Bracing for Nonlinear Impact

Optimal execution with limit price and nonlinear market impact

In a previous post, we examined how to buy Q shares in the least costly way, assuming we have: i) a maximum price s_bar we’re willing to pay ( s_bar > s_0, the current spot price); ii) an infinite time horizon; and iii) a temporary market impact function that’s linear in the speed of trading (number of shares bought per unit of time). A natural question is what happens if the market impact is, instead, a power function. Can we still explicitly compute the optimal speed of trading?

The answer, as we’ll see below, turns out to be yes. On top of that, there’s a robustness in the result, in that the optimal speed we find is simply a constant times the linear speed. This is due to the homogeneity of the problem (which would no longer hold true if one were to assume, say, a time limit, or other constraints or variations in the objective function).

As usual, the stock’s mid price is denoted by s_t, our “negative inventory” is q_t = Q minus our current inventory (so that we start at q_0 = Q and end up at 0), and our cash account x_t. These stochastic processes satisfy

\(\begin{equation*} (1) \left\{ \begin{split} ds_t &= \sigma dW_t\\ dq_t &= -\lambda_t dt\ \ \ q_0=Q,\ dq_t\le 0,\ \lambda_t\ge 0,\\ d x_t &= -(s_t + \delta + f (\lambda_t))\lambda_t dt,\\ \end{split} \right. \end{equation*} \)

where σ is the stock’s volatility and δ + f(λ_t) is the temporary market impact function (δ>0 constant), as a function of the speed of selling λ_t. Since we want to buy a fixed number of shares Q, we’ll pay Qδ — a constant— regardless of our trading strategy; therefore we can set δ=0 in the following without loss of generality. We need to maximize our expected PnL

\((2)\ \ \ \ \ \ \begin{split} &u(s_t,x_t,q_t) = \sup_{\{\lambda_t\},\ q_\tau = 0}\mathbb{E}_t\left(x_\tau - q_\tau s_\tau\right),\\ \\ &\hbox{where }\tau \hbox{ is the stopping time }s_\tau=\overline{s}\hbox{ (maximum price)}\\ \end{split} \)

which can be achieved by solving the stationary Hamilton-Jacobi-Bellman problem

\((3)\ \ \ \ \ \ \begin{equation*} \left\{ \begin{split} &\frac{\sigma^2}{2}u_{ss} + \sup_\lambda\left(-\lambda u_q - (s+f(\lambda))\lambda u_x\right) = 0\\ \\ &u(\overline{s},x,q)= -\infty\ \ \hbox{ if } q>0\\ \\ &u(\overline{s},x,0) = x,\\ \end{split} \right. \end{equation*} \)

where we have penalized the terminal constraint that q = Q minus our inventory be zero (that is, we acquired all the shares we had set to buy) if the stock price reaches our limit s_bar. We’ll make the natural ansatz

\((4) \ \ \ u(s,x,q,t) = x - sq +h(s) q^{\alpha + 1},\)

for some α>0 to be determined, which allows us to rewrite (3) as

\((5)\ \ \ \ \ \ \frac{\sigma^2}{2}u_{ss} + \sup_{\lambda}\left\{-(\alpha+1)\lambda h(s)q^\alpha - F(\lambda)\right\} = 0,\)

where F(λ) = λf(λ) (and with the same boundary conditions as in (3)), which can be rewritten

\((6)\ \ \ \ \ \ \begin{equation*} \left\{ \begin{split} &\frac{\sigma^2}{2}h’’(s) q^{\alpha+1} + F^*\left(-(\alpha+1)h(s)q^\alpha\right) = 0\\ \\ &h(\overline{s}) = -\infty, \end{split} \right. \end{equation*}\)

where F* is the Legendre-Fenchel transform of F, namely

\((7)\ \ \ \ \ \ F^*(\xi) = \sup_{\lambda}\left\{\lambda\xi - F(\lambda)\right\}.\)

Let’s now assume that the market impact function is a pure power:

\((8)\ \ \ \ \ \ f(\lambda)=\theta\lambda^a; \ \ \ \ F(\lambda)=f(\lambda)\lambda = \theta\lambda^{a+1}, \ \ \ \ (a>0).\)

With this choice, F* is given by

\((9)\ \ \ \ \ \ F^*(\xi) = a(a+1)^{-(1+1/a)}\theta^{-1/a}\xi^{1+1/a}\)

so that (6) boils down to

\((10)\ \ \ \ \ \ \frac{\sigma^2}{2}h’’(s)q^{\alpha+1} + a\theta^{-1/a} q^{\alpha(1+1/a)}h^{1+1/a} = 0,\)

which clearly forces α = a, so that all we need to solve is the following ODE:

\((11)\ \ \ \ \ \ \ \begin{equation*} \left\{ \begin{split} &\frac{\sigma^2}{2}h’’(s) + a\theta^{-1/a}h^{1+1/a}(s) = 0\\ \\ &h(\overline{s}) = -\infty. \end{split} \right. \end{equation*}\)

It is straightforward to solve (11) by integration (multiply by h’ and integrate twice) to get

\((12)\ \ \ \ \ \ h(s) = -\frac{(2a+1)^a\sigma^{2a}\theta}{\left(\overline{s} - s\right)^{2a}}.\)

In the linear case (a=1), we recover our old friend

\((13)\ \ \ \ \ \ \ h(s) = -\frac{3\sigma^2\theta}{\left(\overline{s}-s\right)^{2}}\)

(see previous post). Finally, from (7) we can compute the optimal speed of trading (which is the λ=λ* for which the sup is achieved) as

\((14)\ \ \ \ \ \ \lambda^* = \frac{(2a+1)\sigma^2}{\left(\overline{s} - s\right)^2} q. \)

Interestingly, this optimal trading rule is simply a constant times the linear case — due to the homogeneity of the problem. Therefore, it is still, in the nonlinear case, optimal to buy stock at a speed that’s proportional to the remaining inventory to acquire, and inversely proportional to the square distance of the stock price to the limit price.

Quantitatively Yours,