Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 29L / b

Table of contents
- i b
  - Solution i
- ii b
  - Solution ii

i

Solution

The log likelihood is

ℓ (θ) = n lo g θ - θ \sum_{i = 1}^{n} X_{i} .

Its strictly concave maximizer is

θ_{n} = \frac{n}{\sum _{i} X _{i}} = \frac{1}{X} .

ii

Solution

Since

E X_{i} = 1/ θ

and

Var (X_{i}) = 1/ θ^{2}

, the central limit theorem gives

n (\overline{X} - 1/ θ) \Rightarrow N (0, 1/ θ^{2}) .

Applying the delta method to

g (x) = 1/ x

yields

n (θ_{n} - θ) \Rightarrow N (0, θ^{2}) .

Thus, with

z_{1 - α /2} = Φ^{- 1} (1 - α /2)

, an asymptotic interval centered at the MLE is

C_{n} = [θ_{n} - z_{1 - α /2} \frac{θ _{n}}{n}, θ_{n} + z_{1 - α /2} \frac{θ _{n}}{n}],

optionally intersected with

(0, \infty)

. Slutsky's theorem gives coverage tending to

1 - α

.

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