 Solution

ID: past-exam-of-the-probability-statistics-and-options-ii-course-of-the-university-of-oxford/2009/s/s2/c/solution

Each

N_{i}

is geometric on

{1, 2, \dots}

with success probability

a = Z_{p} / (M Z_{q})

. Hence

E [N_{i}] = \frac{M Z _{q}}{Z _{p}}, Z_{q} / Z_{p} = \frac{1}{MN} i = 1 \sum N n_{i} .

Linearity of expectation proves the estimator is unbiased. Moreover the joint probability factors:

P (N_{i} = k, X_{i} = x) = (1 - a)^{k - 1} \frac{p ~ ( x )}{M Z _{q}} = [(1 - a)^{k - 1} a] p (x) .

Thus

N_{i} and X_{i} are independent

, not merely uncorrelated. Fresh independent trials make the simulated cycles independent as well.

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