# Stochastic models with multiplicative noise for economic inequality and mobility

• Maria Letizia Bertotti , Amit K Chattopadhyay and Giovanni Modanese

## Abstract

In this article, we discuss a dynamical stochastic model that represents the time evolution of income distribution of a population, where the dynamics develops from an interplay of multiple economic exchanges in the presence of multiplicative noise. The model remit stretches beyond the conventional framework of a Langevin-type kinetic equation in that our model dynamics is self-consistently constrained by dynamical conservation laws emerging from population and wealth conservation. This model is numerically solved and analysed to evaluate the inequality of income in correlation to other relevant dynamical parameters like the mobility M and the total income μ. Inequality is quantified by the Gini index G. In particular, correlations between any two of the mobility index M and/or the total income μ with the Gini index G are investigated and compared with the analogous quantities resulting from an additive noise model.

Corresponding author: Maria Letizia Bertotti, Free University of Bozen-Bolzano, Faculty of Science and Technology, 39100 Bolzano, Italy, E-mail:

1. Author contribution: All the authors have accepted responsibility for the entire content of this submitted manuscript and approved submission.

2. Research funding: This research was funded in part by the Free University of Bozen-Bolzano through the project DEFENSSE.

3. Conflict of interest statement: The authors declare no conflicts of interest regarding this article.

Appendix

Proof of i , j D i j ( 2 ) ( x ) ξ j = 0 for D i j ( 2 ) as in (7).

The solutions of Eq. (5) are found based on a discretization of time, i.e., considering times t k with k = 0, 1, 2, … and

x i ( t k + 1 ) = x i ( t k ) + D i ( 1 ) ( x ( t k ) ) ( t k + 1 t k ) + j = 1 n D i j ( 2 ) ( x ( t k ) ) ξ j Γ ( t k + 1 t k ) ,

for i = 1, …, n.

If t 0 = 0 and i = 1 n x i ( t 0 ) = 1 , recalling that i = 1 n D i ( 1 ) ( x ) d t = 0 for any x we have,

i = 1 n x i ( t 1 ) = 1 + i = 1 n j = 1 n D i j ( 2 ) ( x ( t 0 ) ) ξ j Γ ( t 1 t 0 ) .

We show next that in fact i = 1 n x i ( t 1 ) = 1 .

Indeed, if the operator D i j ( 2 ) is as in (7) in Subsection 2.1, applying it to any random vector ξ, namely for any choice of {ξ j }, we get

i , j D i j ( 2 ) ( x ) ξ j = j i j D i j ( 2 ) ( x ) ξ j + j D j j ( 2 ) ( x ) ξ j = j i j x i x j ξ j + j x j 1 x j ξ j = j x j ξ j i j x i + j x j 1 x j ξ j = j x j ξ j i x i x j + j x j ξ j j x j 2 ξ j = j x j ξ j i x i + j x j 2 ξ j + j x j ξ j j x j 2 ξ j = j x j ξ j 1 i x i .

Then, in particular,

i , j D i j ( 2 ) ( x ( t 0 ) ) ξ j = j x j ( t 0 ) ξ j 1 i x i ( t 0 ) = 0 ,

the last equality being true because by assumption the quantity in parentheses vanishes.

At this point iteration of this procedure shows that i = 1 n x i ( t k ) = 1 and i , j D i j ( 2 ) ( x ( t k ) ) ξ j = 0 hold true for any k = 1, 2, … too. The claim then follows in view of the arbitrariness of the t k ′s.

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