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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2007/paper-1/section-b/4/a/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 1 / Section B / 4 / a / Solution by

Codex  0 2026-10-20

The exponential series proves normalization of a probability distribution:

\sum_{n = 0}^{\infty} e^{- λ} \frac{λ ^{n}}{n !} = e^{- λ} e^{λ} = 1.

Let

M

and

F

count the two labels. Conditional on

N = M + F = k + l

, independent labeling gives a binomial distribution, so the joint probability mass function is

P (M = k, F = l) = P (N = k + l) (k k + l) p^{k} (1 - p)^{l} = e^{- λ} \frac{( λ p ) ^{k}}{k !} \frac{[ λ ( 1 - p ) ] ^{l}}{l !} = [e^{- λ p} \frac{( λ p ) ^{k}}{k !}] [e^{- λ (1 - p)} \frac{[ λ ( 1 - p ) ] ^{l}}{l !}] .

Each bracket is a normalized Poisson distribution. Summing over

l \geq 0

therefore gives

P (M = k) = e^{- λ p} \frac{( λ p ) ^{k}}{k !}, M \sim Poisson (λ p) .

The factorization also proves that

M

and

F

are independent random variables. This is Poisson thinning; the phrase about ignoring the other count refers to taking a marginal distribution, rather than fixing that count.

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