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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2020

Paper 1

Because $0\le U\le1$, for fixed $t$ and sufficiently large $n$,
$$
\mathbb P\bigl(n(1-U)\ge t\bigr)
=\begin{cases}
1,&t<0,\\
\mathbb P(U\le1-t/n)=(1-t/n)^n,&t\ge0.
\end{cases}
$$
Therefore
$$
\boxed{h(t)=
\begin{cases}
1,&t<0,\\
e^{-t},&t\ge0.
\end{cases}}
$$
Equivalently, $n(1-\widehat\theta/\theta)$ converges in distribution to an <exponential distribution> of rate one.

Put $c=\log(1/\alpha)$. The limiting probability
$$
\mathbb P\!\left(n\left(1-\frac{\widehat\theta}{\theta}\right)\le c\right)
\longrightarrow1-e^{-c}=1-\alpha
$$
and the certain inequality $\widehat\theta\le\theta$ suggest the approximate <confidence interval>
$$
\boxed{\left[\widehat\theta,
\frac{n\widehat\theta}{n-\log(1/\alpha)}\right]}.
$$
Since $\mathbb E[\widehat\theta]=n\theta/(n+1)$, its expected length is
$$
\boxed{
\left(\frac{n\theta}{n+1}\right)
\left(\frac{\log(1/\alpha)}{n-\log(1/\alpha)}\right)},
$$
as required.

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ib / Paper 1 / 19H / d / Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ib / Paper 1 / 19H / d / Solution

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