Past exam of the computer science course of the University of Oxford / 2015 / Probability and computing / 1 / e

Solution

P (X = k) = \frac{4}{k ( k + 1 ) ( k + 2 )}, k = 1, 2, \dots .

These masses are positive. They sum to one because the partial fraction decomposition

\frac{4}{k ( k + 1 ) ( k + 2 )} = 2 (\frac{1}{k ( k + 1 )} - \frac{1}{( k + 1 ) ( k + 2 )})

makes their sum a telescoping series with value

2/ (1 \cdot 2) = 1

E X = 4 \sum_{k = 1}^{\infty} \frac{1}{( k + 1 ) ( k + 2 )} = 4 \sum_{k = 1}^{\infty} (\frac{1}{k + 1} - \frac{1}{k + 2}) = 2.

By contrast, since

k + 1 \leq 2 k

and

k + 2 \leq 3 k

for

k \geq 1

, its second moment satisfies

E X^{2} = 4 \sum_{k = 1}^{\infty} \frac{k}{( k + 1 ) ( k + 2 )} \geq \frac{2}{3} \sum_{k = 1}^{\infty} \frac{1}{k} = \infty.

The comparison uses the divergent harmonic series. To check the variance directly, for

k \geq 4

we have

(k - 2)^{2} \geq k^{2} /4

, so the corresponding tail of

E [(X - 2)^{2}]

also diverges. Thus

E X = 2 < \infty, Var (X) = \infty.

Here “unbounded” means infinite variance for this single probability distribution.

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