Trading strategy with stochastic volatility in a limit order book market

Abstract

In this paper, we employ the Heston stochastic volatility model to describe the stock’s volatility and apply the model to derive and analyze trading strategies for dealers in a security market with price discovery. The problem is formulated as a stochastic optimal control problem, and the controlled state process is the dealer’s mark-to-market wealth. Dealers in the security market can optimally determine their ask and bid quotes on the underlying stocks continuously over time. Their objective is to maximize an expected profit from transactions with a penalty proportional to the variance of cumulative inventory cost. We provide an approximate, analytically tractable solution to the stochastic control problem. Numerical experiments are given to illustrate the effects of various parameters on the performances of trading strategies.

Appendix

1.1 A. Proof of Proposition 2

Proof

• (a) Let $\Big (\delta _t^a,\delta _t^b\Big )_{t\ge 0}\in {\mathcal {A}}$. Then by Dynkin’s formula we have

  $$\begin{aligned}&\displaystyle {\mathbb {E}}_t\left[ \phi \Big (q^{t,q,(\delta _t^a,\delta _t^b)_{t\ge 0}}_{T_R}, \nu ^{t,\nu }_{T_R},T_R\Big )\right] \\&\quad =\displaystyle \phi (q,\nu ,t)+{\mathbb {E}}_t\bigg [ \int _t^{T_R} \Big (\phi _t+\theta (\alpha -\nu ^{t,\nu }_u)\phi _{\nu }+\frac{1}{2}\xi ^2\nu ^{t,\nu }_u \phi _{\nu \nu } \\&\qquad +\left[ \phi \Big (q^{t,q,(\delta _t^a,\delta _t^b)_{t\ge 0}}_u-1,\nu ^{t,\nu }_u,u\Big )- \phi \Big (q^{t,q,(\delta _t^a,\delta _t^b)_{t\ge 0}}_u,\nu ^{t,\nu }_u,u\Big )\right] \lambda ^a(\delta _u^a)\\&\qquad \displaystyle +\left[ \phi \Big (q^{t,q,(\delta _t^a,\delta _t^b)_{t\ge 0}}_u+1,\nu ^{t,\nu }_u,u\Big )- \phi \Big (q^{t,q,(\delta _t^a,\delta _t^b)_{t\ge 0}}_u,\nu ^{t,\nu }_u,u\Big )\right] \lambda ^b(\delta _u^b)\Big ) \mathrm{d}u\bigg ]\\&\qquad \displaystyle +\,{\mathbb {E}}_t\left[ \int _t^{T_R}\xi \sqrt{\nu _u^{t,\nu }}\phi _\nu (q_u^{t,q,(\delta _t^a,\delta _t^b)_{t\ge 0}}, \nu _u^{t,\nu },u)\mathrm{d}B_u\right] , \end{aligned}$$

  where $T_R{=}\min \left\{ R, T, \inf \{s>t; {\mathbb {E}}_t\left[ \int _t^s \nu _u^{t,\nu } \phi ^2_\nu \left( q_u^{t,q,(\delta _t^a,\delta _t^b)_{t\ge 0}},\nu _u^{t,\nu },u\right) \mathrm{d}u\right] {\ge } R\}\right\}$. Note that $T_R \rightarrow T$, as $R\rightarrow \infty$ ( by condition (iii)). The stopped process

  $$\begin{aligned} \left\{ \int _t^{T_R}\xi \sqrt{\nu _u^{t,\nu }}\phi _\nu \Big (q_u^{t,q,(\delta _t^a,\delta _t^b)_{t\ge 0}}, \nu _u^{t,\nu },u\Big )\mathrm{d}B_u\right\} _{R\ge 0} \end{aligned}$$

  is then a martingale under the probability measure $\mathcal{P}$.

  From Condition (i), we have

  $$\begin{aligned}&{\mathbb {E}}_t\left[ \phi \Big (q^{t,q,(\delta _t^a,\delta _t^b)_{t\ge 0}}_{T_R}, \nu ^{t,\nu }_{T_R},T_R\Big )\right] \nonumber \\&\quad =\displaystyle \phi (q,\nu ,t)+{\mathbb {E}}_t\bigg [ \int _t^{T_R} \Big (\phi _t+\theta (\alpha -\nu ^{t,\nu }_u)\phi _{\nu }+\frac{1}{2}\xi ^2 \nu ^{t,\nu }_u\phi _{\nu \nu } \nonumber \\&\qquad \displaystyle +\left[ \phi \Big (q^{t,q,(\delta _t^a,\delta _t^b)_{t\ge 0}}_u-1,\nu ^{t,\nu }_u,u\Big ) -\phi \Big (q^{t,q,(\delta _t^a,\delta _t^b)_{t\ge 0}}_u,\nu ^{t,\nu }_u,u\Big )\right] \lambda ^a(\delta _u^a)\nonumber \\&\qquad \displaystyle +\left[ \phi \Big (q^{t,q,(\delta _t^a,\delta _t^b)_{t\ge 0}}_u+1,\nu ^{t,\nu }_u,u\Big )- \phi \Big (q^{t,q,(\delta _t^a,\delta _t^b)_{t\ge 0}}_u,\nu ^{t,\nu }_u,u\Big )\right] \lambda ^b(\delta _u^b)\Big ) \mathrm{d}u\bigg ]\nonumber \\&\quad \le \displaystyle \phi (q,\nu ,t)-{\mathbb {E}}_t\Bigg [\int _t^{T_R}\bigg (-\frac{\gamma }{2} \left( q_u^{t,q,(\delta _t^a,\delta _t^b)_{t\ge 0}}\right) ^2\nu ^{t,\nu }_u\nonumber \\&\qquad \displaystyle +\Big (\delta _u^a+\beta -\frac{\gamma \eta ^2}{2}\Big (q_u^{t,q, (\delta _t^a,\delta _t^b)_{t\ge 0}}-1\Big )^2\Big )\lambda ^a(\delta _u^a) \nonumber \\&\qquad \displaystyle +\Big (\delta _u^b-\beta -\frac{\gamma \eta ^2}{2}\Big (q_u^{t,q,(\delta _t^a,\delta _t^b)_{t\ge 0}} +1\Big )^2\Big )\lambda ^b(\delta _u^b)\bigg )\mathrm{d}u\Bigg ]\nonumber \\&\quad \le \displaystyle \phi (q,\nu ,t)-\Bigg \{{\mathbb {E}}_t\bigg [\int _t^{T_R} \Big (\delta _u^a+\beta -\frac{\gamma \eta ^2}{2}\Big (q_u^{t,q,(\delta _t^a,\delta _t^b)_{t\ge 0}} -1\Big )^2\Big )\mathrm{d}N^a_u\nonumber \\&\qquad \displaystyle +\Big (\delta _u^b-\beta -\frac{\gamma \eta ^2}{2}\Big (q_u^{t,q,(\delta _t^a,\delta _t^b)_{t\ge 0}} +1\Big )^2\Big )\mathrm{d}N_u^b\nonumber \\&\qquad \displaystyle -\frac{\gamma }{2}\int _t^{T_R}\left( q_u^{t,q,(\delta _t^a,\delta _t^b)_{t\ge 0}} \right) ^2\nu ^{t,\nu }_u\mathrm{d}u\bigg ]\Bigg \}. \end{aligned}$$

  (4.1)

  Meanwhile, we have

  $$\begin{aligned} \displaystyle {\mathbb {E}}_t\left[ \int _t^{T_R}\left( q_u^{t,q,(\delta _t^a,\delta _t^b)_{t\ge 0}}\right) ^2 \nu ^{t,\nu }_u\mathrm{d}u\right]\le & {} \displaystyle {\mathbb {E}}_t\left[ \int _t^{T}\left( q_u^{t,q, (\delta _t^a,\delta _t^b)_{t\ge 0}}\right) ^2\nu ^{t,\nu }_u\mathrm{d}u\right] \\= & {} \displaystyle \int _t^{T}{\mathbb {E}}_t\left[ \left( q_u^{t,q, (\delta _t^a,\delta _t^b)_{t\ge 0}}\right) ^2\nu ^{t,\nu }_u\right] \mathrm{d}u\\< & {} \infty , \end{aligned}$$

  and

  $$\begin{aligned}&\displaystyle \left| {\mathbb {E}}_t\left[ \int _t^{T_R} \left( \delta _u^a+\beta -\frac{\gamma \eta ^2}{2}\Big (q_u^{t,q,(\delta _t^a,\delta _t^b)_{t\ge 0}} -1\Big )^2\right) \mathrm{d}N^a_u\right. \right. \\&\qquad \left. \left. +\left( \delta _u^b-\beta -\frac{\gamma \eta ^2}{2}\Big (q_u^{t,q, (\delta _t^a,\delta _t^b)_{t\ge 0}}+1\Big )^2\right) \mathrm{d}N_u^b\right] \right| \\&\quad \le \displaystyle {\mathbb {E}}_t\left[ \int _t^{T} |\delta _u^a+\beta - \frac{\gamma \eta ^2}{2}\Big (q_u^{t,q,(\delta _t^a,\delta _t^b)_{t\ge 0}}-1\Big )^2|\mathrm{d}N^a_u\right. \\&\qquad \left. +|\delta _u^b-\beta -\frac{\gamma \eta ^2}{2}\Big (q_u^{t,q,(\delta _t^a,\delta _t^b)_{t\ge 0}} +1\Big )^2|\mathrm{d}N_u^b\right] \\&\quad <\displaystyle \infty . \end{aligned}$$

  Since $\phi$ satisfies a polynomial growth condition, we have⁷

  $$\begin{aligned} \left| \phi \Big (q^{t,q,(\delta _t^a,\delta _t^b)_{t\ge 0}}_{T_R},\nu ^{t,\nu }_{T_R},T_R\Big )\right| \le C\left( 1+\max _{u\in [t,T]} |q_u^{t,q,(\delta _t^{*,a},\delta _t^{*,b})_{t\ge 0}}|^n+ |\nu _{T_R}^{t,\nu }|^m\right) .\nonumber \\ \end{aligned}$$

  (4.2)

  Directly applying the dominated convergence theorem and sending R in Eq. (4.1) to infinity yields

  $$\begin{aligned} \phi (q,\nu ,t)\ge & {} \displaystyle {\mathbb {E}}_t\bigg [\int _t^{T} \Big (\delta _u^a+\beta -\frac{\gamma \eta ^2}{2}(q_u^{t,q,(\delta _t^a, \delta _t^b)_{t\ge 0}}-1)^2\Big )\mathrm{d}N^a_u\\&\displaystyle +\left( \delta _u^b-\beta -\frac{\gamma \eta ^2}{2}(q_u^{t,q, (\delta _t^a,\delta _t^b)_{t\ge 0}}+1)^2\right) \mathrm{d}N_u^b\\&\displaystyle -\frac{\gamma }{2}\int _t^{T}\left( q_u^{t,q,(\delta _t^a, \delta _t^b)_{t\ge 0}}\right) ^2\nu ^{t,\nu }_udu\bigg ]\\\ge & {} V^{(\delta _t^a,\delta _t^b)_{t\ge 0}}(q,\nu ,t). \end{aligned}$$

  Since $\Big (\delta _t^a,\delta _t^b\Big )_{t\ge 0}\in {\mathcal {A}}$ is arbitrary, we conclude that

  $$\begin{aligned} \phi (q,\nu ,t)\ge V(q,\nu ,t) \quad \text{ for } \text{ all } (q,\nu ,t)\in {\mathbb {Z}}\times \mathfrak {R}_+\times [0,T]. \end{aligned}$$

  (4.3)

• (b) Apply the above argument to $\Big (\delta _t^a,\delta _t^b\Big )_{t\ge 0}=\Big (\delta _t^{*,a},\delta _t^{*,b}\Big )_{t\ge 0}$. The calculations above give equality and hence

  $$\begin{aligned} \phi (q,\nu ,t)=V^{\Big (\delta _t^{*,a},\delta _t^{*,b}\Big )_{t\ge 0}}(q, \nu , t)\le V(q,\nu ,t) \end{aligned}$$

  (4.4)

  for all $(q,\nu ,t)\in {\mathbb {Z}}\times \mathfrak {R}_+\times [0,T]$. Combining (4.3) and (4.4), we get the results in (b). $\square$

1.2 B. Proof of Theorem 2

Proof

Due to our choice of “mean-variance” objective function, we are able to simplify the problem with an ansatz on the form of the value function:

$$\begin{aligned} V(q,\nu ,t)=f(\nu ,t)+g(\nu ,t)q+h(\nu ,t)q^2. \end{aligned}$$

(4.5)

Then, we have

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle \delta _t^{*,b}=1/k+b-g(\nu ,t)-h(\nu ,t)(2q+1)\\ \displaystyle \delta _t^{*,a}=1/k-b+g(\nu ,t)+h(\nu ,t)(2q-1). \end{array} \right. \end{aligned}$$

The ask–bid spread $\delta _t^{*,a} + \delta _t^{*,b} = \frac{2}{k}-2h(\nu ,t)$ is independent of the inventory. Substituting Eq. (4.5) and the linear approximation of the order arrival terms into Eq. (2.9) and grouping terms of q yields

$$\begin{aligned} \left\{ \begin{array}{lll} g_t+\theta (\alpha -\nu )g_\nu +\frac{1}{2}\xi ^2\nu g_{\nu \nu }=0\\ g(\nu ,T)=0. \end{array} \right. \end{aligned}$$

(4.6)

By the Feynman–Kac formula, $g(\nu ,t)=0$. Grouping terms in the coefficients of $q^2$ yields

$$\begin{aligned} \left\{ \begin{array}{lll} h_t+\theta (\alpha -\nu )h_\nu +\frac{1}{2}\xi ^2\nu h_{\nu \nu }-\frac{\gamma }{2}\nu =0\\ h(\nu ,T)=0, \end{array} \right. \end{aligned}$$

whose solution can be directly obtained using the Feynman–Kac formula

$$\begin{aligned} h(\nu ,t)= & {} \displaystyle E_t\left[ \int _t^T -\frac{1}{2}\gamma \nu _u\mathrm{d}u\right] \nonumber \\= & {} -\displaystyle \frac{1}{2}\gamma \int _t^T E_t[ \nu _u]\mathrm{d}u\nonumber \\= & {} \displaystyle - \frac{\gamma }{2\theta }(\nu -\alpha )\left[ 1-e^{-\theta (T-t)}\right] - \frac{\gamma }{2}\alpha (T-t). \end{aligned}$$

(4.7)

Grouping terms in the coefficients of $q^0$ yields

$$\begin{aligned} \left\{ \begin{array}{lll} f_t+\theta (\alpha -\nu )f_\nu +\frac{1}{2}\xi ^2\nu f_{\nu \nu }+2Ah(\nu ,t)=0\\ f(\nu ,T)=0. \end{array} \right. \end{aligned}$$

Thus,

$$\begin{aligned} \begin{array}{lll} f(\nu ,t)&{}=&{}\displaystyle {\mathbb {E}}_t\left[ \int _t^T2Ah(\nu _r,r)\mathrm{d}r\right] \\ &{}=&{}\displaystyle -\frac{\gamma A}{\theta }(\nu -\alpha )\left[ \frac{1}{\theta }\Big (1-e^{-\theta (T-t)}\Big )-e^{-\theta (T-t)}(T-t)\right] -\frac{\gamma \alpha A}{2}(T-t)^2. \end{array} \end{aligned}$$

By now, we have got an approximation to the solution of the HJB equation, which is given by

$$\begin{aligned} {\tilde{V}}(q,\nu ,t)= & {} \displaystyle -\frac{\gamma A}{\theta }(\nu -\alpha ) \left[ \frac{1}{\theta }\Big (1-e^{-\theta (T-t)}\Big )-e^{-\theta (T-t)}(T-t) \right] -\frac{\gamma \alpha A}{2}(T-t)^2\\&\displaystyle -\displaystyle \frac{\gamma q^2}{2\theta }(\nu -\alpha )\left[ 1-e^{-\theta (T-t)}\right] -\frac{\gamma q^2}{2}\alpha (T-t). \end{aligned}$$

We now analyze the difference between the approximate and the exact solutions under the Heston stochastic volatility model. Let

$$\begin{aligned} V^i(q,\nu ,t)=-\displaystyle \frac{\gamma q^2}{2\theta }(\nu -\alpha )\left[ 1-e^{-\theta (T-t)}\right] -\frac{\gamma q^2}{2}\alpha (T-t). \end{aligned}$$

Suppose that $V(q,\nu ,t)=w^q(t,\nu )+V^i(q,\nu ,t)$, where $(w^q)_{q\in {\mathbb {N}}}$ is a family of functions in ${{\mathcal {C}}}^{1,2}(t,\nu )$, and ${\mathbb {N}}$ is the set of natural numbers. Substituting the above expression into Eq. (2.9) yields

$$\begin{aligned} \left\{ \begin{array}{lll} \displaystyle w_t^q+\theta (\alpha -\nu )w_\nu ^q+\frac{1}{2}\xi ^2\nu w_{\nu \nu }^q+\epsilon ^q(\nu ,t)=0\\ \displaystyle w^q(\nu ,T)=0, \end{array} \right. \end{aligned}$$

(4.8)

where

$$\begin{aligned} \epsilon ^q(\nu ,t)= & {} \displaystyle \frac{A}{k}\Big (e^{-k\delta _t^{*,a}}+e^{-k\delta _t^{*,b}}\Big )\nonumber \\= & {} \displaystyle \frac{A}{k}\Big [e^{\left( -k(w^q-w^{q-1}+\frac{1}{k}-b+\left( \frac{\gamma }{2\theta } (\nu -a)(1-e^{-\theta (T-t)})+\frac{\gamma }{2}\alpha (T-t)\right) (-2q+1)\right) }\nonumber \\&\displaystyle + e^{\left( -k(w^q-w^{q+1}+\frac{1}{k}+b+\left( \frac{\gamma }{2\theta }(\nu -a) (1-e^{-\theta (T-t)})+\frac{\gamma }{2}\alpha (T-t)\right) (2q+1)\right) }\Big ] \end{aligned}$$

(4.9)

We first note that

$$\begin{aligned} \left\{ \begin{array}{l} \underbrace{\delta _t^{*,a}}_\mathrm{{exact}}-\underbrace{\hat{\delta }_t^{*,a}}_\mathrm{{approx}}=w^q-w^{q-1} = E_t \left[ \int _t^T(\epsilon ^q-\epsilon ^{q-1})(\nu _r,r)\mathrm{d}r \right] \\ \underbrace{\delta _t^{*,b}}_\mathrm{{exact}}-\underbrace{\hat{\delta }_t^{*,b}}_\mathrm{{approx}}=w^q-w^{q+1} = E_t \left[ \int _t^T(\epsilon ^q-\epsilon ^{q+1})(\nu _r,r)\mathrm{d}r \right] , \end{array} \right. \end{aligned}$$

and that

$$\begin{aligned} \epsilon ^q(\nu ,t)=\frac{A}{k}\left( 2-k(\delta _t^{*,a}+\delta _t^{*,b})+ \frac{k^2}{2}\left( (\delta _t^{*,a})^2+(\delta _t^{*,b})^2\right) +o((\delta _t^{*,a})^2+ (\delta _t^{*,b})^2)\right) . \end{aligned}$$

Since $\delta _t^{*,a}$ and $\delta _t^{*,b}$ are relatively small and $\delta _t^{*,a}+\delta _t^{*,b}$ is independent of q, we have

$$\begin{aligned} \displaystyle |\delta _t^{*,a}-\hat{\delta }_t^{*,a}|= & {} \displaystyle \left| {\mathbb {E}}_t\left[ \int _t^T\Big (\epsilon ^q-\epsilon ^{q-1}\Big )(\nu _r,r)\mathrm{d}r\right] \right| \\\le & {} \displaystyle {\mathbb {E}}_t\left[ \int _t^T\left| \Big (\epsilon ^q-\epsilon ^{q-1}\Big )(\nu _r,r)\right| \mathrm{d}r\right] \\= & {} \displaystyle O\Big (\left| \delta _r^{*,a}(q)\delta _r^{*,b}(q)-\delta _r^{*,a}(q-1)\delta _r^{*,b}(q-1)\right| \Big ]. \end{aligned}$$

That is, the differences between the exact and the approximate ask quotes can be very small. (Similar arguments can be directly applied to the bid side.) $\square$