Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability / 7 / b

Table of contents
- i b
  - Solution i
- ii b
  - Solution ii
- iii b
  - Solution iii

i

Solution

By independence, the joint probability density is the product of the two uniform distribution densities:

f_{U, V} (u, v) = 1_{{0 \leq u \leq 1, 0 \leq v \leq 1}} .

Changing its values on the boundary does not affect any probability or expected value.

ii

Solution

Split the unit square into

v \leq u

and

v > u

. The maximum is

u

on the first triangle and

v

on the second, so

E [U max {U, V}] = \int_{0}^{1} [\int_{0}^{u} u^{2} d v + \int_{u}^{1} uv d v] d u = \int_{0}^{1} [u^{3} + \frac{u}{2} (1 - u^{2})] d u = \frac{1}{8} + \frac{1}{4} = \frac{3}{8} .

The shared boundary has zero probability, so it is not counted twice in this joint density calculation.

iii

Solution

For

W = max {U, V}

and

0 \leq w \leq 1

, independence gives the cumulative distribution function

F_{W} (w) = P (U \leq w, V \leq w) = w^{2} .

The cumulative distribution function is zero below

0

and one above

1

. Differentiating on the interior gives

f_{W} (w) = 2 w (0 < w < 1), f_{W} (w) = 0 otherwise,

up to irrelevant endpoint values. There are no endpoint atoms. This maximum is an order statistic.

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