 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-1/29j/c/solution

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

X_{1} \sim N (θ, 1)

and, conditionally on

X_{i - 1}

X_{i} ∣ X_{i - 1} \sim N (θ (1 - γ) + γ X_{i - 1}, 1 - γ) .

Ignoring constants independent of

θ

ℓ_{n} (θ) = - \frac{1}{2} (X_{1} - θ)^{2} - \frac{1}{2 ( 1 - γ )} \sum_{i = 2}^{n} [X_{i} - γ X_{i - 1} - θ (1 - γ)]^{2} .

Its score is

S_{n} (θ) = (X_{1} - θ) + \frac{1 - γ}{1 - γ} \sum_{i = 2}^{n} [X_{i} - γ X_{i - 1} - θ (1 - γ)] .

In terms of the independent innovations,

S_{n} (θ) = ε_{1} + \frac{1 - γ}{1 - γ} \sum_{i = 2}^{n} ε_{i} .

I_{n} (θ) = 1 + (n - 1) \frac{( 1 - γ ) ^{2}}{1 - γ} = 1 + (n - 1) \frac{1 - γ}{1 + γ} .

Since

I_{1} (θ) = 1

, the equality

I_{n} = n I_{1}

holds for all

n

exactly when

\frac{1 - γ}{1 + γ} = 1,

namely when

γ = 0 .

Thus the information tensorizes only in the independent case.

θ

the lower bound

Var_{θ} (θ) \geq \frac{1}{I _{n} ( θ )} = [1 + (n - 1) \frac{1 - γ}{1 + γ}]^{- 1} .

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