# Correction to: Automatic threshold and run parameter selection: a climatology for extreme hourly precipitation in Switzerland

• S. Fukutome
• M. A. Liniger
• M. Süveges
Correction

Correction to: Theor Appl Climatol (2015) 120:403–416

The original version of this article unfortunately contained a mistake. Please find below the relevant part of the Appendix, with changes in the last 2 formulae.

Let i, ji, ii, di, denote for a single observation i: the log-likelihood, the score function, the expected information, and the difference between score function and expected information, respectively. Let the derivative with respect to θ be denoted by a prime. Let $$\mathbb{I}(A)$$ be the indicator function for the set A. Then, for a given (u, K) pair,
$${\ell}_i^{\hbox{'}}\left(\theta \right)=-\frac{\mathbb{I}\left({c}_i^{\left(u,K\right)}=0\right)}{\left(1-\theta \right)}+\frac{2\mathbb{I}\left({c}_i^{\left(u,K\right)}>0\right)}{\theta }-{c}_i^{\left(u,K\right)},$$
$${j}_i\left(\theta \right)=\frac{\mathbb{I}\left({c}_i^{\left(u,K\right)}=0\right)}{{\left(1-\theta \right)}^2}+\frac{4\mathbb{I}\left({c}_i^{\left(u,K\right)}>0\right)}{\theta^2}+{\left({c}_i^{\left(u,K\right)}\right)}^2-\frac{4{c}_i^{\left(u,K\right)}}{\theta },$$
$${i}_i\left(\theta \right)=\frac{\mathbb{I}\left({c}_i^{\left(u,K\right)}=0\right)}{{\left(1-\theta \right)}^2}+\frac{2\mathbb{I}\left({c}_i^{\left(u,K\right)}>0\right)}{\theta^2},$$
$${d}_i\left(\theta \right)=\frac{2\mathbb{I}\left({c}_i^{\left(u,K\right)}>0\right)}{\theta^2}+{\left({c}_i^{\left(u,K\right)}\right)}^2-\frac{4{c}_i^{\left(u,K\right)}}{\theta },$$
$${d_i}^{\hbox{'}}\left(\theta \right)=-\frac{4\mathbb{I}\left({c}_i^{\left(u,K\right)}>0\right)}{\theta^3}+\frac{4{c}_i^{\left(u,K\right)}}{\theta^2}.$$
Let $$D\left(\theta \right)={\left(N-1\right)}^{-1}\sum \limits_{k=1}^{N-1}{d}_k\left(\theta \right)$$ and $$I\left(\theta \right)={\left(N-1\right)}^{-1}\sum \limits_{k=1}^{N-1}{i}_k\left(\theta \right)$$ denote the sample means of di and ii. The sample variance of D(θ) is:
$$V\left(\theta \right)={\left(N-1\right)}^{-1}\sum \limits_{k=1}^{N-1}\left\{{\left[{d}_k\left(\theta \right)-{D}^{\hbox{'}}\left(\theta \right)I{\left(\theta \right)}^{-1}{\ell_k}^{\hbox{'}}\left(\theta \right)\right]}^2\right\}.$$
The Information Matrix Test (IMT) Statistic is then:
$$IMT\left(\widehat{\theta}\right)=\left(N-1\right)D{\left(\widehat{\theta}\right)}^2V{\left(\widehat{\theta}\right)}^{-1}$$
where θ has been replaced by the estimated value of $$\widehat{\theta}$$.