Scale construction from a decisional viewpoint

Abstract

Many quantitative scales are constructed using cutoffs on a continuum with scores assigned to the cutoffs. This paper develops a framework for using or constructing such scales from a decision-making standpoint. It addresses questions such as: How many distinct thresholds or cutoffs on a scale (i.e., what levels of granularity) are useful for a rational agent? Where should these thresholds be placed given a rational agent’s preferences and risk-orientation? Do scale score assignments have any bearing on decision-making and if so, how should scores be assigned? Given two possible states of nature $\{A, \sim A\}$, an ordered collection of alternatives $\{R_{0}, R_{1},{\ldots}, R_{K}\}$ from which one is to be selected depending on the probability that A is the case, a simple expected utility condition stipulates when adjacent alternatives are distinguishable and determines the threshold odds separating them. Threshold odds and utilities are mapped onto scale scores via a simple distance model. The placement of the thresholds reflects relative concern over decisional consequences given A versus consequences given ∼ A. Likewise, it is shown that scale scores reflect risk-aversion or risk-seeking not only with respect to A versus ∼ A but also with respect to the rank of the R _j . Connections are drawn between this framework and rank-dependent expected utility (RDEU) theory. Implications are adumbrated for both machine and human decision-making.

Appendix

Proof of Theorem 3

Suppose j ≤  (K + 1)/2 and m  ≤  (K + 1)/2.

Then for j < m, an algebraic rearrangement yields

$$ \frac{{B}_{j}}{{B}_{K}}-\frac{{B}_{ j}^\circ}{{B}^\circ}=\frac{{B}_{j} \Delta{B}_{ mm-1}({Q}_{m}+{Q}_{K-m})}{{B}_{ K}{B}_{K}^\circ} $$

and

$$ \frac{{B}_{K-j}}{{B}_{K}}-\frac{{B}^\circ_{ K-j}}{{B}^\circ}=\frac{-{B}_{j} \Delta{B}_{ mm-1}({Q}_{m}+{Q}_{K-m})}{{B}_{ K}{B}_{K}^\circ} $$

so

$$ \left[{\frac{{{B}}_{j}} {{B}_{ K}}}-\frac{{B}^\circ_{j}}{{B}^\circ}\right]+\left[\frac{{B}_{K-j}}{{B}_{K}}- \frac{{B}^\circ_{K-j}} {{B}^\circ}\right]=0. $$

For $m\leq j \leq (K+1)/2$,

$$ \frac{{B}_{{j}}}{{B}_{{ K}}}-\frac{{B}^\circ_{{j}}} {{B}^\circ}-\frac{- {B}_{{K-j}}\Delta {B}_{{mm-1}}{Q}_{ m}+{B}_{{j}} \Delta {B}_{{K-mK-m+1}}{Q}_{{ K-m}}}{{B}_{K} {B}^\circ_{K}} $$

and

$$ \frac{{B}_{{K-j}}}{{B}_{{ K}}}-\frac{{B}^\circ_{{K-j}}} {{B}^\circ}=\frac{- {B}_{{j}}\Delta {B}_{{mm-1}}{Q}_{ m}+{B}_{{K-j}} \Delta {B}_{{ K-mK-m}+1}{Q}_{{K-m}}}{{B}_{K} {B}^\circ_{K}} $$

From the ESROS property we know that $\Delta {B}_{{ K-mK-m}+1}=\Delta {B}_{{mm}-1}$, so

$$\left[{\frac{{B}_{{j}}}{{B}_{{K}}} -\frac{{B}^\circ_{{ j}}}{{B}^\circ}}\right]+\left[{\frac{{B}_{{ K-j}}}{{{B}_{{K}}}}-\frac{{B}^\circ_{{ K-j}}}{{B}^\circ}}\right]=\frac{\Delta {B}_{{ mm}-1}({B}_{{j}}+{B}_{{K-j}})({Q}_{{ K-m}}-{Q}_{{m}})}{{B}_{{K}}{B}^\circ_{{ K}}}$$

From the Cases I and II assumptions it immediately follows that this quantity is nonnegative for +q and nonpositive for  − q.

Therefore,

${w}^\circ_{{m-1mj}} > ( < ){w}_{{m-1m}}$

or

${w}^\circ_{{K-mK-m}+1} > ( < )\,{w}_{{K-mK-m}+1}$ implies $\Sigma\mu_{{j}} < <$> <$>( > )\Sigma\mu^\circ_{{j}}$. □

Proof of Corollary 3

Assume an ESROS and choose some $m\leq (K+1)/2$.

Consider the BROS family defined by

$$\Delta {H}_{{jj}-1}=({w}_{{K-jK-j}+1}) ^{({a}+1)/2}({w}_{{j}-1{j}})^{{ a}/2} \hbox{ and }\Delta {G}_{{j}-1{j}}= ({w}_{{j}-1{j}})^{1 + {a}/2}({w}_{{ K-jK-j}+1})^{({a}+1)/2}.$$

We consider the two cases:

• I.
  $${w}^\circ_{{m}-1{m}}={w}_{{m}-1{ m}}\pm {q \hbox{ where } w}^\circ_{{m}-1{m}}{w}_{{ m}-1{m}} \leq1,$$

  and

• II.
  $${w}^\circ_{{K-mK-m}+1}={w}_{{K-mK-m}+1} \pm {q \hbox{ where } w}^\circ_{{K-mK-m}+1}{w}_{{K-mK-m}+1} > 1.$$

Case I

We have

$$\Delta {H}^\circ_{{mm}-1}=({w}_{{K-mK-m}+1}) ^{({a}+1)/2}({w}_{{m}-1{m}}\pm {q})^{{ a}/2}=\Delta {H}_{{mm}-1}[({w}_{{m}-1{m}} \pm {q})/{w}_{{m}-1{m}}]^{{ a}/2}$$

and

$$\eqalign{\Delta {H}^\circ_{{ K-mK-m}+1} =({w}_{{m}-1{m}} \pm {q})^{({ a}+1)/2}({w}_{{K-mK-m}+1})^{{a}/2}\cr =\Delta {H}_{{K-mK-m}+1}[({w}_{{m}-1{m}}\pm {q})/{w}_{{m}-1{m}}]^{({a}+1)/2}.}$$

Therefore,

$$\eqalign{\Delta {B}^\circ_{{mm}-1} =\Delta {H}^\circ_{{mm}-1} [1+({w}_{{m}-1{m}}\pm {q})^{2}]^{1/2}\cr =\Delta {B}_{{mm}-1}[({w}_{{m}-1{m}}\pm {q})/{w}_{{m}-1{m}}]^{{ a}/2}\{[1+({w}_{{m}-1{m}}\pm {q})^{2}]/[1 + {w}^{2}_{{m}-1{m}}]\}^{1/2}}$$

and

$$\eqalign{\Delta {B}^\circ_{{K-mK-m}+1} =\Delta {H}^\circ_{{ K-mK-m}+1}[1 + ({w}_{{K-mK-m}+1})^{2}]^{1/2}\cr =\Delta {B}_{{K-mK-m}+1}[({w}_{{m}-1{m}}\pm {q})/{w}_{{m}-1{m}}]^{({a}+1)/2}.}$$

Now, let

$${Q}_{{m}}=[({w}_{{m}-1{m}}\pm {q})/ {w}_{{m}-1{m}}]^{{a}/2}\{[1 + ({w}_{{ m}-1{m}}\pm {q})^{2}]/[1 +{w}^{2}_{{m}-1{ m}}]\}^{1/2}-1$$

and

$${Q}_{{ K-m}}=[({w}_{{m}-1{m}}\pm {q})/{w}_{{ m}-1{m}}]^{({a}+1)/2}- 1.$$

A straightforward algebraic argument shows that ${w}^\circ_{{m}-1{m}}{w}_{{m}-1{m}} \leq1$ implies ${Q}_{{K-m}}-{Q}_{{m}}\geq(\leq)0$ if  + (−)q. The argument is as follows:

Assume  + q. Then

$$\eqalign{{w}^\circ_{{m}-1{m}}{w}_{{m}-1{m}} \leq1 &\Leftrightarrow {w}_{{m}-1{m}} + {q} + ({w}_{{m}-1{m}} + {q})({w}_{{m}-1{ m}})^{2} > {w}_{{m}-1{m}} + ({w}_{{m}-1{m}} + {q})^{2}\cr &\Leftrightarrow({w}_{{m}-1{m}} + {q})/{w}_{{m}-1{m}} > [1 + ({w}_{{m}-1{ m}} + {q})^{2}]/[1 + {w}^{2}_{{m}-1{m}}] \cr &\Leftrightarrow {Q}_{{K-m}}-{Q}_{{m}}\geq0. }<!endaligned> $$

A similar argument starting with  − q yields ${Q}_{{K-m}}- {Q}_{{m}}\leq 0$.

Case II

We have

$$\eqalign{\Delta {H}^\circ_{{mm}-1} =({w}_{{K-mK-m}+1} \pm {q})^{({a}+1)/2}({w}_{{m}-1{m}})^{{ a}/2}\cr =\Delta {H}_{{mm}-1}[({w}_{{K-mK-m}+1}\pm {q})/{w}_{{K-mK-m}+1}]^{({ a}+1)/2}}$$

and

$$\eqalign{ \Delta {H}^\circ_{{K-mK-m}+1} = ({w}_{{m}-1{m}})^{({a}+1)/2}({w}_{{ K-mK-m}+1}\pm {q})^{{a}/2}\cr =\Delta {H}_{{ K-mK-m}+1}[({w}_{{K-mK-m}+1}\pm {q})/{w}_{{ m}-1{m}}]^{{a}/2}.}$$

Therefore,

$$\eqalign{\Delta {B}^\circ_{{mm}-1} =\Delta {H}^\circ_{{mm}-1} [1+({w}_{{m}-1{m}}\pm {q})^{2}]^{1/2}\cr =\Delta {B}_{{mm}-1}[({w}_{{K-mK-m}+1}\pm {q})/{w}_{{K-mK-m}+1}]^{({a}+1)/2}}$$

and

$$ \eqalign{ \Delta {B}^\circ_{{K-mK-m}+1}&=\Delta {H}^\circ_{{K-mK-m}+1}[1+({w}_{{K-mK-m}+1}\pm {q})^{2}]^{1/2}\cr &=\Delta {B}_{{ K-mK-m}+1}[({w}_{{K-mK-m}+1}\pm {q})/{w}_{{ K-mK-m}+1}]^{{a}/2}\cr \quad\{[1 + ({w}_{{K-mK-m}+1}\pm {q})^{2}]/[1 + {w}^{2}_{{K-mK-m}+1}]\}^{1/2}. }<!endaligned> $$

Now, let

$${Q}_{{m}}=[({w}_{{K-mK-m}+1}\pm {q})/ {w}_{{K-mK-m}+1}]^{({a}+1)/2}-1$$

and

$$ \eqalign{{Q}_{{K-m}}= [({w}_{{K-mK-m}+1}\pm {q})/{w}_{{K-mK-m}+1}]^{{a}/2}\cr \{[1 + ({w}_{{K-mK-m}+1}\pm {q})^{2}]/[1 + {w}^{2}_{{ K-mK-m}+1}]\}^{1/2}-1.}$$

A straightforward algebraic argument similar to that in Case I shows that

$${w}^\circ_{{K-mK-m}+1}{w}_{{K-mK-m}+1} > 1 \hbox{ implies }{Q}_{{K-m}}-{Q}_{{m}} > ( < ) 0\quad\hbox{if} +(-){q}.$$

Thus, the BROS defined in Eq. 24 satisfied the requirements of Theorem 3 and is L-congruent. □