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 Incoming links: Independent Poisson conditioning

Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 46 / 1 / b / Solution 2026-10-03

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Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 1 / 13K / b / Solution Created 2026-09-24 Updated 2026-09-25

For nonnegative

y_{i}

summing to

m

, divide the joint Poisson mass by

P (M = m) = e^{- λ_{+}} λ_{+}^{m} / m!

:

P (Y = y ∣ M = m) = \frac{m !}{\prod y _{i} !} \prod_{i} (\frac{λ _{i}}{λ _{+}})^{y_{i}} .

This is independent Poisson conditioning.

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