Shorter prediction intervals for anonymous individual assessments in group decision-making via pairwise comparisons

Abstract

With reference to pairwise comparisons in group decision-making, consider the problem of how a decision-maker, who knows his/her own assessment and the aggregate assessment of all decision-makers, can predict the assessments of the other decision-makers. We address this problem from a statistical perspective and propose statistical prediction intervals. These are seen to be often significantly shorter and hence appreciably more powerful than their deterministic counterparts available in the literature, accounting for the confidence level as well. In the process, analytical closed form solutions for these deterministic intervals are also obtained. Extensive computations and simulations show that our statistical approach remains very robust to departure from underlying assumptions, frequently being even more efficient. Moreover, this approach is found to be useful generally in predicting an unknown individual component of a given total, thus having much wider applicability beyond the immediate context of pairwise comparisons in group decision-making.

Appendix: Proofs

Proof of Proposition 1. First, note that b_m ≥ $b_{m}^{ - }$, because by (2) and (3), b_m ≥ L and b_m = $(b - \Sigma_{k = 1}^{m - 1} \lambda_{k} b_{k} )/\lambda_{m}$ ≥ $(b - \lambda_{1} b_{1} - U\Sigma_{k = 2}^{m - 1} \lambda_{k} )/\lambda_{m}$ = $\tilde{L}$. To see the existence of a feasible solution attaining this lower bound, first let L > $\tilde{L}$. Then $b_{m}^{ - }$ = L, which is attained when b₂ = … = b_m–1 = $(b - \lambda_{1} b_{1} - \lambda_{m} L)/\Sigma_{k = 2}^{m - 1} \lambda_{k}$, b_m = L. This solution is feasible because it obviously meets (3); moreover, it meets (2) as

$$ \left\{ {(b - \lambda_{1} b_{1} - \lambda_{m} L)/\sum\limits_{k = 2}^{m - 1} {\lambda_{k} } } \right\} - L = \left( {b - \lambda_{1} b_{1} - L\sum\limits_{k = 2}^{m} {\lambda_{k} } } \right)/\sum\limits_{k = 2}^{m - 1} {\lambda_{k} \ge 0,} $$

by (4), while

$$ U - \left\{ {{{(b - \lambda_{1} b_{1} - \lambda_{m} L)} \mathord{\left/ {\vphantom {{(b - \lambda_{1} b_{1} - \lambda_{m} L)} {\sum\limits_{k = 2}^{m - 1} {\lambda_{k} } }}} \right. \kern-\nulldelimiterspace} {\sum\limits_{k = 2}^{m - 1} {\lambda_{k} } }}} \right\} = {{\left( {\lambda_{m} L - b + \lambda_{1} b_{1} + U\sum\limits_{k = 2}^{m - 1} {\lambda_{k} } } \right)} \mathord{\left/ {\vphantom {{\left( {\lambda_{m} L - b + \lambda_{1} b_{1} + U\sum\limits_{k = 2}^{m - 1} {\lambda_{k} } } \right)} {{{\sum\limits_{k = 2}^{m - 1} {\lambda_{k} } = \lambda_{m} (L - \tilde{L})} \mathord{\left/ {\vphantom {{\sum\limits_{k = 2}^{m - 1} {\lambda_{k} } = \lambda_{m} (L - \tilde{L})} {\sum\limits_{k = 2}^{m - 1} {\lambda_{k} > 0.} }}} \right. \kern-\nulldelimiterspace} {\sum\limits_{k = 2}^{m - 1} {\lambda_{k} > 0.} }}}}} \right. \kern-\nulldelimiterspace} {{{\sum\limits_{k = 2}^{m - 1} {\lambda_{k} } = \lambda_{m} (L - \tilde{L})} \mathord{\left/ {\vphantom {{\sum\limits_{k = 2}^{m - 1} {\lambda_{k} } = \lambda_{m} (L - \tilde{L})} {\sum\limits_{k = 2}^{m - 1} {\lambda_{k} > 0.} }}} \right. \kern-\nulldelimiterspace} {\sum\limits_{k = 2}^{m - 1} {\lambda_{k} > 0.} }}}} $$

Next, if $\tilde{L}$ ≥ L, then $b_{m}^{ - }$ = $\tilde{L}$, which is attained when b₂ = … = b_m–1 = U, b_m = $\tilde{L}$. This is again a feasible solution because it obviously meets (3); it also satisfies (2) as $\tilde{L}$ ≥ L, and by (4),

$$ U{-}\tilde{L} = (U\Sigma_{k = 2}^{m} \lambda_{k} - b + \lambda_{1} b_{1} )/\lambda_{m} \ge 0. $$

Thus, the claim about $b_{m}^{ - }$ is proved. The claim about $b_{m}^{ + }$ follows similarly.

Proof of Theorem 1

Let $\tilde{S}$ = $\Sigma_{k = 1}^{n - 1} Y_{k}$. Following Olds (1952), with a more compact notation, the pdfs of $\tilde{S}$ and S are given, respectively, by

$$ g_{n - 1} {{(\tilde{s};c_{1} , \ldots ,c_{n - 1} )} \mathord{\left/ {\vphantom {{(\tilde{s};c_{1} , \ldots ,c_{n - 1} )} {\left\{ {(n - 2)!\prod\limits_{k = 1}^{n - 1} {c_{k} } } \right\}}}} \right. \kern-\nulldelimiterspace} {\left\{ {(n - 2)!\prod\limits_{k = 1}^{n - 1} {c_{k} } } \right\}}},\;0 \, \le \tilde{s} \le \sum\limits_{k = 1}^{n - 1} {c_{k} ,} $$

$$ {\text{and}}\;g_{n} {{(s;c_{1} , \ldots ,c_{n} )} \mathord{\left/ {\vphantom {{(s;c_{1} , \ldots ,c_{n} )} {\left\{ {(n - 1)!\prod\limits_{k = 1}^{n} {c_{k} } } \right\}}}} \right. \kern-\nulldelimiterspace} {\left\{ {(n - 1)!\prod\limits_{k = 1}^{n} {c_{k} } } \right\}}},\;0 \, \le s \le \sum\limits_{k = 1}^{n} {c_{k} .} $$

Because Y_n and $\tilde{S}$ are independent and Y_n is uniform on [0, c_n], their joint pdf equals

$$ g_{n - 1} {{(\tilde{s};c_{1} , \ldots ,c_{n - 1} )} \mathord{\left/ {\vphantom {{(\tilde{s};c_{1} , \ldots ,c_{n - 1} )} {\left\{ {(n - 2)!\prod\limits_{k = 1}^{n} {c_{k} } } \right\}}}} \right. \kern-\nulldelimiterspace} {\left\{ {(n - 2)!\prod\limits_{k = 1}^{n} {c_{k} } } \right\}}},\;0 \le y_{n} \le c_{n} ,\;0 \, \le \tilde{s} \le \sum\limits_{k = 1}^{n - 1} {c_{k} .} $$

Since $\tilde{S}$ = S − Y_n, the transformation (Y_n,$\tilde{S}$) → (Y_n, S) yields the joint pdf of Y_n and S as

$${g_{n-1} (s-y_{n}; c_{1}, \ldots , c_{n-1})} / { \{ (n-2) ! \mathop \Pi \limits _{k=1} ^{n} c_{k}}, \quad y_{n} \in D, \quad 0 \le s \le \mathop \Sigma \limits _{k=1} ^{n} c_{k},$$

recalling (7). Dividing this by the pdf of S as stated above, the result follows.

Proof of Theorem 2

We first present the following lemma.

Lemma A1

Let n ≥ 2. Then (a) $g_{n} (s)$ = $g_{n} (n - s)$, 0 ≤ s ≤ n, (b) $g_{n} (s)$ is increasing in s on [0, n/2] and decreasing in s on [n/2, n].

Proof. (a) This is known in the literature on Irwin–Hall distribution that concerns the sum of independent random variables, each uniform on [0, 1], and can be deduced from the fact that

$$ \Sigma_{j = 0}^{n} ( - 1)^{j} (_{j}^{n} )(s - j)^{n - 1} = 0. $$

(b) This also seems to be known as a folklore. A formal proof is presented here as we could not find one in the literature. By (18), (b) holds for n = 2, for then $g_{2} (s)$ equals s if 0 ≤ s ≤ 1, and 2 − s if 1 ≤ s ≤ 2. To apply induction, suppose (b) holds for n = p (≥ 2), and consider n = p + 1. By (18), $g_{p + 1} (s)$ is differentiable, with derivative

$$ \begin{gathered} g^{\prime}_{p + 1} (s) = p\Sigma_{j = 0}^{p + 1} ( - 1)^{j} (_{\;j}^{p + 1} )(s - j)^{p - 1} \varepsilon (s - j) \hfill \\ = p[s^{p - 1} \varepsilon (s) + \Sigma_{j = 1}^{p} ( - 1)^{j} \{ (_{j}^{p} ) + (_{j - 1}^{\,\;p} )\} (s - j)^{p - 1} \varepsilon (s - j) \hfill \\ + ( - 1)^{p + 1} (s - p - 1)^{p - 1} \varepsilon (s - p - 1)] \hfill \\ = p\{ g_{p} (s) - g_{p} (s - 1)\} , \hfill \\ \end{gathered} $$

(A.1)

using (18) and the fact that

$$ \begin{gathered} \Sigma_{j = 1}^{p} ( - 1)^{j} (_{j - 1}^{\;\,p} )(s - j)^{p - 1} \varepsilon (s - j) + ( - 1)^{p + 1} (s - p - 1)^{p - 1} \varepsilon (s - p - 1) \hfill \\ = \Sigma_{j = 0}^{p} ( - 1)^{j + 1} (_{j}^{p} )(s - 1 - j)^{p - 1} \varepsilon (s - 1 - j) = - g_{p} (s - 1). \hfill \\ \end{gathered} $$

By (18), $g_{p} (s - 1)$ = 0 if $s - 1$ ≤ 0. Hence, if 0 < s ≤ p/2, then by (A.1) and induction hypothesis, $g^{\prime}_{p + 1} (s)$ > 0. Moreover, if p/2 < s < (p + 1)/2, then s − 1 < p − s < p/2, so that by (A.1), part (a) and induction hypothesis,

$$ g^{\prime}_{p + 1} (s) = p\{ g_{p} (p - s) - g_{p} (s - 1)\} > 0. $$

Thus, $g^{\prime}_{p + 1} (s)$ > 0 for 0 < s < (p + 1)/2, i.e., $g_{p + 1} (s)$ is increasing in s on [0, (p + 1)/2]. Hence, by part (a), $g_{p + 1} (s)$ is decreasing in s on [(p + 1)/2, p + 1], and the result follows by induction.

Proof of Theorem 2.

(I) Clearly, s − y_n decreases from s − $Y_{n}^{ - }$ to s − $Y_{n}^{ + }$ as y_n increases from $Y_{n}^{ - }$ to $Y_{n}^{ + }$. From (16), observe that the following hold for n ≥ 3:

• (a) If 0 < s ≤ (n − 1)/2, then $Y_{n}^{ - }$ = 0, s − $Y_{n}^{ - }$ = s ≤ (n − 1)/2 and s − $Y_{n}^{ + }$ = max(s − 1, 0) ≥ 0.

• (b) If (n + 1)/2 ≤ s < n, then $Y_{n}^{ + }$ = 1, s − $Y_{n}^{ - }$ = min(s, n − 1) ≤ n − 1, and s − $Y_{n}^{ + }$ = s − 1 ≥ (n − 1)/2.

• (c) If (n − 1)/2 < s < (n + 1)/2, then $Y_{n}^{ - }$ = 0, $Y_{n}^{ + }$ = 1, s − $Y_{n}^{ - }$ = s > (n − 1)/2, s − $Y_{n}^{ + }$ = s − 1 < (n − 1)/2, and s − y_n equals (n − 1)/2 when y_n equals s − (n − 1)/2 = s*.

The result now follows if one invokes Lemma A1(b) with n there replaced by n − 1, and recalls (17).

(II) This is intuitively expected from (I) but a formal constructive proof will help. For ease in notation, write y for y_n and h(y) for $h(y_{n} |s)$, the cumulative distribution function associated with h(y) being denoted by H(y). As h(y) is continuous and hence bounded on the compact interval D, it follows that H(y) is continuous on D. So, there exist d₁ and d₂ in the interior of D such that H(d₁) = $\alpha$ and H(d₂) = $1 - \alpha$. Under (a) or (b) of (I), take q(s) = h(d₂) or q(s) = h(d₁), for then h(y) ≥ q(s) if and only if y ≤ d₂ or y ≥ d₁, respectively.

Consider now (c) of (I). Then D = [0, 1]. Let $h_{1}^{ - 1} (.)$ and $h_{2}^{ - 1} (.)$ be the piecewise inverses of h(.) on [0, s*] and [s*, 1], respectively. These inverse functions are well defined and continuous on their respective domains, as h(.) is continuous and strictly monotone on [0, s*] and [s*, 1]. Let h_min = min{h(0), h(1)} and h_max = h(s*). For h_min ≤ q ≤ h_max, define

$$ \begin{gathered} y^{(1)} (q) = \left\{ \begin{gathered} \;h_{1}^{ - 1} (q),\;{\text{if}}\;h(0) < q \le h_{\max } \hfill \\ \;0,\quad \quad \,\,\,{\text{if}}\;h_{\min } \le q \le h(0) \hfill \\ \end{gathered} \right., \hfill \\ y^{(2)} (q) = \left\{ \begin{gathered} \;h_{2}^{ - 1} (q),\;{\text{if}}\;h(1) < q \le h_{\max } \hfill \\ \;1,\quad \quad \,\,\;{\text{if}}\;h_{\min } \le q \le h(1) \hfill \\ \end{gathered} \right., \hfill \\ \end{gathered} $$

(A.2)

and for any y $\in$ D, by (c) of (I), note that h(y) ≥ q if and only if $y^{(1)} (q)$ ≤ y ≤ $y^{(2)} (q)$, so that the integral of h(y) over {y: h(y) ≥ q} equals $H(y^{(2)} (q)) - H(y^{(1)} (q))$ = ψ(q), say. Now, by (A.2), ψ(h_min) = 1 and ψ(h_max) = 0. Since ψ(q) is continuous in q on [h_min, h_max] due to the continuity of $h_{1}^{ - 1} (.)$, $h_{2}^{ - 1} (.)$ and H(.), it follows that there exists q(s) in (h_min, h_max) such that ψ(q(s)) = $1 - \alpha$, thus completing the proof.