or equal to $n-1$. Let us show that the dimension of this subspace is $n-1$.

Indeed, let us assume that the dimension is smaller than $n-1$. Since the pre-ordering is non-trivial, there exist tuples $y=(y_1,\ldots ,y_n)$ and $y'=(y'_1,\ldots ,y'_n)$ for which $y\succ y'$ and thus, $Z=(Z_1,\ldots ,Z_n)\in S_\succ$, where $Z_i=\ln \left( \displaystyle \frac{y_i}{y'_i}\right)$. From $Z\in S_\succ$, we conclude that $-Z\in S_\prec$.

Since the linear space $S_\sim$ is a less than $(n-1)$-dimensional subspace of an $n$-dimensional linear space, there is a path connecting $Z\in S_\succ$ and $-Z\in S_\prec$ which avoids $S_\sim$. In mathematical terms, this path is a continuous mapping $\gamma :[0,1]\rightarrow R^n$ for which $\gamma (0)=Z$ and $\gamma (1)=-Z$. Since this path avoids $S_\sim$, every point $\gamma (t)$ on this path belongs either to $S_\succ$ or to $S_\prec$.

Let $\overline{t}$ denote the supremum (least upper bound) of the set of all the values $t$ for which $\gamma (t)\in S_\succ$. By definition of the supremum, there exists a sequence $t^{(k)}\rightarrow \overline{t}$ for which $\gamma \left( t^{(k)}\right) \in S_\succ$. Similarly to Part 11, we can use continuity to prove that in the limit, $\gamma \left( \overline{t}\right) \in S_\succ$ or $\gamma \left( \overline{t}\right) \in S_\sim$. Since the path avoids the set $S_\sim$, we thus get $\gamma \left( \overline{t}\right) \in S_\succ$.

Similarly, since $\gamma (1)\not \in S_\succ$, there exists a sequence $t^{(k)}\downarrow \overline{t}$ for which $\gamma \left( t^{(k)}\right) \in S_\prec$. We can therefore conclude that in the limit, $\gamma \left( \overline{t}\right) \in S_\succ$ or $\gamma (\overline{t})\in S_\sim$—a contradiction with our previous conclusion that $\gamma \left( \overline{t}\right) \in S_\succ$.

This contradiction shows that the linear space $S_\sim$ cannot have dimension smaller than $n-1$ and thus, that this space have dimension $n-1$.

$14^\circ$. Every $(n-1)$-dimensional linear subspace of an $n$-dimensional superspace separates the superspace into two half-spaces. Let us show that one of these half-spaces is $S_\succ$ and the other is $S_\prec$.

Indeed, if one of the subspaces contains two tuples $Z$ and $Z'$ for which $Z\in S_\succ$ and $Z'\in S_\prec$, then the line segment $\gamma (t)=t\cdot Z+(1-t)\cdot Z'$ containing these two points also belongs to the same subspace, i.e., avoids the set $S_\sim$. Thus, similarly to Part 13, we would get a contradiction.

So, if one point from a half-space belongs to $S_\succ$, all other points from this subspace also belong to the set $S_\succ$. Similarly, if one point from a half-space belongs to $S_\prec$, all other points from this subspace also belong to the set $S_\prec$.

$15^\circ$. Every $(n-1)$-dimensional linear subspace of an $n$-dimensional space has the form

$$\begin{aligned} \alpha _1\cdot Z_1+\ldots +\alpha _n\cdot Z_n=0 \end{aligned}$$

(39)

for some real values $\alpha _i$, and the corresponding half-spaces have the form

$$\begin{aligned} \alpha _1\cdot Z_1+\ldots +\alpha _n\cdot Z_n>0 \end{aligned}$$

(40)

and

$$\begin{aligned} \alpha _1\cdot Z_1+\ldots +\alpha _n\cdot Z_n<0. \end{aligned}$$

(41)

The set $S_\succ$ coincides with one of these subspaces. If it coincides with the set of all tuples $Z$ for which $\alpha _1\cdot Z_1+\ldots +\alpha _n\cdot Z_n<0$, then we can rewrite it as

$$\begin{aligned} (-\alpha _1)\cdot Z_1+\ldots +(-\alpha _n)\cdot Z_n>0, \end{aligned}$$

(42)

i.e., as $\alpha '_1\cdot Z_1+\ldots +\alpha '_n\cdot Z_n>0$ for $\alpha '_i=-\alpha _i$.

Thus, without losing generality, we can conclude that the set $S_\succ$ coincides with the set of all the tuples $Z$ for which $\alpha _1\cdot Z_1+\ldots +\alpha _n\cdot Z_n>0$. We have mentioned that

$$\begin{aligned} y'=(y'_1,\ldots ,y'_n)\succ y=(y_1,\ldots ,y_n) \Leftrightarrow (Z_1,\ldots ,Z_n)\in S_\succ , \end{aligned}$$

(43)

where $Z_i=\ln \left( \displaystyle \frac{y'_i}{y_i}\right)$. So,

$$\begin{aligned} y'\succ y\Leftrightarrow \end{aligned}$$

$$\begin{aligned} \alpha _1\cdot Z_1+\ldots +\alpha _n\cdot Z_n= \alpha _1\cdot \ln \left( \frac{y'_1}{y_1}\right) +\ldots +\alpha _n\cdot \ln \left( \frac{y'_n}{y_n}\right) >0. \end{aligned}$$

(44)

Since $\ln \left( \displaystyle \frac{y'_i}{y_i}\right) =\ln (y'_i)-\ln (y_i)$, the last inequality in the formula (44) is equivalent to

$$\begin{aligned} \alpha _1\cdot \ln (y'_1)+\ldots +\alpha _n\cdot \ln (y'_n)>\alpha _1\cdot \ln (y_1)+\ldots +\alpha _n\cdot \ln (y_n). \end{aligned}$$

(45)

Let us take $\exp$ of both sides of the formula (45); then, due to the monotonicity of the exponential function, we get an equivalent inequality

$$\begin{aligned} \exp (\alpha _1\cdot \ln (y'_1)+\ldots +\alpha _n\cdot \ln (y'_n))>\exp (\alpha _1\cdot \ln (y_1)+\ldots +\alpha _n\cdot \ln (y_n)). \end{aligned}$$

(46)

Here,

$$\begin{aligned} \exp (\alpha _1\cdot \ln (y'_1)+\ldots +\alpha _n\cdot \ln (y'_n))= \exp (\alpha _1\cdot \ln (y'_1))\cdot \ldots \cdot \exp (\alpha _n\cdot \ln (y'_n)), \end{aligned}$$

where for every $i$, $e^{\alpha _i\cdot z_i}=\left( e^{z_i}\right) ^{\alpha _i}$, with $z_i\mathop {=}\limits ^{\mathrm{{def}}}\ln (y'_i)$, implies that

$$\begin{aligned} \exp (\alpha _i\cdot \ln (y'_i))=(\exp (\ln (y'_i)))^{\alpha _i}=(y'_i)^{\alpha _i}, \end{aligned}$$

(47)

so

$$\begin{aligned} \exp (\alpha _1\cdot \ln (y'_1)+\ldots +\alpha _n\cdot \ln (y'_n))= (y'_1)^{\alpha _1}\cdot \ldots \cdot (y'_n)^{\alpha _n} \end{aligned}$$

(48)

and similarly,

$$\begin{aligned} \exp (\alpha _1\cdot \ln (y_1)+\ldots +\alpha _n\cdot \ln (y_n))= y_1^{\alpha _1}\cdot \ldots \cdot y_n^{\alpha _n}. \end{aligned}$$

(49)

Thus, due to (44), (45), (46), (48) and (49), the condition $y'\succ y$ is equivalent to:

$$\begin{aligned} \prod _{i=1}^n y_i^{\alpha _i}>\prod _{i=1}^n (y'_i)^{\alpha _i}. \end{aligned}$$

(50)

Similarly, we prove that

$$\begin{aligned} y_1,\ldots ,y_n)\sim y'=(y'_1,\ldots ,y'_n)\Leftrightarrow \prod _{i=1}^n y_i^{\alpha _i}=\prod _{i=1}^n (y'_i)^{\alpha _i}. \end{aligned}$$

(51)

The condition $\alpha _i>0$ follows from our assumption that the pre-ordering is monotonic.

The theorem is proven.