Relative performance evaluation and peer-performance summarization errors

Abstract

In tests of the relative performance evaluation (RPE) hypothesis, empiricists rarely aggregate peer performance in the same way as a firm’s board of directors. Framed as a standard errors-in-variables problem, a commonly held view is that such aggregation errors attenuate the regression coefficient on systematic firm performance towards zero, which creates a bias in favor of the strong-form RPE hypothesis. In contrast, we analytically demonstrate that aggregation differences generate more complicated summarization errors, which create a bias against finding support for strong-form RPE (potentially inducing a Type-II error). Using simulation methods, we demonstrate the sensitivity of empirical inferences to the bias by showing how an empiricist can conclude erroneously that boards, on average, do not apply RPE, simply by selecting more, fewer, or different peers than the board does. We also show that when the board does not apply RPE, empiricists will not find support for RPE (that is, precluding a Type-I error).

Appendix: Proofs

Proof of Proposition 1

As outlined, we relax the individual rationality constraint, CE _i  = 0, and solve for each firm i the relaxed problem (8). Note that the relaxed problem is well behaved. Using the Lagrange-parameter λ _i for the incentive compatibility constraint, (4), the Lagrangian for firm i is given by

$$ L_{i} = a_{i} - \kappa_{i} (a_{i} ) - \frac{{r_{i} }}{2}(v_{ii}^{2} {\text{Var}}[y_{i} ] + 2v_{ii} v_{ip} {\text{Cov}}[y_{i} ,y_{ip} (\varvec{\delta}_{i} )] + v_{ip}^{2} {\text{Var}}[y_{ip} (\varvec{\delta}_{i} )] - \lambda_{i} (\kappa_{i}^{\prime } (a_{i} ) - v_{ii} ) $$

(18)

where Var[y _i ] = σ ² _i  + μ ² _i σ ² _m , ${\text{Cov}}[y_{i} ,y_{ip} (\varvec{\delta}_{i} )] = \mu_{i} \sum\nolimits_{j = 1}^{n} {\delta_{ij} \mu_{j} \sigma_{m}^{2} }$, and ${\text{Var}}[y_{ip} (\varvec{\delta}_{i} )] = \sum\nolimits_{j = 1}^{n} {\delta_{ij}^{2} (\sigma_{j}^{2} + \mu_{j}^{2} \sigma_{m}^{2} )}$.

From (18), the Karush–Kuhn–Tucker conditions are given by

$$ \partial L_{i} /\partial a_{i} = 1 - \kappa_{i}^{\prime } (a_{i} ) - \lambda_{i} \kappa_{i}^{\prime \prime } (a_{i} ) = 0, $$

(19)

$$ \partial L_{i} /\partial v_{ii} = - r_{i} (v_{ii} {\text{Var}}[y_{i} ] + v_{ip} {\text{Cov}}[y_{i} ,y_{ip} (\varvec{\delta}_{i} )]) + \lambda_{i} = 0, $$

(20)

$$ \partial L_{i} /\partial v_{ip} = - r_{i} (v_{ii} {\text{Cov}}[y_{i} ,y_{ip} (\varvec{\delta}_{i} )] + v_{ip} {\text{Var}}[y_{ip} (\varvec{\delta}_{i} )]) = 0, $$

(21)

$$ \partial L_{i} /\partial \lambda_{i} = \kappa_{i}^{\prime } (a_{i} ) - v_{ii} = 0, $$

(22)

and for k = 1,…,n, k ≠ i,

$$ \begin{aligned} \frac{{\partial L_{i} }}{{\partial \delta_{ik} }} = & - \frac{{r_{i} }}{2}\left( {2v_{ii} v_{ip} \frac{{\partial {\text{Cov}}[y_{i} ,y_{ip} ({\varvec{\delta}}_{i} )]}}{{\partial \delta_{ik} }} + v_{ip}^{2} \frac{{\partial {\text{Var}}[y_{ip} ({\varvec{\delta}}_{i} )]}}{{\partial \delta_{ik} }}} \right) \\ = & - r_{i} v_{ip} (v_{ii} \mu_{i} \mu_{k} \sigma_{m}^{2} + v_{ip} \delta_{ik} (\sigma_{k}^{2} + \mu_{k}^{2} \sigma_{m}^{2} )) = 0. \\ \end{aligned} $$

(23)

Normalizing, $\sum\nolimits_{k = 1}^{n} {\varvec{\delta}_{ik} = 1}$, and solving (19)–(23) yields (9a), (9b), and (9c). The rest of the proof follows as outlined in the text. □

Proof of Corollary 1

The partial correlation, $\rho_{ij{\cdot}\dag }$, is defined by (Kendall 1991)

$$ \rho_{ij{\cdot}\dag } = \frac{{\rho_{ij} - \rho_{i\dag }\rho_{j\dag } }}{{\sqrt {(1 - \rho_{i\dag }^{2} )(1 - \rho_{j\dag}^{2} )} }}, $$

with $\rho_{ij} = {\text{Corr}}\left[ {y_{i} ,y_{j} } \right]$; $\rho_{i\dag } = {\text{Corr}}[y_{i} ,y_{ip} (\varvec{\delta}_{i}^{\dag } )];$ and $\rho_{j\dag } = {\text{Corr}}[y_{j} ,y_{jp} (\varvec{\delta}_{i}^{\dag } )]$ denoting the associated correlation coefficients. From (22) and (9b), we get for the risk-minimized aggregation, $\varvec{\delta}_{i}^{\dag }$:

$$ 2\frac{{\partial {\text{Cov}}[y_{i} ,y_{ip}(\varvec{\delta}_{i}^{\dag } )]}}{{\partial {{\updelta}}_{ij} }} =- \frac{{v_{ip}^{\dag } }}{{v_{ii}^{\dag } }}\frac{{\partial{\text{Var}}[y_{ip} (\varvec{\delta}_{i}^{\dag } )]}}{{\partial{{\updelta}}_{ij} }} = \frac{{{\text{Cov}}[y_{i} ,y_{ip}(\varvec{\delta}_{i}^{\dag } )]}}{{{\text{Var[}}y_{ip}(\varvec{\delta}_{i}^{\dag } )]}}\frac{{\partial{\text{Var}}[y_{ip} (\varvec{\delta}_{i}^{\dag } )]}}{{\partial{{\updelta}}_{ij} }}\quad {\text{for}}\,{\text{all}}\quad j \ne i.$$

(24)

Using $\frac{{\partial {\text{Cov}}[ {y_{i} ,\sum\nolimits_{j = 1}^{n} {\delta_{ij} y_{j} } }]}}{{\partial \delta_{ij} }} = {\text{Cov}}[y_{i} ,y_{j} ]$ and $\frac{{\partial {\text{Var}}[y_{ip} (\varvec{\delta}_{i} )]}}{{\partial \delta_{ij} }} = \frac{{\partial {\text{Cov}}\left[ {\sum\nolimits_{k = 1}^{n} {\delta_{ik} y_{k} }, \sum\nolimits_{k = 1}^{n} {\delta_{ik} y_{k} } } \right]}}{{\partial \delta_{ij} }} = 2{\text{Cov}}\left[ {\sum\nolimits_{k = 1}^{n} {\delta_{ik} y_{k} ,y_{j} } } \right] = 2{\text{Cov}}[y_{ip} (\varvec{\delta}_{i} ),y_{j} ]$, we can rewrite (24) as

$$ {\text{Cov}}[y_{i} ,y_{j} ] = \frac{{{\text{Cov}}[y_{i} ,y_{ip} (\varvec{\delta}_{i}^{\dag } )]{\text{Cov}}[y_{ip} (\varvec{\delta}_{i}^{\dag } ),y_{j} ]}}{{{\text{Var}}[y_{ip} (\varvec{\delta}_{i}^{\dag } )]}} = \rho_{i\dag } \rho_{j\dag } {\text{SD}}[y_{i} ]{\text{SD}}[y_{j} ]\quad {\text{for}}\,{\text{all}}\quad j \ne i. $$

(25)

Noting Cov[y _i , y _j ] = ρ _ij SD[y _i ]SD[y _j ], (25) implies ρ _ij  = ρ _i† ρ _j† for all j ≠ i and, given the above definition of ρ _ij·†, then it follows that ρ _ij·† = 0, for all j ≠ i. □

Proof of Lemma 1

Due to the orthogonal regression in (12a) and (12b), the variables $y_{u} (\varvec{\delta}^{e} )$ and $y_{s} (\varvec{\delta}^{e} )$ are not correlated. The regression coefficient in (15) is given by

$$ \hat{\beta }_{s} (\varvec{\delta}^{b} ,\varvec{\delta}^{e} ) = {\text{Cov}}[w(\varvec{\delta}^{b} ),y_{s} (\varvec{\delta}^{e} )]/{\text{Var}}[y_{s} (\varvec{\delta}^{e} )]. $$

(26)

Substituting $y_{s} (\varvec{\delta}^{e} ) = ({\text{Cov}}[y_{f} ,y_{p} (\varvec{\delta}^{e} )]/{\text{Var}}[y_{p} (\varvec{\delta}^{e} )])y_{p} (\varvec{\delta}^{e} )$ as per (12a), $w (\varvec{\delta}^{b} ) = f^{b} + v_{f}^{b} y_{f} + v_{p}^{b} y_{p} (\varvec{\delta}^{b} ),$ and the ratio $v_{p}^{b} /v_{f}^{b} = - {\text{Cov}}[y_{f} ,y_{p} (\varvec{\delta}^{b} )]/{\text{Var}}[y_{p} (\varvec{\delta}^{b} )]$ into (26), and simplifying yields

$$ \hat{\beta }_{s} (\varvec{\delta}^{b} ,\varvec{\delta}^{e} ) = \left( {1 - \frac{{{\text{Cov}}[y_{f} ,y_{p} (\varvec{\delta}^{b} )]{\text{Cov}}[y_{p} (\varvec{\delta}^{e} ),y_{p} (\varvec{\delta}^{b} )]}}{{{\text{Cov}}[y_{f} ,y_{p} (\varvec{\delta}^{e} )]{\text{Var}}[y_{p} (\varvec{\delta}^{b} )]}}} \right)v_{f}^{b} . $$

(27)

Using the correlation coefficients, ρ _fb , ρ _be , and ρ _fe , and simplifying (27) results in (15). Finally, using the partial correlation coefficient, ρ _fe·b  = (ρ _fe  − ρ _fb ρ _be )[(1 − ρ ² _fb )(1 − ρ ² _be )]^−1/2, yields the right hand side of (15). □

Proof of Proposition 2

Given risk-minimized aggregated peer performance, $y_{p}^{b} = y_{p} (\varvec{\delta}^{\dag } )$, following Corollary 1, the performance of the focal firm, y _f , is conditionally independent from the performance of any peer firm, y _j , that is, ρ _fj  − ρ _fb ρ _jb  = 0 for any peer firm j ≠ f. Accordingly, the performance of the focal firm is also conditionally independent from any aggregation of peer performance, implying that

$$ \rho_{fe} - \rho_{fb} \rho_{be} = 0,\quad {\text{for}}\,{\text{any}}\,{\text{aggregation}}\,\varvec{\delta}^{e} \ne 0. $$

(28)

Substituting (28) in (15) completes the proof. □

Proof of Proposition 3

From Lemma 1, $\hat{\beta }_{s} (\varvec{\delta}^{b} ,\varvec{\delta}^{e} ) = 0$ if and only if ρ _fe  = ρ _fb ρ _be . As shown in (25), ρ _fe  = ρ _fb ρ _be can be equivalently restated as

$$ {\text{Cov}}[y_{f} ,y_{p} (\varvec{\delta}^{e} )]/{\text{Cov}}[y_{f} ,y_{p} (\varvec{\delta}^{b} )] = {\text{Cov}}[y_{p} (\varvec{\delta}^{e} ),y_{p} (\varvec{\delta}^{b} )]/{\text{Var}}[y_{p} (\varvec{\delta}^{b} )]. $$

(29)

The left side of (29) does not depend on firm specific risk, since ${\text{Cov}}[y_{f} ,y_{p} (\varvec{\delta}^{e} )] = \mu_{f} \sum\nolimits_{j} {} \delta_{j}^{e} \mu_{j} \sigma_{m}^{2} ,\,{\text{Cov}}[y_{f} ,y_{p} (\varvec{\delta}^{b} )] = \mu_{f} \sum\nolimits_{j} {} \delta_{j}^{b} \mu_{j} \sigma_{m}^{2} ,$ and thus ${\text{Cov}}[y_{f} ,y_{p} (\varvec{\delta}^{e} )]/{\text{Cov}}[y_{f} ,y_{p} (\varvec{\delta}^{b} )] = (\sum\nolimits_{j} {} \delta_{j}^{e} \mu_{j} )/(\sum\nolimits_{j} {} \delta_{j}^{b} \mu_{j} ).$

The right side of (29), ${\text{Cov}}[y_{p} (\varvec{\delta}^{e} ),y_{p} (\varvec{\delta}^{b} )]/{\text{Var}}[y_{p} (\varvec{\delta}^{b} )]$, is the regression coefficient when regressing the empiricist’s peer-group performance, $y_{p} (\varvec{\delta}^{e} )$, on the board’s peer-group performance, $y_{p} (\varvec{\delta}^{b} )$. In general, this regression coefficient depends on each peer j’s firm-specific risk, σ ² _j , that is,

$$ {\text{Cov}}[y_{p} (\varvec{\delta}^{e} ),y_{p} (\varvec{\delta}^{b} )] = \sum\limits_{j} {\delta_{j}^{e} \delta_{j}^{b} \sigma_{j}^{2} } + \left( {\sum\limits_{j} {\delta_{j}^{e} \mu_{j} } } \right)\left( {\sum\limits_{j} {\delta_{j}^{b} \mu_{j} } } \right)\sigma_{m}^{2} ,\,{\text{and}} $$

(30)

$$ {\text{Var}}[y_{p} (\varvec{\delta}^{b} )] = \sum\limits_{j} {(\delta_{j}^{b} )^{2} \sigma_{j}^{2} } + \left( {\sum\limits_{j} {\delta_{j}^{b} \mu_{j} } } \right)^{2} \sigma_{m}^{2} . $$

(31)

Since the left side of (29) is independent of firm-specific risk, the condition $\hat{\beta }_{s} (\varvec{\delta}^{b} ,\varvec{\delta}^{e} ) = 0$ is only satisfied if the right side of (29) is not affected by the peers’ firm-specific risk. This result, however, requires that either the board’s peer-group performance is a sufficient statistic for the empiricist’s peer-group performance, implying $\varvec{\delta}^{b} =\varvec{\delta}^{\dag }$, or that the board’s peer-group performance and the empiricist’s peer-group performance coincide, that is, $\varvec{\delta}^{b} =\varvec{\delta}^{e} .$ □

Proof of Proposition 4

Using (16), the regression coefficient is characterized by

$$ \hat{\beta }_{s} (0,\varvec{\delta}^{e} ) = {\text{Cov}}[w(0),y_{s} (\varvec{\delta}^{e} )]/{\text{Var}}[y_{s} (\varvec{\delta}^{e} )]. $$

(32)

Substituting into (32) the agent’s compensation, (16), plus the definition for $y_{s} (\varvec{\delta}^{e} )$, (12a), and simplifying yields

$$ \hat{\beta }_{s} (0,\varvec{\delta}^{e} ) =\frac{{{\text{Cov}}\left[ {v_{f}^{0} y_{f},\frac{{{\text{Cov}}[y_{f} ,y_{p} (\varvec{\delta}^{e})]}}{{{\text{Var}}[y_{p} (\varvec{\delta}^{e} )]}}y_{p}(\varvec{\delta}^{e} )} \right]}}{{{\text{Var}}\left[{\frac{{{\text{Cov}}[y_{f} ,y_{p} (\varvec{\delta}^{e})]}}{{{\text{Var}}[y_{p} (\varvec{\delta}^{e} )]}}y_{p}(\varvec{\delta}^{e} )} \right]}} = \frac{{{\frac{{{\text{Cov}}[y_{f} ,y_{p} (\varvec{\delta}^{e})]}}{{{\text{Var}}[y_{p} (\varvec{\delta}^{e} )]}}} ^{2} }}{{{\frac{{{\text{Cov}}[y_{f} ,y_{p} (\varvec{\delta}^{e})]}}{{{\text{Var}}[y_{p} (\varvec{\delta}^{e} )]}}} ^{2}}}v_{f}^{0} = v_{f}^{0} \ne 0. $$

□

Proof of Proposition 5

From Lemma 1, we get

$$ \begin{aligned} \hat{\beta }_{s} (\varvec{\delta}^{b} ,\varvec{\delta}^{e} ) = & \left( {1 - \frac{{\rho_{fb} \rho_{be} }}{{\rho_{fe} }}} \right)v_{f}^{b} = \left( {1 - \frac{{{\text{Cov}}[y_{f} ,y_{p} (\varvec{\delta}^{b} )]}}{{{\text{Cov}}[y_{f} ,y_{p} (\varvec{\delta}^{e} )]}}\frac{{{\text{Cov}}[y_{p} (\varvec{\delta}^{b} ),y_{p} (\varvec{\delta}^{e} )]}}{{{\text{Var}}[y_{p} (\varvec{\delta}^{b} )]}}} \right)v_{f}^{b} \\ = & \left( {1 - \frac{{\mu_{f} \mu_{b} \sigma_{m}^{2} }}{{\mu_{f} \mu_{e} \sigma_{m}^{2} }}\frac{{{\text{Cov}}[y_{p} ({{\updelta}}^{b} ),y_{p} ({{\updelta}}^{e} )]}}{{\mu_{b}^{2} \sigma_{m}^{2} + \sigma_{b}^{2} }}} \right)v_{f}^{b} . \\ \end{aligned} $$

Noting that in case (i):

$$ {\text{Cov}}[y_{p} (\varvec{\delta}^{b} ),y_{p} (\varvec{\delta}^{e} )] = \mu_{b} \mu_{e} \sigma_{m}^{2} + {\text{Cov}}\left[ {\sum\limits_{{j \in I_{b} }} {\delta_{j}^{e} \varepsilon_{j} } ,\sum\limits_{{j \in I_{b} }} {\delta_{j}^{b} \varepsilon_{j} } } \right] = \mu_{b} \mu_{e} \sigma_{m}^{2} + \sum\limits_{{j \in I_{b} }} {\delta_{j}^{e} \delta_{j}^{b} \sigma_{j}^{2} } , $$

in Case (ii), ${\text{Cov}}[y_{p} (\varvec{\delta}^{b} ),y_{p} (\varvec{\delta}^{e} )] = \mu_{b} \mu_{e} \sigma_{m}^{2} + {\text{Cov[}}\sum\nolimits_{{j \in I_{e} }} {\delta_{j}^{e} \varepsilon_{j} } ,\sum\nolimits_{{j \in I_{b} }} {\delta_{j}^{b} \varepsilon_{j} } ]= \mu_{b} \mu_{e} \sigma_{m}^{2} + \sum\nolimits_{{j \in I_{e} }} {\delta_{j}^{e} \delta_{j}^{b} \sigma_{j}^{2} } ,$ and in Case (iii), ${\text{Cov}}[y_{p} (\varvec{\delta}^{b} ),y_{p} (\varvec{\delta}^{e} )] = \mu_{b} \mu_{e} \sigma_{m}^{2} ,$ proves our result. □

Proof of Proposition 6

Using (12b), (13), and (14), we get:

$$ \begin{aligned} \hat{\beta }_{u} (\varvec{\delta}^{b} ,\varvec{\delta}^{e} ) = & {\text{Cov}}[w(\varvec{\delta}^{b} ),y_{u} (\varvec{\delta}^{e} )]/{\text{Var}}[y_{u} (\varvec{\delta}^{e} )] \\ = & {\text{Cov}}\left[ {v_{f}^{b} \left( {y_{f} + \frac{{v_{p}^{b} }}{{v_{f}^{b} }}y_{p} (\varvec{\delta}^{b} )} \right),y_{f} - \frac{{{\text{Cov}}[y_{f} ,y_{p} (\varvec{\delta}^{b} )]y_{p} (\varvec{\delta}^{e} )}}{{{\text{Var}}[y_{p} (\varvec{\delta}^{b} )]}}} \right]/{\text{Var}}\left[ {y_{f} - \frac{{{\text{Cov}}[y_{f} ,y_{p} (\varvec{\delta}^{e} )]y_{p} (\varvec{\delta}^{e} )}}{{{\text{Var}}[y_{p} (\varvec{\delta}^{b} )]}}} \right] \\ = & \frac{{1 - \rho_{fe}^{2} - \rho_{fb}^{2} - \rho_{fb} \rho_{fe} \rho_{be} }}{{1 - \rho_{fe}^{2} }}v_{f}^{b} . \\ \end{aligned} $$

Using $\hat{\beta }_{s} (\varvec{\delta}^{b} ,\varvec{\delta}^{e} ) = [1 - (\rho_{fb} \rho_{be} /\rho_{fe} )]v_{f}^{b}$ from Lemma 1 and rearranging yields (17), which is zero when ρ _be·f  = 0. □

Proof of Proposition 7

Using (16), the regression coefficient is characterized by

$$ \begin{aligned} \hat{\beta }_{u} (0,\varvec{\delta}^{e} ) = & {\text{Cov}}[w(0),y_{u} (\varvec{\delta}^{e} )]/{\text{Var}}[y_{u} (\varvec{\delta}^{e} )] \\ = & {\text{Cov}}\left[ {v_{f}^{0} y_{f} ,y_{f} - \frac{{{\text{Cov}}[y_{f} ,y_{p} (\varvec{\delta}^{e} )]y_{p} (\varvec{\delta}^{e} )}}{{{\text{Var}}[y_{p} (\varvec{\delta}^{e} )]}}} \right]/{\text{Var}}\left[ {y_{f} - \frac{{{\text{Cov}}[y_{f} ,y_{p} (\varvec{\delta}^{e} )]y_{p} (\varvec{\delta}^{e} )}}{{{\text{Var}}[y_{p} (\varvec{\delta}^{e} )]}}} \right] \\ = & v_{f}^{0} \left( {{\text{Var}}[y_{f} ] - \frac{{{\text{Cov}}[y_{f} ,y_{p} (\varvec{\delta}^{e} )]^{2} }}{{{\text{Var}}[y_{p} (\varvec{\delta}^{e} )]}}} \right)/\left( {{\text{Var}}[y_{f} ] - \frac{{{\text{Cov}}[y_{f} ,y_{p} (\varvec{\delta}^{e} )]^{2} }}{{{\text{Var}}[y_{p} (\varvec{\delta}^{e} )]}}} \right) = v_{f}^{0} . \\ \end{aligned} $$

Recalling from Proposition 4, $\hat{\beta }_{s} (0,\varvec{\delta}^{e} ) = v_{f}^{0} ,$ yields $\hat{\beta }_{u} (0,\varvec{\delta}^{e} ) - \hat{\beta }_{s} (0,\varvec{\delta}^{e} ) = 0$ for any aggregation $\varvec{\delta}^{e} \ne 0.$ □