Almgren-Chriss Mean-Variance Optimal Execution
A trader needs to sell $X_0$ shares over the time interval $[0, T]$. The unaffected mid-price follows arithmetic Brownian motion:
$dS_t = \sigma \, dW_t$
Trading incurs two types of market impact:
- Temporary impact: linear with coefficient $\eta$. When you trade at rate $\dot{x}(t)$ shares per unit time, you pay an additional $\eta \, |\dot{x}(t)|$ per share in slippage.
- Permanent impact: linear with coefficient $\gamma$. Your trading permanently shifts the mid-price by $\gamma \, \dot{x}(t)$.
Let $x(t)$ denote the number of shares remaining at time $t$, so $x(0) = X_0$ and $x(T) = 0$. The execution cost $C$ is the total slippage paid relative to the initial mid-price $S_0$.
- Write down the expected cost $E[C]$ and the variance $\text{Var}(C)$ of the execution cost as functionals of the trajectory $x(t)$.
- Formulate the Almgren-Chriss objective: minimize
$E[C] + \lambda \, \text{Var}(C)$
over admissible trajectories $x(t)$, where $\lambda > 0$ is the risk-aversion parameter.
- In the no-permanent-impact case ($\gamma = 0$), derive the Euler-Lagrange equation and show that the optimal trajectory takes the form
$x(t) = A \sinh\!\bigl(\kappa(T - t)\bigr) + B \cosh\!\bigl(\kappa(T - t)\bigr)$
where $\kappa = \sqrt{\lambda \sigma^2 / \eta}$. Determine the constants $A$ and $B$ from the boundary conditions.
Hints
- Write the total cost as an integral over the trading rate and the inventory. Which part of the cost depends on the path, and which part is fixed regardless of how you trade?
- The objective functional has the form $\int (\eta \dot{x}^2 + \lambda \sigma^2 x^2) \, dt$. Apply the Euler-Lagrange equation to get a second-order ODE for $x(t)$.
- The ODE $\ddot{x} = \kappa^2 x$ has solutions in terms of $\sinh$ and $\cosh$. Write the general solution centered at $T$ and use $x(T) = 0$ to kill one term, then $x(0) = X_0$ to fix the other.
Worked Solution
How to Think About It: This is the foundational model for optimal execution -- how to liquidate a large position while balancing two competing costs. If you trade too fast, you pay heavy temporary impact (slippage). If you trade too slowly, the price drifts randomly while you still hold inventory, and that unhedged inventory creates variance in your total cost. The risk-aversion parameter $\lambda$ controls the trade-off: high $\lambda$ means you hate variance and want to front-load the selling; $\lambda = 0$ means you only care about expected cost and trade at a constant rate (TWAP).
Quick Estimate: Consider the extreme cases. When $\lambda = 0$, you minimize expected cost alone; with only temporary impact, the optimal strategy is to sell at a constant rate $X_0/T$ (TWAP). When $\lambda \to \infty$, you are infinitely risk-averse and want to dump everything immediately -- a block trade at $t = 0$. For finite $\lambda$, you expect something in between: front-loaded selling that tapers off, controlled by how large $\lambda \sigma^2 / \eta$ is relative to
Approach: We write the cost functional, apply the calculus of variations (Euler-Lagrange equation), and solve the resulting ODE.
Formal Solution:
Part 1 -- Cost functionals. Let $v(t) = -\dot{x}(t) \geq 0$ be the selling rate. The execution price at time $t$ is $S_t - \eta v(t)$ (temporary impact) and the mid-price evolves as $dS_t = -\gamma v(t) \, dt + \sigma \, dW_t$ (permanent impact shifts the price against you).
The total implementation shortfall (cost relative to $S_0$) is:
$C = \gamma \int_0^T v(t)\left(\int_0^t v(s) \, ds\right) dt + \eta \int_0^T v(t)^2 \, dt - \sigma \int_0^T x(t) \, dW_t$
The permanent-impact piece can be simplified. Since $x(0) = X_0$ and $x(T) = 0$, integrating by parts gives the permanent impact contribution as $\tfrac{1}{2}\gamma X_0^2$, which is path-independent. So:
$E[C] = \frac{1}{2}\gamma X_0^2 + \eta \int_0^T v(t)^2 \, dt$
For the variance, only the stochastic integral contributes:
$\text{Var}(C) = \sigma^2 \int_0^T x(t)^2 \, dt$
This is the Ito isometry applied to the stochastic integral $\int_0^T x(t) \, dW_t$.
Part 2 -- Objective functional. We minimize:
$J[x] = \eta \int_0^T \dot{x}(t)^2 \, dt + \lambda \sigma^2 \int_0^T x(t)^2 \, dt$
The $\frac{1}{2}\gamma X_0^2$ term is a constant independent of the path and drops out of the optimization. The boundary conditions are $x(0) = X_0$ and $x(T) = 0$.
Part 3 -- Euler-Lagrange equation ($\gamma = 0$). The integrand (Lagrangian) is $L(x, \dot{x}) = \eta \dot{x}^2 + \lambda \sigma^2 x^2$. The Euler-Lagrange equation is:
$\frac{\partial L}{\partial x} - \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} = 0$
$2\lambda \sigma^2 x - 2\eta \ddot{x} = 0$
$\ddot{x}(t) = \kappa^2 x(t), \quad \kappa = \sqrt{\frac{\lambda \sigma^2}{\eta}}$
This is a second-order linear ODE with constant coefficients. The general solution is:
$x(t) = C_1 e^{\kappa t} + C_2 e^{-\kappa t}$
or equivalently in hyperbolic form centered at $T$:
$x(t) = A \sinh\!\bigl(\kappa(T - t)\bigr) + B \cosh\!\bigl(\kappa(T - t)\bigr)$
Boundary conditions:
- $x(T) = 0$: Setting $t = T$ gives $A \sinh(0) + B \cosh(0) = B = 0$.
- $x(0) = X_0$: Setting $t = 0$ gives $A \sinh(\kappa T) = X_0$, so $A = X_0 / \sinh(\kappa T)$.
The optimal trajectory is:
$\boxed{x(t) = X_0 \, \frac{\sinh\!\bigl(\kappa(T - t)\bigr)}{\sinh(\kappa T)}, \quad \kappa = \sqrt{\frac{\lambda \sigma^2}{\eta}}}$
The corresponding optimal trading rate is:
$v(t) = -\dot{x}(t) = X_0 \kappa \, \frac{\cosh\!\bigl(\kappa(T - t)\bigr)}{\sinh(\kappa T)}$
which is largest at $t = 0$ and smallest at $t = T$, confirming the front-loaded pattern.
Answer: The optimal execution trajectory in the no-permanent-impact case is $x(t) = X_0 \sinh(\kappa(T-t)) / \sinh(\kappa T)$ with $\kappa = \sqrt{\lambda \sigma^2 / \eta}$. Expected cost under temporary impact is $E[C] = \eta \int_0^T v(t)^2 \, dt$ and cost variance is $\text{Var}(C) = \sigma^2 \int_0^T x(t)^2 \, dt$. The permanent impact contributes a fixed cost $\frac{1}{2}\gamma X_0^2$ regardless of trajectory. The key dimensionless parameter $\kappa T$ controls how aggressively front-loaded the execution is: $\kappa T \to 0$ recovers TWAP, $\kappa T \to \infty$ approaches an immediate block trade.
Intuition
The Almgren-Chriss model captures the core tension in every execution algorithm: trading fast reduces your exposure to random price moves (variance) but increases your market impact cost. The optimal trajectory is a smooth curve that front-loads execution -- you sell faster at the start when your inventory (and therefore your risk) is largest, then taper off. The shape is controlled by a single dimensionless number $\kappa T$, which measures how "urgent" the liquidation is. When $\kappa T$ is small (low risk aversion, short horizon, or cheap impact), you barely deviate from TWAP. When $\kappa T$ is large, you dump most of the position immediately.
This model is the starting point for virtually every execution algo on a real trading desk. In practice, you layer on more realistic features -- nonlinear impact, stochastic volatility, signals about short-term alpha -- but the core insight remains: optimal execution is about balancing the deterministic cost of impact against the stochastic cost of holding inventory. The permanent impact being path-independent is a key structural result: it means permanent impact does not affect the shape of the optimal trajectory, only temporary impact and variance do.