 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-iii/7/b/iv/solution

k = 3

gives

I_{5} = 2 θ I_{3} = 2 θ^{2}

. Consequently

var (X_{i}^{2}) = E [X_{i}^{4}] - (E [X_{i}^{2}])^{2} = I_{5} - I_{3}^{2} = 2 θ^{2} - θ^{2} = θ^{2} .

var (θ) = \frac{n θ ^{2}}{n ^{2}} = \frac{θ ^{2}}{n} .

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