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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2025

Paper 1

The change of variable $x=(y/\lambda)^k$ has <derivative> $kx/y$, so
$$
f_\lambda(y)=e^{-(y/\lambda)^k}\frac{k}{\lambda}\left(\frac y\lambda\right)^{k-1},\qquad y>0.
$$
For fixed $k$ this is
$$
k y^{k-1}\exp\left\{\eta y^k-A(\eta)\right\},\qquad
\eta=-\lambda^{-k}<0,\quad A(\eta)=-\log(-\eta).
$$
Thus the sample's <sufficient statistic> is $T=\sum_iY_i^k$.

Since $(Y/\lambda)^k\sim\operatorname{Exp}(1)$,
$$
\mathbb E(Y^k)=\lambda^k.
$$
The log likelihood, up to constants, is $-kn\log\lambda-T/\lambda^k$. Differentiating gives the unique maximum
$$
\boxed{\widehat\lambda=\left(\frac1n\sum_{i=1}^nY_i^k\right)^{1/k}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 5K / Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 5K / Solution

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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 5K / Solution

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