Interpreting Trade Half-Life in the Almgren–Chriss Execution Schedule

Summary

The document asks how to interpret the trade half-life parameter in the Almgren–Chriss optimal execution model. It compares the paper’s definition of the parameter as the inverse of the decay rate with the remaining-inventory formula, which uses a ratio of hyperbolic sine terms. Substituting the proposed half-life into that finite-horizon expression does not generally leave exactly one exponential fraction of the original inventory.

The key issue raised is that the schedule depends on both the decay rate and the fixed execution horizon. The stated half-life interpretation may describe the characteristic decay timescale of an exponential component, rather than an exact depletion point for the full finite-horizon schedule. The document provides the model equation and identifies this apparent mismatch, but does not include an answer resolving it. Readers should therefore distinguish the parameter’s timescale interpretation from a direct claim about the inventory ratio at a particular time.

Key ideas

• The remaining inventory in the stated schedule depends on the decay parameter and the execution horizon.
• Substituting the inverse decay rate into the finite-horizon formula does not generally produce an exact exponential fraction of the initial inventory.
• The post raises an interpretation question but supplies no resolution or empirical evidence.
• The half-life label should be interpreted in the context of the model’s finite-horizon schedule.

Tags

# Explanation of trade half life in Almgren-Chriss paper


# Explanation of trade half life in Almgren-Chriss paper












In Section 2.3 of Almgren, Robert, and Neil Chriss. "Optimal execution of portfolio transactions." Journal of Risk 3 (2001): 5-40 the authors define the half-life of a trade as $\theta \equiv 1/\kappa$ and note that it is exactly the amount of time it takes to deplete the portfolio by a factor of $e$.

Eq(17) derives the amount of inventory remaining at a trading time $t$ to be $$ \frac{\sinh(\kappa(T - t))}{\sinh(\kappa T)} X. $$

However, setting $t = \theta \equiv 1 / \kappa$ in the above does not yield $X / e$.

I suspect I am misinterpreting the claim and seek some guidance to set me straight. Thank you!

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.