The <Markov inequality> applied to the first exponential-moment bound gives
$$
\mathbb P\left\{
\frac1\theta\sum_{i=1}^n
(\psi(\theta X_i)-\theta\mu)\geq t
\right\}
\leq
\exp\left(-\theta t+\frac{\theta^2\sigma^2n}{2}\right).
$$
The second bound controls the lower tail. The <union bound> therefore gives
$$
\mathbb P\left\{
\left|\frac1\theta\sum_{i=1}^n
(\psi(\theta X_i)-\theta\mu)\right|\geq t
\right\}
\leq
2\exp\left(-\theta t+\frac{\theta^2\sigma^2n}{2}\right).
$$
Since
$$
\widehat\mu_\theta-\mu
=\frac1{n\theta}\sum_{i=1}^n
(\psi(\theta X_i)-\theta\mu),
$$
putting $L=\log(2/\delta)$,
$$
\theta=\sqrt{\frac{2L}{\sigma^2n}},
\qquad
t=\sqrt{2\sigma^2nL}
$$
makes the exponent equal to $-L$. It follows that the <Catoni mean estimator> satisfies
$$
\boxed{
\mathbb P\left(
|\widehat\mu_\theta-\mu|
\geq\sigma\sqrt{\frac{2\log(2/\delta)}n}
\right)\leq\delta}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 223 / 4 / b / Solution

{c}

A Catoni mean estimator replaces each scaled observation by a nondecreasing influence function whose exponential envelope is controlled by a quadratic. This yields sub-Gaussian deviation bounds assuming only a finite variance.


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