Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 5K (Statistical Modelling)

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- Solution 5K

Solution

The change of variable

x = (y / λ)^{k}

k x / y

, so

f_{λ} (y) = e^{- (y / λ)^{k}} \frac{k}{λ} (\frac{y}{λ})^{k - 1}, y > 0.

For fixed

k

this is

k y^{k - 1} exp {η y^{k} - A (η)}, η = - λ^{- k} < 0, A (η) = - lo g (- η) .

Thus the sample's sufficient statistic is

T = \sum_{i} Y_{i}^{k}

.

Since

(Y / λ)^{k} \sim Exp (1)

,

E (Y^{k}) = λ^{k} .

The log likelihood, up to constants, is

- kn lo g λ - T / λ^{k}

. Differentiating gives the unique maximum

λ = (\frac{1}{n} i = 1 \sum n Y_{i}^{k})^{1/ k} .

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