Past exam of the mathematics course of the University of Oxford / Undated / Prelims / Mathematics III resit / 6 / a

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

i

Solution

A joint probability density function integrates to one. Here

1 = C \int_{0}^{1} \int_{0}^{1} u (1 + 2 v) d v d u = C (\frac{1}{2}) (2),

so

C = 1.

This value also makes the probability density function nonnegative everywhere.

ii

Solution

Obtain each marginal density by integrating the joint probability density function over the other variable:

f_{U} (u) = \int_{0}^{1} u (1 + 2 v) d v = 2 u, 0 \leq u \leq 1,

f_{V} (v) = \int_{0}^{1} u (1 + 2 v) d u = \frac{1}{2} + v, 0 \leq v \leq 1.

Each marginal density is zero off its displayed interval and has total integral one.

iii

Solution

On the product support

[0, 1]^{2}

, the joint probability density function factors as

f_{U, V} (u, v) = u (1 + 2 v) = (2 u) (1/2 + v) = f_{U} (u) f_{V} (v) .

The equality also holds outside the support with both sides zero. The probability density function factorization criterion proves that $U$ and $V$ are independent random variables.

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