Optimal filter rules for selling stocks in the emerging stock markets

Abstract

With the application of the optimal stopping techniques, this paper proposes a filter rule for investors in emerging stock markets. In a bull market, once the stock price falls down to the optimal filter size, investors should sell the stock to avoid massive losses. We show that the optimal filter size is a function of the historical highest price, the weights of the future returns and the current drawdown in the investor’s utility function, the characteristics of the underlying stochastic price process, and the discount rate. Out-of-sample tests verify that this filter rule is valid, and the selling signals generated by the filter rule are at the beginning of the downtrend in the most emerging stock markets.

Appendices

Appendix A: Proof of proposition 3.1

The following lemma is useful in the proof of proposition 3.1

Lemma A.1

Consider a real-valued continuous function $f(x) = ax^{c_1} + bx^{c_2}$ defined on ${\mathbb {R}}^{+}$. If $a > 0$, $b < 0$, and $c_2 > c_1$, then $f(x) \ge 0$ on $(0,(-\frac{a}{b})^{\frac{1}{c_2-c_1}}]$ and $f(x) < 0$ on $((-\frac{a}{b})^{\frac{1}{c_2-c_1}},+\infty )$.

Proof

Lemma A.1 Let $f(x) \ge 0$, then we have

$$\begin{aligned} ax^{c_1} + bx^{c_2} \ge 0 \Rightarrow ax^{c_1} \ge -bx^{c_2}. \end{aligned}$$

Because $b < 0$, and $c_2 > c_1$, we can divide both sides by b and $x^{c_1}$ to derivate the following inequations

$$\begin{aligned} ax^{c_1} \ge -bx^{c_2} \Rightarrow -\frac{a}{b} \ge x^{c_2 - c_1} \Rightarrow \left( -\frac{a}{b}\right) ^{\frac{1}{c_2-c_1}} \ge x. \end{aligned}$$

Lemma A.1 is proved. $\square$

Proof of Proposition3.1

We start by the infinitesimal generator ${\mathbb {L}}$ defined by

$$\begin{aligned} {\mathbb {L}} = \frac{\sigma ^2x^2}{2}\frac{d^2}{dx^2}+\mu x\frac{d}{dx} \end{aligned}$$

and the ordinary differential equation defined by

$$\begin{aligned} {\mathbb {L}}f(x) = rx \end{aligned}$$

The positive increasing solution $\psi (x)$ and the decreasing solution $\varphi (x)$ of the ODE are following

$$\begin{aligned} \psi (x) = x^{\gamma _1} \ \ ,\ \ \varphi (x) = x^{\gamma _0} \end{aligned}$$

where

$$\begin{aligned} \gamma _{0},\gamma _{1} \triangleq \frac{-\left( \frac{2\mu }{\sigma ^2}-1\right) \mp \sqrt{\left( \frac{2\mu }{\sigma ^2}-1\right) ^2+\frac{8r}{\sigma ^2}}}{2} \end{aligned}$$

It is clear that $\gamma _1 > 0 , \gamma _0 < 0$.

For the geometric Brownian motion with the drift term $\mu$ and the diffusion term $\sigma$, these two solutions can represent the first passage time $\tau _k = \inf \{t \ge : X_t = k\}$ by

$$\begin{aligned} {\mathbb {E}}^{x}(e^{-r \tau _k}) = \left\{ \begin{aligned} \frac{\psi (x)}{\psi (k)} \ \ ,&\ \ if \ \ x \le k\\ \frac{\varphi (x)}{\varphi (k)} \ \ ,&\ \ if \ \ x \ge k \end{aligned} \right. \end{aligned}$$

(1.1)

We consider a general optimal stopping problem defined by

$$\begin{aligned} V(x) = \sup _{\tau \in \Gamma }{\mathbb {E}}^{x}[e^{-r\tau }g(X_{\tau })] \end{aligned}$$

We set the interval [m, n] which the initial value x of the process $X_t$ belongs to. If $X_t$ leaves the interval [m, n], the process will stop. The expectation ${\mathbb {E}}^{x}[e^{-r\tau }g(X_{\tau })]$can be represented as the average of escaping from the upper bound n and escaping from the lower bound m.

$$\begin{aligned} \begin{aligned} {\mathbb {E}}^{x}(e^{-r(\tau _m \wedge \tau _n)}g_{\tau _m \wedge \tau _n})&= g(m){\mathbb {E}}^{x}(e^{-r\tau _m}1_{\tau _m<\tau _n}) + g(n){\mathbb {E}}^{x}(e^{-r\tau _n}1_{\tau _m>\tau _n})\\&= g(m)\frac{\psi (x)\varphi (n)-\psi (n)\varphi (x)}{\psi (m)\varphi (n)-\psi (n)\varphi (m)}+g(n)\frac{\psi (m)\varphi (x)-\psi (x)\varphi (m)}{\psi (m)\varphi (n)-\psi (n)\varphi (m)} \end{aligned}\nonumber \\ \end{aligned}$$

(1.2)

And if $m = x = n$, ${\mathbb {E}}^{x}(e^{-r(\tau _m \wedge \tau _n)}g_{\tau _m \wedge \tau _n}) = g(x)$ The optimal stopping problem is considering to find the optimal interval $[m^*(x), n^*(x)]$ to maximize ${\mathbb {E}}^{x}(e^{-r(\tau _m \wedge \tau _n)}g_{\tau _m \wedge \tau _n})$.

Before giving the solution, we firstly define the continuous and strictly function F(x) by

$$\begin{aligned} F(x) \triangleq \frac{\psi (x)}{\varphi (x)} = x^{\gamma _1 - \gamma _0} \end{aligned}$$

(1.3)

The Eq. 1.2 can be expressed as

$$\begin{aligned} {\mathbb {E}}^{x}(e^{-r(\tau _m \wedge \tau _n)}g_{\tau _m \wedge \tau _n}) = \varphi (x)\bigg [ \frac{g(m)}{\varphi (m)}\cdot \frac{F(m)-F(x)}{F(n)-F(m)}+\frac{g(n)}{\varphi (n)}\cdot \frac{F(x)-F(m)}{F(n)-F(m)}\bigg ] \end{aligned}$$

(1.4)

Furthermore, we set function $G(\cdot )$ as

$$\begin{aligned} G(\cdot ) := \frac{g}{\varphi }\circ F^{-1}(\cdot ). \end{aligned}$$

(1.5)

and

$$\begin{aligned} Y_t := F(X_t). \end{aligned}$$

(1.6)

We replace $X_t$ in Eq. 1.2 by $Y_t$, then

$$\begin{aligned} {\mathbb {E}}^{x}(e^{-r(\tau _m \wedge \tau _n)}g_{\tau _m \wedge \tau _n})= \left\{ \begin{aligned}&\varphi (F^{-1}(y))\bigg [ G(y_m)\frac{y_m-y}{y_n-y_m}+G(y_n)\frac{y-y_n}{y_n-y_m}\bigg ]&m < n \\&\varphi (F^{-1}(y))G(y)&m = n \end{aligned} \right. \nonumber \\ \end{aligned}$$

(1.7)

where $y_m = F(m)$, $y_n = F(n)$.

It is natural that the value function V(x) is $V(F^{-1}(y))$, and it can represented under the optimal interval $[m^*(x), n^*(x)]$ as

$$\begin{aligned} V(F^{-1}(y)) = \left\{ \begin{aligned}&\varphi (F^{-1}(y))\bigg [ G(y_m^*)\frac{y_m^*-y}{y_n^*-y_m^*}+G(y_n^*)\frac{y-y_n^*}{y_n^*-y_m^*}\bigg ]&m^*(x) < n^*(x)\\&\varphi (F^{-1}(y))G(y)&m ^*(x) = n^*(x) \end{aligned} \right. \nonumber \\ \end{aligned}$$

(1.8)

where $y_m^* = F(m^*(x))$, $y_n^* = F(n^*(x))$ and $y = F(x)$. Since $V(x) \ge {\mathbb {E}}^{x}(e^{-r(\tau _m \wedge \tau _n)}g_{\tau _m \wedge \tau _n})$ for any [m, n], we have

$$\begin{aligned} G(y_m^*)\frac{y_m^*-y}{y_n^*-y_m^*}+G(y_n^*)\frac{y-y_n^*}{y_n^*-y_m^*} \ge G(y) \end{aligned}$$

So the value function is the smallest nonnegative concave majorant of G(y). And the continue region ${\mathcal {S}}$ and the optimal stopping time $\tau ^*$ is defined as

$$\begin{aligned} {\mathcal {S}} = \{x :V(x) = g(x)\} \qquad and \qquad \tau ^* = \inf \{t \ge 0: X_t \in {\mathcal {S}}\} \end{aligned}$$

In our optimal stopping problem, the reward function G(y) is

$$\begin{aligned} G_S(y) = \frac{\frac{\phi _1}{\phi _2} \sqrt{S} - (S - y^{\frac{1}{\gamma _1 - \gamma _0}})}{y^{\frac{\gamma _0}{\gamma _1 - \gamma _0}}} \end{aligned}$$

(1.9)

where the subscript S means that the highest price is fixed at S. The method to find the smallest concave majorant is making an tangent line of the curve $G_S(y)$ from the origin point. Hence, the concavity of $G_S(y)$ is necessary to indicate the existence of the tangent line. The first-order and second-order derivative of $G_S(y)$ are presented as followed.

$$\begin{aligned} G^{\prime }_S(y)= & {} -\frac{\gamma _0}{\gamma _1-\gamma _0}\left( \frac{\phi _1}{\phi _2}\sqrt{S} - S\right) y^{-\frac{\gamma _1}{\gamma _1-\gamma _0}} + \phi _2\frac{1-\gamma _0}{\gamma _1-\gamma _0}y^{\frac{1-\gamma _1}{\gamma _1-\gamma _0}} \end{aligned}$$

(1.10)

$$\begin{aligned} G^{\prime \prime }_S(y)= & {} \frac{\gamma _0\gamma _1}{ (\gamma _1-\gamma _0)^2}\left( \frac{\phi _1}{\phi _2}\sqrt{S} - S\right) y^{-\frac{\gamma _1}{\gamma _1-\gamma _0}-1} + \phi _2\frac{(1-\gamma _0)(1-\gamma _1)}{(\gamma _1-\gamma _0)^2}y^{\frac{1-\gamma _1}{\gamma _1-\gamma _0}-1}\qquad \end{aligned}$$

(1.11)

$G^{\prime \prime }_S(y)$ has the same formula as f(x) in lemma A.1. We can take $\frac{\gamma _0\gamma _1}{ (\gamma _1-\gamma _0)^2}(\frac{\phi _1}{\phi _2}\sqrt{S} - S)$ as a, $\phi _2\frac{(1-\gamma _0)(1-\gamma _1)}{(\gamma _1-\gamma _0)^2}$ as b, $-\frac{\gamma _1}{\gamma _1-\gamma _0}-1$ as $c_1$ and $\frac{1-\gamma _1}{\gamma _1-\gamma _0}-1$ as $c_2$.

Owing to $\gamma _1 > 1$ and $\gamma _0 < 0$, we have

$$\begin{aligned} \phi _2\frac{(1-\gamma _0)(1-\gamma _1)}{(\gamma _1-\gamma _0)^2}< 0 ,\quad -\frac{\gamma _1}{\gamma _1-\gamma _0}-1 < \frac{1-\gamma _1}{\gamma _1-\gamma _0}-1. \end{aligned}$$

We define the lower bound for S by

$$\begin{aligned} \frac{\phi _1}{\phi _2}\sqrt{S} - S \le 0 \quad \Rightarrow \quad S\ge {\underline{S}} = (\frac{\phi _1}{\phi _2})^2. \end{aligned}$$

When $\frac{\phi _1}{\phi _2}\sqrt{S} - S < 0$, $\frac{\gamma _0\gamma _1}{ (\gamma _1-\gamma _0)^2}(\frac{\phi _1}{\phi _2}\sqrt{S} - S) > 0$. Based on lemma A.1, $G^{\prime \prime }_S(y)$ is positive around the origin point and negative with y increasing to infinity. The curve of $G_S(y)$ is convex with y around zero and becomes concave with y increasing, which guarantees the existence of the tangent line from the origin point.

We can compute the tangent point $x^*(S)$ that

$$\begin{aligned} \begin{aligned} x^*(S) = \frac{\frac{\phi _1}{\phi _2}\sqrt{S}-S}{\phi _2}\frac{\gamma _1}{1 - \gamma _1}. \end{aligned} \end{aligned}$$

(1.12)

Since the process $X_t$ is below S, we have the upper bound that

$$\begin{aligned} S \ge x^*(S) = \frac{\frac{\phi _1}{\phi _2}\sqrt{S}-S}{\phi _2}\frac{\gamma _1}{1 - \gamma _1}\quad \Rightarrow \quad {\overline{S}} = \left( \frac{\phi _1\gamma _1}{\phi _2}\right) ^2. \end{aligned}$$

(1.13)

$\square$

Proof of Theorem 3.4

We consider to use mean value theorems for definite integrals in Eq. 3.13 but we just take the approximate value.

$$\begin{aligned} \int ^{S+\epsilon }_{S} P(u)\times P_f(u)\times \left( \frac{\phi _1}{\phi _2}\sqrt{u} - e(u)\right) du \approx P(S) \times P_f(S)\times \left( \frac{\phi _1}{\phi _2}\sqrt{S} - e(S)\right) \epsilon \\ \int ^{S+\epsilon }_{S}\frac{F'(n)dn}{F(n)-F(n-e(n))} \approx \epsilon \frac{F'(S)}{F(S)-F(S-e(S))} \end{aligned}$$

Then we define a function $v(S,\epsilon )$ by

$$\begin{aligned} \begin{aligned} v(S,\epsilon ) = \sup _{e} \frac{\varphi (x)}{\varphi (S-e(S))} \frac{F'(S)}{F(S)-F(S-e(S))}(\frac{\phi _1}{\phi _2}\sqrt{S} - e(S))\epsilon \\ + \frac{\varphi (S)}{\varphi (S+\epsilon )}\times \exp (-\epsilon \frac{F'(S)}{F(S)-F(S-e(S))}) V(S+\epsilon ). \end{aligned} \end{aligned}$$

(6.14)

Lemma 4 from Egami and Oryu (2017) shows that

$$\begin{aligned} \frac{v(S,\epsilon )}{\varphi (S)} \overset{\epsilon \rightarrow 0}{=} \frac{\varphi '(S+\epsilon )}{\varphi '(S)}\frac{V(S+\epsilon )}{\varphi (S+\epsilon )} \end{aligned}$$

(6.15)

From Eq. 6.14, It can be useful to notice that

$$\begin{aligned} \begin{aligned} v(S,\epsilon )-\frac{\varphi (S)}{\varphi (S+\epsilon )}\times \exp (-\epsilon \frac{F'(S)}{F(S)-F(S-e^*(S))}) V(S+\epsilon ) \\ = \frac{\varphi (x)}{\varphi (S-e^*(S))} \frac{F'(S)}{F(S)-F(S-e^*(S))}\left( \frac{\phi _1}{\phi _2}\sqrt{S} - e^*(S)\right) \epsilon . \end{aligned} \end{aligned}$$

(6.16)

where $e^*(S)$ is the maximizer of Eq. 6.14. Equations 6.15 and 6.16 solve $v(S, \epsilon )$. It is clear that $\lim _{\epsilon \rightarrow 0}v(S,\epsilon ) = V(S)$, and we have the expression of the value function V(S) that

$$\begin{aligned} \begin{aligned} V(S) = \sup _{e}&\frac{\varphi (S)}{\varphi (S - e(S))}\frac{F'(S)\varphi '(S)}{\varphi ''(S)(F(S)-F(S-e(S)))+F'(S)\varphi (S)}\\&\times \left( \frac{\phi _1}{\phi _2}\sqrt{S} - e(S)\right) . \end{aligned} \end{aligned}$$

(6.17)

Hence, the optimal stopping problem is to find the maximizer $e^*(S)$ of Eq. 6.17$\square$

Appendix B: Robust test

1.1 B.1: Using weekly data

We apply the weekly data to check the robustness of our filter rule. Figures 6, 7, 8, 9 and 10 are the results. we can find that during the 2008 global financial crisis in Chinese stock market, the selling signal based on daily data is around November 2007, but the selling signal based on weekly data is around December 2007. It results in an additional loss of 1.5%, other periods and markets have similar results. However, compared with the entire crisis, the filter rule based on weekly data is still effective to control losses.

Fig. 6

The Threshold for the Chinese Stock Market(Weekly)

Fig. 7

The Threshold for the Indian Stock Market(Weekly)

Fig. 8

The Threshold for the Brazilian Stock Market(Weekly)

Fig. 9

The Threshold for the Russian Stock Market(Weekly)

Fig. 10

The Threshold for the South African Stock Market(Weekly)

1.2 B.2: Using mean absolute deviation

Considering the standard deviation is not the only proxy of the optimal drawdown for investors, we also use the mean absolute deviation(MAD) for estimating ${\hat{\phi }}_1/{\hat{\phi }}_2$ to conduct the robust test. The new esitmated results is presented in Table 5

Table 5 The estimations of $\phi _1/\phi _2$ based on the mean absolute deviation and the standard deviation

We can find that the estimated $\phi _1/\phi _2$ based on the mean absolute deviation is smaller than our original results. We also apply the estimated $\phi _1/\phi _2$ based on the mean absolute deviation to our empirical tests. The results are shown by Figs. 11, 12, 13, 14, and 15. The thresholds becomes smaller, and the filter rule is more likely to deliver selling signals, especially when the price drop a lot during a uptrend. It makes investor sell stocks too early to get the potential profits in the subsequent price rising. Hence, the standard deviation would be better to estimate $\phi _1/\phi _2$.

Fig. 11

The Threshold for the Chinese Stock Market(based on MAD)

Fig. 12

The Threshold for the Indian Stock Market(basd on MAD)

Fig. 13

The Threshold for the Brazilian Stock Market(basd on MAD)

Fig. 14

The Threshold for the Russian Stock Market(based on MAD)

Fig. 15

The Threshold for the South African Stock Market(based on MAD)