Past exam of the year 1 of the computer science course of the University of Oxford / 2013 / Discrete mathematics, probability and continuous mathematics / 8 / b

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

i

Solution

Independence of random variables factors the joint probability mass function as

P (X = x, Y = y) = P (X = x) P (Y = y)

. Therefore

E [f (X) g (Y)] = x \sum y \sum f (x) g (y) P (X = x, Y = y) = x \sum y \sum f (x) P (X = x) g (y) P (Y = y) = (x \sum f (x) P (X = x)) (y \sum g (y) P (Y = y)) .

With finite expected values, these rearrangements are justified by absolute convergence: the analogous sum with absolute values equals

E [∣ f (X) ∣] E [∣ g (Y) ∣] < \infty

. Thus

E [f (X) g (Y)] = E [f (X)] E [g (Y)] .

ii

Solution

Assume the variances are finite. Expand the square and apply linearity of expectation:

var (X + Y) = E [X^{2}] + 2 E [X Y] + E [Y^{2}] - (E [X] + E [Y])^{2} = var (X) + var (Y) + 2 (E [X Y] - E [X] E [Y]) .

Part (i), applied to the identity functions, makes the bracketed covariance zero by independence of random variables. Hence

var (X + Y) = var (X) + var (Y) .

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