A combined Brass-random walk approach to probabilistic household forecasting: Denmark, Finland, and the Netherlands, 2011–2041

Abstract

Probabilistic household forecasts to 2041 are presented for Denmark, Finland, and the Netherlands. Future trends in fertility, mortality and international migration are taken from official population forecasts. Time series of shares of the population in six different household positions are modelled as random walks with drift. Brass’ relational model preserves the age patterns of the household shares. Probabilistic forecasts for households are computed by combining predictive distributions for the household shares with predictive distributions of the populations, specific for age and sex. If current trends in the three countries continue, we will witness a development towards more and smaller households, often driven by increasing numbers of persons who live alone. We can be quite certain that by 2041, there will be between two and four times as many persons aged 80 and over who live alone when compared with the situation in 2011.

Appendix: Back transformation from ξ to α

In “Modelling household shares” section the shares α _j are transformed into fractions ξ _k . In this “Appendix” we outline the back transformation from ξ _k to α _j . We suppress indices for age, sex, time, and country. The starting point is the set of expressions that transform the shares α _j into fractions ξ _k .

$$\begin{aligned} \xi_{2} & = {\text{logit}}\left( {\left( {\alpha_{3} + \alpha_{4} } \right)/\left( {1 \, - \alpha_{1} } \right)} \right) \\ \xi_{3} & = {\text{logit}}\left( {\alpha_{4} /\left( {\alpha_{3} + \alpha_{4} } \right)} \right) \\ \xi_{4} & = {\text{logit}}\left( {\left( {\alpha_{2} + \alpha_{7} } \right)/\left( {\alpha_{2} + \alpha_{5} + \alpha_{6} + \alpha_{7} } \right)} \right) \\ \xi_{5} & = {\text{logit}}\left( {\left( {\alpha_{2} } \right)/\left( {\alpha_{2} + \alpha_{7} } \right)} \right) \\ \xi_{6} & = {\text{logit}}\left( {\alpha_{5} /\left( {\alpha_{5} + \alpha_{6} } \right)} \right) \\ \end{aligned}$$

There are many equivalent expressions for the α _j written as functions of the ξ _k . One of these is the following set

$$\begin{aligned} \alpha_{2} & = {{\left( {1 \, {-}\alpha_{1} } \right)\exp (\xi_{4} )\exp (\xi_{5} )} \mathord{\left/ {\vphantom {{\left( {1 \, {-}\alpha_{1} } \right)\exp (\xi_{4} )\exp (\xi_{5} )} {\left\{ {\left( {1 \, + \, \exp \left( {\xi_{2} } \right)} \right)\left( {1 \, + \, \exp \left( {\xi_{4} } \right)} \right) \, \left( {1 \, + \, \exp \left( {\xi_{5} } \right)} \right)} \right\}}}} \right. \kern-0pt} {\left\{ {\left( {1 \, + \, \exp \left( {\xi_{2} } \right)} \right)\left( {1 \, + \, \exp \left( {\xi_{4} } \right)} \right) \, \left( {1 \, + \, \exp \left( {\xi_{5} } \right)} \right)} \right\}}} \\ \alpha_{3} & = {{\left( {1 \, - \alpha_{1} } \right)\exp \left( {\xi_{2} } \right)} \mathord{\left/ {\vphantom {{\left( {1 \, - \alpha_{1} } \right)\exp \left( {\xi_{2} } \right)} {\left\{ {\left( {1 \, + \, \exp \left( {\xi_{2} } \right)} \right)(1 \, + \, \exp \left( {\xi_{3} } \right)} \right\}}}} \right. \kern-0pt} {\left\{ {\left( {1 \, + \, \exp \left( {\xi_{2} } \right)} \right)(1 \, + \, \exp \left( {\xi_{3} } \right)} \right\}}} \\ \alpha_{4} & = \alpha_{3} \exp \left( {\xi_{3} } \right) \\ \alpha_{6} & = {{\left( {1 \, {-}\alpha_{1} {-}\alpha_{3} {-}\alpha_{4} } \right)} \mathord{\left/ {\vphantom {{\left( {1 \, {-}\alpha_{1} {-}\alpha_{3} {-}\alpha_{4} } \right)} {\left\{ {\left( {1 \, + \, \exp \left( {\xi_{4} } \right)} \right)(1 \, + \, \exp \left( {\xi_{6} } \right)} \right\}}}} \right. \kern-0pt} {\left\{ {\left( {1 \, + \, \exp \left( {\xi_{4} } \right)} \right)(1 \, + \, \exp \left( {\xi_{6} } \right)} \right\}}} \\ \alpha_{5} & = \alpha_{6} \exp \left( {\xi_{6} } \right) \\ \alpha_{7} & = {{\alpha_{6} \exp \left( {\xi_{4} } \right)\left( {1 \, + \, \exp \left( {\xi_{6} } \right)} \right)} \mathord{\left/ {\vphantom {{\alpha_{6} \exp \left( {\xi_{4} } \right)\left( {1 \, + \, \exp \left( {\xi_{6} } \right)} \right)} {\left( {1 \, + \, \exp \left( {\xi_{5} } \right)} \right)}}} \right. \kern-0pt} {\left( {1 \, + \, \exp \left( {\xi_{5} } \right)} \right)}} \\ \end{aligned}$$

By assumption, α ₁ is independent of ξ _k (k = 2, 3,…6).