Past exam of the year 1 of the computer science course of the University of Oxford / 2017 / Discrete mathematics, continuous mathematics and probability / 9 / a

e^{- t}

for

t \geq 0

E T = 1

and

E [T^{2}] = 2

, so

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

ii

Because

T

and

U

E [T U] = E T E U = 1.

The original independence assumption is essential to this factorization.

For the overlapping products, factor using the independence of

T, U, V

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

U

contributes its second moment. In contrast,

T U

and

VW

use disjoint independent factors and are themselves independent, giving

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

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