An MCMC based Bayesian inference approach to parameter estimation of distributed lag models for forecasting used product returns for remanufacturing

Abstract

Forecasting the quantity and timing of used product returns is one of the major challenges faced by remanufacturers. Distributed lag model has been proposed in recent years to forecast used product returns based on past sales data. However, the forecasting accuracy of a distributed lag model is affected to a large extent by the estimation of the parameters of the lag function. The Bayesian inference approach for parameter estimation which has been proposed by previous studies requires solving the marginal likelihood function which is often difficult even for a slightly complex lag function. In this research, a Markov Chain Monte Carlo (MCMC) based Bayesian inference approach is proposed to estimate the parameters of a lag function which provides an efficient way to sample parameter values from a posterior distribution regardless of the complexity of a lag function. An example case study of forecasting used product returns was undertaken to illustrate the proposed parameter estimation approach which was validated by comparing it with the maximum likelihood estimate (MLE) method. The validation results show that the forecasting accuracy of the number of used product returns based on the parameters estimated by using the proposed MCMC based Bayesian approach is better than that estimated by using the MLE method in terms of mean absolute percent errors and variance of errors.

APPENDIX: Derivation of covariance matrix of the error vector of a DLM

In this section, the derivation of covariances of the error vector, u ∈ (u₃, u₄, u₅, …u_T)′ where u_t = ε_t − 2(1 − q)ε_t − 1 + (1 − q)²ε_t − 2 for t = 3,4,5 … T is presented. We assume the error terms, i.e. ε_t, ε_t − 1, ε_t − 1 are normally distributed with zero mean and σ² variance. Hence, the covariance between any two like terms is equivalent to its variance; i.e. Cov(ε_t, ε_t) = Cov(ε_t − 1, ε_t − 1) = Cov(ε_t − 1, ε_t − 1)= σ² whereas covariance between unlike terms is equivalent to zero.

The covariances of the error vector denoted by Σ_u is represented by a (T − 2) × (T − 2) matrix as shown below.

$$ {\sum}_u=\left[\begin{array}{cccccc}\mathrm{Cov}\left({u}_t,{u}_t\right)& \mathrm{Cov}\left({u}_t,{u}_{t+1}\right)& \mathrm{Cov}\left({u}_t,{u}_{t+2}\right)& \cdots & \cdots & \mathrm{Cov}\left({u}_{\mathrm{t}},{u}_T\right)\\ {}\mathrm{Cov}\left({u}_{t+1},{u}_t\right)& \mathrm{Cov}\left({u}_{t+1},{u}_{t+1}\right)& \mathrm{Cov}\left({u}_{t+2},{u}_{t+3}\right)& \cdots & \cdots & \mathrm{Cov}\left({u}_{t+1},{u}_T\right)\\ {}\mathrm{Cov}\left({u}_{t+2},{u}_t\right)& \mathrm{Cov}\left({u}_{t+3},{u}_{t+2}\right)& \mathrm{Cov}\left({u}_{t+2},{u}_{t+2}\right)& \cdots & \cdots & \mathrm{Cov}\left({u}_{t+2},{u}_T\right)\\ {}\vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ {}\vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ {}\mathrm{Cov}\left({u}_T,{u}_t\right)& \mathrm{Cov}\left({u}_T,{u}_{t+1}\right)& \mathrm{Cov}\left({u}_T,{u}_{t+2}\right)& \cdots & \cdots & \mathrm{Cov}\left({u}_T,{u}_T\right)\end{array}\right] $$

Detailed computation of the elements of Σ_u is presented below.

$$ {\displaystyle \begin{array}{l}\mathrm{Cov}\left({u}_t,{u}_t\right)=\mathrm{E}\left[\left({u}_t-\mathrm{E}\left({u}_t\right)\right)\left({u}_t-\mathrm{E}\left({u}_t\right)\right)\right]=\mathrm{E}\left[{u_t}^2\right]\\ {}\kern6.25em =\mathrm{E}\left({\left({\varepsilon}_t-2\left(1-q\right){\varepsilon}_{t-1}+{\left(1-q\right)}^2\ {\varepsilon}_{t-2}\right)}^2\right)\kern0.5em \mathrm{for}\ \mathrm{t}=3,4,5\dots \mathrm{T}\\ {}\begin{array}{l}\kern5em =\mathrm{E}\Big({\varepsilon}_t^2-4\left(1-q\right){\varepsilon}_t{\varepsilon}_{t-1}+2{\left(1-q\right)}^2{\varepsilon}_t{\varepsilon}_{t-2}+4{\left(1-q\right)}^2{\left({\varepsilon}_{t-1}\right)}^2-4{\left(1-q\right)}^3{\varepsilon}_{t-1}{\varepsilon}_{t-2}+\\ {}\kern6.5em {\left(1-q\right)}^4{\varepsilon}_{t-2}^2\Big)\\ {}\kern6em =\mathrm{E}\left({\varepsilon}_t^2\right)-\mathbf{4}\left(1-q\right)\boldsymbol{E}\left({\varepsilon}_t{\varepsilon}_{t-1}\right)+2{\left(1-q\right)}^2E\left({\varepsilon}_t{\varepsilon}_{t-2}\right)+4{\left(1-q\right)}^2E\left({\left({\varepsilon}_{t-1}\right)}^2\right)-\\ {}\kern6em 4{\left(1-q\right)}^3\mathrm{E}\left({\varepsilon}_{t-1}{\varepsilon}_{t-2}\right)+{\left(1-q\right)}^4\mathrm{E}\left({\varepsilon}_{t-2}^2\right)\\ {}\kern5em ={\sigma}^2+4{\left(1-q\right)}^2{\sigma}^2+{\left(1-q\right)}^4\ {\sigma}^2\kern0.75em ={\sigma}^2\ \left(1+4{\left(1-q\right)}^2+{\left(1-q\right)}^4\ \right)\end{array}\end{array}} $$

$$ {\displaystyle \begin{array}{l}\begin{array}{l}\mathrm{Cov}\left({u}_t,{u}_{t+1}\right)=\mathrm{E}\left[\left({u}_t-\mathrm{E}\left({u}_t\right)\right)\left({u}_{t+1}-\mathrm{E}\left({u}_{t+1}\right)\right)\right]=\mathrm{E}\left[{u}_t{u}_{t+1}\right]\\ {}\kern6.5em =\mathrm{E}\left[{\varepsilon}_t-2\left(1-q\right){\varepsilon}_{t-1}+{\left(1-q\right)}^2\ {\varepsilon}_{t-2}\ \right)\left({\varepsilon}_{t+1}-2\left(1-q\right){\varepsilon}_t+{\left(1-q\right)}^2\ {\varepsilon}_{t-1}\right]\\ {}\kern6.5em =\mathrm{E}\Big[{\varepsilon}_t{\varepsilon}_{t+1}-2\left(1-q\right){\varepsilon}_t^2+{\left(1-q\right)}^2{\varepsilon}_t{\varepsilon}_{t-1}-2\left(1-q\right)\left({\varepsilon}_{t-1}{\varepsilon}_{t+1}\right)+4{\left(1-q\right)}^2{\varepsilon}_t{\varepsilon}_{t-1}-\\ {}\kern7.75em 2{\left(1-q\right)}^3{\varepsilon}_{t-1}^2+{\left(1-q\right)}^2{\varepsilon}_{t+1}{\varepsilon}_{t-2}-2{\left(1-q\right)}^3\left(\kern0.5em {\varepsilon}_t{\varepsilon}_{t-2}\right)+{\left(1-q\right)}^4{\varepsilon}_{t-1}{\varepsilon}_{t-2}\Big]\end{array}\\ {}\kern6.5em =\mathrm{E}\left({\varepsilon}_t{\varepsilon}_{t+1}\Big)-2\left(1-q\right)E\left({\varepsilon}_t^2\right)+{\left(1-q\right)}^2E\left({\varepsilon}_t{\varepsilon}_{t-1}\right)-2\left(1-q\right)E\left({\varepsilon}_{t-1}{\varepsilon}_{t+1}\right)+4\right)\\ {}\kern6.5em {\left(1-q\right)}^2E\left({\varepsilon}_t{\varepsilon}_{t-1}\right)-2{\left(1-q\right)}^3E\left({\varepsilon}_{t-1}^2\right)+{\left(1-q\right)}^2E\left({\varepsilon}_{t+1}{\varepsilon}_{t-2}\right)-2{\left(1-q\right)}^3E\left(\kern0.5em {\varepsilon}_t{\varepsilon}_{t-2}\right)+\\ {}\kern7.75em {\left(1-q\right)}^4E\left({\varepsilon}_{t-1}{\varepsilon}_{t-2}\right)\\ {}\kern6em =-2\left(1-q\right){\sigma}^2-2{\left(1-q\right)}^3{\sigma}^2=-2{\sigma}^2\left(\left(1-q\right)+{\left(1-q\right)}^3\right)=-2{\sigma}^2\left(1-q\right)\left(1+{\left(1-q\right)}^2\right)\end{array}} $$

$$ {\displaystyle \begin{array}{l}\begin{array}{l}\mathrm{Cov}\left({u}_t,{u}_{t+2}\right)=\mathrm{E}\left({\varepsilon}_t-2\left(1-q\right){\varepsilon}_{t-1}+{\left(1-q\right)}^2\ {\varepsilon}_{t-2}\ \left)\right({\varepsilon}_{t+2}-2\left(1-\mathrm{q}\right){\varepsilon}_{t+1}+{\left(1-q\right)}^2\ {\varepsilon}_t\ \right)\\ {}\kern6.25em =\mathrm{E}\Big({\varepsilon}_t{\varepsilon}_{t+2}-2\left(1-q\right)\ {\varepsilon}_t{\varepsilon}_{t+1}+{\left(1-q\right)}^2{\varepsilon}_t^2-2\left(1-q\right){\varepsilon}_{t-1}{\varepsilon}_{t+2}+4{\left(1-q\right)}^2{\varepsilon}_{t-1}{\varepsilon}_{t+1}+\\ {}\kern6.25em 2{\left(1-q\right)}^3\ {\varepsilon}_t{\varepsilon}_{t-1}+{\left(1-q\right)}^2\ {\varepsilon}_{t-2}{\varepsilon}_{t+2}-2{\left(1-q\right)}^3{\varepsilon}_{t-2}{\varepsilon}_{t+1}+{\left(1-q\right)}^4{\varepsilon}_t{\varepsilon}_{t-2}\ \Big)\end{array}\\ {}\kern6.25em =\mathrm{E}\left({\varepsilon}_t{\varepsilon}_{t+2}\right)-2\left(1-q\right)\mathrm{E}\left({\varepsilon}_t{\varepsilon}_{t+1}\right)+{\left(1-q\right)}^2E\left({\varepsilon}_t^2\right)-2\left(1-q\right)\mathrm{E}\left({\varepsilon}_{t-1}{\varepsilon}_{t+2}\right)+4{\left(1-q\right)}^2\\ {}\kern6.25em E\left({\varepsilon}_{t-1}{\varepsilon}_{t+2}\right)++2{\left(1-q\right)}^3\mathrm{E}\left({\varepsilon}_t{\varepsilon}_{t-1}\right)+{\left(1-q\right)}^2\mathrm{E}\left({\varepsilon}_{t-2}{\varepsilon}_{t+2}\right)-2{\left(1-q\right)}^3E\left({\varepsilon}_{t-2}{\varepsilon}_{t+1}\right)+\\ {}\kern6.25em {\left(1-q\right)}^4E\left({\varepsilon}_t{\varepsilon}_{t-2}\right)\\ {}\kern6.25em ={\left(1-q\right)}^2{\sigma}^2\end{array}} $$

$$ {\displaystyle \begin{array}{l}\begin{array}{l}\mathrm{Cov}\left({u}_t,{u}_{t+3}\right)=\mathrm{E}\left({\varepsilon}_t-2\left(1-q\right){\varepsilon}_{t-1}+{\left(1-q\right)}^2\ {\varepsilon}_{t-2}\ \left)\right({\varepsilon}_{t+3}-2\left(1-q\right){\varepsilon}_{t+2}+{\left(1-q\right)}^2\ {\varepsilon}_{t+1}\ \right)\\ {}\kern6.5em =\mathrm{E}\Big({\varepsilon}_t{\varepsilon}_{t+3}-2\left(1-q\right)\ {\varepsilon}_t{\varepsilon}_{t+2}+{\left(1-q\right)}^2{\varepsilon}_t{\varepsilon}_{t+1}-2\left(1-q\right){\varepsilon}_{t-1}{\varepsilon}_{t+3}+4{\left(1-q\right)}^2\ {\varepsilon}_{t-1}{\varepsilon}_{t+2}-\\ {}\kern6.5em 2{\left(1-q\right)}^3{\varepsilon}_{t-1}{\varepsilon}_{t+1}+{\left(1-q\right)}^2\ {\varepsilon}_{t-2}{\varepsilon}_{t+3}-2{\left(1-q\right)}^3{\varepsilon}_{t-2}{\varepsilon}_{t+2}+{\left(1-q\right)}^4\ {\varepsilon}_{t-2}{\varepsilon}_{t+1}\Big)\end{array}\\ {}\kern6.5em =\mathrm{E}\left({\varepsilon}_t{\varepsilon}_{t+3}\right)-2\left(1-q\right)E\left({\varepsilon}_t{\varepsilon}_{t+2}\right)+{\left(1-q\right)}^2E\left({\varepsilon}_t{\varepsilon}_{t+1}\right)-2\left(1-q\right)E\left({\varepsilon}_{t-1}{\varepsilon}_{t+3}\right)+\\ {}\kern7em 4{\left(1-q\right)}^2\mathrm{E}\left({\varepsilon}_{t-1}{\varepsilon}_{t+2}\right)-2{\left(1-q\right)}^3E\left({\varepsilon}_{t-1}{\varepsilon}_{t+1}\right)+{\left(1-q\right)}^2\mathrm{E}\left({\varepsilon}_{t-2}{\varepsilon}_{t+3}\right)-\\ {}\kern7em 2{\left(1-q\right)}^3E\left({\varepsilon}_{t-2}{\upvarepsilon}_{t+2}\right)+{\left(1-q\right)}^4\ \mathrm{E}\left({\varepsilon}_{t-2}{\varepsilon}_{t+1}\right)\\ {}\kern6.25em =0\end{array}} $$

From the above computation, the values for the rest of non-diagonal entries of the matrix, i.e. Cov(u_t, u_t + 3), for t = 3,4,5… T will be zeros. Therefore, the covariance matrix Σ_u can be represented in terms p & q parameters as shown below.

$$ {\sum}_u=\left[\begin{array}{cccccc}1+4{\left(1-\mathrm{q}\right)}^2+{\left(1-\mathrm{q}\right)}^4& -2\left(1-\mathrm{q}\right)\left(1+{\left(1-\mathrm{q}\right)}^2\right)& {\left(1-\mathrm{q}\right)}^2& 0& \cdots & 0\\ {}-2\left(1-\mathrm{q}\right)\left(1+{\left(1-\mathrm{q}\right)}^2\right)& 1+4{\left(1-\mathrm{q}\right)}^2+{\left(1-\mathrm{q}\right)}^4& -2\left(1-\mathrm{q}\right)\left(1+{\left(1-\mathrm{q}\right)}^2\right)& {\left(1-\mathrm{q}\right)}^2& \cdots & 0\\ {}{\left(1-\mathrm{q}\right)}^2& -2\left(1-\mathrm{q}\right)\left(1+{\left(1-\mathrm{q}\right)}^2\right)& 1+4{\left(1-\mathrm{q}\right)}^2+{\left(1-\mathrm{q}\right)}^4& -2\left(1-\mathrm{q}\right)\left(1+{\left(1-\mathrm{q}\right)}^2\right)& \cdots & 0\\ {}0& {\left(1-\mathrm{q}\right)}^2& -2\left(1-\mathrm{q}\right)\left(1+{\left(1-\mathrm{q}\right)}^2\right)& 1+4{\left(1-\mathrm{q}\right)}^2+{\left(1-\mathrm{q}\right)}^4& \cdots & \vdots \\ {}\vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ {}0& 0& 0& 0& \cdots & 1+4{\left(1-\mathrm{q}\right)}^2+{\left(1-\mathrm{q}\right)}^4\end{array}\right] $$