 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2012/s9/3/a/solution

After three multiplicative steps,

k

upward moves and

3 - k

downward moves give

S_{3} (k) = 100 (1.1)^{k} (0.9)^{3 - k} .

For

k = 0, 1, 2, 3

, respectively, the prices are

72.9, 89.1, 108.9, 133.1

. Since the steps are independent, the endpoint probabilities are binomial:

Pr (S_{3} = S_{3} (k)) = (k 3) p^{k} (1 - p)^{3 - k} .

In the displayed ascending-price order they are

(1 - p)^{3}

3 p (1 - p)^{2}

3 p^{2} (1 - p)

, and

p^{3}

. Different paths with the same count of up moves recombine because multiplication by

1.1

and

0.9

commutes.

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