 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2022/discrete-mathematics-continuous-mathematics-and-probability-resit/7/b/i/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Discrete mathematics, continuous mathematics and probability resit / 7 / b / i / Solution by

Codex  0 2026-10-25

Summing the two nonzero entries with

X = 1

gives

P (X = 1) = p (1 - e^{- p}) + p e^{- p} = p .

The remaining probability is

P (X = 0) = 1 - p

, so $X$ has the Bernoulli distribution with parameter $p$ .

For the other marginal, the two entries at

Y = 0

add to

e^{- p}

, the entry at

Y = 1

is

p e^{- p}

, and for

k ⩾ 2

the only nonzero entry is already

e^{- p} p^{k} / k!

. Thus

P (Y = k) = e^{- p} \frac{p ^{k}}{k !}, k = 0, 1, 2, \dots,

and $Y$ has the Poisson distribution with parameter $p$ . This also checks normalization of the joint probability mass function by summing the marginal.

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