 Solution

ID: past-exam-of-the-information-theory-course-of-the-university-of-oxford/2023/3/c/ii/solution

For

0 \leq q < 1

, the number of attempts has the geometric distribution with success probability

1 - q

Pr (M_{1} = k) = q^{k - 1} (1 - q), k = 1, 2, \dots .

E M_{1} = k = 0 \sum \infty Pr (M_{1} > k) = k = 0 \sum \infty q^{k} = \frac{1}{1 - q} .

q = 0

M_{1} = 1

deterministically. At

q = 1

M_{1} = + \infty

and its expectation is

+ \infty

, rather than a finite waiting time.

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