Almgren-Chriss Optimal Execution

Hard·25 min read

Market MicrostructureOptimal ExecutionMarket ImpactStochastic ControlAlmgren-Chriss

Quick Quiz

1. In the Almgren-Chriss model, the variance of the implementation shortfall for a given trading trajectory $x(t)$ is $\mathrm{Var}[\mathrm{IS}] = \sigma^2 \int_0^T x(t)^2\, dt$ . Why does the variance depend on the remaining inventory $x(t)$ and not on the trading rate $v(t)$ ?

The variance depends on

v(t)

because trading faster increases price uncertaintyThe variance arises from mid-price diffusion while the position is open: a larger remaining position

x(t)

means more exposure to price moves

dS_t = \sigma\, dW_t

at each instantThe variance is caused by permanent market impact, which is proportional to total shares soldThe

x(t)^2

term comes from the quadratic utility function, not from price risk

2. The Almgren-Chriss optimal trajectory is $x^*(t) = X \sinh(\kappa(T-t))/\sinh(\kappa T)$ where $\kappa = \sqrt{\lambda\sigma^2/(2\eta)}$ . This trajectory is front-loaded relative to TWAP. Which of the following correctly explains why?

Front-loading reduces temporary market impact by selling at lower rates early when the book is thinEarly in the horizon, the remaining inventory

x(t)

is large, so the variance penalty

\lambda\sigma^2 x(t)^2

is large. The optimal strategy sells more shares early to reduce this term fasterFront-loading captures price appreciation by selling before temporary impact revertsThe hyperbolic sine is an approximation to exponential decay; front-loading is an artifact of the linear approximation

3. TWAP is the optimal Almgren-Chriss strategy when $\lambda = 0$ . This follows from minimising $\eta\int_0^T v(t)^2\,dt$ subject to $\int_0^T v(t)\,dt = X$ . Which mathematical inequality demonstrates that constant $v = X/T$ achieves the minimum?

Jensen's inequality:

\mathbb{E}[v^2] \geq (\mathbb{E}[v])^2

, with equality iff

v

is constantCauchy-Schwarz:

\int v^2\,dt \cdot T \geq (\int v\,dt)^2 = X^2

, with equality iff

v

is constantThe triangle inequality applied to the

L^2

norm of

v

Grönwall's inequality applied to the ODE for remaining inventory

4. In the Almgren-Chriss model, increasing the risk-aversion parameter $\lambda$ always increases the expected implementation shortfall while decreasing the variance. Therefore, a risk-neutral trader ( $\lambda = 0$ ) always achieves the lowest expected cost.

truefalse

5. The efficient frontier in the Almgren-Chriss model is the set of (Expected IS, Variance) pairs achieved by optimal strategies for all $\lambda \geq 0$ . A point on the frontier is dominated if another point has both lower expected cost and lower variance. Which of the following strategies is always dominated (not on the efficient frontier)?

TWAP (constant rate over the full horizon)VWAP (trading at the rate proportional to historical volume participation)Selling half the position immediately and TWAP the restAn arbitrary non-monotone schedule (e.g., buying back shares during execution)

6. In the Almgren-Chriss model, permanent impact shifts the mid-price permanently while temporary impact affects only the fill price in that period. Why is permanent impact a sunk cost that does not affect the optimal trading strategy?

Permanent impact is constant regardless of the trading schedule, so it cannot be minimised by changing the scheduleThe total permanent impact cost depends only on

X

(total shares sold) and

\gamma

, not on the schedule

\{n_k\}

, making it independent of the optimisation variablePermanent impact is zero in the Almgren-Chriss model; only temporary impact is modelledPermanent impact affects future periods, so it is hedged by the subsequent trades in the schedule

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