Past exam of the year 1 of the computer science course of the University of Oxford / 2013 / Discrete mathematics, probability and continuous mathematics / 9 / a

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

i

Solution

On

0 < x < 2

,

2 x - x^{2} = x (2 - x) > 0

. A valid probability density function must integrate to one, so

1 = α \int_{0}^{2} (2 x - x^{2}) d x = α [x^{2} - \frac{x ^{3}}{3}]_{0}^{2} = \frac{4}{3} α .

Thus

α = 3/4

, which is positive and ensures nonnegativity. Its expected value is

E [X] = \frac{3}{4} \int_{0}^{2} (2 x^{2} - x^{3}) d x = \frac{3}{4} [\frac{2 x ^{3}}{3} - \frac{x ^{4}}{4}]_{0}^{2} = 1.

Therefore

Valid with α = \frac{3}{4}, E [X] = 1.

The probability density function is symmetric around

1

, providing a second check on the expected value.

ii

Solution

The factor

2 x - x^{3} = x (2 - x^{2})

is positive on

(0, 2)

and negative on

(2, 2)

. For every nonzero

α

, the proposed probability density function is therefore negative on an interval. The choice

α = 0

has total integral zero. In fact,

\int_{0}^{2} (2 x - x^{3}) d x = [x^{2} - \frac{x ^{4}}{4}]_{0}^{2} = 0,

so no scaling could normalise it even without the sign obstruction. Consequently

No value of α gives a probability density function; no corresponding expectation exists .

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