Diversification, gambling and market forces

Abstract

Though simple and appealing, mean-variance portfolio choice theory does not describe actual diversification choices by investors, especially their propensity to gamble and the solvency constraints they face. Using 8 million trades realized by 90,000 individual investors, we show that diversification choices are in fact strongly driven by the skewness of returns, especially in bull markets, but also by the amount to be invested in risky assets. Increasing this amount by 10 % leads to increase by 3.8 % the number of stocks in investors’ portfolios, controlling for portfolio skewness. An important contribution of this paper is to show that the strength of the relationship between diversification and the skewness of returns is shaped by market forces. A strong negative relationship exists in bull markets but disappears in bear markets, a result not found in the literature. Our results survive several robustness checks, including controlling for individual heterogeneity and time-variability of stock price co-movements.

Appendix: Moments of equally-weighted portfolios of Arrow–debreu securities

Let $\Omega =\left\{ \omega _{1},\ldots ,\omega _{n}\right\}$ denote a finite state-space with $n$ equally-likely states of nature and assume that all Arrow–Debreu securities, denoted $X_{1},\ldots, X_{n}$, are traded. $X_{i}$ pays 1 in state $\omega _{i}$ and 0 elsewhere. $(p_{1},\ldots,p_{n})$ stands for a sequence of equally-weighted portfolios containing respectively $1,2,\ldots, n,$ AD securities. Without loss of generality, we assume that $p_{k}$ contains $1/k$ units of each of the first $k$ securities.¹⁶

Before analyzing portfolios, we briefly recall the elementary properties of the moments of AD securities.

Proposition 1

For any $1\le k\le n$,

$$\begin{aligned} E(X_{k})=m_{k}=\frac{1}{n}\end{aligned}$$

(9)

$$\begin{aligned} V(X_{k})=\frac{n-1}{n^{2}}\end{aligned}$$

(10)

$$\begin{aligned} E\left[ (X_{k}-m_{k})^{3}\right]=\frac{(n-1)(n-2)}{n^{3}}\end{aligned}$$

(11)

$$\begin{aligned} cov(X_{k},X_{k^{*}})=-\frac{1}{n^{2}}\text { if }k\ne k^{*} \end{aligned}$$

(12)

Proof

The first point is obvious since states are equally-likely. $\sigma _{i}^{2}=E(X_{i}^{2})-E(X_{i})^{2}=\frac{1}{n}- \frac{1}{n^{2}}$ since $X_{i}^{m}=X_{i}$ for any positive integer $m$. The third central moment is calculated as follows

$$\begin{aligned} E\left[ (X_{i}-\mu _{i})^{3}\right]&=E(X_{i}^{3})-3\mu _{i} E(X_{i}^{2})+3\mu _{i}^{2}E(X_{i})-\mu _{i}^{3}\\ &=\frac{1}{n}-3\frac{1}{n^{2}}+3\frac{1}{n^{3}}-\frac{1}{n^{3}} =\frac{1}{n}-\frac{3}{n^{2}}+\frac{2}{n^{3}} \\ &=\frac{(n-1)(n-2)}{n^{3}} \end{aligned}$$

(13)

Finally, we get $cov(X_{i},X_{j})=E(X_{i}X_{j})- E(X_{i}) E(X_{j})=-1/n^{2}$ since $X_{i}X_{j}\equiv 0$ when $i\ne j$.

Consider now a portfolio $p_{k}$ invested in the first $k$ AD securities and denote $\mu _{k}$ ($\sigma _{k}^{2})$ the expectation (variance) of payoffs of $p_{k}$.

Proposition 2

$$\begin{aligned} \forall k,\mu _{k}&=\frac{1}{n}\\ \sigma _{k}^{2}&=\frac{1}{n}\left( \frac{1}{k}- \frac{1}{n}\right) \end{aligned}$$

(14)

Proof

Proposition 1 allows to write the covariance matrix of the $n$ AD securities payoffs as

$$\begin{aligned} {\mathbf {V}}_{n}=\frac{1}{n}{\mathbf {I}}_{n}-\frac{1}{n^{2}} {\mathbf {1}}_{(n,n)} \end{aligned}$$

(15)

where ${\mathbf {I}}_{n}$ is the $(n,n)$ identity matrix and ${\mathbf {1}}_{(n,n)}$ is a $(n,n)$ matrix containing only ones. As $p_{k}=\frac{1}{k} { \sum \nolimits _{i=1}^{k}} X_{i}$, we get

$$\begin{aligned} \sigma _{k}^{2}=\frac{1}{k^{2}}{\mathbf {1}}_{(k)}^{\prime } {\mathbf {V}}_{k}{\mathbf {1}}_{(k)} \end{aligned}$$

(16)

where ${\mathbf {1}}_{(k)}$ denotes a column vector of ones with $k$ components and ${\mathbf {V}}_{k}$ the square matrix of the first $k$ rows and columns of ${\mathbf {V}}_{n}$.

Equation (15) implies ${\mathbf {V}}_{k}=\frac{1}{n} {\mathbf {I}}_{k}-\frac{1}{n^{2}}{\mathbf {1}}_{(k,k)}.$ We then write

$$\begin{aligned} \sigma _{k}^{2}&=\frac{1}{k^{2}}{\mathbf {1}}_{(k)}^{\prime } {\mathbf {V}}_{k}{\mathbf {1}}_{(k)}=\frac{1}{k^{2}}{\mathbf {1}}_{(k)}^{\prime } \left( \frac{1}{n}{\mathbf {I}}_{k}-\frac{1}{n^{2}}{\mathbf {1}}_{(k,k)} \right) {\mathbf {1}}_{(k)} \\ &=\frac{1}{k^{2}n}{\mathbf {1}}_{(k)}^{\prime }{\mathbf {I}}_{k} {\mathbf {1}}_{(k)}-\frac{1}{k^{2}n^{2}}{\mathbf {1}}_{(k)}^{\prime } {\mathbf {1}}_{(k,k)}{\mathbf {1}}_{(k)} \\ &=\frac{1}{kn}-\frac{1}{n^{2}}=\frac{1}{n}\left( \frac{1}{k}-\frac{1}{n}\right) \end{aligned}$$

(17)

As expected, the variance of the equally-weighted portfolio decreases with the number of AD securities in the portfolio. The case $k=n$ gives $\sigma _{n}^{2}=0$ which is consistent withe fact that $p_{n}$ is a risk-free portfolio paying $1/n$ in each state.

The inverse of the number of stocks in portfolios is often considered as a measure of diversification (denoted $D_{1}$ by Mitton and Vorkink (2007)). Proposition 2 shows that the variance of returns increases linearly with $D_{1}$. When $k=n$, the portfolio is risk-free and the variance of payoffs is equal to 0.

Denote now $s_{k}^{3}$ the third central moment of $p_{k}$ defined by:

$$\begin{aligned} s_{k}^{3}=E\left( \left( \frac{1}{k} {\sum \limits _{j=1}^{k}} X_{j}-\mu _{k}\right) ^{3}\right) =\frac{1}{k^{3}}E\left( \left( {\sum \limits _{j=1}^{k}} X_{j}-\frac{k}{n}\right) ^{3}\right) \end{aligned}$$

(18)

Denoting $Y_{k}={ \sum \nolimits _{j=1}^{k}} X_{j}$ gives $s_{k}^{3}=\frac{1}{k^{3}}E((Y_{k}-\frac{k}{n})^{3})$. The specificities of AD securities imply that $E(Y_{k}^{3})=E(Y_{k}^{2})=\frac{k}{n}$. In fact, these relations simply come from the fact that $X_{j}^{m}X_{j^{*}}^{t}=0$ for any pair $(m,t)$ of strictly positive integers and different indices $j$ and $j^{*}$. We now get easily $s_{k}^{3}$.

Proposition 3

The central third moment of $p_{k}$ is valued:

$$\begin{aligned} s_{k}^{3}=\frac{1}{n^{3}}\left[ \left( \frac{n}{k}-1\right) \left( \frac{n}{k}-2\right) \right] \end{aligned}$$

(19)

Proof

$$\begin{aligned} s_{k}^{3}&=\frac{1}{k^{3}}E\left[ (Y_{k}^{3}- \left( \frac{k}{n}\right) ^{3}-3\left( \frac{k}{n}\right) Y_{k}^{2}+3\left( \frac{k}{n}\right) ^{2}Y_{k}\right] \\ &=\frac{1}{k^{3}}\left[ \frac{k}{n}-\left( \frac{k}{n} \right) ^{3}-3\left( \frac{k}{n}\right) ^{2}\,+\,3 \left( \frac{k}{n}\right) ^{3}\right] \end{aligned}$$

(20)

Rearranging terms leads to

$$\begin{aligned} s_{k}^{3}&=\frac{1}{n^{3}}\left[ \left( \frac{n}{k}\right) ^{2} -3\left( \frac{n}{k}\right) +2\right] \\ &=\frac{1}{n^{3}}\left[ \left( \frac{n}{k}-1\right) \left( \frac{n}{k}\,-2\right) \right] \end{aligned}$$

(21)

We know that $k<n;$ consequently an equally weighted portfolio has a positive skewness as long as the number of AD securities it contains is lower than $n/2$. Beyond this threshold, skewness becomes negative. When $n$ is even, the distribution of returns is symmetric for $k=n/2$, leading to a zero skewness for the portfolio return. Using the above diversification measure $D_{1}$, we get that the third order moment increases quadratically in $D_{1}$. In fact, we have:

$$\begin{aligned} s_{k}^{3}=\frac{1}{n}\left[ \left( D_{1}-\frac{1}{n}\right) \left( D_{1}-\frac{2}{n}\right) \right] \end{aligned}$$

(22)

We can also establish a very simple relationship between $s_{k}^{3}$ and $\sigma _{k}^{2}$ using Eqs. (14) and (19).

$$\begin{aligned} s_{k}^{3}=\sigma _{k}^{2}\left( D_{1}-\frac{2}{n}\right) \end{aligned}$$

(23)