Estimating age-dependent performance in paired comparisons competitions: application to snooker

Rose D. Baker; Ian G. McHale

doi:10.1515/jqas-2023-0082

Article

Estimating age-dependent performance in paired comparisons competitions: application to snooker

Rose D. Baker and Ian G. McHale

Published/Copyright: February 1, 2024

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From the journal Journal of Quantitative Analysis in Sports Volume 20 Issue 2

Abstract

We first present a model for the outcome of snooker matches in which player strengths are allowed to vary deterministically with time. The results allow us to identify the greatest players of all time, and to examine the relationship between age and performance. Second, we present a random effects model which uses the estimated strengths from our first model, to forecast player performance, and to assess the extent to which early promise has been maintained. Ronnie O’Sullivan and Stephen Hendry are the two candidates for the title of the greatest of all time. We find that peak performance occurs between the ages of 25 and 30, younger than would be expected when compared to findings in other sports. Outside sport, these findings contribute to the general literature on variation of performance with age.

Keywords: paired comparisons; snooker; age-dependence; barycentric interpolation; random-effects model

Corresponding author: Ian G. McHale, University of Liverpool Management School, Liverpool, L69 7ZH, UK, E-mail: ian.mchale@liverpool.ac.uk

Research ethics: Not applicable.
Author contributions: The authors have accepted responsibility for the entire content of this manuscriptand approved its submission.
Competing interests: The authors state no conflict of interest.
Research funding: None declared.
Data availability: The raw data can be obtained on request from the corresponding author.

Appendix

Details of the ageing model

This appendix gives details of model fitting for the individual player ageing model.

The likelihood function

Using equation (4), the likelihood for a player is

(6) L = F ( 2 π ) ( m + 1 ) / 2 ∫ − ∞ ∞ exp ( − ψ / 2 ) d ϵ d η ,

where

(7) F = ( 2 π σ 2 ) − n / 2 λ − 1 ( ϕ ) − m .

With Δ_t = y_t − h_t,

(8) ψ = ∑ t = 1 n Δ t − ∑ j = 1 m r t j ϵ j − η 2 σ 2 + ∑ j = 1 m ϵ j 2 ϕ 2 + η 2 λ 2 .

Writing v = ( ϵ 1 … ϵ m , η ) T , we have

(9) ψ = A − 2 B T v + v T M v ,

where M is symmetric, A = ∑ t = 1 n Δ t 2 / σ 2 and

(10) B i = ∑ t = 1 n Δ t r t i / σ 2 if i ≤ m ∑ t = 1 n Δ t / σ 2 if i = m + 1

Also,

(11) M i j = ∑ t = 1 n r t i r t j / σ 2 + δ i j / ϕ 2 if i , j ≤ m ∑ t = 1 n r t i / σ 2 if i ≤ m , j = m + 1 n / σ 2 + 1 / λ 2 if i = j = m + 1

Model fitting

To fit the model to data by likelihood-based methods, the likelihood function must be integrated over the m + 1 normal random variates ϵ , η.

We can evaluate the integral, allowing for random effects at each node, by completing the square in the exponent. Then

(12) ψ = A − 2 B T v + v T M v = C + ( v − δ ) T M ( v − δ ) ,

from which we read off Mδ = B, C = A − δ ^TB. The vector δ is found by solving the m + 1 linear equations Mδ = B. The distribution of v given y₁ … y_n is multivariate normal with mean δ and covariance matrix M⁻¹.

Maximum likelihood estimators are known to underestimate scale parameters, such as σ, ϕ and λ, but because of the large sample size, this bias will be negligible.

Note that one can think of this random-effects model in Bayesian terms; the normal distribution of the errors would be the prior pdf, and our likelihood would then become the posterior probability. This approach would then be empirical Bayes, based on maximum posterior probability.

The vector of realized random effects, δ , is found by solving the m + 1 linear equations. The method of computation used here is efficient for this example, and differs from methods commonly used, such as iterative generalized least squares. This is often used to estimate the model parameters, using the EM algorithm, but the approach described here is more direct. Doing a Cholesky decomposition M = LL^T, where L is lower-diagonal, the m + 1 linear equations for δ can be solved, and the determinant | M | = ∏ i = 1 m + 1 L i i 2 calculated. There is no need to invert the matrix M. Thus finally

(13) L = F ⁡ exp ( − A / 2 + δ T B / 2 ) / | M | 1 / 2 .

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Received: 2023-09-05

Accepted: 2023-12-22

Published Online: 2024-02-01

Published in Print: 2024-06-25

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Articles in the same Issue

https://doi.org/10.1515/jqas-2023-0082

Keywords for this article

paired comparisons; snooker; age-dependence; barycentric interpolation; random-effects model