A trade gap longer than two seconds

Question

On a lightly traded contract, the time between consecutive trades is exponentially distributed with a mean of 2 seconds. A trade just printed.

What is the probability that the next trade takes more than 2 seconds to arrive?

Accepted Answer

For an exponential gap, the chance the wait exceeds time t is P(T t) = e^{-\lambda t}, where \lambda is the trade rate. The rate is one trade per average gap, so \lambda = 	frac{1}{2} per second. For t = 2 seconds, P(T 2) = e^{-\frac{1}{2}\cdot 2} = e^{-1} \approx 0.368. Notice a general fact hiding here: asking whether the wait exceeds exactly one mean always gives e^{-1} \approx 0.37, no matter the units. So a bit more than a third of the time, the gap runs longer than average. Because the exponential is skewed, the average gap is not the typical or "middle" gap; long gaps drag the mean up…

A trade gap longer than two seconds

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