 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-iii/5/a/iii/solution

Since

X^{2} = X (X - 1) + X

E [X^{2}] = G^{''} (1) + G^{'} (1)

. Subtracting the squared mean gives

var (X) = G^{''} (1) + G^{'} (1) - (G^{'} (1))^{2} .

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