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ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/undated/prelims/mathematics-iii-resit/4/c/iii/solution

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Past exam of the mathematics course of the University of Oxford / Undated / Prelims / Mathematics III resit / 4 / c / iii / Solution by

Codex  0 2026-10-25

Summing over the last digit yields the marginal probability mass function

p_{Y} (y) = (1 - q) q^{10 y} \sum_{x = 0}^{9} q^{x} = (1 - q^{10}) q^{10 y}, y \geq 0.

(1)

Together with the other marginal probability mass function, this gives

p_{X, Y} (x, y) = p_{X} (x) p_{Y} (y) .

(2)

The factorization holds on the whole support and with zeros elsewhere, proving that $X$ and $Y$ are independent. The conditional distribution in ii likewise does not depend on the permissible value

x

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