Can fat-tail create the momentum and reversal?

ABSTRACT

We suggest that fat-tail variations can cause both short-term momentum and long-term reversal simultaneously, in both the time series and cross-sectional returns of securities. The fat-tail of the distribution is known to explain many anomalies in the financial market, but not momentum. To support our argument, we adopt widely accepted models in the literature, which generate reversal only, and revise a single assumption: Each random variable follows a non-normal stable distribution rather than a normal distribution. This single difference generates additional short-term return momentum. This finding shows that 1) investor irrationality is not essential to generate both phenomena, and 2) we must be cautious not to overuse normal distributions in the models.

KEYWORDS:

• Stable distribution
• momentum
• reversal
• market efficiency

JEL CLASSIFICATION:

• G12
• G14

Disclosure statement

This paper extends chapter 2 of Bae's doctoral dissertation.

Proof of Proposition 1

Equation (7)(7) $d{f}_{i,t}=-{\theta }_i{f}_{i,t}dt+d{L}_{f,i,t},i=1,\dots ,I$(7) is equivalent to

(A4) $f_{i,t}=e^{-{\theta }_{i}t}{\int }_{-\mathrm{\infty }}^t{e}^{{\theta }_{i}s}d{L}_{f,i,s},i=1,\dots ,I.$(A4)

For convenience, we define the following five variables:

(A5) $X_{i,1}={\int }_0^T{e}^{{\theta }_{i}s}d{L}_{f,i,t},$(A5)

(A6) $X_{i,2}={\int }_{-\tau }^0{e}^{{\theta }_{i}s}d{L}_{f,i,t},$(A6)

(A7) $X_{i,3}={\int }_{-\mathrm{\infty }}^{-\tau }{e}^{{\theta }_{i}s}d{L}_{f,i,t},$(A7)

(A8) $Y_{j,1}=L_{g,j,T}-L_{g,j,0},$(A8)

(A9) $Y_{j,2}=L_{g,j,0}-L_{g,j,-\tau }.$(A9)

Then, the scale parameters of $X_{i,2}$, $X_{i,3}$, and $Y_{j,2}$ are given as

(A10) ${\left(\frac{1}{\alpha {\theta }_i}\left(1-e^{-\alpha {\theta }_i\tau }\right)\right)}^{1/\alpha },$(A10)

(A11) ${\left(\frac{1}{\alpha {\theta }_i}\right)}^{1/\alpha }{e}^{-{\theta }_i\tau },$(A11)

(A12) ${\tau }^{1/\alpha },$(A12)

respectively. From these representations, we can decompose each factor at T, 0, and $-\tau$, as follows:

(A13) $f_{i,T}=e^{-{\theta }_{i}T}\left(X_{i,1}+X_{i,2}+X_{i,3}\right),$(A13)

(A14) $f_{i,0}=X_{i,2}+X_{i,3},$(A14)

(A15) $f_{i,-\tau }=e^{{\theta }_i\tau }{X}_{i,3},$(A15)

(A16) $g_{j,T}=Y_{j,1}+Y_{j,2}+L_{g,j,-\tau ,}$(A16)

(A17) $g_{j,0}=Y_{j,2}+L_{g,j,-\tau },$(A17)

(A18) $g_{j,-\tau }=L_{g,j,-\tau }.$(A18)

Then, from these forms, the returns can be written as

(A19) $r_k\left(0,T\right)=\sum _{i=1}^I{\beta }_{f,i,k}\left(e^{-{\theta }_{i}T}{X}_{i,1}-\left(1-e^{-{\theta }_{i}T}\right)X_{i,2}\text{break}-\left(1-e^{-{\theta }_{i}T}\right)X_{i,3}\right)+\sum _{j=1}^J{\beta }_{g,j,k}{Y}_{j,1},$(A19)

(A20) $r_k\left(-\tau ,0\right)=\sum _{i=1}^I{\beta }_{f,i,k}\left(X_{i,2}-\left(e^{{\theta }_i\tau }-1\right)X_{i,3}\right)+\sum _{j=1}^J{\beta }_{g,j,k}{Y}_{j,2}.$(A20)

Then, according to Lemma 1, we have the relation

(A21) $\mathrm{E}\left[r_k\left(0,T\right)|r_k\left(-\tau ,0\right)\right]=A_k\left(\tau ,T\right)r_k\left(-\tau ,0\right),$(A21)

where

(A22) $A_k\left(\tau ,T\right)\equiv \begin{array}{c}\underset{\_}{\sum _{i=1}^I\frac{{\left|{\beta }_{f,i,k}\right|}^{\alpha }}{\alpha {\theta }_i}\left(1-e^{-{\theta }_{i}T}\right)\left(-\left(1-e^{-\alpha {\theta }_i\tau }\right)+e^{-{\theta }_i\tau }{\left(1-e^{-{\theta }_i\tau }\right)}^{\alpha -1}\right)}\\ \sum _{i=1}^I\frac{{\left|{\beta }_{f,i,k}\right|}^{\alpha }}{\alpha {\theta }_i}\left(\left(1-e^{-\alpha {\theta }_i\tau }\right)+{\left(1-e^{-{\theta }_i\tau }\right)}^{\alpha }\right)+\sum _{j=1}^J{\left|{\beta }_{g,j,k}\right|}^{\alpha }\tau \end{array}.$(A22)

Therefore, we can observe continuation or reversal when $A_k\left(\tau ,T\right)$ is positive or reversal, respectively. The sign of $A_k\left(\tau ,T\right)$ depends on its numerator because its denominator is always positive.

To investigate the sign of $A_k\left(\tau ,T\right)$, we define a function h such that

(A23) $h\left(\tau \right)=-\left(1-e^{-\alpha \theta \tau }\right)+e^{-\theta \tau }{\left(1-e^{-\theta \tau }\right)}^{\alpha -1}.$(A23)

In the case of $\alpha =2$, we have $h\left(0\right)=0$ and $h^{\prime }\left(\tau \right)=-\theta e^{-\theta \tau }<0$. Therefore, $A_k\left(\tau ,T\right)<0$ for all $\tau >0$, $T>0$, and k. This means that there is always reversal under the normal distribution.

On the contrary, in the case of $1<\alpha <2$, we have

(A24) $\underset{\tau \to \mathrm{\infty }}{lim}h\left(\tau \right)=-1.$(A24)

This implies that there is a $\bar{\tau }>0$ such that $A_k\left(\tau ,T\right)<0$ for all $\tau >\bar{\tau }$. In other words, there is reversal for long-term returns. In addition, we have $h\left(0\right)=0$ and

(A25) $\underset{\tau \to 0^{+}}{lim}{h}^{\prime }\left(\tau \right)=+\mathrm{\infty }$(A25)

because

(A26) $h^{\prime }\left(\tau \right)=-\alpha \theta e^{-\alpha \theta \tau }-\theta e^{-\theta \tau }{\left(1-e^{-\theta \tau }\right)}^{\alpha -1}+\left(\alpha -1\right)\theta e^{-2\theta \tau }{\left(1-e^{-\theta \tau }\right)}^{\alpha -2}.$(A26)

This implies that there is a $\underset{-}{\tau }>0$ such that $A_k\left(\tau ,T\right)>0$ for all $0<\tau <\underset{-}{\tau }$. In other words, there is continuation for short-term returns. We have shown short-term momentum and long-term reversal only in time series. However, this is sufficient to show the cross-sectional relation because the prices of zero-cost portfolios are also given by Equation (3)(3) $p_{k,t}=\sum _{i=1}^I{\beta }_{f,i,k}{f}_{i,t}+\sum _{j=1}^J{\beta }_{g,j,k}{g}_{j,t}$(3) .

Proof of Proposition 2

Let ${\hat{d}}_t=d_t-\delta \sim i.i.d.S_{\alpha }\left(\sigma ,0,0\right)$. Then

(A27) $r\left(t,t+m\right)=\frac{1}{r_f\left(t+m\right)}\left(\sum _{i=t+1}^{t+m}{\hat{d}}_i-\frac{m}{t}\sum _{i=1}^t{\hat{d}}_i\right).$(A27)

Hence,

(A28) $E\left[r\left(t,t+m\right)|r\left(t-n,t\right)=x\right]=\frac{1}{r_f\left(t+m\right)}E\left[\left.\sum _{i=t+1}^{t+m}{\hat{d}}_i-\frac{m}{t}\sum _{i=1}^t{\hat{d}}_i\right|\sum _{i=t-n+1}^t{\hat{d}}_i-\frac{n}{t-n}\sum _{i=1}^{t-n}{\hat{d}}_i=r_{f}tx\right]=-\frac{m}{r_{f}t\left(t+m\right)}E\left[\left.\sum _{i=1}^t{\hat{d}}_i\right|\sum _{i=t-n+1}^t{\hat{d}}_i-\frac{n}{t-n}\sum _{i=1}^{t-n}{\hat{d}}_i=r_{f}tx\right]=-\frac{m}{r_{f}t\left(t+m\right)}\frac{n-{\left(\frac{n}{t-n}\right)}^{\alpha -1}\left(t-n\right)}{n+{\left(\frac{n}{t-n}\right)}^{\alpha }\left(t-n\right)}{r}_{f}tx=\frac{m{n}^{\alpha -1}}{\left(t+m\right)\left(n+n^{\alpha }{\left(t-n\right)}^{1-\alpha }\right)}\left({\left(t-n\right)}^{2-\alpha }-n^{2-\alpha }\right)x.$(A28)

If $\alpha =2$, then ${\left(t-n\right)}^{2-\alpha }-n^{2-\alpha }=0$, which yields $E\left[r\left(t,t+m\right)|r\left(t-n,t\right)\right]=0$. On the other hand, under $1<\alpha <2$, $E\left[r\left(t,t+m\right)|r\left(t-n,t\right)\right]$ and $r\left(t-n,t\right)$ have the same and opposite signs when $0<n<t/2$ and $t/2<n<t$, respectively.

Proof of Corollary 1

Let

(A29) $\hat{R}\left(t_1,t_2\right)\equiv R\left(t_1,t_2\right)-\frac{\delta }{r_f}\left({\left(1+r_f\right)}^{t_2-t_1}-1\right)$(A29)

Then

(A30) $\hat{R}\left(t,t+m\right)=\sum _{i=t+1}^{t+m}\left(\frac{1}{r_f\left(t+m\right)}+{\left(1+r_f\right)}^{t+m-i}\right){\hat{d}}_i-\frac{m}{r_{f}t\left(t+m\right)}\sum _{i=1}^t{\hat{d}}_i.$(A30)

Hence, when $1<\alpha \le 2$,

(A31) $E\left[R\left(t,t+m\right)|R\left(t-n,t\right)\right]=\frac{\delta }{r_f}\left({\left(1+r_f\right)}^m-1\right)+E\left[\hat{R}\left(t,t+m\right)|\hat{R}\left(t-n,t\right)\right]=C_t\left(n,m\right)\hat{R}\left(t-n,t\right)+\frac{\delta }{r_f}\left({\left(1+r_f\right)}^m-1\right)=C_t\left(n,m\right)R\left(t-n,t\right)+D_t\left(n,m\right),$(A31)

where

(A32) $C_t\left(n,m\right)\equiv \frac{m}{r_{f}t\left(t+m\right)}\text{break}\begin{array}{c}\underset{\_}{\left(t-n\right){\left(\frac{n}{r_{f}t\left(t-n\right)}\right)}^{\alpha -1}-\sum _{i=t-n+1}^t{\left(\frac{1}{r_{f}t}+{\left(1+r_f\right)}^{t-i}\right)}^{\alpha -1}}\\ \left(t-n\right){\left(\frac{n}{r_{f}t\left(t-n\right)}\right)}^{\alpha }+\sum _{i=t-n+1}^t{\left(\frac{1}{r_{f}t}+{\left(1+r_f\right)}^{t-i}\right)}^{\alpha }\end{array},$(A32)

(A33) $D_t\left(n,m\right)\equiv \frac{\delta }{r_f}\left({\left(1+r_f\right)}^m-1\right)-C_t\left(n,m\right)\frac{\delta }{r_f}\left({\left(1+r_f\right)}^n-1\right).$(A33)

Note that, if $\alpha =2$,

(A34) $\left(t-n\right){\left(\frac{n}{r_{f}t\left(t-n\right)}\right)}^{\alpha -1}-\sum _{i=t-n+1}^t{\left(\frac{1}{r_{f}t}+{\left(1+r_f\right)}^{t-i}\right)}^{\alpha -1}=-\frac{1}{r_f}\left({\left(1+r_f\right)}^m-1\right)<0.$(A34)

Therefore, $C_t\left(n,m\right)<0$ if $\alpha =2$. Furthermore,

(A35) $\left(t-n\right){\left(\frac{n}{r_{f}t\left(t-n\right)}\right)}^{\alpha -1}-\sum _{i=t-n+1}^t{\left(\frac{1}{r_{f}t}+{\left(1+r_f\right)}^{t-i}\right)}^{\alpha -1}\to t-2n$(A35)

as $\alpha \to 1^{+}.$ Therefore, when $\alpha \to 1^{+}$,$C_t\left(n,m\right)$ is positive and negative for $0<n<t/2$ and$t/2<n<t$, respectively.

Proof

Because the conditional expectation is given by Equations (A21)(A21) $\mathrm{E}\left[r_k\left(0,T\right)|r_k\left(-\tau ,0\right)\right]=A_k\left(\tau ,T\right)r_k\left(-\tau ,0\right),$(A21) and (A22) with I = 1, it is sufficient to show the existence of $\hat{\tau }$, such that

(B1) $h\left(\hat{\tau }\right)=0,$(B1)

(B2) $h\left(\tau \right)>0,\mathrm{i}\mathrm{f}0<\tau <\hat{\tau },$(B2)

(B3) $h\left(\tau \right)<0,\mathrm{i}\mathrm{f}\tau >\hat{\tau },$(B3)

for a function h in Equation (A23)(A23) $h\left(\tau \right)=-\left(1-e^{-\alpha \theta \tau }\right)+e^{-\theta \tau }{\left(1-e^{-\theta \tau }\right)}^{\alpha -1}.$(A23) . When we define a function $\varphi$ as

(B4) $\varphi \left(x\right)=-\left(1-x^{\alpha }\right)+x{\left(1-x\right)}^{\alpha -1},$(B4)

then our problem becomes equivalent to the existence of $\hat{x}$ such that

(B5) $\varphi \left(\hat{x}\right)=0,$(B5)

(B6) $\varphi \left(x\right)>0,\mathrm{f}\mathrm{o}\mathrm{r}\hat{x}<x<1,$(B6)

and

(B7) $\varphi \left(x\right)<0,\mathrm{f}\mathrm{o}\mathrm{r}0<x<\hat{x}.$(B7)

Then, the function $\varphi$ satisfies

(B8) $\varphi \left(0\right)=-1,$(B8)

(B9) $\varphi \left(0.5\right)=2^{1-\alpha }-1<0,$(B9)

(B10) $\varphi \left(x\right)=-{\left(1-x\right)}^{\alpha -2}+\left(1-\alpha x\right)+\alpha x^{\alpha -1},\text{ >}0\text{if}0\text{ <}x\text{<}\mathrm{0.5.}$(B10)

This implies that

(B11) $\varphi \left(x\right)<0,\mathrm{f}\mathrm{o}\mathrm{r}0<\rho <\mathrm{0.5.}$(B11)

In addition, note that

(B12) $\varphi \left(0.5\right)=2^{1-\alpha }-1<0,$(B12)

(B13) $\varphi \left(1\right)=0,$(B13)

(B14) $\underset{x\to 1^{-}}{lim}{f}^{\prime }\left(x\right)=-\mathrm{\infty },$(B14)

(B15) ${\varphi }^{″}\left(x\right)=-x{\left(1-x\right)}^{\alpha -3}\left(\alpha -1\right)\left(2-\alpha \right)-\left(\alpha -1\right)\left(2{\left(1-x\right)}^{\alpha -2}-\alpha x^{\alpha -2}\right)\text{ <}0,\text{if}\text{0}.\text{5<x<1}.$(B15)

Therefore, there is a unique number $\hat{x}$ such that

(B16) $\varphi \left(x\right)<0,\mathrm{f}\mathrm{o}\mathrm{r}0<x<\hat{x},$(B16)

(B17) $\varphi \left(x\right)>0,\mathrm{f}\mathrm{o}\mathrm{r}\hat{x}<x<1.$(B17)

Notes

¹ For example, Carr and Wu (2003) develop option pricing formulas under the assumption that the underlying asset follows a stable distribution. Ibragimov and Walden (2007) use a stable distribution to show the limits of diversification for heavy-tailed risk. Anand et al. (2016) analyse the asset allocation problem under the assumption of a normal tempered stable distribution. Gabaix (2012) shows heavy-tailed distributions can explain various financial anomalies although the author does not utilize stable distributions. Chang et al. (2015) develop early warning signal models by considering heavy-tailed distributions.

² Moskowitz, Ooi, and Pedersen (2012) mention that Barberis, Shleifer, and Vishny (1998), Daniel, Hirshleifer, and Subrahmanyam (1839-1885), and Hong and Stein (1999) focus only on the time series relations of a single risky asset.

³ Although they do not explicitly represent a mean-reverting noise process, the authors indicate that their representation is equivalent to AR(1), the first-order autoregressive model.

⁴ Moreover, Sarwar, Mateus, and Todorovic (2017) show that size and value premium as well as the momentum premium is related to the cyclical variations.

⁵ We use the symbol $\stackrel{\mathrm{\Delta }}{=}$ for equivalence in distribution.

⁶ When $\alpha =2$, $\beta$ vanishes. Thus, uniqueness excludes $\beta$ when $\alpha =2$.

⁷ According to the original definition of Samoradnitsky and Taqqu (1994), $X\left(0\right)=0$ almost surely and the process is defined on non-negative times. However, for convenience, we do not impose these assumptions.

⁸ For example, when $f_{1,t}$is an asset-specific risk of the first asset, ${\beta }_{f,1,k}=0$ for all $k\ne 1$.

⁹ This assumption is equivalent to an autoregressive model, AR(1). However, our assumption allows t to be a real number rather than an integer.

¹⁰ This result is obtained by Summers (1986), Fama and French (1988), and Arnott et al. (2015).

¹¹ It is equivalent to an assumption that investors regard dividends as normal variables and they use Bayesian updating with a diffuse prior (Lewellen and Shanken 2002). Accordingly, this subsection allows misconception about the distribution because we assume that dividends follow stable variables later.

¹² This is result of Lewellen and Shanken (2002).

¹³ The logic is valid for any $n$ such that $t/2<n<t$.