 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ib/paper-2/6h/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

[0, 1]

and

[0, θ]

. Hence

E_{θ} X = \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{θ}{2} = \frac{1 + θ}{4},

E_{θ} (4 \overline{X} - 1) = θ .

Thus

θ

is unbiased.

Also,

E_{θ} X^{2} = \frac{1 + θ ^{2}}{6}, Var_{θ} X = \frac{5 θ ^{2} - 6 θ + 5}{48} .

n (θ - θ) d N (0, \frac{5 θ ^{2} - 6 θ + 5}{3}) .

Replacing

θ

θ

θ \pm z_{1 - α /2} \frac{5 θ ^{2} - 6 θ + 5}{3 n} .

It may be intersected with the parameter space

(0, 1)

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