Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 6 / a

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

i

Solution

For the rate-one exponential distribution, the probability density function is

e^{- t}

for

t \geq 0

. By integration by parts,

E T = \int_{0}^{\infty} t e^{- t} d t = 1, E (T^{2}) = \int_{0}^{\infty} t^{2} e^{- t} d t = 2.

Thus

E T = 1, Var (T) = 2 - 1^{2} = 1.

ii

Solution

By independence, the expected value of the product factors:

E (T U) = E T E U = 1.

iii

Solution

The products

T U

and

U V

share

U

, so they are not independent. Using the definition of covariance and the independence of

T, U, V

,

Cov (T U, U V) = E (T U^{2} V) - E (T U) E (U V) = (E T) (E U^{2}) (E V) - 1 = 2 - 1 = 1.

By contrast,

T U

and

VW

depend on disjoint sets of independent random variables, so they are independent and have zero covariance. Hence

Cov (T U, U V) = 1, Cov (T U, VW) = 0.

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