 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/discrete-mathematics-continuous-mathematics-and-probability/9/a/ii/solution

E (X^{p}) = \int_{0}^{1} x^{p} d x = lim_{ε ↓ 0} \frac{1 - ε ^{p + 1}}{p + 1} = \frac{1}{p + 1} (p > - 1) .

In particular

E (X) = 1/2

and

E (X^{2}) = 1/3

, so

Var (X) = \frac{1}{3} - \frac{1}{4} = \frac{1}{12} .

For

p = - 1

\int_{ε}^{1} x^{- 1} d x = - lo g ε \to + \infty

; for

p < - 1

, the power integral also tends to

+ \infty

. Thus $E (X^{p}) = + \infty$ for $p \leq - 1$ , so there is no finite expected value. Negative powers need only be defined almost surely on

X > 0

; the null event

X = 0

has no effect on this conclusion.

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