 Solution

ID: past-exam-of-the-probability-statistics-and-options-ii-course-of-the-university-of-oxford/2011/m/m5/a/vi/solution

X_{2}

, is uniform on

0, \dots, N - 1

. Given

X_{2} = j \geq 1

, the second child takes an average of

(j + 1) /2

stickers; when

j = 0

the amount is zero. Hence, for

N \geq 1

E Y_{2} = \frac{1}{N} j = 1 \sum N - 1 \frac{j + 1}{2} = \frac{( N - 1 ) ( N + 2 )}{4 N} .

The expression gives zero for

N = 1

, as expected.

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