Stochastic dominance and thick-tailed wealth distributions

Abstract

Right-skewed and thick-tailed wealth distributions have been documented as an empirical regularity across space and time. A key mechanism for explaining these distributional features is proportional random growth. We investigate the comparative statics of a well-defined class of random growth models when allowing for stochastically ordered shifts in the wealth return process. An order-contingent monotone comparative statics property is identified, according to which pure increases in risk (e.g. higher volatility of capital returns) foster top wealth concentration whereas first-order stochastically dominated shifts in the return process (induced by e.g. proportional capital income taxation) rather lower inequality at the upper end of the distribution. Our analysis points to the potentially ambiguous effects on top wealth inequality of introducing or modifying capital income tax treatments in the presence of stochastic returns.

Appendix

Thick-tailed distributions

Let L(x) be a real-valued function. Then L(x) is said to vary regularly with index $\mu \in (0, \infty )$ if

$$\begin{aligned} \lim _{x \rightarrow \infty } \frac{L(tx)}{L(x)}=t^{-\mu },\quad \forall t>0 \end{aligned}$$

A probability distribution with a differentiable CDF F(x) is a power-law with tail index $\mu$ if $L(x):=\left( 1-F(x) \right)$ is regularly varying with index $\mu >0$. The distribution F(x) is then said to be thick-tailed with tail index $\mu$.

Regularity conditions for Kesten processes (Kesten 1973).

Consider the stochastic Eq. (1) where $(A_t, B_t) \in \mathbb {R}^2_+$ are i.i.d. random variables. For some $\mu >1$, let the the following hold:

$$\begin{aligned} E[A^\mu ]=1;\quad E\left[ (A)^\mu \max \left\{ ln(A), 0 \right\} \right] <\infty ;\quad E[B]\in (0, \infty ) \end{aligned}$$

Also, assume $\left( B/(1-A) \right)$ is non-degenerate and the distribution of ln(A) is non-lattice (i.e. its support is not contained in $\lambda \mathbb {Z}$ for some $\lambda$). Then there exist constants $(c_{-}, c_+)$—at least one positive—such that for $x \rightarrow \infty$:

$$\begin{aligned} \begin{aligned} x^{\mu } Pr(W>x) \quad \rightarrow \quad c_+ \\ x^{\mu } Pr(W<-x) \quad \rightarrow \quad c_{-} \end{aligned} \end{aligned}$$

where W is the stationary solution to (1), i.e. $W\overset{d}{=} A W+B$. The solution to (1) then converges in probability to W as $t \rightarrow \infty$.

Proof of Proposition 1

Recall that we restrict attention to sequences $(A_t)$ and $(A_t')$ complying with Assumption 1, and therefore associated with tail indexes $\mu _A >1$ and $\mu _{A'}>1$, respectively. Let a denote the generic realization of the random variable A. We start noticing that the function $g: a \mapsto a^{\mu }$ is strictly increasing and strictly convex for all $a \in \mathbb {R}_+$.

• i) Consider $F_{A} \succcurlyeq _{1} F_{A'}$ with $F_{A} \ne F_{A'}$. Then from Theorem 1, part (i) we have

  $$\begin{aligned} 1=E[(A')^{\mu _{A'}}]=\int _{\mathbb S} (a')^{\mu _{A'}} d F_{A'} < \int _{\mathbb S} (a)^{\mu _{A'}} d F_{A} \end{aligned}$$

  Suppose that $\mu _{A'} \le \mu _{A}$. Then by Holder’s inequality we have

  $$\begin{aligned} \int _{\mathbb S}(a)^{\mu _{A'}}dF_{A} \le \left[ \int _{\mathbb S}\left( (a)^{\mu _{A'}} \right) ^{\frac{\mu _A}{\mu _{A'}}}dF_{A} \right] ^{\frac{\mu _{A'}}{\mu _A}}=\left[ \int _{\mathbb S} (a)^{\mu _A}dF_{A} \right] ^{\frac{\mu _{A'}}{\mu _A}}=1 \end{aligned}$$

  which is a contradiction. Hence it must be $\mu _{A'}> \mu _A$.

• ii) Consider $F_{A} \succcurlyeq _{2} F_{A'}$ with $\int _{\mathbb S} a dF_{a} = \int _{\mathbb S} a' dF_{A'}$ $F_{A} \ne F_{A'}$ From Theorem 1, part (ii) we have

  $$\begin{aligned} 1=E[(A)^{\mu _A}]=\int _{\mathbb S} (a)^{\mu _A} d F_{a} < \int _{\mathbb S} (a')^{\mu _A} d F_{A'}=E[(A')^{\mu _A}] \end{aligned}$$

  and the second assertion follows by applying again Holder’s inequality.

$\square$

Proof of Corollary 1

Since the composite function $g: z \mapsto (\phi (z))^{\mu }$—with $\mu >1$—is strictly increasing if and only if $d \phi /d z >0$ for all z, and strictly convex if and only if

$$\begin{aligned} (\mu -1) \phi (z)^{\mu -2} \left( d \phi / d z \right) ^2+\phi (z)^{\mu -1}(d^2 \phi / d z^2) >0\quad \forall z \end{aligned}$$

which is equivalent to requiring

$$\begin{aligned} \frac{d^2 \phi }{d z^2}>-(\mu -1)\frac{\left( d \phi / d z \right) ^2}{\phi (z)}=\Lambda (z)\quad \forall z \end{aligned}$$

(11)

both points follow readily from Proposition 1. $\square$

Proof of Proposition 2

In the following, we consider the transformation $R'=(1-\tau (R))R$ of the random variable $R \sim F_R$ (a copy of $R_t$), with $F_{R'}=F_{(1-\tau (R))R}$.

• i) The assertion follows immediately if the mapping $(1-\tau (r_t))r_t$ is monotone in $r_t$ (and hence invertible). More generally, let $g: \mathbb {R}_+ \mapsto \mathbb {R}_+$ be a function such that for any $x \in \mathbb {R}_{+}$ it holds $g(x) \le x$. Let also M denote the set of all random variables $X: \Omega \mapsto \mathbb {R}_+$ defined on a properly filtered probability space $(\Omega , \Sigma , P)$. Then there always exists a function $g^*: M \mapsto M$ such that for any $\omega \in \Omega$ the function $g^*(X): \Omega \rightarrow \mathbb {R}_+$ is given by $g^*(X)(\omega )=g(X(\omega ))$.

  Consider now a random variable X with density $f_X$, and define $\Gamma _{y}:=\left\{ x= X(\omega ): g(x)\le y \right\}$ for any real y. By construction $y\le x$ for all x belonging to the border of $\Gamma _y$, i.e for all $x\in \partial \Gamma _y:=\left\{ x: g(x)= y \right\}$. Let $\underline{x}=\min \partial \Gamma _y$, then it holds

  $$\begin{aligned} Pr(g^*(X)<y) = \int _{\Gamma _{y}}f(x) dx \ge \int _{0}^{\underline{x}} f(x) dx \ge \int _{0}^{y} f(x) dx = Pr(X< y) \end{aligned}$$

  (12)

  Since the after-tax rate of return $r'_t=(1-\tau (r_t))r_t$ satisfies the restriction $r'_t \le r_t$, it plays the role of the $g^*(\cdot )$ function in the above. Using the first and the last terms in (12), we can therefore write $Pr(R_t<y) \le Pr(R'_t <y)$ for any real y, from which it follows $F_R \succcurlyeq _{1} F_{R'}$.

• ii) Let $\tau (r_t)=\tau \in (0,1)$, such that $r'_t=(1-\tau )r_t$. By the previous point $F_R$ first-order stochastically dominates $F_{R'}$. However, since the imposition of a capital income tax distorts households’ saving decisions, a behavioral response arises—i.e. a change in the share $\psi (\theta , \delta , F_R)$—whose strength depends on the underlying probability distribution shift. Hence, Corollary 1 cannot be applied. In order to resort to Proposition 1, we need to show that $F_R \succcurlyeq _{1} F_{R'}$ implies $F_{A(R)} \succcurlyeq _{1} F_{A(R')}$. Under CRRA preferences, the effective random return on wealth (i.e. the multiplicative coefficient of the wealth accumulation equation) is

  $$\begin{aligned} A(X)=\frac{X}{1+\left( \theta E[X^{1-\delta }] \right) ^{-\frac{1}{\delta }}},\quad X \in \left\{ R, R' \right\} \end{aligned}$$

  (13)

$\square$

Let $R'=(1-\tau )R$. Suppose, by contradiction, that there exists some $r \in \mathbb {S} \subset \mathbb {R}_+$ such that

$$\begin{aligned} Pr(A(R)<r)>Pr(A(R')<r) \end{aligned}$$

which is equivalent to requiring

$$\begin{aligned} r \cdot \left[ 1+\left( \theta E[R^{1-\delta }] \right) ^{-\frac{1}{\delta }} \right] > \frac{r}{1-\tau } \cdot \left[ 1+\left( \theta E[((1-\tau )R)^{1-\delta }] \right) ^{-\frac{1}{\delta }} \right] \end{aligned}$$

(14)

or

$$\begin{aligned} 1+\left( \theta E[R^{1-\delta }] \right) ^{-\frac{1}{\delta }} > \frac{1}{1-\tau } +\frac{1}{(1-\tau )^{1/\delta }}\left( \theta E[R^{1-\delta }] \right) ^{-\frac{1}{\delta }} \end{aligned}$$

(15)

Since $\tau \in (0,1)$, by simple algebra a contradiction is obtained. Hence, for all $r \in \mathbb {S}$ it must be that $Pr(A(R)<r)<Pr(A(R')<r)$, or $F_{A(R)} \succcurlyeq _{1} F_{A(R')}$.

Proof of Corollary 2

We first show that, for any tax schedule $\tau (\cdot )$, it holds $F_{A(R)} \succcurlyeq _{1} F_{A(R')}$ if $\delta \le 1$ (sufficiency). This is trivially true in the knife-edge case $\delta =1$, as the imposition of a capital income tax does not change the fraction of current wealth devoted to savings, Corollary 1 part i) therefore applies.

Consider now the case $\delta <1$. Since $g: z \mapsto (z)^{1-\delta }$ is an increasing function, $R \succcurlyeq _{1} R'$ implies $E[R^{1-\delta }]>E[R'^{1-\delta }]$. Recalling the saving decision rule (9), this in turn implies $\psi (\theta , \delta , F_R)>\psi (\theta , \delta , F_{R'})$. According to the invariance property of the stochastic dominance ordering, if $R \succcurlyeq _{1} R'$ then $c\cdot R \succcurlyeq _{1} c\cdot R'$ for any constant $c>0$. The assertion then readily follows from noticing that $R' \succcurlyeq _{1} \frac{\psi (\theta , \delta , F_{R'})}{\psi (\theta , \delta , F_R)} R'$.

To prove the necessity part, we show that, when $\delta >1,$ there exist tax schedules $\tau (\cdot )$ for which $F_R \succcurlyeq _{1} F_{R'}$ does not imply $F_{A(R)} \succcurlyeq _{1} F_{A(R')}$. To this end, for a fixed (realized) rate of return $\underline{r} \in \mathbb {S}$, let us consider a progressive capital income tax schedule which takes a simple two-bracket form:

$$\begin{aligned} \tau (r)= \left\{ \begin{array}{rl} \tau _{min}, &{}\text{ if } \, r < \underline{r} \\ \\ \tau _{max}, &{}\text{ if } \, r \ge \underline{r}\\ \end{array} \right. \end{aligned}$$

(16)

where $0\le \tau _{min}< \tau _{max}<1$. For all random variables $R \in \mathbb {S}$ the following stochastic dominance ordering holds:

$$\begin{aligned} (1-\tau _{min}) R \succcurlyeq _1(1-\tau (R)) R \succcurlyeq _1 (1-\tau _{max}) R \end{aligned}$$

(17)

Let us now consider the random variable defined by (13). For $r < \underline{r}$ we have

$$\begin{aligned} Pr(A((1-\tau (r)) R)< r) = Pr\left( R < r \left( \frac{1+\left( \theta E[((1-\tau (r)) R)^{1-\delta }] \right) ^{-\frac{1}{\delta }}}{1-\tau _{min}}\right) \right) \end{aligned}$$

(18)

and

$$\begin{aligned} Pr(A((1-\tau _{min}) R)< r) = Pr\left( R < r \left( \frac{1+\left( \theta E[((1-\tau _{min} R)^{1-\delta }] \right) ^{-\frac{1}{\delta }}}{1-\tau _{min}}\right) \right) \end{aligned}$$

(19)

For $\delta > 1$, we get $E[((1-\tau (r)) R)^{1-\delta }] > E[((1-\tau _{min}) R)^{1-\delta }]$, from which it follows

$$\begin{aligned} \left( \frac{1+\left( \theta E[((1-\tau (r)) R)^{1-\delta }] \right) ^{-\frac{1}{\delta }}}{1-\tau _{min}}\right) < \left( \frac{1+\left( \theta E[((1-\tau _{min} R)^{1-\delta }] \right) ^{-\frac{1}{\delta }}}{1-\tau _{min}}\right) \end{aligned}$$

(20)

Hence, it must be that $Pr(A((1-\tau _{min}) R)< r) > Pr(A((1-\tau (r)) R) < r)$, and the first-order stochastic dominance ordering between the random variables $(1-\tau _{min}) R$ and $(1-\tau (R)) R$ is not inherited by the random variables $A((1-\tau _{min}) R)$ and $A((1-\tau (r)) R)$. $\square$

Piece-wise linear taxation.

The CDF of the after-tax return on wealth $(1-\tau )r_t$ under the flat rate $\tau$—with the pre-tax $r_t \sim U[0, \bar{r}]$ – simply reads as $F(r)=\frac{r}{\bar{r}(1-\tau )}$, with mean $r_{M}:=E[(1-\tau )r_t]=(1-\tau )\frac{\bar{r}}{2}$. Any mean-preserving spread on $r_t$—call it $r'_t$, with associated CDF $F'(\cdot )$—obtained through a change in the tax system, must be such that $\int _{0}^{\bar{r}(1-\tau )}r'_t dF'=r_{M}$ and $\int _0^{y} F(r)dr \le \int _0^{y} F'(r)dr$ for all $y \le \bar{r}(1-\tau )$.

A straightforward instance of such risk transformation is a piece-wise linear tax scheme according to which (i) a lower flat rate $\tau '<\tau$ is imposed on all taxpayers, and (ii) a fixed tax T (fulfilling some conditions) is also raised on higher-than-average returns on wealth $r > r_M$. Specifically, let $1-\tau '=\frac{1-\tau }{\gamma }$ for a given $\gamma \in (0,1)$ and T such that $T<r_M$ (all taxpayers gaining $r>r_M$ on their wealth can afford paying this lump sum) and $T=(\tau -\tau ')\bar{r}>0$ (the lump sum is strictly positive). Combining these requirement yields $T=\frac{1-\gamma }{\gamma }(1-\tau )\bar{r}$ with $\gamma >\frac{2}{3}$. It is easy to see that the CDF of the after-tax rates of return on wealth in the presence of the new flat rate $\tau '$ writes as

$$\begin{aligned} F'(r) = \left\{ \begin{array}{ll} \frac{r}{\bar{r} (1-\tau ')} &{} \quad \text{ if } \quad r \le r_M \\ \\ \frac{r}{\bar{r} (1-\tau ')} +(1-\gamma ) &{} \quad \text{ if } \quad r>r_{M} \end{array} \right. \end{aligned}$$

(21)

and hence the associated mean $E[r'_t]$ is equal to $r_M$. Since $F(0)=F'(0)=0$ and $F\left( (1-\tau )\bar{r} \right) =F'\left( (1-\tau )\bar{r}\right) =1$, by the equality of the means we have

$$\begin{aligned} \int _{0}^{(1-\tau ) \bar{r}}\left( F(r)-F'(r) \right) dr= & {} \left[ - \int _0^{(1-\tau ) \bar{r}} r (f( r ) - f'( r ) ) d r \right] \\&+\,\left[ F\left( (1-\tau )\bar{r} \right) -F'\left( (1-\tau )\bar{r} \right) \right] (1-\tau )\bar{r}=0 \end{aligned}$$

from which it follows $\int _0^{y} F(r)dr \le \int _0^{y} F'(r)dr$ for all y. Hence, $F'$ constitutes a mean-preserving second-order dominant change in risk.