Suppose, there is a HF strategy (agent) that is based on order book microstructure, and it is able to make good executions locally. More formally, in average its execution price is better than asset price $\tau$ sec. after the execution. Suppose, we manage two such agents: one for long orders, another for short orders.

The question is how to develop a controller that synchronizes between two and manages their mutual position given position limit N on each side, and maximal order size n.

I assume, this is a very broadly studied problem, especially among market makers. Can you please recommend relevant articles and ideas that provide overview of this topic and most sophisticated approaches. I'm especially interested in the very details such as: 1) Timing. is it prudent to generate random time intervals between last execution and new orders placement? 2) Pricing. How many orders can be executed on the same price? Thank you.


2 Answers 2


This paper Dealing with the Inventory Risk. A solution to the market making problem, has a full bibliography and explains the intra day market making mechanism.

The model is made of two components:

  • a diffusion of the fair price (to model the market risk)
  • a point process (with an intensity in $A \exp -k \delta$ (where $\delta$ is the distance to the fair price) to model the probability to be hit once you choose quote price

Then a stochastic control framework is set up to continuously adjust the quotes: distances (bid / ask) to the faire price.

Thanks to a tricky change of variable, the problem is solved.

  • $\begingroup$ I like the referenced paper thus far, still need to dig a bit deeper as it seems to make a lot of assumptions re price dynamics and arrival times. You may want to point out that this is partly your paper unless the name of one of the authors is coincidental. But nonetheless +1. $\endgroup$
    – Matt Wolf
    Feb 4, 2013 at 1:58
  • $\begingroup$ yes I am one of the authors of the paper, @Freddy, the assumptions we made are realistic in the sense that when the market is not "stressed", they are reasonable. When the market is stressed, no model will work... $\endgroup$
    – lehalle
    Feb 4, 2013 at 22:37
  • $\begingroup$ @lehalle, thanks for the disclaimer. Regarding your statement that no models work in stressed markets, I beg to disagree. You would essentially disqualify jump models with your statements. Discontinuous jumps are clearly a situation in which some external "stress" factor induced prices to discontinuously jump 20%, for example. Or to stick to this particular area, in many markets a stock can suddenly stop trading due to pending announcements or external events which generally precludes a larger jump, even when considering intraday hft markets. $\endgroup$
    – Matt Wolf
    Feb 5, 2013 at 3:42
  • $\begingroup$ @lehalle, one remark. I understand that you didn't put much effort for simulation, but this assumption is way too optimistic: "we assumed that our orders were entirely filled when a trade occurred at or above the ask price quoted by the agent". $\endgroup$
    – Serg
    Feb 5, 2013 at 11:56
  • $\begingroup$ @lehalle, did you test this model in the ideal setup, i.e. where price, time, and trades process are continuous and behave exactly as the model assumes? Would it generate any non-zero p&l? If not, why would it became profitable when used in discrete world? $\endgroup$
    – Serg
    Feb 5, 2013 at 12:01

You first need to clearly define your constraints first:

  • max single position size
  • max net exposure

I am not sure why you want to limit order size. The whole idea of hft strategies is to maximize turnover. As long as your strategy generates alpha you should allow it to trade as often as the strategy prescribes. All you need to then do is to constrain the strategy and OMS to adhere to your position and exposure limits.

  • $\begingroup$ Yes, in general, the max. order size can be twice the max. exposure. I can of course let both sides to trade freely just withing exposure limits, but there might be better techniques, such as answered by Lehalle. $\endgroup$
    – Serg
    Feb 4, 2013 at 0:30

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