Risk management for crude oil futures: an optimal stopping-timing approach

Abstract

Timing the selling of crude oil futures to control risk is a worth studying question given the swift fall of their prices. This paper proposes an optimal stopping model to find the optimal selling time at the beginning of the downtrend. The model depends on the crude oil futures prices drawdown and the boundary to identify the occurrence of downtrend in real-time. The numerical simulation and empirical analyses help verify the effectiveness of the proposed optimal stopping time model, especially, in 2007, when the model can effectively avoid losses. The conclusions of the paper provide a new perspective for investors to control risk.

Proof of Theorem 1

According to Peskir and Shiriaev (2006), $E|\tau - \theta |$ can be represented as

$$\begin{aligned} E|\tau -\theta |=\mathrm {E}\left[ \int _{0}^{\tau }F(\frac{S_{t}-B_{t}}{\sqrt{1-t}})dt\right] +\frac{1}{2} \end{aligned}$$

(1.1)

where $F(x) = 4\Phi (x) - 3$. Our optimal stopping problem 3.5 can be transformed as

$$\begin{aligned} V = \inf _{\tau \in \mathfrak {M}} E\left( \int ^{\tau }_0 F\left( \frac{S_t - B_t}{\sqrt{1 - t}}dt\right) \right) + \frac{1}{2}. \end{aligned}$$

(1.2)

Because $S_t-B_t$ and $|B_t|$ have the same distribution, we can replace $S_t - B_t$ with $|B_t|$. Then our optimal stopping problem can be simplified as

$$\begin{aligned} V = \inf _{\tau \in \mathfrak {M}} E\left( \int ^{\tau }_0 F\left( \frac{|B_t|}{\sqrt{1 - t}}dt\right) \right) + \frac{1}{2}\text {.} \end{aligned}$$

(1.3)

Let $1-t = e^{-2s}$, then $\frac{B_t}{\sqrt{1 - t}} = e^sB_{1-e^{-2s}}$. Its differential is

$$\begin{aligned} de^sB_{1-e^{-2s}} = e^{s}B_{1-e^{-2s}}ds + e^{s}dB_{1-e^{-2s}} = e^{s}B_{1-e^{-2s}}ds + \sqrt{2}d\tilde{B}_s \end{aligned}$$

(1.4)

where $\tilde{B}_s = \frac{1}{\sqrt{2}}\int _{0}^{s}e^{t}dB_{1-e^{-2t}}$. Because $\tilde{B}_s$ is the Itô integral, its mean value and variance are $E\tilde{B}_s = 0$ and $Var(\tilde{B}_s) = s$, and it is also a martingale. Hence, $\tilde{B}_s$ can be defined as a Brownian motion. We set process $Y_s$ as

$$\begin{aligned} Y_s&= e^sB_{1-e^{-2s}}\text {, } \quad Y_0 = y \quad \text {,} \end{aligned}$$

(1.5)

$$\begin{aligned} dY_s&= Y_sds + \sqrt{2}d\tilde{B}_s\text {.} \end{aligned}$$

(1.6)

Then the infinitesimal generator of $Y_s$ is

$$\begin{aligned} \mathcal {L}_{Y}=Y\frac{d}{dY}+\frac{d^{2}}{dY^{2}}\text {.} \end{aligned}$$

(1.7)

We replace $\frac{|B_t|}{\sqrt{1 - t}}$ with $Y_s$ to obtain

$$\begin{aligned} V=2\inf _{\tau \in \mathfrak {M}}\mathrm {E}\left[ \int _{0}^{s_{\tau }}e^{-2s}F(|Y_{s}|)ds\right] +\frac{1}{2}\text {,} \end{aligned}$$

(1.8)

where $s_{\tau }=\log (1/\sqrt{1-\tau })$. $\tau$ on the filtration $\mathcal {F}^B_t$and $s_{\tau }$ on the filtration $\mathcal {F}^{\tilde{B}_s}$ correspond one to one. So the optimal stopping problem can be simplified as

$$\begin{aligned} H=\inf _{\rho \ge 0}\mathrm {E}\left[ \int _{0}^{\rho }e^{-2s}F(|Y_{s}|)ds\right] \text {.} \end{aligned}$$

(1.9)

which is a general optimal stopping problem. We assume that the optimal stopping time $\rho$ satisfies

$$\begin{aligned} \rho =\inf \{s>0\ :\ |Y_{s}|\ge Y^{*}\}\text {.} \end{aligned}$$

(1.10)

According to Peskir and Shiriaev (2006), the optimal stopping problem is equivalent to a Dirichlet differential equation with the free boundary:

$$\begin{aligned} \left\{ \begin{aligned}&(\mathcal {L}_{Y}-2)H(y)=-F(|y|)\ \text {when} \ y\in (-y^{*},\ y^{*})\ \\&H(\pm y^{*})=0 \\&H'(\pm y^{*})=0 \end{aligned} \right. \end{aligned}$$

(1.11)

The solution is

$$\begin{aligned} \hat{H}(y)=\Phi (y^{*})(1+y^{2})-y\varphi (y)+(1-y^{2})\Phi (y))-\frac{3}{2}, y\in [0,\ y^{*}] \end{aligned}$$

(1.12)

where $y^*$ is the unique root of the following equation.

$$\begin{aligned} 4\Phi (y) - 2y\phi (y) - 3 = 0 \end{aligned}$$

(1.13)

Next, we need to verify the function $\hat{H}(y)$ is the minimum value function and $\rho$ is the optimal stopping time. From Itô integral, we have

$$\begin{aligned} e^{-2s}\hat{H}(Y_{s})=\hat{H}(y)+\int _{0}^{s}e^{-2t}(\mathcal {L}_{Y}\hat{H}(Y_{t})-2\hat{H}(Y_{t}))dt +\sqrt{2}\int _{0}^{s}e^{-2t}\hat{H}'(Y_{s})d\tilde{B}_{t} \end{aligned}$$

(1.14)

Because when $y\in (-y^{*},\ y^{*})$, $\mathcal {L}_{Y}\hat{H}(y)-2\hat{H}(y)=-F(|y|)$, and when $y\notin (-y^{*},\ y^{*})$ , $\mathcal {L}_{Y}\hat{H}(y)-2\hat{H}(y) >-F(|y|)$, we obtain

$$\begin{aligned} e^{-2s}\hat{H}(Y_{s})\ge \hat{H}(y)-\int _{0}^{s}e^{-2t}F(|Y_{t}|)dt+\sqrt{2}\int _{0}^{s}e^{-2t}\hat{H}'(Y_{s})d\tilde{B}_{t}\text {, } \end{aligned}$$

(1.15)

We set s equals $\rho$, then $\hat{H}(Y_{\rho }) = \hat{H}(y^*) = 0$. We have

$$\begin{aligned}&0\ge \hat{H}(y)-\int _{0}^{\rho }e^{-2t}F(|Y_{t}|)dt+\sqrt{2}\int _{0}^{\rho }e^{-2t}\hat{H}'(Y_{s})d\tilde{B}_{t}\text {, } \end{aligned}$$

(1.16)

$$\begin{aligned}&\int _{0}^{\rho }e^{-2t}F(|Y_{t}|)dt\ge \hat{H}(y)+\sqrt{2}\int _{0}^{\rho }e^{-2t}\hat{H}'(Y_{s})d\tilde{B}_{t}\text {.} \end{aligned}$$

(1.17)

Take the expectation of both sides of the above equation, we have

$$\begin{aligned} E\int _{0}^{\rho }e^{-2t}F(|Y_{t}|)dt\ge \hat{H}(y)\text {.} \end{aligned}$$

(1.18)

It is noted that the left-hand side of the inequality $E\int _{0}^{\rho }e^{-2t}F(|Y_{t}|)dt$ is the minimum of the object function H(y). Because of the definition of the minimum value $H(y) \le \hat{H}(y)$, we have$H(y) = \hat{H}(y)$.