Does Demography Change Wealth Inequality?

Abstract

In this article, we investigate the effect of demography on wealth inequality. We propose an economic growth model with overlapping generations in which individuals are altruistic towards their children and differ with respect to the age of their parent. We denote the age gap between the parent and their child as generational gap. The introduction of the generational gap allows us to analyze wealth inequality not only across cohorts but also within cohorts. Our model predicts that a decline in fertility raises wealth inequality within cohorts and, simultaneously, it reduces inequality at the population level (across cohorts). In contrast, increases in life expectancy result in a non-monotonic effect on wealth inequality by age and across cohorts.

Appendices

Appendix 1: Number of Offspring and the Fraction of Wealth Received

Number of Offspring

Similar to the foundation of the stable population theory by Lotka (1939), we will consider the total number of heirs of an individual at age x is a convolution of the fertility function with the survival probability. To simplify the notation and without loss of generality, in this proof, we get rid of the time at birth.

Let 𝜗(l) be the number of children born from an individual of age l. Let us assume that 𝜗(l) is an independently distributed random variable defined on the non-negative integers. Further let z _i (a), with i = 0, …, 𝜗(l), denote the ith child of age a. Let z _i (a) be a Bernoulli distributed independent random variable with probability S(a) (i.e. S(a) denotes the probability that the child survives up to age a). Then $\mathcal {Z} (a,l)$ defining the total number of children of exact age a born from an individual of age l can be defined as

$$\displaystyle \begin{aligned} \mathcal{Z} (a,l)= \sum\nolimits_{i=0}^{\vartheta (l)} z_{i}(a). \end{aligned} $$

(38)

Assuming 𝜗(l) is distributed at the population level according to a Poisson with parameter m(l), where m(l) is the age-specific fertility rate at the exact age l. Then, the distribution of $\mathcal {Z}(a,l)$ is a convolution of the fertility function with the survival probably

$$\displaystyle \begin{aligned} P\left\{\mathcal{Z}(a,l)=\sigma\right\}&=P\left\{\sum\nolimits_{i=1}^{\vartheta(l)}z_{i}(a)=\sigma\right\}\\ &=\sum\nolimits_{n=1}^{\infty}P\left\{\sum\nolimits_{i=1}^{n}z_{i}(a)=\sigma\right\}P\left\{\vartheta(l)=n\right\}. \end{aligned} $$

(39)

To find the distribution that results from (39) we use the characteristic function of $\mathcal {Z}(a,l)$ as follows

$$\displaystyle \begin{aligned} \varphi_{\mathcal{Z}(a,l)}(t)&=\operatorname{E}\left[e^{it \mathcal{Z}(a,l)}\right]=\operatorname{E}\left[e^{it\sum\nolimits_{j=0}^{\vartheta(l)}z_{j}(a)}\right]\\ &=\sum_{n=0}^{\infty}\operatorname{E}\left[e^{it\sum\nolimits_{j=0}^{n}z_{j}(a)}\right]\operatorname{P}\left\{\vartheta(l)=n\right\}\\ &=\sum_{n=0}^{\infty}\left(\prod_{j=0}^{n}\operatorname{E}\left[e^{it z_{j}(a)}\right]\right)\operatorname{P}\left\{\vartheta(l)=n\right\}\\ &=\sum_{n=0}^{\infty}\left(\varphi_{z_{j}(a) }(t)\right)^n\operatorname{P}\left\{\vartheta(l)=n\right\}=\sum_{n=0}^{\infty}\frac{\left(\varphi_{z_{j}(a) }(t)m(l)\right)^ne^{-m(l)}}{n!}\\ &=e^{-m(l)}\sum_{n=0}^{\infty}\frac{\left(\varphi_{z_{j}(a) }(t)m(l)\right)^n}{n!}=e^{-m(l)}e^{\varphi_{z_{j}(a) }(t)m(l)}=e^{m(l)\left(\varphi_{z_{j}(a) }(t)-1\right)}.{} \end{aligned} $$

(40)

Given that the characteristic function of the Bernoulli distribution is

$$\displaystyle \begin{aligned} \varphi_{z_{j}(a) }(t)=\operatorname{E}\left[e^{it z_{j}(a) }\right]=e^{it}S(a)+1-S(a).{} \end{aligned} $$

(41)

Substituting (41) in (40) gives

$$\displaystyle \begin{aligned} \varphi_{\mathcal{Z}(a,l)}(t)&=e^{m(l) S(a)\left(e^{it}-1\right)},{} \end{aligned} $$

(42)

which is the characteristic function of a Poisson process. Thus, we have

$$\displaystyle \begin{aligned} \mathcal{Z}(a,l){\stackrel {d}{\sim}} \operatorname{Po}(m(l) S(a)).{} \end{aligned} $$

(43)

If we define the random variable total number of children of an individual with exact age x as

$$\displaystyle \begin{aligned} \mathcal{N}(x)=\int_0^x\mathcal{Z}(x-l,l)dl.{} \end{aligned} $$

(44)

Then, from probability theory we have that the sum of Poisson processes is also a Poisson process. Hence, from (43) and (44) it follows

$$\displaystyle \begin{aligned} \mathcal{N}(x){\stackrel {d}{\sim}} \operatorname{Po}\left(\int_0^xm(l) S(x-l)dl\right), \end{aligned} $$

(45)

where the mean of $\mathcal {N}(x)$ is equal to the total number of heirs, see Eq. (2), given that $n(x)=\int _0^xm(l) S(x-l)dl$.

Fraction of Wealth Received

In order to derive (6) we must answer the question: what will be the expected fraction of wealth that corresponds to an individual if the parent dies at the exact age x? If the parent leaves one monetary unit as bequest, the wealth will be equally split between our individual and all the remaining siblings. From (45) we know that the number of offspring from a parent of x is a random number distributed according to a Poisson process. Therefore, the expected number of offspring from a parent of exact age x is, in this case, given by

$$\displaystyle \begin{aligned} E\left[\mathcal{N}(x)\middle|\mathcal{N}(x)\geq 1\right]&=\frac{1}{1-P\left\{\mathcal{N}(x)=0\right\}}\sum_{\sigma=1}^\infty \sigma P\left\{\mathcal{N}(x)=\sigma\right\}\\ &=\frac{1}{1-e^{-n(x)}}\sum_{\sigma=1}^\infty \sigma \frac{[n(x)]^\sigma e^{-n(x)}}{\sigma!}\\ &=\frac{n(x)}{1-e^{-n(x)}}.{} \end{aligned} $$

(46)

Given that each unit of wealth will be split equally among the expected number of offspring, the fraction of wealth received by any offspring from a parent of age x becomes the inverse of (46),

$$\displaystyle \begin{aligned} \eta(x)=\frac{1-e^{-n(x)}}{n(x)}, \end{aligned} $$

(47)

which coincides with (6).

Appendix 2: Total Bequest Given Equals Total Bequest Received

For convenience we integrate with respect to age and generational gaps. Let us define all wealth transfers given at time t as

$$\displaystyle \begin{aligned} \int_{0}^\omega \mu(x,t-x)\int_0^{\omega} N(x,t-x,l) k(x,t-x,l)\left[1-\theta(x,t-x)\right]dldx.{} \end{aligned} $$

(48)

Equation (48) is the integral over all ages and generational gaps of the capital left as bequest at each age x in year t by individuals whose parents were l years older at the time of birth, i.e. $k(x,t-x,l)\left [1-\theta (x,t-x)\right ]$, times the total number of people dying with those demographic characteristics, μ(x, t − x)N(x, t − x, l). Also, let us define all wealth transfers received as

$$\displaystyle \begin{aligned} \int_{0}^\omega\int_{0}^{\omega-x} B(x,t-x,l)N(x,t-x,l)dl dx. {}\end{aligned} $$

(49)

Equation (49) is the integral over all ages and generational gaps of the average bequest received at age x in year t from the death of their parent at age x + l. From (5) we have that the bequest received only takes non zero values for l ∈ [0, w − x).

In order to prove that all wealth transfers given equal all wealth transfers received, we show that by substituting terms in (49) we get (48). The inverse is completely analogous and we leave it to the reader.

Proof

First, by substituting (19) and (21) in (49), we get

$$\displaystyle \begin{aligned} \int_{0}^\omega\int_{0}^{\omega-x} B(x,t-x,l)S(x,t-x)m(l,t-x-l)N(l,t-x-l)dl dx.{}\end{aligned} $$

(50)

Using (5), defining a = x + l, and rearranging terms in (50), we have

$$\displaystyle \begin{aligned} \int_{0}^\omega\int_{x}^{\omega} \mu(a,t-a)N(a,t-a)k(a,t-a)\eta(a,t-a)S(x,t-x)m(a-x,t-a)da dx.{}\end{aligned} $$

(51)

Changing the order of integration in (51) and leaving outside of the inner integral those variables that do not depend on x gives

$$\displaystyle \begin{aligned} \int_{0}^\omega \mu(a,t-a)N(a,t-a)k(a,t-a)\eta(a,t-a)\int_{0}^{a} S(x,t-x)m(a-x,t-a)dx da.{}\end{aligned} $$

(52)

Expressing the inner integral of (52) in terms of the generational gap l = a − x gives

$$\displaystyle \begin{aligned} \int_{0}^\omega \mu(a,t-a)N(a,t-a)k(a,t-a)\eta(a,t-a)\int_{0}^{a} S(a-l,t-a+l)m(l,t-a)dl da.{}\end{aligned} $$

(53)

According to (2), the inner integral of (53) equals the average number of births of an individual born in year t − a at age a. Then, using (6) in (53), we have

$$\displaystyle \begin{aligned} \int_{0}^\omega \mu(a,t-a)N(a,t-a)k(a,t-a)[1-\theta(a,t-a)] da.{}\end{aligned} $$

(54)

Next, using the fact that $N(a,t-a)k(a,t-a)=\int _0^{\omega }N(a,t-a,l)k(a,t-a,l)dl$ in (54), and assuming a = x, we have

$$\displaystyle \begin{aligned} \int_{0}^\omega \mu(x,t-x)\int_0^{\omega}N(x,t-x,l)k(x,t-x,l)[1-\theta(x,t-x)]dl dx, \end{aligned} $$

(55)

which is equivalent to (48). □