 Solution

ID: past-exam-of-the-probability-statistics-and-options-ii-course-of-the-university-of-oxford/2010/s/s2/a/ii/solution

For one trial and any measurable set

A

P (Y \in A, accept) = \int_{A} q (y) \frac{p ( y )}{Mq ( y )} d y = \frac{1}{M} \int_{A} p (y) d y .

Since the acceptance probability is

1/ M

, conditioning gives target probability

\int_{A} p

. Repeating failed independent trials does not alter this conditional law: summing over the trial number multiplies by the geometric sum

\sum_{j \geq 0} (1 - 1/ M)^{j} / M = 1

. Taking

A = (- \infty, x)

proves

P (X < x) = \int_{- \infty}^{x} p (y) d y .

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