How Deep (Inside the Book) Is Your Love?

An update to the Bee Gees’ hit using stochastic optimal control

Say we want to liquidate a position of q shares by the end of the trading session T, using only limit orders (so as to limit market impact). What is the optimal depth δ>0 inside the book at which we should place our orders?

That’s the kind of question which can easily be answered using a standard stochastic optimal control framework (see for instance my previous post, as well as my course on Udemy). Let’s first assume that the intensity of the Poisson-type market order flow that can eat through depth δ in the book (s + δ being the limit price of the sell order, s = mid price of the stock) is given by

\((1)\ \ \ \ \ \ A\, \frac{1}{\delta^k}, \ \ \ (k>1, \ A>0)\)

(We use a power function here for computational ease, but the reasoning below would work for the usual exponentially decreasing function.) Our trader state variables are: his cash amount x, the mid price of the stock s, his inventory q, and time. He needs to maximize his utility function (expected PnL)

\((2)\ \ \ \ \ \ u(s,x,q,t) = \sup_{\delta}\mathbb{E}_{s,x,q,t}\left(x_T + q_T s_T\right).\)

Under the standard assumption that the stock’s mid price s follows a diffusion with volatility σ, and assuming that all trades happen in increments of q_0, u can be shown to solve the Hamilton-Jacobi-Bellman problem

\((3)\ \ \ \ \ \ \begin{equation*} \left\{ \begin{split} &u_t + \frac{\sigma^2}{2}u_{ss} + A\sup_{\delta}\left(\frac{1}{\delta^k}\left(u(s,x + (s+\delta)q_0,q-q_0,t) - u(s,x,q,t)\right)\right) = 0\\ \\ &u(s,x,q=0,t) = x\\ \\ &u(s,x,q,t=T) = x + qs. \end{split} \right. \end{equation*}\)

We can now make the following Ansatz

\((4)\ \ \ \ \ \ u(s,x,q,t) = x + qs + h(q,\tau),\ \ \tau = T-t,\)

so that, to first order (assuming q_0 is small) (3) reduces to:

\((5)\ \ \ \ \ \ \begin{equation*} \left\{ \begin{split} &h_\tau= Aq_0 \sup_{\delta} \frac{1}{\delta^k}\left(\delta - h_q\right) \\ \\ &h(0,\tau) = 0\\ \\ &h(q,0) = 0. \end{split} \right. \end{equation*}\)

The optimal δ = δ* at which the sup is achieved in (5) is given by

\((6)\ \ \ \ \ \ \delta^* = \frac{k}{k-1} h_q,\)

so that (5) reduces to the first-order Hamilton-Jacobi equation

\((7)\ \ \ \ \ \ h_\tau = \frac{A q_0}{k}\left(1 - \frac{1}{k}\right)^{k-1} \frac{1}{h_q^{k-1}},\)

along with the same boundary conditions as in (5). Let’s look for the solution to (7) in a homogeneous form

\((8)\ \ \ \ \ \ h(q,\tau) = c q^\alpha \tau^\beta,\ \ \ (c,\,\alpha,\ \beta = \hbox{const}). \)

Substituting (8) into (7), we find

\((9)\ \ \ \ \ \ \begin{equation*} \begin{split} &\alpha = 1 - \frac{1}{k},\ \ \ \beta = \frac{1}{k}\\ \\ &c = \left(Aq_0\right)^{1/k}, \end{split} \end{equation*}\)

so that the optimal depth at which to place our limit orders is given by

\(\boxed{(10)\ \ \ \ \ \ \delta^* = c\left(\frac{\tau}{q}\right)^{1/k}}\)

Let’s plot this optimal depth as a function of time = T - τ (T=1 below) and inventory q:

As we can see, we find that, consistent with our intuition, the optimal depth δ*

• is decreasing as a function of inventory: indeed, with a large inventory, we are willing to sacrifice a bit of liquidity premium δ in order to increase the probability of getting filled so as to reduce our inventory

• is decreasing in time (increasing w.r.t. τ), since we need to liquidate faster as maturity T approaches.

(Python code below for paying subscribers.)

Quantitatively Yours,

---

from __future__ import print_function
import numpy as np
import matplotlib.pyplot as plt

DisplayPlots = True
k = 5
c = 0.167
t0 = 0
t1 = 1
T = 300
q0 = 500
q1 = 10000
Q = 100
t_pts = np.linspace(t0,t1,T)
q_pts = np.linspace(q0,q1,Q)
delta = np.zeros((T,Q))
for i in range(T):
    for j in range(Q):
        tau_val = t1 - t_pts[i]
        q_val = q_pts[j]
        delta[i,j] = c * (tau_val/q_val)**(1/k)

if DisplayPlots: 
    t, q = np.meshgrid(t_pts, q_pts)

    # Plot the contour
    plt.figure()
    lvls = 0.01*np.array([0.,1.,2.,3.,4.,5.])
    contour_plot = plt.contourf(t, q, delta.T, cmap='coolwarm',levels=lvls)
    plt.colorbar(contour_plot, label='depth ($)')
    plt.title('optimal depth of order placement', fontsize=14)
    plt.xlabel('time', fontsize=14)
    plt.ylabel('inventory', fontsize=14)

    plt.show()