Thinning a Poisson order flow

Question

Orders arrive at a matching engine as a Poisson process at $\lambda = 10$ per minute. Independently, each order is a "buy" with probability $0.4$ and a "sell" otherwise.

(a) What is the probability of exactly $3$ buys in the next minute? (b) What is the probability of no orders at all in the next $6$ seconds?

Accepted Answer

(a) Randomly keeping each event of a Poisson process with a fixed probability p, independently, gives another Poisson process at rate \lambda p (this is thinning). So buys form a Poisson process at 10 	imes 0.4 = 4 per minute. In one minute, \mu = 4: P(N = 3) = e^{-4}\frac{4^3}{3!} = e^{-4}\frac{64}{6} = 10.667\,e^{-4} \approx 10.667 	imes 0.01832 \approx 0.195. (b) For "no orders at all," use the full process (both buys and sells), rate 10 per minute. Over 6 seconds = 0.1 minute, \mu = 10 	imes 0.1 = 1: P(N = 0) = e^{-1} \approx 0.368. Takeaway. Thinning a Poisson process by an independen…

Thinning a Poisson order flow

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