A Becker–Tomes model with investment risk

Abstract

Recent studies find that sufficiently volatile idiosyncratic investment risk plays an important role in generating wealth inequality. I introduce idiosyncratic investment risk into the Becker and Tomes (J Polit Econ 87:1153–1189, 1979) model and find an explicit expression of the stationary wealth distribution in this simple model. This explicit expression brings us new insights of how bequest motives and estate taxes influence wealth distributions. I find that inheritance increases wealth inequality in models with idiosyncratic investment risk through exaggerating labor earnings uncertainty, while inheritance decreases wealth inequality in the Becker and Tomes (1979) model through mitigating labor earnings uncertainty. This causes estate taxes to have different impacts on wealth inequality in my model and the Becker and Tomes (1979) model.

Appendix

1.1 The tail of the stationary wealth distribution

In the steady state of the aggregate economy, Eq. (5) implies

$$\begin{aligned} L_{t+1}=d_{t+1}L_{t}+\eta _{t+1}, \end{aligned}$$

(A.1)

where

$$\begin{aligned} d_{t+1}=\frac{\tilde{R}_{t+1}(1-b)}{1+\left[ \tilde{R}_{t+1}(1-b)\right] ^{ \frac{\gamma -1}{\gamma }}\chi ^{-\frac{1}{\gamma }}}, \end{aligned}$$

(A.2)

and

$$\begin{aligned} \eta _{t+1}=\frac{1}{1+\left[ \tilde{R}_{t+1}(1-b)\right] ^{\frac{\gamma -1}{ \gamma }}\chi ^{-\frac{1}{\gamma }}}\left( H_{t+1}+G\right) . \end{aligned}$$

(A.3)

In order to investigate the stationary wealth distribution with serially correlated $\{H_{t}\}$ and $\{\tilde{R}_{t}\}$, we need the following definition.

Definition 2

Let $(\ell ,\mathcal {F})$ be a measurable space and let $\left\{ x_{n}\right\}$ be a stationary Markov process with transition kernel $Q(x,\cdot )$ defined on it. A Markov-modulated process (MMP) associated with $\left\{ x_{n}\right\}$ is a stationary Markov process $\left\{ (x_{n},\zeta _{n})\right\}$ defined on a product space $(\ell \times \Upsilon ,\mathcal {F}\otimes \Xi )$, whose transitions depend only on the position of $x_{n}$. That is, for any $n\ge 0$, $A\in \mathcal {F}$, $B\in \Xi$,

$$\begin{aligned} Pr \left( x_{n}\in A,\zeta _{n}\in B\mid \sigma \left( (x_{i},\zeta _{i}):i<n\right) \right) =\int _{A}Q(x,dy)\Gamma (x,y,B)|_{x=x_{n-1}}, \end{aligned}$$

where $\Gamma (x,y,\cdot )=Pr (\zeta _{1}\in \cdot \mid x_{0}=0,x_{1}=y)$ is a kernel on $(\ell \times \ell \times \Xi )$.

Lemma 1

Let

$$\begin{aligned} m(x)=\log E(d_{t+1})^{x}. \end{aligned}$$

Then m(x) is a convex function of $x>0$.

Proof

See page 158 of Loève (1977). $\square$

Lemma 2

m(x) is a continuous function of $x>0$.

Proof

By Proposition 17 of Chapter 5 in Royden (1988), Lemma 1 implies Lemma 2. $\square$

Proof of Theorem 1

Note that the process $\left\{ (H_{t},v_{t})\right\}$, where $v_{t}=\left( d_{t},\eta _{t}\right)$, is a Markov-modulated process associated with $\{H_{t}\}$.

In order to apply Theorem 1.5 of Rointershtein (2007) to the process $\{L_{t}\}$, we will verify (A1)–(A7) of Assumption 1.2 in Rointershtein (2007).

(A1) is obviously satisfied since the Borel sigma-algebra is countably generated.

By Assumption 1, $\{H_{t}\}$ is irreducible. Thus (A2) is satisfied.

By Assumption 2, we have

$$\begin{aligned} Pr \left( H_{t+1}\le h\mid H_{t}=x\right) =\int _{0}^{h}f(x,y)\hbox {d}y=\int _{0}^{h}\bar{H}f(x,y)\frac{1}{\bar{H}}\hbox {d}y, \end{aligned}$$

for $h\in (0,\bar{H})$. Let $\mu ^{Leb}$ be the Lebesgue measure. We construct a probability measure $\lambda$ on $(0,\bar{H})$, such that $\lambda (A)=\frac{1}{\bar{H}}\mu ^{Leb}(A)$ for any Borel set A. Since f(x, y) is uniformly bounded above on $(0,\bar{H})\times (0,\bar{H})$, $\bar{ H}f(x,y)$ is also uniformly bounded above on $(0,\bar{H})\times (0,\bar{H})$ . Thus the family of functions $\{\bar{H}f(x,\cdot ):(0,\bar{H})\rightarrow [0,\infty )\}_{x\in (0,\bar{H})}$ is uniformly integrable with respect to the measure $\lambda$. Then (A3) is satisfied for $m_{1}=1$ and the measure $\lambda$ we construct.

From Assumption 2, we know that $Pr (H_{t+1}\in (0,\bar{H}))=1$. And from Assumption 4 we know that $Pr (\tilde{R}_{t+1}\in [$ Ṟ,$\bar{ R}])=1$. Thus from Eq. (A.3) we know that there exists an $\bar{\eta }>0$ such that $Pr (\eta _{t+1}<\bar{\eta })=1$. Thus (A4) is satisfied.

From Assumption 4, we know that $Pr (\tilde{R}_{t+1}\in [$ Ṟ, $\bar{R}])=1$. Thus $d_{t+1}$ is bounded, since $d_{t+1}$ is a continuous function of $\tilde{R}_{t+1}$ [see Eq. (A.2)]. We also know that $d_{t+1}$ is bounded away from zero since Ṟ$>0$. Thus there exists an $c_{\rho }>1$ such that $Pr (\frac{1}{c_{\rho }}<d_{t+1}<c_{\rho })=1$. Thus (A5) is satisfied.

For $x>0$, we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{1}{n}\log E\left[ \prod \limits _{i=1}^{n-1} \left( d_{i}\right) ^{x}\right] =\log E(d_{t+1})^{x}=m(x), \end{aligned}$$

since, by Assumption 3, $\left\{ \tilde{R}_{t}\right\}$ is i.i.d. along generations. From Lemmas 1 and 2, we know that m(x) is convex and continuous. From Assumption 5, we have $E\left( d_{t+1}\right) <1$. Thus $m(1)<0$. By Assumption 6, $E(d_{t+1})^{2}>1$. Thus $m(2)>0$. Then we know that there exists a unique $\mu \in (1,2)$ such that $m(\mu )=0$, i.e.

$$\begin{aligned} E(d_{t+1})^{\mu }=1. \end{aligned}$$

Thus (A6) is satisfied.

By Assumption 4, $\tilde{R}_{t+1}$ has a probability density function $l(\cdot )$ on [Ṟ,$\bar{R}]$. Thus the distribution of $\log d_{t+1}$ is nonarithmetic. Thus (A7) is satisfied.

We have verified (A1)–(A7) of Assumption 1.2 in Rointershtein (2007). Applying Theorem 1.5 of Rointershtein (2007) to the process $\{L_{t}\}$, we have

$$\begin{aligned} \lim _{x\rightarrow \infty }\frac{Pr (L_{\infty }>x)}{x^{-\mu }}=c, \end{aligned}$$

with $c>0$. $\square$

1.2 Proof of Theorem 2

Proof

Let $\mu ^{\prime }\in (1,2)$ solves

$$\begin{aligned} E(d_{t+1}^{\prime })^{\mu ^{\prime }}=1. \end{aligned}$$

Since $d_{t+1}\succeq _\mathrm{FSD}d_{t+1}^{\prime }$ and $f(d)=(d)^{\mu ^{\prime }}$ is an increasing function of d, applying Theorem 1.2.8 of Müller and Stoyan (2002), we have

$$\begin{aligned} E(d_{t+1})^{\mu ^{\prime }}\ge E(d_{t+1}^{\prime })^{\mu ^{\prime }}=1. \end{aligned}$$

Thus

$$\begin{aligned} \log E(d_{t+1})^{\mu ^{\prime }}\ge 0. \end{aligned}$$

By Assumption 5, we have $\log E(d_{t+1})<0$. From Lemmas 1 and 2, we know that $\log E(d_{t+1})^{x}$ is convex and continuous in x. Thus there exists a $\mu >1$ such that $\log E\left( d_{t+1}\right) ^{\mu }=0$, and $\mu \le \mu ^{\prime }$. $\square$

1.3 Proof of Proposition 2

Proof

Suppose that $\chi >\chi ^{\prime }$. Thus by Eq. (A.2) we know that

$$\begin{aligned} d_{t+1}=\frac{\tilde{R}_{t+1}(1-b)}{1+\left[ \tilde{R}_{t+1}(1-b)\right] ^{ \frac{\gamma -1}{\gamma }}\chi ^{-\frac{1}{\gamma }}}>d_{t+1}^{\prime }= \frac{\tilde{R}_{t+1}(1-b)}{1+\left[ \tilde{R}_{t+1}(1-b)\right] ^{\frac{ \gamma -1}{\gamma }}\left( \chi ^{\prime }\right) ^{-\frac{1}{\gamma }}}. \end{aligned}$$

For $x\in \mathbb {R}$, $d_{t+1}\le x$ implies $d_{t+1}^{\prime }\le x$. Thus we have

$$\begin{aligned} Pr (d_{t+1}\le x)\le Pr (d_{t+1}^{\prime }\le x), \end{aligned}$$

for $x\in \mathbb {R}$. Then we know that $d_{t+1}\succeq _\mathrm{FSD}d_{t+1}^{\prime }$. Applying Theorem 2, we know that the Pareto exponent $\mu$ of the wealth distribution under $\chi$ is smaller than under $\chi ^{\prime }$. $\square$

1.4 Proof of Proposition 4

Proof

By Eq. (A.2), we have

$$\begin{aligned} d_{t+1}= & {} \frac{\tilde{R}_{t+1}(1-b)}{1+\left[ \tilde{R}_{t+1}(1-b)\right] ^{\frac{\gamma -1}{\gamma }}\chi ^{-\frac{1}{\gamma }}} \\= & {} \frac{1}{\left[ \tilde{R}_{t+1}(1-b)\right] ^{-1}+\left[ \tilde{R} _{t+1}(1-b)\right] ^{-\frac{1}{\gamma }}\chi ^{-\frac{1}{\gamma }}}. \end{aligned}$$

Thus $d_{t+1}$ decreases with b.

Suppose that $b<b^{\prime }$. Thus we have $d_{t+1}>d_{t+1}^{\prime }$. For $x\in \mathbb {R}$, $d_{t+1}\le x$ implies $d_{t+1}^{\prime }\le x$. Thus we have

$$\begin{aligned} Pr (d_{t+1}\le x)\le Pr (d_{t+1}^{\prime }\le x), \end{aligned}$$

for $x\in \mathbb {R}$. Then we know that $d_{t+1}\succeq _\mathrm{FSD}d_{t+1}^{\prime }$. Applying Theorem 2, we know that the Pareto exponent $\mu$ of the wealth distribution under b is smaller than under $b^{\prime }$. $\square$

1.5 Proof of Proposition 5

Proof

Suppose that $g<g^{\prime }$. Thus we have $\frac{d_{t+1}}{g}>\frac{ d_{t+1}}{g^{\prime }}$. For $x\in \mathbb {R}$, $\frac{d_{t+1}}{g}\le x$ implies $\frac{d_{t+1}}{g^{\prime }}\le x$. Thus we have

$$\begin{aligned} Pr \left( \frac{d_{t+1}}{g}\le x\right) \le Pr \left( \frac{d_{t+1}}{ g^{\prime }}\le x\right) , \end{aligned}$$

for $x\in \mathbb {R}$. Then we know that $\frac{d_{t+1}}{g}\succeq _\mathrm{FSD}\frac{d_{t+1}}{ g^{\prime }}$. Applying Theorem 2 to the process $\left\{ \hat{L} _{t}\right\}$, we know that the Pareto exponent $\mu$ of the wealth distribution under g is smaller than under $g^{\prime }$. $\square$

1.6 Serially correlated $\left\{ \tilde{R}_{t}\right\}$

We introduce a stationary Markov process $\left\{ x_{t}\right\}$ into the model such that the process $\left\{ (x_{t},\psi _{t})\right\}$, where $\psi _{t}=\left( \tilde{R}_{t},H_{t}\right)$, is a Markov-modulated process associated with $\{x_{t}\}$. Thus the Markov process $\left\{ x_{t}\right\}$ is the underlying process.

Assumption 1$^{\prime }$.$\{x_{t}\}$ is on the measurable space $( \mathbb {R} ,\mathbf {B})$, where $\mathbf {B}$ is the Borel sigma-algebra.

Assumption 2$^{\prime }$.$\{x_{t}\}$ is irreducible.

Assumption 3$^{\prime }$. Let $Q(x,\cdot )$ be the transition kernel of $\{x_{t}\}$. There exist a probability measure $\lambda$ on $( \mathbb {R} ,\mathbf {B})$, a number $m_{1}\in \mathbb {N}$, and a measurable density kernel $f(x,y): \mathbb {R} ^{2}\rightarrow [0,\infty )$ such that

$$\begin{aligned} Q^{m_{1}}(x,A)=\int _{A}f(x,y)\lambda (dy), \end{aligned}$$

and the family of functions $\{f(x,\cdot ): \mathbb {R} \rightarrow [0,\infty )\}_{x\in \mathbb {R} }$ is uniformly integrable with respect to the measure $\lambda$.

Assumption 4$^{\prime }$.$H_{t}\in (0,\bar{H})$.

Assumption 5$^{\prime }$.$\tilde{R}_{t}\in [$Ṟ$,\bar{R}]$ with Ṟ$>0$.

Assumption 6$^{\prime }$. Let $\Lambda (x)=\lim \sup _{n\rightarrow \infty }\frac{1}{n}\log E\left[ \prod \nolimits _{i=1}^{n-1} \left( d_{i}\right) ^{x}\right]$. There exist $\mu _{1}>1$ and $\mu _{2}>1$ such that $\Lambda (\mu _{1})\ge 0$ and $\Lambda (\mu _{2})<0$.

Assumption 7$^{\prime }$. There do not exist a constant $\alpha >0$ and a measurable function $\beta : \mathbb {R} \rightarrow [0,\alpha )$ such that²⁹

$$\begin{aligned} Pr \left( \log (d_{1})\in \beta (x_{0})-\beta (x_{1})+\alpha \mathbb {Z} \right) =1. \end{aligned}$$

Proof of Proposition 7

The process $\left\{ (x_{t},v_{t})\right\}$, where $v_{t}=\left( d_{t},\eta _{t}\right)$, is a Markov-modulated process associated with $\{x_{t}\}$ since the process $\left\{ (x_{t},\psi _{t})\right\}$, where $\psi _{t}=\left( \tilde{R} _{t},H_{t}\right)$, is a Markov-modulated process associated with $\{x_{t}\}$.

In order to apply Theorem 1.5 of Rointershtein (2007) to the process $\{L_{t}\}$, we will verify (A1)–(A7) of Assumption 1.2 in Rointershtein (2007).

(A1) is obviously satisfied since the Borel sigma-algebra is countably generated.

By Assumptions 2$^{\prime }$ and 3$^{\prime }$, (A2) and (A3) are satisfied.

From Assumption 4$^{\prime }$, we know that $Pr (H_{t+1}\in (0,\bar{H}))=1$. And from Assumption 5$^{\prime }$ we know that $Pr (\tilde{R}_{t+1}\in [$ Ṟ,$\bar{R}])=1$. Thus from Eq. (A.3) we know that there exists an $\bar{\eta }>0$ such that $Pr (\eta _{t+1}<\bar{\eta } )=1$. Thus (A4) is satisfied.

From Assumption 5$^{\prime }$, we know that $Pr (\tilde{R}_{t+1}\in [$ Ṟ, $\bar{R}])=1$. Thus $d_{t+1}$ is bounded, since $d_{t+1}$ is a continuous function of $\tilde{R}_{t+1}$ [see Eq. (A.2)]. We also know that $d_{t+1}$ is bounded away from zero since Ṟ$>0$. Thus there exists an $c_{\rho }>1$ such that $Pr (\frac{1}{c_{\rho }}<d_{t+1}<c_{\rho })=1$. Thus (A5) is satisfied.

By Assumptions 6$^{\prime }$ and 7$^{\prime }$, (A6) and (A7) are satisfied.

We have verified (A1)–(A7) of Assumption 1.2 in Rointershtein (2007). By Lemma 2.3 of Rointershtein (2007) we know that, for $x>0$, the following limit exists,

$$\begin{aligned} \Lambda (x)=\lim _{n\rightarrow \infty }\frac{1}{n}\log E\left[ \prod \limits _{i=1}^{n-1}\left( d_{i}\right) ^{x}\right] . \end{aligned}$$

We then show the following lemma.

Lemma 3

$\Lambda (x)$ is a convex function of $x>0$.

Proof

Note that $\log E\left[ \prod \nolimits _{i=1}^{n-1}\left( d_{i}\right) ^{x}\right] =\log E\left[ \left( \prod \nolimits _{i=1}^{n-1}d_{i}\right) ^{x} \right]$. Viewing $\left( \prod \nolimits _{i=1}^{n-1}d_{i}\right)$ as a random variable, we know, from Lemma 1, that $\log E\left[ \left( \prod \nolimits _{i=1}^{n-1}d_{i}\right) ^{x}\right]$ is a convex function of $x>0$. Thus $\frac{1}{n}\log E\left[ \prod \nolimits _{i=1}^{n-1}\left( d_{i}\right) ^{x}\right]$ is a convex function of $x>0$. Then we know that $\Lambda (x)$ is a convex function of $x>0$. $\square$

Thus we have Lemma 4 as a corollary to Lemma 3.

Lemma 4

$\Lambda (x)$ is a continuous function of $x>0$.

Proof

By Proposition 17 of Chapter 5 in Royden (1988), Lemma 3 implies Lemma 4. $\square$

Lemmas 3 and 4 imply that $\Lambda (x)$ is convex and continuous. From Assumption 6$^{\prime }$, we know that there exist $\mu _{1}>1$ and $\mu _{2}>1$ such that $\Lambda (\mu _{1})\ge 0$ and $\Lambda (\mu _{2})<0$. Thus there exists a unique $\mu >1$ such that

$$\begin{aligned} \Lambda (\mu )=0. \end{aligned}$$

Applying Theorem 1.5 of Rointershtein (2007) to the process $\{L_{t}\}$, we have

$$\begin{aligned} \lim _{x\rightarrow \infty }\frac{Pr (L_{\infty }>x)}{x^{-\mu }}=c, \end{aligned}$$

with $c>0$. $\square$

1.7 Proof of Proposition 8

Proof

Suppose that $\chi >\chi ^{\prime }$. Thus by Eq. (A.2) we know that

$$\begin{aligned} d_{t+1}=\frac{\tilde{R}_{t+1}(1-b)}{1+\left[ \tilde{R}_{t+1}(1-b)\right] ^{ \frac{\gamma -1}{\gamma }}\chi ^{-\frac{1}{\gamma }}}>d_{t+1}^{\prime }= \frac{\tilde{R}_{t+1}(1-b)}{1+\left[ \tilde{R}_{t+1}(1-b)\right] ^{\frac{ \gamma -1}{\gamma }}\left( \chi ^{\prime }\right) ^{-\frac{1}{\gamma }}}. \end{aligned}$$

Let

$$\begin{aligned} \Lambda _{1}(x)=\lim _{n\rightarrow \infty }\frac{1}{n}\log E\left[ \prod \limits _{i=1}^{n-1}\left( d_{i}^{\prime }\right) ^{x}\right] , \end{aligned}$$

for $x>0$. Suppose that $\Lambda _{1}(\mu ^{\prime })=0$. Thus we have

$$\begin{aligned} \Lambda (\mu ^{\prime })=\lim _{n\rightarrow \infty }\frac{1}{n}\log E\left[ \prod \limits _{i=1}^{n-1}\left( d_{i}\right) ^{\mu ^{\prime }}\right] \ge \lim _{n\rightarrow \infty }\frac{1}{n}\log E\left[ \prod \limits _{i=1}^{n-1} \left( d_{i}^{\prime }\right) ^{\mu ^{\prime }}\right] =0. \end{aligned}$$

From Assumption 6$^{\prime }$, we know that there exists $\mu _{2}>1$ such that $\Lambda (\mu _{2})<0$. Then we know that there exists a $\mu >1$ such that $\Lambda (\mu )=0$, and $\mu \le \mu ^{\prime }$. $\square$

1.8 Proof of Proposition 9

Proof

I apply Theorem 1.8 of Mirek (2011) to prove Proposition 9 . I need to verify Assumptions 1.6 and 1.7 of Mirek (2011).

Let $\vartheta =\left( \tilde{R},H\right)$ and

$$\begin{aligned} \psi _{\vartheta }(L)=\left\{ \begin{array}{ll} H-(1-b)\tilde{R}\theta L+G, &{} \text { if }L\le \phi \\ dL+\eta , &{} \hbox {otherwise} \end{array} \right. . \end{aligned}$$

The process $\left\{ L_{t}\right\}$ is generated by $L_{t+1}=\psi _{\vartheta }(L_{t})$. Thus $\psi _{\vartheta }(L)$ is Lipschitz continuous.

Verification of Assumption 1.6 of Mirek (2011). For every $z>0$, let

$$\begin{aligned} \psi _{\vartheta ,z}(L)=z\psi _{\vartheta }\left( \frac{1}{z}L\right) . \end{aligned}$$

$\psi _{\vartheta ,z}$ are called dilatations of $\psi _{\vartheta }$. Let

$$\begin{aligned} \bar{\psi }_{\vartheta }(L)=\lim _{z\rightarrow 0}\psi _{\vartheta ,z}(L). \end{aligned}$$

Thus we have

$$\begin{aligned} \bar{\psi }_{\vartheta }(L)=\lim _{z\rightarrow 0}\psi _{\vartheta ,z}(L)=\lim _{z\rightarrow 0}\left[ z\psi _{\vartheta }\left( \frac{1}{z} L\right) \right] =dL,\text { for }\forall L\ge 0. \end{aligned}$$

Since $\psi _{\vartheta }(L)$ is piecewise linear, It is easy to find a random variable $N_{\vartheta }$ with bounded support such that

$$\begin{aligned} |\psi _{\vartheta }(L)-dL|\le N_{\vartheta },\text { for }\forall L\ge 0. \end{aligned}$$

Assumption 1.6 of Mirek (2011) is satisfied.

Verification of Assumption 1.7 of Mirek (2011). As for Assumption 1.7 of Mirek (2011), condition (H3) is satisfied since

$$\begin{aligned} d_{t+1}=\frac{\tilde{R}_{t+1}(1-b)}{1+\left[ \tilde{R}_{t+1}(1-b)\right] ^{ \frac{\gamma -1}{\gamma }}\chi ^{-\frac{1}{\gamma }}}, \end{aligned}$$

by Eq. (A.2). $\left\{ \tilde{R}_{t}\right\}$ is i.i.d. along time and the support of $\tilde{R}_{t}$ is closed.

The law of $\log d$ is nonarithmetic since $\tilde{R}_{t}$ has a probability density function $l(\cdot )$ on [Ṟ,$\bar{R}]$ by Assumption 3$^{\prime \prime }$. Thus (H4) in Assumption 1.7 of Mirek (2011) is satisfied.

Let $m(x)=\log E(d)^{x}$. From Assumption 4$^{\prime \prime }$, we have $E\left( d\right) <1$. Thus $m(1)<0$. By Assumption 5$^{\prime \prime }$, $E(d)^{2}>1$. Thus $m(2)>0$. From Lemmas 1 and 2, we know that m(x) is convex and continuous. Thus there exists a unique $\mu \in (1,2)$ such that $m(\mu )=0$, i.e.

$$\begin{aligned} E(d)^{\mu }=1. \end{aligned}$$

We also know that $E\left( d^{\mu }|\log d|\right) <\infty$, since d is bounded.

$E\left[ \left( N_{\vartheta }\right) ^{\mu }\right] <\infty$, since $N_{\vartheta }$ is bounded.

Assumption 1.7 of Mirek (2011) is satisfied.

Applying Theorem 1.8 of Mirek (2011), we find that the stationary distribution of the process $\left\{ L_{t}\right\}$, $L_{\infty }$, has an asymptotic Pareto tail, i.e.

$$\begin{aligned} \lim _{x\rightarrow \infty }\frac{Pr (L_{\infty }>x)}{x^{-\mu }}=c, \end{aligned}$$

with $c>0$. $\square$