Futures Exchange Data Feed Delay

Worked Solution

How to Think About It: You are sitting on a trading desk with two feeds -- one fast, one slow. The slow feed lags by some random amount that is memoryless (exponential). The moment you see a price move on the fast feed, the consolidated feed is showing a stale quote. That stale quote is a free option: you know the new price, but the rest of the market (those on the slow feed) is still trading at the old one. The question is how long that window stays open, and how much you can extract before it closes.

The key parameter is the rate $\lambda = 1/50 = 0.02$ per microsecond. Everything flows from the exponential survival function $P(\delta > t) = e^{-\lambda t}$.

Quick Estimate: With mean delay $50\mu s$, the consolidated feed takes twice the mean ($100\mu s$) to catch up only $e^{-2} \approx 13.5\%$ of the time. If your order latency is $\tau = 10\mu s$, you are racing to exploit the stale quote, and $e^{-0.2} \approx 82\%$ of the time you get there before it updates. That is a big edge. If it takes you $50\mu s$ (the mean delay itself), you still catch it $e^{-1} \approx 37\%$ of the time. So speed really matters -- every microsecond shifts your capture rate by roughly $\lambda = 2\%$.

Approach: Use the exponential CDF for part (1), then combine with price impact for parts (2) and (3).

Formal Solution:

(a) Probability the consolidated feed is still stale after $100\mu s$:

The delay is $\delta \sim \text{Exp}(\lambda)$ with $E[\delta] = 1/\lambda = 50\mu s$, so $\lambda = 0.02$ per $\mu s$.

$$P(\delta > 100) = e^{-\lambda \cdot 100} = e^{-2} \approx 0.1353$$

About $13.5\%$ of the time, the consolidated feed has not yet updated after $100\mu s$.

(b) Information advantage:

Suppose the true price moved by $\Delta p$ at time $t_0$. During the delay $\delta$, the consolidated feed shows the old (stale) price. A trader with the fast feed can: - Buy at the stale (low) price if $\Delta p > 0$ - Sell at the stale (high) price if $\Delta p < 0$

If your order takes $\tau$ microseconds to reach the exchange, you capture the stale price only if the consolidated feed has not yet updated by the time your order arrives:

$$P(\text{stale at time } \tau) = P(\delta > \tau) = e^{-\lambda \tau}$$

So the expected profit per opportunity is:

$$E[\Pi] = E[|\Delta p|] \cdot e^{-\lambda \tau}$$

This is your information advantage: you earn $E[|\Delta p|]$ on each opportunity, discounted by the probability that the window is still open when your order hits.

(c) Connection to latency arbitrage:

This is exactly the mechanism behind latency arbitrage. Faster traders exploit the gap between when they observe new information and when slower participants react. The exponential model makes the value of speed transparent:

• $\tau = 10\mu s$: $P(\text{stale}) = e^{-0.2} \approx 81.9\%$
• $\tau = 50\mu s$: $P(\text{stale}) = e^{-1} \approx 36.8\%$
• $\tau = 100\mu s$: $P(\text{stale}) = e^{-2} \approx 13.5\%$

The marginal value of a microsecond reduction in latency is:

$$\frac{dE[\Pi]}{d\tau} = -\lambda \cdot E[|\Delta p|] \cdot e^{-\lambda \tau}$$

At $\tau = 0$, each microsecond is worth $\lambda \cdot E[|\Delta p|] = 0.02 \cdot E[|\Delta p|]$, i.e., roughly $2\%$ of the expected absolute price move per microsecond. This explains the arms race for co-location, microwave links, and FPGA-based trading systems -- the payoff to shaving microseconds is directly proportional to this exponential survival probability.

Answer: (a) $P(\delta > 100) = e^{-2} \approx 13.5\%$. (b) Expected profit per opportunity is $E[\Pi] = E[|\Delta p|] \cdot e^{-\lambda \tau}$, where $\lambda = 0.02/\mu s$ and $\tau$ is your order latency. (c) Each microsecond of latency improvement increases your capture rate by approximately $2\%$, which is why firms invest heavily in speed infrastructure.