text:
Mental Illness in America
The news is full of crazy people. Grafton Thomas, the man who attacked a house full of Hasidic Jews with a machete, has been charged with "Hate Crimes," but evidence is mounting that he is insane:
But on Sunday afternoon, a family friend, Taleea Collins, and Mr. Thomas's pastor, the Rev. Wendy Paige, said that he had struggled with mental illness for two decades and had repeatedly sought help over the years. . . . Ms. Paige said Mr. Thomas had been diagnosed with schizophrenia.
I have written before that the intersection of mental illness with extreme ideology is a major threat to our safety, and here is another example.
But what should we do about it? The Rev. Paige says she
had struggled for years to understand why Thomas wasn't institutionalized. "There hasn't been anyone who has given a real solution to deal with a grown man who is dealing with schizophrenia, other than 'Go home and call us if something happens.'"
Mitchell Silber, a former director of intelligence for the NYC police who has lately made himself an expert on antisemitic hate crimes, writes that
Failing to treat individuals with documented mental health issues is not an acceptable solution.
But what would an "acceptable solution" be? Our laws do not allow locking up mentally ill people unless they present a clear danger to themselves or others, and many people who knew Grafton Thomas thought he was mild-mannered and anything but violent; before this attack, could a case have been made that he was a danger to society? He did once have a tense confrontation with the police, which led to his being charged with "second-degree reckless endangerment and menacing a police or peace officer," but those sorts of confrontations happen every day. Are we going to lock up all the homeless people who have ever had a run in with the cops?
Meanwhile there has been lots of news about the ongoing homelessness crisis in California, with everyone from Donald Trump to the NY Times weighing in. The problem of shit, needles and trash on the streets of San Francisco is so bad that big tech conferences that used to be held there are moving elsewhere.
The Times rather predictably takes the tone that this is an economic problem, and they explicitly address it under the heading of inequality:
The path to becoming homeless can start with a large medical bill that causes someone to fall behind on their rent payments, which leads to eventual eviction. More than half of the people surveyed in Los Angeles cited economic hardship as the primary reason that they fell into homelessness. In San Francisco, 26 percent of the homeless surveyed cited the loss of a job as the primary cause.
But really in California as everywhere else homelessness is very much enmeshed with mental illness. Statistics on homelessness are a mess because there are two kinds of homeless people: the kind who lose their homes through eviction or fleeing from abuse and then very quickly find new housing (the median length of stay in a US homeless shelter is one night), and the long-term homeless. The statistics above, from Kevin Drum at Mother Jones and therefore not likely to be a conservative hack job, show that most long-term homeless people have serious non-economic problems. Fully 78% have diagnosed mental health problems, and 75% have substance abuse problems; 50% have the trifecta of physical health problems, mental health problems, and substance abuse. Notice that alcoholics and drug abusers are especially unlikely to be found in shelters, because in a shelter they have a hard time getting high.
Take a look at the Times photo essay on a miserable homeless camp in Oakland; tell me that seems like a gathering of sane people.
This is not to say that economics does not play a role here; we have probably all known people with mental health or drug problems that might well have landed them on the street if they did not have middle class families and other resources. I certainly have. It may well be true that some people end up homeless after losing a job; losing a job is traumatic for anyone, but for mentally ill people just holding it together it might be the shove that sends them over the edge. Housing costs and the shortage of subsidized housing may also be factors; again, struggles that people with good mental health might be able to manage might be too much for the mentally ill.
But to write about homelessness without mentioning mental health is just absurd.
That being said, what can we do about it? Better community health care is obviously thing one, and it might be enough to make a big dent in the problem. If you've ever tried to find good mental health providers who are taking new patients and your insurance, you know what a problem that can be even for middle class people. It must be ten times worse for poor people. California has spent billions over the past decade "fighting homelessness," but I think they would have done better to invest in mental health care. My only hesitation is that since the need is so vast, it is hard to imagine how much money we might need to spend to provide decent mental health care for everyone.
And what about people who refuse care? Because this is a serious problem. "He was on medication for a while but didn't like it and stopped taking it" is a line in many mental health tragedies, including that of Grafton Thomas. Is there a solution to this beyond "Go home and call us if something happens"?
Some people have talked about re-opening state mental hospitals for long-term care, but that is very expensive and the last time we tried it the system was full of brutality and neglect. Several commentators have mentioned the number of mentally ill people in prison, with the idea, stated or not, that maybe they would be better off in mental hospitals. Maybe so, but most mentally ill people have not committed crimes that would get them long prison sentences, and we are a long way from being able to predict which ones are likely to. Scott Alexander, a psychiatrist, has written a long piece about why putting the mentally ill homeless in hospitals is a bad idea, and I recommend it for the curious.
So what is the answer? I honestly don't know. But it seems to me that if we want a world in which everyone who wants one has a home, we first need to invest in mental health care, and if we want to protect our public spaces from being overrun by crazy people who yell at us, we may need to violate a few rights.
Labels: psychology, society
The obvious middle ground, as you allude to later in your post, is to make mental healthcare available to anyone who wants it. We need to solve our massive homelessness crisis. We need to get people out of poverty. These negative conditions promote and perpetuate mental illness, so we must alleviate or eliminate them to whatever extent is possible.
We can't have a reactionary mindset, we have to be preventative. The best way to prevent crimes by the mentally ill is to prevent mental illness itself. If we're squeemish about locking up people who only might be a threat, then we need to be doing everything we can to ensure people never get that far along - that their problems are handled before there is even a chance of them being a threat.
You have a homeless man who is schizophrenic, and you think he might potentially be a danger to people, but you don't want to lock him up until it's unquestionably plain and obvious that he's a threat? Then give him a decent house; give him a decent job; give him decent food; give him all the medical attention he could possibly need; give him assisted living; help him to be stable and healthy, and surround him with people who can monitor him for worsening of his condition and intervene before he does become a threat to himself or others.
Until we start investing in our own society, it's going to keep slowly falling apart.
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Mercedes SLC 180 Is Brand's Most Affordable Cabrio In U.K.
Will 156 horsepower be enough for open-top thrills?
The SLC's days might be numbered if we were to rely on a couple of recent reports indicating there's not going to be a next generation. Should you be interested in owning one of the few remaining hardtops while it's still available, the cabriolet is now cheaper than ever, at least in United Kingdom. Kicking off at £32,039 ($41,530), the newly added SLC 180 AMG Line sits at the bottom of the range and can be had for as low as £276 on a 36-month plan with a £3,999 customer deposit.
You'll have to settle with just 156 horsepower (116 kilowatts) and 184 pound-feet (250 Newton-meters) of torque provided by a turbocharged 1.6-liter gasoline engine. Despite the lesser engine, it's still a decent performer as the droptop Mercedes will cover the 0-62 mph (0-100 kph) in 7.9 seconds and max out at a respectable 140 mph (225 kph).
Power is channeled to the wheels through a standard six-speed manual gearbox, but you can spend an extra £1,600 and get the SLC 180 AMG Line with Mercedes' silky smooth nine-speed automatic. Although we're dealing with an entry-level model, the car still comes generously equipped as it features standard cruise control, sports suspension, 18-inch alloy wheels, air conditioning, Artico seats, sports braking system, and Active Brake Assist system.
The addition of the 180 AMG Line model is not the only change Mercedes U.K. has made to its SLC lineup considering the SLC Sport model line has been discontinued while all AMG Line cars are now £500 cheaper than before. The most expensive SLC money can buy would have to be the hot biturbo V6-powered 43 model from £47,600 ($61,713) coming with 362 hp (270 kW) and 384 lb-ft (520 Nm) for a sprint in 4.7 seconds and a top speed capped at 155 mph (250 kph).
Note: Raw conversion from British pounds to U.S. dollars. Does not take into account differences in tax.
The SLC 180 AMG Line has been added to the SLC range, with prices starting from £32,039 OTR.
With monthly payments starting from as low as £276, the new SLC 180 makes the Mercedes-Benz roadster range more accessible than ever before.
The SLC 180 features a turbo-charged 1.6-litre petrol engine producing 156 hp and 250 Nm. It can accelerate from 0 to 62 mph in 7.9 seconds and has a top speed of 140 mph. It emits a low 134 g/km of CO2 and delivers up to 48.7 mpg combined. A six-speed manual transmission is standard, or for £1,600 customers can opt for a 9G-Tronic automatic gearbox.
The SLC 180 AMG Line comes as standard with 18-inch alloy wheels; Artico seats; Active Brake Assist; cruise control; air conditioning; sports suspension; and sports braking system.
The SLC Sport model line has also been discontinued to make the range even simpler, with AMG Line now the only option. The price of AMG Line trim has also been reduced by £500, making the SLC line-up even more attractive.


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text:
\section{Introduction}

	Suppose that we have a sample of observations $(p_i,y_i)_{i=1}^n$ from $(P_i,Y_i)_{i=1}^n$ such that $(P_i,Y_i)$ has the same distribution as $(P,Y) \in [0,1]\times \{0,1\}$, $i=1,\dots,n$.
	The interpretation is that $P_i$ is a prediction for the probability that $Y_i = 1$.

	The random variables are defined on some underlying probability space $(\Omega,\mathcal{F})$ and  $\mathcal{P}$ denotes all probability measures on $(\Omega,\mathcal{F})$. \citet{HosmerLemeshow1980} propose a test for the null hypothesis of perfect calibration
	\[
	\mathcal{H}_{\text{HL},n} = \{\mathbb{P} \in \mathcal{P}\;|\; \mathbb{E}_{\mathbb{P}}(Y_i | P_i) = P_i\; \text{ $\mathbb{P}$-almost surely},\; i=1,\dots,n\}.
	\]

	The Hosmer-Lemeshow (henceforth HL) test is based on partitioning the interval $ [0,1] $ in $ g \in \mathbb{N} $ bins and counting the  observed numbers of events, $ o_{1g} $, and no event occurrences, $ o_{0g} $, in each bin. Based on that binning and counting procedure, the HL test statistic  to test for perfect calibration of the probability predictions is
	\begin{align}
		\label{eq:HLTest}
		\widehat C = \sum_{k=1}^{g} \left[ \dfrac{(o_{1k} - \widehat e_{1k})^2}{\widehat e_{1k}} + \dfrac{(o_{0k} - \widehat e_{0k})^2}{\widehat e_{0k}} \right] ,
	\end{align}
	where $ \widehat{e}_{1k} $ and $ \widehat{e}_{0k} $ are the expected event and no event occurrences in bin $ k $, respectively \citep{HosmerLemeshowSturdivant2013}.
	Under the null hypothesis, $\widehat C$ asymptotically follows a $\chi^2$-distribution with $g$ degrees of freedom given that the sample $(P_i,Y_i)_{i=1}^n$ was not used for model estimation (and $g-2$ degrees of freedom otherwise).

	Technically, the choice of the binning procedure is up the user of the HL test and is conventionally implemented via quantile based binning strategies with $g=10$, resulting in equally populated bins (decile-of-risk). Less commonly, the test is based on equidistantly spaced bins, where the unit interval (or the range of prediction values) is divided  into $g$ equidistant bins.
	While little attention is devoted to the binning procedure in practical applications, it implicitly determines the set of alternatives the test has power against \citep[Section 5]{Dimitriadis2022Honest}, such that the test result is often highly sensitive to the exact implementation of the binning; see e.g., \citet{Hosmeretal1997, Bertolini2000, Kuss2002} and our empirical application in Section \ref{sec:application}.

	Nevertheless, the HL test is still the literature's favorite for checking the calibration of binary prediction models and commonly used in current and highly influential medical and epidemiological studies; see amongst many others \citet{HLappl1, HLappl3, HLappl4}.

	In this article, we suggest a safe and stable HL test based on e-values (that we describe below) and isotonic regression \citep{Ayer1955, Brunk1965}.
	The test is henceforth called eHL test.
	\cite{Dimitriadise2016191118} recently propose the use of isotonic regression to resolve the closely related instability issue stemming from binning approaches in so-called reliability diagrams in forecast evaluation.
	While feasible inference on the isotonic regression for classical testing procedures is hampered by complicated asymptotic distributions and an inconsistency of the bootstrap, the e-values adopted here prove to be an appealing alternative in this setting.
	Based on online isotonic regression studied by \citet{Kotlowski2017}, we show that (an ideal version of) our eHL test has power against essentially all deviations from calibration, which makes it theoretically superior to the classical HL test.

	E-values, where `e' abbreviates the word `expectation',  were proposed recently as an alternative to p-values in testing problems. In a nutshell, an e-value is a realization of a non-negative random variable whose expected value is at most one under a given null hypothesis. This already signals that an e-value itself allows for meaningful interpretations since an e-value greater than one provides evidence against the null hypothesis.
	Additionally, the multiplicative inverse of an e-value is a conservative p-value by Markov's inequality. From a game-theoretic perspective, the e-value has a simple financial meaning in the sense that the e-value can be seen as the factor by which a skeptic multiplies her money when betting against the null hypothesis; see \citet{ShaferVovk2019,Shafer2021}.

	An important advantage of e-values over p-values is their uncomplicated behavior in combinations: the arithmetic average of e-values also is an e-value, likewise the product of independent or successive e-values; see \citet{Shafer2021,GrunwaldHeideETAL2019,WangRamdas2020}. In practice, this appeals because more evidence can be added later, i.e.~evidence across studies can easily be combined.

	The proposed eHL test offers a safe alternative to a fragile state-of-the-art approach by avoiding ad-hoc choices and software instabilities.

	It can be regarded as an application of the Universal Inference approach of \citet{Wasserman2020}. While this method allows to construct valid tests under only weak assumptions, it has been observed that this validity often comes at the price of a diminished power \citep{Strieder2022, Tse2022}. In Section \ref{sec:construction_eHL}, we show that an ideal -- but computationally infeasible -- variant of the eHL test does have guaranteed power to detect essentially all violations of calibration. Our proof relies on connections between the proposed e-value and the regret in random permutation online isotonic regression, which is studied by \citet{Kotlowski2017}. It has been observed that power guarantees for anytime-valid tests can be obtained by means of regret bounds of online prediction methods, see for example the discussion in \citet{Casgrain2022}. Our result demonstrates that such a connection also exists in the batch case of $e$-values for a fixed sample size $n$ due to connections with the online random permutation setting.

	In Section \ref{sec:simulation}, we compare a feasible version of the eHL test to the classical HL test in a simulation study.
	As expected, we find that the eHL test has conservative rejection rates under the null hypothesis and quickly develops power under model mispecification.
	While its empirical test power is lower than the one of the classical HL test, we do not consider this to be problematic as HL tests are often carried out in cases of vast data sets and are even criticized as being ``too powerful'' in that they reject essentially all, even acceptably well calibrated models \citep{Paul2013, Nattino2020}.
	See \cite{Dimitriadis2022Honest} for an alternative solution to this problem based on confidence bands.

	We apply the eHL test in Section \ref{sec:application} to predictions of a logistic regression model for the binary event of credit card defaults in Taiwan in 2005, where over-issuing of credit cards lead to many default payments and a subsequent credit card crisis \citep{YehLien2009}.
	The eHL test provides clear evidence against calibration of the logistic model predictions, and further illustrates that recalibration methods work well.
	In contrast, the classical HL test based on different natural binning choices delivers equivocal results with p-values ranging from 0 to 0.91 for a single prediction method, implying that a researcher could have cherry-picked the binning specification and hence the test result to her will.

	\section{Construction of HL e-values \label{sec:construction_eHL}}

	\subsection{Preliminaries}

	An e-variable for $\mathcal{H}_{\text{HL},n}$ is a non-negative random variable $E$ (that is allowed to take the value $+\infty$) such that $\mathbb{E}_{\mathbb{P}}(E) \le 1$ for all $\mathbb{P} \in \mathcal{P}$. An e-value is a realization of an e-variable. An e-variable $E$ always yields a valid p-variable $1/E$ (a p-value is a realized p-variable) by Markov's inequality, since
	\begin{equation}\label{eq:Markov}
		\mathbb{P}\Big(\frac{1}{E} \le \alpha\Big) = \mathbb{P}\Big(E \ge \frac{1}{\alpha}\Big) \le \alpha \mathbb{E}_{\mathbb{P}}(E) \le \alpha, \quad \text{for all $\mathbb{P} \in \mathcal{H}_{\text{HL},n}$}.
	\end{equation}
	We reject the null hypothesis $\mathcal{H}_{\text{HL},n}$ if we observe a large value of $E$. If we want to ensure a classical p-guarantee then we have to determine the rejection region for a given $\alpha$ by \eqref{eq:Markov}. \citet{VovkWang2021} show that this is essentially the only way to transform an e-variable into a p-variable.
	We say that an e-variable has the alternative hypothesis $\mathcal{H'} \subset \mathcal{P}$ if $\mathbb{E}_{\mathbb{Q}}(E) > 1$ for all $\mathbb{Q} \in \mathcal{H}'$.

	\subsection{Sample size one}\label{sec:HL1lambda}
	We first construct e-variables for the sample size one Hosmer-Lemeshow null hypothesis
	\[
	\mathcal{H}_{\text{HL},1} = \{\mathbb{P} \in \mathcal{P}\;|\; \mathbb{E}_{\mathbb{P}}(Y | P) = P \}.
	\]
	In the special case here, e-variables are likelihood ratios conditional on $P$. Indeed, if $q \in [0,1]$, an e-variable for $\mathcal{H}_{\text{HL},1}$ is given by
	\[
	E_q(P,Y) = \frac{q^Y(1-q)^{1-Y}}{P^Y(1-P)^{1-Y}} = \begin{cases}q/P, & \text{if $Y = 1$,}\\ (1-q)/(1-P), &\text{if $Y=0$.} \end{cases}
	\]
	The variable $E_q(P,Y)$ is clearly non-negative, and for $\mathbb{P} \in \mathcal{H}_{\text{HL},1}$,
	\begin{align*}
		\mathbb{E}_{\mathbb{P}}(E_q(P,Y)) & = \mathbb{E}_{\mathbb{P}}\left(\mathbb{E}_{\mathbb{P}}(Y \mid P)\frac{q}{P}  + \mathbb{E}_{\mathbb{P}}(1-Y \mid P)\frac{1-q}{1-P} \right) \\
		& = \mathbb{E}_{\mathbb{P}}\left(P\frac{q}{P} + (1-P) \frac{1-q}{1-P}\right) = 1.
	\end{align*}
	To find alternative hypotheses for the e-variable $E_q$, let $\bar{\pi} = \mathbb{E}_{\mathbb{Q}}(Y \mid P)$. Then,
	\begin{align*}
		\mathbb{E}_{\mathbb{Q}}(E_q(P, Y) \mid P) & = \bar{\pi} \frac{q}{P} + (1 - \bar{\pi})\frac{1-q}{1-P}

	\end{align*}
	is strictly larger one if and only if, $\bar{\pi} > P$ and $q > P$, or, $\bar{\pi} < P$ and $q < P$, i.e., if $\pi$ and $q$ are to the same side of $P$.
	This shows that if $q < P$, $E_q$ has the alternative
	\begin{equation}\label{eq:alternative_q<P}
		\mathcal{H}' = \{\mathbb{Q} \in \mathcal{P} \mid \mathbb{E}_{\mathbb{Q}}(Y\mid P) < P\},
	\end{equation}
	and if $q > P$, $E_q$ has the alternative
	\begin{equation}\label{eq:alternative_q>P}
		\mathcal{H}' = \{\mathbb{Q} \in \mathcal{P} \mid \mathbb{E}_{\mathbb{Q}}(Y\mid P) > P\}.
	\end{equation}

	It is possible to show that basically any e-variable for $\mathcal{H}_{\text{HL},1}$ is of the form $E = E_q(P, Y)$ for some $q$ (depending on $P$) but this requires some more arguments; it follows by the construction in \citet{HenziZiegel2021}, see also \citet{Waudby-SmithRamdas2020}.
	The connection of $E_q(P,Y)$ to the e-variables in \citet{HenziZiegel2021} of type $E = 1 + \lambda D$ with $D \geq -1$ such that $\mathbb{E}_{\mathbb{P}}(D) = 0$ for $\mathbb{P} \in \mathcal{H}_{\text{HL},1}$, follows from the fact that $\lambda$ in this representation can be bijectively mapped to $q$. In this context,
	\begin{equation}\label{eq:Elambda}
		E = 1 + \lambda (P - Y)
	\end{equation} is an  e-variable for $\mathcal{H}_{\text{HL},1}$ for any $\lambda$ that is $\sigma(P)$-measurable with $-(1/P) \le \lambda \le 1/(1-P)$. If $P = 1$, there is no restriction on $\lambda$ from above, and analogously if $P = 0$, there is no restriction from below. By choosing $\lambda = (P-q)/(P(1-P))$, we obtain that $E = E_q(P,Y)$.

	Clearly, the e-variable $E_q(P,Y)$ may take the value infinity if either $P = 0$ and $Y = 1$ or $P = 1$ and $Y = 0$ occurs; a single observation $Y = 1$ or $Y = 0$ is sufficient to reject the hypothesis of calibration with certainty if the predicted probabilities are in $\{0,1\}$. For the remainder of the theoretical part of this paper, we will always make the assumption $\mathbb{P}(P \in \{0,1\}) = 0$ to exclude these special but uninteresting cases.

	%

	\subsection{Combining e-values in the iid case \label{sec:comb_evals}}

	We assume now that $(P_i,Y_i)_{i=1}^n$ are independent and identically distributed (iid).
	For testing $\mathcal{H}_{\text{HL},n}$, we suggest the e-variable
	\begin{equation}\label{eq:EHLn}
		E_{\text{HL},n}^{\text{id}} = \prod_{i=1}^n E_{q_i}(P_i,Y_i),
	\end{equation}
	where $q_i$ is $\sigma(P_1,\dots,P_i,Y_1,\dots,Y_{i-1})$-measurable. For $\mathbb{P}\in \mathcal{H}_{\text{HL},n}$, we have
	\begin{align*}
		\mathbb{E}_{\mathbb{P}}E_{\text{HL},n}^{\text{id}} &= \mathbb{E}_{\mathbb{P}}\Big(\mathbb{E}_{\mathbb{P}}\Big(\prod_{i=1}^n E_{q_i}(P_i,Y_i)|P_1,\dots,P_n,Y_1,\dots,Y_{n-1}\Big)\Big)\\
		&= \mathbb{E}_{\mathbb{P}}\Big(\prod_{i=1}^{n-1} E_{q_i}(P_i,Y_i)\mathbb{E}_{\mathbb{P}}\Big(E_{q_n}(P_n,Y_n)|P_1,\dots,P_n,Y_1,\dots,Y_{n-1}\Big)\Big)\\
		&= \mathbb{E}_{\mathbb{P}}\Big(\prod_{i=1}^{n-1} E_{q_i}(P_i,Y_i)\Big(1 + \frac{P_n - q_n}{P_n(1-P_n)} \mathbb{E}_{\mathbb{P}}(P_n - Y_n |P_1,\dots,P_n,Y_1,\dots,Y_{n-1})\Big)\Big)\\
		&= \mathbb{E}_{\mathbb{P}}\Big(\prod_{i=1}^{n-1} E_{q_i}(P_i,Y_i)\Big(1 + \frac{P_n - q_n}{P_n(1-P_n)}  \mathbb{E}_{\mathbb{P}}(P_n - Y_n |P_n)\Big)\Big)\\
		& = \mathbb{E}_{\mathbb{P}}\Big(\prod_{i=1}^{n-1} E_{q_i}(P_i,Y_i)\Big) = \mathbb{E}_{\mathbb{P}}E_{\text{HL},n-1}^{\text{id}} = \dots =  1,
	\end{align*}
	where we used the equivalent representation of $E_q(P,Y)$ in \eqref{eq:Elambda}.
	In particular, from the above derivation it is easy to see that $(E_{\text{HL},n}^{\text{id}})_{n \in \mathbb{N}}$ is a test martingale.

	The e-variable $E_{\text{HL},n}^{\text{id}}$ depends on the ordering of $(P_i,Y_i)_{i=1}^n$ through the choice of $q_i$. Let $S_n$ denote all permutations of $\{1,\dots,n\}$, and for $\sigma \in S_n$ define $E_{\text{HL},n}^{\sigma}$ as $E_{\text{HL},n}^{\text{id}}$ for the random variables $(P_{\sigma(i)},Y_{\sigma(i)})_{i=1}^n$ instead of $(P_i,Y_i)_{i=1}^n$. Generally,
	\[
	\sup_{\sigma \in S_n} E_{\text{HL},n}^{\sigma}
	\]
	is not an e-variable for $\mathcal{H}_{\text{HL},n}$, so one would guess that there are opportunities to fish for (spurious) significance by choosing some specific ordering of a sample of observations $(p_i,y_i)_{i=1}^n$. If there is a natural ordering of the observations such as a time stamp then the problem usually does not occur in applications since a different ordering of the observations is hard to justify. Indeed, when the observations are sequential (and possibly dependent), the e-variable defined at \eqref{eq:EHLn} is also an e-variable for the hypothesis
	\[
	\mathcal{H}_{\text{HL},n,seq} = \{\mathbb{P} \in \mathcal{P}\;|\; \mathbb{E}_{\mathbb{P}}(Y_i | P_1,\dots,P_{i},Y_1,\dots,Y_{i-1}) = P_i\; \text{$\mathbb{P}$-almost surely},\; i=1,\dots,n\}.
	\]
	Contrary to classical theory, the sequential case is easier to treat than the iid case and has been the focus of many works employing e-values including for example \citet{Waudby-SmithRamdas2020,HenziZiegel2021}.

	Coming back to our situation with iid data, an alternative to \eqref{eq:EHLn} could be
	\[
	E_{\text{HL},n,\text{sym}} = \frac{1}{n! } \sum_{\sigma \in S_n} E_{\text{HL},n}^{\sigma}.
	\]
	This strategy is essentially the merging technique for independent e-values in Section 4 of \citet{VovkWang2021}, and the object of interest in this article.

	\subsection{An ideal test with power guarantees}
	\label{sec:IdealTest}

	The statistic $E_{\text{HL},n,\text{sym}}$ is an e-variable solely under the requirement that for $i = 1, \dots, n$ and all permutations $\sigma$, the probabilities $q_{\sigma(i)}$ in $E_{\text{HL},n}^{\sigma}$ are a measurable function of $(P_{\sigma(j)}, Y_{\sigma(j)})$, $j = 1, \dots, i - 1$, and of $P_{\sigma(i)}$. In the following, we write
	\[
	q_{\sigma, \sigma(i)} = f_i(P_{\sigma(1)}, \dots, P_{\sigma(i)}, Y_{\sigma(1)}, \dots, Y_{\sigma(i-1)}),
	\]
	using the same algorithm $f_i$ for constructing $q_{\sigma, \sigma(i)}$ based on $P_{\sigma(1)}, \dots, P_{\sigma(i)}, Y_{\sigma(1)}, \dots, Y_{\sigma(i-1)}$ for all permutations $\sigma$. The challenge is then how to choose the functions $f_1, \dots, f_n$ such that the test has power. As argued by \citet{GrunwaldHeideETAL2019,Shafer2021}, a suitable measure of power for e-values is the growth rate $\mathbb{E}_{\mathbb{Q}}[\log(E)]$ under an alternative distribution $\mathbb{Q}$, so that ideally, $E$ grows exponentially fast in the sample size if the null hypothesis is violated.

	Our algorithm for choosing $f_1, \dots, f_n$ is inspired by permutation online isotonic regression, studied extensively by \citet{Kotlowski2017}. In machine learning applications, isotonic regresison is an established method for the recalibration of binary classifiers; see e.g.\ \citet{ZadroznyElkan2002}  or \citet{Flach2012}. Recently, \citet{Dimitriadise2016191118} related the isotonic regression approach to reliability diagrams, which are a key diagnostic tool in evaluating probability forecast for binary events, especially in meteorology. Our results demonstrate that isotonic regression is also suitable for constructing universal tests of calibration.

	To introduce the algorithm for constructing our e-variable, let $p_1, \dots, p_i \in [0,1]$ be probability predictions and $y_1, \dots, y_i \in \{0,1\}$ be the corresponding outcomes. Then the isotonic regression of $y_1, \dots, y_i$ on $p_1, \dots, p_i$ can be described as the maximizer of
	\begin{equation} \label{eq:isoreg}
		\hat{R}_n(g_1, \dots, g_i) = \hat{R}(g_1, \dots, g_i; \, p_1, \dots, p_i, y_1, \dots, y_i) = \sum_{j=1}^i \log \, \left( \left(\frac{g_j}{p_j}\right)^{y_j}\left(\frac{1-g_j}{1-p_j}\right)^{1-y_j} \right),
	\end{equation}
	over all $g_1, \dots, g_i$ such that $g_k \leq g_l$ if $p_k \leq p_l$.
	Notice that the quantity in \eqref{eq:isoreg} is simply a normalized version of the logarithmic score, and the maximizer does not depend on the fact that we normalize by $p_j^{y_j}(1-p_j)^{1-y_j}$.
	Moreover notice that up to rescaling by $1/i$, this criterion also equals the sample version of $\mathbb{E}_{\mathbb{Q}}[\log(E)]$ when the e-variable $E$ is the likelihood ratio between the probabilities $g_j$ and $p_j$. A unique maximizer exists --- unique since we exclude the cases $p_j = 0$ and $y_j = 1$ or $p_j = 1$ and $y_j = 0$ for some $j$ --- and can be computed efficiently with the PAV-Algorithm \citep{Ayer1955}. This estimator only defines a recalibrated version of $p_1, \dots, p_i$, and a method is required to define the regression at a $p_{i+1} \in [0,1]$ not contained in the sample. To obtain out-of-sample predictions with small regret in terms of log-loss, we rely on a strategy originally proposed by \citet{Vovk2015} and applied by \citet{Kotlowski2017} to derive regret bounds for isotonic regression in an online setting. The out-of-sample value at $p_{j+1}$ is defined as follows,

	\begin{equation}  \label{eq:isoreg_oos}
		f_{i+1}(p_1, \dots, p_i, p_{i+1}, y_1, \dots, y_i) = \frac{g_{i+1,1}}{g_{i+1,1}+1-g_{i+1,0}},
	\end{equation}
	where $g_{i+1,1}$ and $g_{i+1,0}$ are the $(i+1)$-th component the isotonic regression of $p_i, \dots, p_i, p_{i+1}$ with observations $y_1, \dots, y_i, 1$ or $y_1, \dots, y_i, 0$, respectively.
	That is, to define the isotonic regression at the unseen $p_{i+1}$, we fit two isotonic regression in which we include $p_{i+1}$ in the sample with artificial observations of $1$ and of $0$ respectively, and take the ratio \eqref{eq:isoreg_oos} as recalibrated probability. The definition \eqref{eq:isoreg_oos} is extended to the case $i = 0$ by setting $g_{1,1} = g_{1,0} = 0.5$. The workflow to construct $E_{\text{HL},n,\text{sym}}$ is then described in Algorithm \ref{alg:esym}.

	\begin{algorithm}
		\caption{Construction of $E_{\text{HL},n,\text{sym}}$}\label{alg:esym}
		\begin{algorithmic}[1]
			\State $E_{\text{HL},n,\text{sym}} \gets 0$
			\For{all permutations $\sigma$ of $\{1, \dots, n\}$}
			\State $E_{\text{HL},n}^{\sigma} \gets (0.5/P_{\sigma(1)})^{Y_{\sigma(1)}}(0.5/(1-P_{\sigma(1)}))^{1-Y_{\sigma(1)}}$
			\For{$i=1,\dots,n-1$}
			\State $q_{\sigma, \sigma(i+1)} \gets f_{i}(P_{\sigma(1)}, \dots, P_{\sigma(i+1)}, Y_{\sigma(1)}, \dots, Y_{\sigma(i)})$ as defined in \eqref{eq:isoreg_oos}
			\State $E_{\text{HL},n}^{\sigma} \gets E_{\text{HL},n}^{\sigma} \cdot (q_{\sigma, \sigma, \sigma(i+1)}/P_{\sigma(i+1)})^{Y_{\sigma(i+1)}}((1-q_{\sigma, \sigma(i+1)})/(1-P_{\sigma(i+1)}))^{1-Y_{\sigma(i+1)}}$
			\EndFor
			\State $E_{\text{HL},n,\text{sym}} \gets E_{\text{HL},n,\text{sym}} + E_{\text{HL},n}^{\sigma}/n!$
			\EndFor
			\State \textbf{return} $E_{\text{HL},n,\text{sym}}$
		\end{algorithmic}
	\end{algorithm}

	To state a result about the power of $E_{\text{HL},n,\text{sym}}$, we need a population version of the isotonic regression estimator. For a function $\pi \colon [0,1] \rightarrow [0,1]$, let
	\[
	R_{\mathbb{Q}}(\pi) = \mathbb{E}_{\mathbb{Q}}\left[\log \left( (\pi(P)/P)^Y((1-\pi(P))/(1-P))^{1-Y}\right)\right]
	\]
	if this expectation exists. Let $\mathcal{F}_{\uparrow, [0,1]}$ be the set of nondecreasing functions $\pi \colon [0,1] \rightarrow [0,1]$. If $\mathbb{Q}$ is the empirical distribution of $(P_1, Y_1), \dots, (P_n, Y_n)$, then it is easy to see that $R_{\mathbb{Q}}$ coincides with the target function $\hat{R}$ of the usual isotonic regression in finite samples. With these definitions, we can state the following result about the power of $E_{\text{HL},n,\text{sym}}$.

	\begin{thm} \label{thm:power}
		Let $(P_1, Y_1), \dots, (P_n, Y_n), (P, Y)$ be iid with distribution $\mathbb{Q}$ such that
		\begin{equation} \label{eq:integrability}
			\mathbb{E}_{\mathbb{Q}}[\log(P)^2 + \log(1-P)^2] < \infty.
		\end{equation}
		Then,
		\begin{itemize}
			\item[(i)] there exists a $\mathbb{Q}$-almost-surely unique maximizer $\pi^* \in \mathcal{F}_{\uparrow, [0,1]}$ of $R_{\mathbb{Q}}$;
			\item[(ii)] for a version of $\pi^*$ from part (i), let
			\[
			D(\mathbb{Q}) = R_{\mathbb{Q}}(\pi^*) = \mathbb{E}_{\mathbb{Q}}[\log \, (\pi^*(P)/P)^Y((1-\pi^*(P))/(1-P))^{1-Y}];
			\]
			then $D(\mathbb{Q}) \geq 0$, with equality if and only if $\mathbb{Q} \in \mathcal{H}_{\text{HL},1}$;
			\item[(iii)] the e-value $E_{\text{HL},n,\text{sym}}$ constructed with Algorithm \ref{alg:esym} satisfies
			\[
			E_{\text{HL},n,\text{sym}} \geq \exp\left(\sum_{i=1}^{n}\log\, \left(\frac{\pi^*(P_i)}{P_i}\right)^{Y_i} \left(\frac{1-\pi^*(P_i)}{1-P_i}\right)^{1-Y_i} -C\sqrt{n\log(n)^2} \right)
			\]
			for an universal constant $C > 0$, and hence
			\[
			\mathbb{E}_{\mathbb{Q}}[\log(E_{\text{HL},n,\text{sym}})] \geq nD(\mathbb{Q}) - C\sqrt{n\log(n)^2}.
			\]
		\end{itemize}
	\end{thm}

	The integrability assumption \eqref{eq:integrability} is solely required to prove parts (i) and (ii) of the theorem, and the lower bound on $E_{\text{HL},n,\text{sym}}$ and the expectation of its logarithm in fact hold for any $\pi \in \mathcal{F}_{\uparrow, [0,1]}$. However, part (iii) only becomes useful in conjunction with (i) and (ii): the fact that $D(\mathbb{Q}) \geq 0$ with equality if and only if $\mathbb{Q} \in \mathcal{H}_{\text{HL},1}$ implies that the test has positive growth rate for all alternative distributions $\mathbb{Q}$ if $n$ is large enough. This is a surprising result, since it might seem that restricting our estimator of $\mathbb{E}_{\mathbb{Q}}[Y|P]$ to isotonic functions in $P$ implies some restriction on the class of alternatives against which the test has power --- which is not the case.

	Part (iii) of Theorem \ref{thm:power} only gives a diverging lower bound on the expected value of $\log(E_{\text{HL},n,\text{sym}})$. However, under assumption \eqref{eq:integrability} the average growth rate
	\[
	\frac{1}{n}\sum_{i=1}^{n}\log\, \left(\frac{\pi^*(P_i)}{P_i}\right)^{Y_i} \left(\frac{1-\pi^*(P_i)}{1-P_i}\right)^{1-Y_i}
	\]
	satisfies the strong law of large numbers, and since $D(\mathbb{Q}) > 0$ for $\mathbb{Q} \not\in \mathcal{H}_{\text{HL},1}$, this implies that $E_{\text{HL},n,\text{sym}} \rightarrow \infty$ almost surely as $n \rightarrow \infty$. In particular, for any Type-I error $\alpha$ and desired power $1-\beta$, there exists a sample size $N$ such that $\mathbb{Q}(E_{\text{HL},N,\text{sym}} \geq 1/\alpha) \geq 1-\beta$ for $N \geq n$.

	\begin{proof}[Proof of Theorem \ref{thm:power}]
		Part (i) is a consequence of the following facts. If $(\pi_n)_{n\in\mathbb{N}}$ is a sequence in $\mathcal{F}_{\uparrow, [0,1]}$ such that $\lim_{n\rightarrow\infty} R_{\mathbb{Q}}(\pi_n) = \sup_{\pi \in \mathcal{F}_{\uparrow, [0,1]}}R(\pi)$, then by Helly's selection theorem, there exists a subsequence $(\pi_{n_k})_{k\in\mathbb{N}}$ converging pointwise to some $\pi^* \in \mathcal{F}_{\uparrow, [0,1]}$. The function $\pi(P) \mapsto \log \left( (\pi(P)/P)^Y((1-\pi(P))/(1-P))^{1-Y}\right)$ inside the expectation in the definition of $R_{\mathbb{Q}}$ is strictly concave, and the set $\mathcal{F}_{\uparrow, [0,1]}$ is convex. Hence $R_{\mathbb{Q}}(\pi^*) \geq R_{\mathbb{Q}}(\pi)$ for all $\pi \in \mathcal{F}_{\uparrow, [0,1]}$, and equality holds if and only if $\pi = \pi^*$ $\mathbb{Q}$-almost-surely, provided that $R_{\mathbb{Q}}(\pi^*)$ is finite, which is shown below.

		The nonnegatitivy in part (ii) holds because $\mathcal{F}_{\uparrow, [0,1]}$ contains the identity function, and we only have to prove that $D(\mathbb{Q}) > 0$ if $\mathbb{Q} \not\in \mathcal{H}_{\text{HL},1}$. For this, it is sufficient to show that there exists some $\pi$ with $R_{\mathbb{Q}}(\pi) > 0$. We start with some results about the existence of certain expected values.
		Since $Y|P$ is Bernoulli with expectation $\bar{\pi}(P)$, we have
		\begin{align*}
			& \mathbb{E}_{\mathbb{Q}}[\log \, (\pi^*(P)/P)^Y((1-\pi^*(P))/(1-P))^{1-Y} \mid P] \\
			& \quad = \bar{\pi}(P)\log(\pi^*(P)/P) + (1-\bar{\pi}(P))\log((1-\pi^*(P))/(1-P))
		\end{align*}
		and the nonnegativity of the Kullback-Leibler divergence implies
		\begin{align*}
			& \bar{\pi}(P)\log(\pi^*(P)/P) + (1-\bar{\pi}(P))\log((1-\pi^*(P))/(1-P)) \\
			& \quad \leq \bar{\pi}(P)\log(\bar{\pi}(P)/P) + (1-\bar{\pi}(P))\log((1-\bar{\pi}(P))/(1-P)),
		\end{align*}
		hence we obtain
		\begin{align}
			0 & \leq \mathbb{E}_{\mathbb{Q}}[\log \, (\pi^*(P)/P)^Y((1-\pi^*(P))/(1-P))^{1-Y}] \nonumber\\
			& \leq \mathbb{E}_{\mathbb{Q}}[\bar{\pi}(P)\log(\bar{\pi}(P)/P) + (1-\bar{\pi}(P))\log((1-\bar{\pi}(P))/(1-P))] \nonumber \\
			& \leq \mathbb{E}_{\mathbb{Q}}[|\log(P)| + |\log(1-P)|] \nonumber \\
			& \leq \sqrt{\mathbb{E}_{\mathbb{Q}}[\log(P)^2]} + \sqrt{\mathbb{E}_{\mathbb{Q}}[\log(1-P)^2]} < \infty. \label{eq:finite_integral}
		\end{align}
		Let now $\tilde{\pi}(P)$ be a version of the conditional expectation of $\bar{\pi}(P)$ with respect to the sigma lattice generated by $P$, which $\tilde{\pi}(P)$ satisfies the following properties:
		\begin{align}
			& \tilde{\pi} \text{ is increasing }; \label{eq:expectation_lattice_incr} \\
			& \mathbb{E}_{\mathbb{Q}}[(\bar{\pi}(P) - \tilde{\pi}(P))h(P)] \leq 0 \text{ for all increasing } h \text{ such that } \mathbb{E}_{\mathbb{Q}}[h(P)^2] < \infty; \label{eq:expectation_lattice_ineq} \\
			& \mathbb{E}_{\mathbb{Q}}[\bar{\pi}(P)\mathbbm{1}_B(P)] = \mathbb{E}_{\mathbb{Q}}[\tilde{\pi}(P)\mathbbm{1}_B(P)] \text{ for all } B \text{ in the } \sigma\text{-field generated by } \tilde{\pi}. \label{eq:expectation_lattice_eq}
		\end{align}
		Equation \eqref{eq:expectation_lattice_incr} holds by definition of the conditional expectation given a sigma lattice, and \eqref{eq:expectation_lattice_ineq} and \eqref{eq:expectation_lattice_eq} are by \citet[Equations (3.9) and (3.11)]{Brunk1965}. By \eqref{eq:expectation_lattice_eq}, we have $\bar{\pi}(P) = P$ almost surely if and only if $\tilde{\pi}(P) = P$ almost surely. By definition of $\pi^*(P)$, we know that
		\[
		\mathbb{E}_{\mathbb{Q}}[\log \, (\tilde{\pi}(P)/P)^Y((1-\tilde{\pi}(P))/(1-P))^{1-Y}] \leq \mathbb{E}_{\mathbb{Q}}[\log \, (\pi^*(P)/P)^Y((1-\pi^*(P))/(1-P))^{1-Y}].
		\]
		The goal is now to prove
		\[
		\mathbb{E}_{\mathbb{Q}}\left[\log \left( (\tilde{\pi}(P)/P)^Y((1-\tilde{\pi}(P))/(1-P))^{1-Y}\right)\right] \geq 0,
		\]
		with equality if and only if $\tilde{\pi}(P) = P$ $\mathbb{Q}$-almost-surely. Notice that $\tilde{\pi}$ is only defined on the support of $P$, but one can assume without loss of generality that it is defined on the whole interval $[0,1]$ by right-continuous constant extrapolation in parts where it is not defined. Since $\tilde{\pi}$ is increasing, there exist at most countably many disjoint intervals $A_i \subseteq [0,1]$, $i \in \mathcal{I}$, on which $\tilde{\pi}$ is constant with some value $c_i \in [0,1]$. Furthermore, there are at most countably many disjoint intervals $B_j$, $j \in \mathcal{J}$, whose union equals $[0,1] \setminus \bigcup_{i \in \mathcal{I}} A_i$. We now make a few case distinctions.

		Fix $i$ with $c_i > 0$ and assume that $A_i = [a_i, b_i]$; the following arguments can be easily modified for the case that $A_i$ is (half-)open. Define the function
		\[
		h_i(x) = \begin{cases}
			\log(1/a_i) + 1, \ & \text{ if }x < a_i, \\
			\log(1/x), & \text{ if } x \in [a_i, b_i], \\
			-1, &\text{ if } x > b_i.
		\end{cases}
		\]
		Then $h_i(P)$ is square integrable due to \eqref{eq:finite_integral}, $h_i$ is decreasing, and constant outside of $[a_i, b_i]$, so that
		\begin{align*}
			0 \overset{\eqref{eq:expectation_lattice_ineq}}{\leq} \mathbb{E}_{\mathbb{Q}}[(\bar{\pi}(P) - \tilde{\pi}(P))h_i(P)]
			& \overset{\eqref{eq:expectation_lattice_eq}}{=} \mathbb{E}_{\mathbb{Q}}[(\bar{\pi}(P) - \tilde{\pi}(P))\log(1/P)\mathbbm{1}_{A_i}(P)] \\
			& \overset{\eqref{eq:expectation_lattice_eq}}{=} \mathbb{E}_{\mathbb{Q}}[(\bar{\pi}(P) - \tilde{\pi}(P))\log(c_i/P)\mathbbm{1}_{A_i}(P)] \\
			& \, \,= \mathbb{E}_{\mathbb{Q}}[(\bar{\pi}(P) - \tilde{\pi}(P))\log(\tilde{\pi}(P)/P)\mathbbm{1}_{A_i}(P)],
		\end{align*}
		where the last step is due to the fact that $\tilde{\pi}(P) = c_i$ for $P \in A_i$. Hence
		\[
		\mathbb{E}_{\mathbb{Q}}[\bar{\pi}(P)\log(\tilde{\pi}(P)/P)\mathbbm{1}_{A_i}(P)] \geq 	\mathbb{E}_{\mathbb{Q}}[\tilde{\pi}(P)\log(\tilde{\pi}(P)/P)\mathbbm{1}_{A_i}(P)].
		\]
		If $c_i = 0$, the above inequality is still true because \eqref{eq:expectation_lattice_eq} implies that $\tilde{\pi}(P) = \bar{\pi}(P) = 0$ for $P \in A_i$ in that case, and we define $0\log(0) := 0$.

		Similarly, for $c_i < 1$ we define
		\[
		h_i(x) = \begin{cases}
			\log(1/(1-b_i)) + 1, \ & \text{ if } x > b_i, \\
			\log(1/(1-x)), & \text{ if } x \in [a_i, b_i], \\
			-1, &\text{ if } x < a_i,
		\end{cases}
		\]
		which is square integrable and increasing. As before,
		\begin{align*}
			0 \overset{\eqref{eq:expectation_lattice_ineq}}{\leq} \mathbb{E}_{\mathbb{Q}}[(\tilde{\pi}(P) - \bar{\pi}(P))h_i(P)]
			& \overset{\eqref{eq:expectation_lattice_eq}}{=} \mathbb{E}_{\mathbb{Q}}[(\tilde{\pi}(P) - \bar{\pi}(P))\log(1/(1-P))\mathbbm{1}_{A_i}(P)] \\
			& \overset{\eqref{eq:expectation_lattice_eq}}{=} \mathbb{E}_{\mathbb{Q}}[(\tilde{\pi}(P) - \bar{\pi}(P))\log((1-c_i)/(1-P))\mathbbm{1}_{A_i}(P)] \\
			& \, \,= \mathbb{E}_{\mathbb{Q}}[(1-\bar{\pi}(P) - (1-\tilde{\pi}(P)))\log((1-\tilde{\pi}(P))/(1-P))\mathbbm{1}_{A_i}(P)],
		\end{align*}
		so we obtain
		\[
		\mathbb{E}_{\mathbb{Q}}[(1-\bar{\pi}(P))\log((1-\tilde{\pi}(P))/(1-P))\mathbbm{1}_{A_i}(P)] \geq 	\mathbb{E}_{\mathbb{Q}}[(1-\tilde{\pi}(P))\log((1-\tilde{\pi}(P))/(1-P))\mathbbm{1}_{A_i}(P)],
		\]
		which also holds if $\tilde{\pi}(P) = 1$ on $A_i$. Hence we have shown that
		\begin{align*}
			0 & \leq \mathbb{E}_{\mathbb{Q}}[\mathbbm{1}_{A_i}(P)(\tilde{\pi}(P)\log(\tilde{\pi}(P)/P) + (1-\tilde{\pi}(P))\log((1-\tilde{\pi}(P))/(1-P)))] \\
			& \leq \mathbb{E}_{\mathbb{Q}}[\mathbbm{1}_{A_i}(P)(\bar{\pi}(P)\log(\tilde{\pi}(P)/P) + (1-\bar{\pi}(P))\log((1-\tilde{\pi}(P))/(1-P)))] \\
			& = \mathbb{E}_{\mathbb{Q}}[\mathbbm{1}_{A_i}(P)\log \, (\tilde{\pi}(P)/P)^Y((1-\tilde{\pi}(P))/(1-P))^{1-Y}],
		\end{align*}
		and equality holds if and only if $\tilde{\pi}(P) = P$ $\mathbb{Q}$-almost-surely on $A_i$, since the Kullback-Leibler divergence is non-negative.

		Consider now an interval $B_j$. Since $\tilde{\pi}$ is strictly increasing on $B_j$, the sigma field generated by $\tilde{\pi}$ contains all Borel sets which are subsets of $B_j$. Then \eqref{eq:expectation_lattice_eq} implies that $\tilde{\pi}(P) = \bar{\pi}(P)$ $\mathbb{Q}$-almost-surely on $B_j$, hence
		\begin{align*}
			& \mathbb{E}_{\mathbb{Q}}\left[\mathbbm{1}_{B_j}(P)\left(\bar{\pi}(P)\log(\tilde{\pi}(P)/P) + (1-\bar{\pi}(P))\log((1-\tilde{\pi}(P))/(1-P))\right)\right] \\
			& \ =  \mathbb{E}_{\mathbb{Q}}\left[\mathbbm{1}_{B_j}(P)\left(\tilde{\pi}(P)\log(\tilde{\pi}(P)/P) + (1-\tilde{\pi}(P))\log((1-\tilde{\pi}(P))/(1-P))\right)\right]\geq 0
		\end{align*}
		with equality if and only if $\tilde{\pi}(P) = P$ $\mathbb{Q}$-almost-surely on $B_j$.

		With the above derivations, we obtain that for any finite number of indices $i_1, \dots, i_n \in \mathcal{I}$, $j_1, \dots, j_n \in \mathcal{J}$ and
		\[
		C_n = \left(\bigcup_{k=1}^n A_{i_k}\right) \cup \left(\bigcup_{l=1}^n B_{j_l}\right),
		\]
		the following inequalities hold,
		\begin{align}
			0 & \leq \mathbb{E}_{\mathbb{Q}}\left[\mathbbm{1}_{C_n}(P)\left(\tilde{\pi}(P)\log(\tilde{\pi}(P)/P) + (1-\tilde{\pi}(P))\log((1-\tilde{\pi}(P))/(1-P))\right)\right] \label{eq:int_lower_bound}\\
			& \leq \mathbb{E}_{\mathbb{Q}}\left[\mathbbm{1}_{C_n}(P)\left(\bar{\pi}(P)\log(\tilde{\pi}(P)/P) + (1-\bar{\pi}(P))\log((1-\tilde{\pi}(P))/(1-P))\right)\right] \label{eq:int_upper_bound}.
		\end{align}
		Since the integrand in \eqref{eq:int_lower_bound} is non-negative and the integrand in \eqref{eq:int_upper_bound} dominated pointwise by
		\[
		M(P) = \bar{\pi}(P)\log(\bar{\pi}(P)/P) + (1-\bar{\pi}(P))\log((1-\bar{\pi}(P))/(1-P))
		\]
		with $\mathbb{E}_{\mathbb{Q}}[M(P)] < \infty$, we can choose index sequences such that $\bigcup_{n=1}^N C_n$ increases to $[0,1]$, and apply Fatou's Lemma and the dominated convergence theorem to obtain
		\begin{align}
			0 & \leq \mathbb{E}_{\mathbb{Q}}\left[\tilde{\pi}(P)\log(\tilde{\pi}(P)/P) + (1-\tilde{\pi}(P))\log((1-\tilde{\pi}(P))/(1-P))\right] \label{eq:int_lower_bound2}\\
			& \leq \mathbb{E}_{\mathbb{Q}}\left[\bar{\pi}(P)\log(\tilde{\pi}(P)/P) + (1-\bar{\pi}(P))\log((1-\tilde{\pi}(P))/(1-P))\right] \nonumber.
		\end{align}
		Equality in \eqref{eq:int_lower_bound2} holds if and only if $\tilde{\pi}(P) = P$ almost surely.

		For part (iii), the inequality of arithmetic and geometric mean implies that
		\[
		E \geq \left(\prod_{\sigma \in \mathcal{S}_n}\prod_{i=1}^{n} \frac{q_{\sigma,\sigma(i)}^{Y_{\sigma(i)}}(1-q_{\sigma,\sigma(i)})^{1-Y_{\sigma(i)}}}{P_{\sigma(i)}^{Y_{\sigma(i)}}(1-P_{\sigma(i)})^{1-Y_{\sigma(i)}}}\right)^{1/n!} = \exp\left(\frac{1}{n!}\sum_{\sigma \in \mathcal{S}_n} \sum_{i=1}^n \log\frac{q_{\sigma,\sigma(i)}^{Y_{\sigma(i)}}(1-q_{\sigma,\sigma(i)})^{1-Y_{\sigma(i)}}}{P_{\sigma(i)}^{Y_{\sigma(i)}}(1-P_{\sigma(i)})^{1-Y_{\sigma(i)}}}\right).
		\]
		The term inside the exponential can be written as
		\[
		L = \mathbb{E}_{\sigma}\left[\sum_{i=1}^n \log\frac{q_{\sigma,\sigma(i)}^{Y_{\sigma(i)}}(1-q_{\sigma,\sigma(i)})^{1-Y_{\sigma(i)}}}{P_{\sigma(i)}^{Y_{\sigma(i)}}(1-P_{\sigma(i)})^{1-Y_{\sigma(i)}}}\right],
		\]
		which is the negative of the entropic loss defined in Section 4.4 of \citet{Kotlowski2017}, and the expectation $\mathbb{E}_{\sigma}[\cdot]$ is with respect to the uniform distribution over all permutations $\sigma$ of $\{1,\dots,n\}$. It follows from Lemma 2.1, Theorem 4.3 and the proof of Theorem 4.1 of \citet{Kotlowski2017} that for all $K \in \mathbb{N}$,
		\[
		L - \sum_{i=1}^n \log\frac{\hat{\pi}_i^{Y_{i}}(1-\hat{\pi}_i)^{1-Y_{i}}}{P_{i}^{Y_{i}}(1-P_{\sigma(i)})^{1-Y_{i}}} \geq -\sum_{k=1}^n \left(\frac{2}{K}+\frac{4K}{k}\log(1+k)\right),
		\]
		where $\hat{\pi}_1, \dots, \hat{\pi}_n$ is the isotonic regression of $Y_1, \dots, Y_n$ on $P_1, \dots, P_n$, i.e.~the maximizer of
		\[
		(g_1, \dots, g_n) \mapsto \hat{R}(g_1, \dots, g_n; \, P_1, \dots, P_n, Y_1, \dots, Y_n).
		\]
		as defined at \eqref{eq:isoreg}. The result now follows because
		\[
		\hat{R}(\hat{\pi}_1, \dots, \hat{\pi}_n; \, P_1, \dots, P_n, Y_1, \dots, Y_n) \geq \hat{R}(\pi^*(P_1), \dots, \pi^*(g_n); \, P_1, \dots, P_n, Y_1, \dots, Y_n)
		\]
		and $\sum_{k=1}^n (2/K + 4K\log(1+k)/k) = \mathcal{O}(\sqrt{n(\log(n))^2})$ for $K$ of order $\sqrt{n/(\log(n))^2}$.
	\end{proof}

	\subsection{A feasible version of the test} \label{sec:feasible}
	The ideal test described in Algorithm \ref{alg:esym} cannot be implemented for practically relevant $n$, as it requires to compute e-values over all $n!$ permutations of $\{1, \dots, n\}$. Even for a single permutation $\sigma$, the inner loop in Algorithm \ref{alg:esym} has computational complexity of $\mathcal{O}(n^2)$: it requires computing $2n$ isotonic regressions to generate out-of-sample predictions. We suggest to address these problems above by the simplified version in Algorithm \ref{alg:feasible}, which can be regarded as a version of the split likelihood ratio test by \citet{Wasserman2020}.

	\begin{algorithm}[tbh]
		\caption{Split LRT version of the e-value}\label{alg:feasible}
		\begin{algorithmic}[1]
			\State \textbf{Parameters:} split fraction $s \in (0,1)$, number of splits $B \in \mathbb{N}$.
			\State $E_{\text{HL},n} \gets 0$
			\For{$b = 1, \dots, B$}
			\State randomly select $\lfloor ns \rfloor$ pairs $(Y_{i}, P_{i})$, $j \in S_b = \{i_1, \dots, i_{\lfloor ns \rfloor}\}$, without replacement
			\State estimate the isotonic of regression of $(Y_{i}, P_{i})$, $i \in S_b$, by maximizing \eqref{eq:isoreg}
			\State generate predictions $q_i$ for $\mathbb{E}[Y|P_i]$, $i \in \{1, \dots, n\} \setminus S_b$, from the isotonic regression
			\State $E_{\text{HL},n} \leftarrow E_{\text{HL},n} + \prod_{i \in \{1, \dots, n\} \setminus S_b} (q_i/P_i)^{Y_i}((1-q_i)/(1-P_i))^{1-Y_i} /B$
			\EndFor
			\State \textbf{return} $E_{\text{HL},n}$
		\end{algorithmic}
	\end{algorithm}

	A delicate point in Algorithm \ref{alg:feasible} is Step 6, where one needs to generate out-of-sample predictions from the isotonic regression fit. Naive extrapolation approaches could lead to predicted probabilities $q_i \in \{0,1\}$ and hence an e-value of zero if either $q_i = 0$ and $Y_i = 1$ or $q_i = 1$ and $Y_i = 0$.

	Let $p_1 < \dots < p_m$ denote the distinct values of $P_i$, $i \in S_b$, and $\hat{\pi}_1 \leq \dots \leq \hat{\pi}_m$ the corresponding values of the isotonic regression. A well known result about isotonic regression states that there exists a partition of $S_b$ into index sets $\mathcal{I}_1, \dots, \mathcal{I}_d$ such that $\hat{\pi}_j$ is the empirical mean of the $Y_i$ with indices in $\mathcal{I}_j$,
	\[
	\hat{\pi}_j = \frac{1}{\#\mathcal{I}_j}\sum_{i \in \mathcal{I}_j} Y_i.
	\]
	To remedy the problem of predictions in $\{0,1\}$, we propose to apply the smoothed Laplace predictor, equivalent to Jeffreys' prior in binomial proportion estimation,
	\[
	\check{\pi}_j = \frac{1}{\# \mathcal{I}_j + 1} \left(0.5 + \sum_{i \in \mathcal{I}_j}Y_i \right) \in (0,1).
	\]
	For out-of-sample predictions at $P_i \not\in \{p_1, \dots, p_m\}$, one can then apply linear interpolation
	\[
	q_i = \begin{cases} \displaystyle
		\frac{p_l - P_i}{p_l - p_k} \check{\pi}_k + \frac{P_i - p_k}{p_l - p_k} \check{\pi}_l, & \ \text{ if } P_i \in [p_k, p_l], \\[0.2em]
		\check{\pi}_1,  & \ \text{ if } P_i < p_1, \\[0.2em]
		\check{\pi}_m,  & \ \text{ if } P_i > p_m., \\
	\end{cases}
	\]
	where it is now guaranteed that $q_i \in (0,1)$.

	\section{Simulations}
	\label{sec:simulation}

	This section evaluates the empirical performance of the feasible version of the proposed test in Section \ref{sec:feasible} together with sensible values of the splitting fraction $s \in (0,1)$.
	We follow the simulation setup of \citet{Hosmeretal1997} with a quadratic misspecification in assessing HL-type tests, which is, if at all, just slightly modified in more recent contributions \citep{HosmerHjort2002, XieEtAl2008, Allison2014, CanaryEtAl2017, Nattino2020}.
	Replication material for the simulations and the application in Section \ref{sec:application} in the statistical software \texttt{R} is available under \href{https://github.com/marius-cp/eHL}{https://github.com/marius-cp/eHL}.

	For $i=1,\dots,2n$ with $n \in \{1024, 2048, 4096, 8192\}$, we simulate the iid covariate $X_i \stackrel{\text{iid}}{\sim} \text{U}(-3,3)$ and let the response variables $Y_i \sim \text{Bernoulli}(\pi_i)$ be independent, where the true conditional event probability $\bar{\pi}_i$ follows a logistic transformation of a quadratic model
	\begin{align}
		\label{Eq:true}
		\bar{\pi}_i =  \bar{\pi} (X_i) = \mathbb{P} (Y_i=1 \mid X_i; \, \beta_0, \beta_1, \beta_2) = \dfrac{\exp( \beta_0 + \beta_1 X_i + \beta_2 X_i^2)}{1+\exp( \beta_0 + \beta_1 X_i + \beta_2 X_i^2)}.
	\end{align}

	We split the simulated data into an estimation set and validation set, both of size $n$.
	Based on the data in the estimation set, we estimate the parameters of a linear, and hence misspecified, logistic regression model by maximum likelihood and denote the parameter estimates by $\big(\widehat{\beta}_0, \widehat{\beta}_1\big)$.
	The probability of a positive outcome is then predicted by
	\begin{align}
		\label{Eq:pred}
		P_i = \dfrac{\exp \big( \widehat{\beta}_0 + \widehat{\beta}_1 X_i \big)}  {1+\exp( \widehat{\beta}_0 + \widehat{\beta}_1 X_i )}.
	\end{align}

	We vary the severity of the misspecification, expressed through the magnitude of $\beta_2$.
	Following \citet{Hosmeretal1997}, we characterize the ``lack of linearity'' through the conditions $\bar{\pi}(-3) = j - 0.00733745$, $\bar{\pi}(-1.5) = 0.05$ and $\bar{\pi}(3)= 0.95$ such that the value $j = 0$ results in the very accurate approximation $\beta_2 \approx 0$, i.e., a linear effect of $X_i$ on the log odds-ratio.
	We consider a sequence of 51 equally spaced values of $j$ in the interval $[0, 0.1]$.
	Notice that for each choice of $j$, the values of $\beta_0$ and $\beta_1$ are also determined by these conditions.

	\begin{table}[tb]
		\caption{Rejection rates in percentage points of the classical HL test and our eHL test under the null hypothesis with $j=0$ and the true regression parameters $(\beta_0, \beta_1)$ in \eqref{Eq:pred} at a significance level of $5\%$.
			We treat an e-value above 20 as a rejection in the eHL test.}

	\label{tab:HLsize}
	\centering
	\begin{tabular}[t]{c c r c rrr}
	\toprule
	&& HL && \multicolumn{3}{c}{eHL} \\
	\cmidrule(l{3pt}r{3pt}){3-3}
	\cmidrule(l{3pt}r{3pt}){5-7}

	$s$ && && $1/3$ & $1/2$ & $2/3$ \\
	\midrule
	$n=1024$ && 6.2 && 0.5 & 1.0 & 0.4  \\
	$n=2048$ && 5.0 &&  0.1 & 0.4 & 0.6  \\
	$n=4096$ && 4.7 && 0.2 & 0.4 & 0.6  \\
	$n=8192$ && 4.5 && 0.0 & 0.1 & 0.5 \\
	\bottomrule
\end{tabular}
\end{table}

Table \ref{tab:HLsize} reports rejection rates of the tests over 1000 simulation replications, where we set $\beta_2 = 0$ (i.e., $j=0$), and use the true regression parameters $(\beta_0, \beta_1)$ in \eqref{Eq:pred} to guarantee that the null hypothesis $\mathcal{H}_{\text{HL},n}$ holds.
For the classical HL test, we use ten equally populated (quantile-spaced) bins, where the exact procedure follows the method \textsf{Q}{$^R$} described in Appendix \ref{sec:HLInstability}.
For the feasible eHL test of Section \ref{sec:feasible}, we use the splitting fractions $s \in \{1/3, 1/2, 2/3\}$.
To limit computation time, we choose a relatively low amount of bootstrap replications $B=10$ in the eHL test as we are mainly interested in rejection rates averaged across simulation replications, and hence, stability of the test is less of a concern as e.g., in the subsequent empirical application.
Here and in the following, we treat e-values above 20 as a test rejection at the $5\%$ significance level.
The table shows that all tests are well sized, where all eHL versions exhibit rejection frequencies much below the nominal value of $5\%$, which is not unusual for tests based on e-values.

\begin{figure}[tb]
\includegraphics[width=\linewidth]{Fig_Simulation.pdf}
\caption{Rejection (left) and e-value growth (right) rates for the classical HL (cHL) test, the feasible eHL test and an oracle eHL test for a range of splitting factors $s$.
	The oracle eHL test is based on the true $\pi_i$.
	The $x$-axis contains the severity of model mispecification, and the vertically aligned plots correspond to different sample sizes.}
\label{fig:Rejection_Rates}
\end{figure}

Figure \ref{fig:Rejection_Rates} analyzes the tests' behaviour under the alternative hypotheses induced by $j>0$.
Notice that we use the true parameters $(\beta_0, \beta_1)$ in \eqref{Eq:pred} for $j=0$ but estimates $\big(\widehat{\beta}_0, \widehat{\beta}_1\big)$ for any $j>0$ as the pseudo-true parameters are unknown under model mispecification. In this analysis, we further include an oracle version of the eHL test, whose e-values are optimal in the sense that they are based on $q_i = \bar{\pi}_i$, i.e., the practically unknown true conditional event probabilities.
The oracle eHL version with $s=1/2$ facilitates a fair comparison with the feasible HL test based on the same splitting factor.
The left panel of the figure shows classical test rejection rates for a nominal significance level of $5\%$.
Following the explanations in Section \ref{sec:IdealTest} together with \citet{GrunwaldHeideETAL2019} and \citet{Shafer2021}, a suitable measure of power for e-values is the growth rate $\mathbb{E} \big[ \log(E_{\text{HL},n}) \big]$, which is shown in the right panel Figure \ref{fig:Rejection_Rates}, where we approximate the expectation by the average log e-value over the simulation replications.
We restrict attention to $n \in \{1024, 4096\}$ as the other sample sizes do not yield further insights.

We find that all tests develop power for increasing mispecification $j$.
E.g., the feasible eHL versions already have substantial power for both sample sizes against an alternative with $j \approx 0.043$, which \citet{Hosmeretal1997} interpret as a value inducing only `slight' mispecification. (Notice that $j \approx 0.043$ equals 0.05 in their parametrization.)
There seems to be little difference among the feasible eHL tests when using different splitting fractions $s$, and hence, we do not find arguments to deviate from the natural choice of $s=1/2$, which we continue to use in the application.

The higher power of the classical HL test can be explained by the required sample split in the eHL test, and the estimation error in assigning suitable values for $q_i$.
The two oracle eHL tests make these steps redundant and hence achieve comparable power to the classical HL test.
Perhaps surprisingly, the difference between the two oracle eHL tests with different $s$ is smaller than the respective difference to the feasible test versions based on estimated $q_i$'s, which means that tuning the test to a specific alternative through the $q_i$'s is the main empirical challenge of the eHL test.

Notice that the often overlooked bin specification in the classical HL test implicitly determines the set of alternatives the test has power against as e.g.\ illustrated in \citet[Section 5]{Dimitriadis2022Honest}. As the sample split in the eHL test allows for \emph{estimating} a suitable alternative, Theorem \ref{thm:power} shows that the (ideal version of the) eHL test has power against all alternatives.

Turning to the growth rates of the feasible eHL tests, we find that larger choices of $s$ perform better for slight model mispecifications (small $j$) while the opposite is true for large mispecifications.
This can be explained since as discussed around \eqref{eq:alternative_q<P}--\eqref{eq:alternative_q>P}, $\bar{\pi}_i$ must be on the `correct' side of $P_i$ to gain power, which might be violated for small $s$ (and $n$) under slight mispecifications.

\section{Application: Credit Card Defaults in Taiwan}
\label{sec:application}

In this application, we analyze (re-)calibration of probability predictions for the binary event of credit card defaults in Taiwan in the year 2005.
In that time period, banks in Taiwan over-issued credit cards, also to unqualified applicants, who at the same time overused the cards for consumption, resulting in severe credit card debts  and damaged consumer finance confidence \citep{YehLien2009, LoHarvey2011}.
This crisis calls for improved and in particular calibrated probability predictions for credit card defaults that can be used for a thorough risk management and improved financial regulations.

For our analysis, we use a data set of $m = 30\,000$ credit card holders from Taiwan in 2005, that is publicly available from the UCI Machine Learning Repository \citep{UCIMLRep, YehLien2009} under \href{https://archive.ics.uci.edu/ml/datasets/default+of+credit+card+clients}{https://archive.ics.uci.edu/ml/datasets/default+of+credit+card+clients}.
Specifically, the binary response variable $Y_i \in \{0,1\}$ contains information on whether a default payment, $Y_i = 1$, occurred for customer $i=1,\dots,m$.
We observe a relatively high rate of $22.12\%$ of default payments in the data set that reflects the above mentioned credit card crisis.
The data set further includes 23 explanatory variables, containing information on the amount of given credit, gender, education, marital status, age, and various historical payment records for the past six months.

\begin{table}[tb]
\centering
\caption{E-values of the eHL and the range of p-values of the classical HL test, the latter stemming from 80 reasonable binning procedures as detailed in Table \ref{tab:HLinstabilities} and Appendix \ref{sec:HLInstability}.}
\label{tab:application_results}
\begin{tabular}[t]{l l l}
	\toprule
	Prediction method & eHL e-values  & Range of HL p-values \\
	\midrule
	Logistic model & \hspace{0.5em}$7.0 \cdot 10^{28}$ & [0.00, 0.00] \\
	Logistic model with increased estimation set & \hspace{0.5em}$9.6 \cdot 10^{22}$ & [0.00, 0.00] \\
	Isotonic recalibration &  20.04 & [0.00, 0.91] \\
	Bagged isotonic recalibration &   \hspace{0.5em}6.14 & [0.00, 0.53] \\
	\bottomrule
\end{tabular}
\end{table}

We randomly split the data into an estimation and a Recalibration set $\mathcal{R}$ with $M=12000$ observations each, and a Validation set $\mathcal{V}$ containing the remaining $n=6000$ observations.
We use the estimation set to fit a standard logistic regression model based on all predictor variables by maximum likelihood and compute the model predictions on the recalibration and validation sets, respectively.
We run all the following tests on the validation set.

Table \ref{tab:application_results} reports the e-values of the feasible version of our calibration test described in Section \ref{sec:feasible} based on $B=10000$ bootstrap replication and with a splitting factor of $s=1/2$ that is motivated by our simulation results.
We further report the range between the smallest and largest p-value of the classical HL test, where the different p-values result from five different, but natural binning procedures using $g=5,\dots,20$ bins, respectively.
We provide further details on these implementation choices in Appendix \ref{sec:HLInstability}.

The predictions from the logistic model result in an e-value far beyond the value of 20 in Table \ref{tab:application_results}, hence implying that these predictions are clearly miscalibrated.
In this setting, all implementation choices of the classical HL test agree and deliver p-values very close to zero.
The second row of the table shows that even when using all observations in the ``increased estimation set'' comprising the estimation and the recalibration set, the situation barely changes and both the eHL and HL tests agree (under all implementation choices).

As a consequence of these clear rejections, we now aim at isotonically recalibrating the probability predictions, a technique that proved valuable in other disciplines \citep{Guo2017, Vannitsem2018}, where it is also called ``post-processing''.
For this, we estimate an isotonic regression on the recalibration set $\mathcal{R}$ and generate recalibrated predictions by transforming the logistic predictions on $\mathcal{V}$ with the estimated isotonic regression function.
Table \ref{tab:application_results} shows that our calibration test has an e-value just above 20, i.e., a weak rejection when interpreted as a (conservative) test at the $5\%$ level.

\begin{figure}[tb]
\centering
\includegraphics[width=\linewidth]{Fig_calibrationcurve.pdf}
\caption{Bagged isotonic recalibration curve of the logit predictions.
	The blue curve shows the mean, and the red band the range of the pointwise $1\%$ and $99\%$ quantiles, over all bagging iterations.}
\label{fig:recalibration_curve}
\end{figure}

As a nonparametric method, the isotonic regression is known to involve substantial estimation noise that might adversely affect the recalibrated predictions.
Hence, we stabilize the estimation through the classical bagging (bootstrap aggregation) method of \citet{Breiman1996bagging}.
In detail, we draw $\widetilde{B} = 100$ bootstrap samples $\mathcal{R}_b, b=1,\dots,\widetilde{B}$  of size $M$ from the recalibration set $\mathcal{R}$ and estimate the isotonic regression on each  bootstrap sample $\mathcal{R}_b$.
The final predictions are obtained by recalibrating with the average of the estimated isotonic regression functions, displayed in Figure \ref{fig:recalibration_curve}.

The last row of Table \ref{tab:application_results} shows an e-value of approximately 6 implying only very weak evidence against the null hypothesis of calibration, once again illustrating the practical strength of both, bagging and recalibration methods.
The estimated re-calibration function displayed in Figure \ref{fig:recalibration_curve} reinforces the importance of recalibrating the logistic model predictions by showing that it substantially deviates from the diagonal.

\begin{figure}[tb]
\centering
\includegraphics[width=\linewidth]{Fig_HLinstabilty.pdf}
\caption{Histograms of $p$-values of the classical out-of-sample HL test based the five binning procedures given in Appendix \ref{sec:HLInstability} based on 5-20 bins, respectively, resulting in a total of 80 test results.}
\label{fig:HLinstabilities}
\end{figure}

For these two recalibration methods, the various natural implementation choices of the HL test, further described in Appendix \ref{sec:HLInstability}, result in p-values ranging between essentially 0 and 0.91 (and 0.53 respectively).
The corresponding p-value histograms in Figure \ref{fig:HLinstabilities} (and the detailed results in Table \ref{tab:HLinstabilities}) show the continuum of p-values, where the null hypothesis is rejected in approximately half of the cases at the $5\%$ level, implying that a researcher can essentially tailor the test decision to her will.
As already noted by \citet{Hosmeretal1997, Bertolini2000, Kuss2002}, this is a disconcerting state of affairs for a commonly used testing procedure and calls for more robust alternatives, such as the eHL test proposed in this paper.
Appendix \ref{sec:HLInstability} further shows that the feasible eHL test version is affected less by such instabilities arising from the repeated sample splits, at least if $B$ is chosen sufficiently large as in this application.

\section{Discussion \label{sec:discussion}}

This article proposes an e-test for perfect calibration, which is a safe testing counterpart to the widely used Hosmer-Lemeshow test. The proposed eHL test follows a simple betting interpretation (see \cite{Shafer2021}) where the e-value can be seen as the factor by which we multiply the bet against the hypothesis of perfect calibration. Intuitively, when accumulating money by the bet, we gain evidence against the null. Here, the e-value depends on the probability prediction, its corresponding realization, and an arbitrary value, which we suggest estimating in a two-step approach by isotonic regression. Furthermore, we assess the empirical performance of the test to detect quadratic model misspecifications. The simulations show that in samples of more than 2000 observations, the eHL test allows to reliably detect levels of quadratic misspecification, which \citet[p.~973]{Hosmeretal1997} denote to be slight. The intrinsic flexibility of the e-values allows the application of stable data-driven methods (here isotonic regression) instead of the typical binning and counting technique in the HL test. However, this flexibility comes at the cost of lower power in small samples of less than 2000 observations.

The feasible version of our test is based on random splits of the training data. Since the null hypothesis $\mathcal{H}_{\text{HL},n}$ requires calibration conditional on the predictions $P_1, \dots, P_n$, one also obtains a valid tests if the splits are performed systematically based on $P_1, \dots, P_n$, for example, by choosing two subsets with similar distribution of the $P_i$ in order for the isotonic regression estimator to extrapolate well. Systematic sampling approaches to increase the power have already been applied by \citet{Duan2022} for testing treatment effects.

Our article focuses on the batch setting where a fixed sample of size $n$ is available, rather than the online setting in which $(P_i, Y_i)$, $i \in \mathbb{N}$, arrive sequentially. However, the fact that powerful tests based on isotonic regression can be constructed in the batch setting suggests that similar approaches may be fruitful for online testing. \citet[Section 7.1]{Kotlowski2016} describe algorithms with sublinear regret for online isotonic regression (without the random permutation setting). We believe that in conjunction with parts (i) and (ii) of our Theorem \ref{thm:power}, it is possible to derive power guarantees for sequential calibration tests where $\mathbb{E}[Y|P]$ is estimated sequentially with isotonic regression. We leave such extensions for future work.

\section*{Acknowledgments}

A.~Henzi and J.~Ziegel gratefully acknowledge financial support from the Swiss National Science Foundation.
T.~Dimitriadis gratefully acknowledges financial support from the German Research Foundation (DFG) through grant number 502572912.

\bibliographystyle{apalike}

redpajama_set_name: RedPajamaArXiv


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The protesters lined a sidewalk space, the place they waved Israeli flags and denounced the freshman congresswoman over latest remarks that some have described as anti-Semitic, the Los Angeles Day by day Information reported. The ambiance was a mixture of dancing and music blended with the vitriolic feedback towards Omar, the report stated.
Omar, a 37-year-old immigrant from Somalia who got here to the U.S. together with her household in 1995, has confronted a storm of criticism from pro-Israel politicians and teams after her February tweet that stated "It's all concerning the Benjamins child," in reference to the assist that some U.S. lawmakers have supplied to Israel.
The freshman Democrat drew scorn from Republicans and a few in her personal celebration. She later apologized and clarified her criticism of the Israeli authorities.
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A handful of counterprotesters additionally appeared exterior the occasion to voice assist for Omar.
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Omar's go to to Southern California is anticipated to proceed Sunday, as she is scheduled to attend a non-public meet-and-greet in Irvine, in response to a flyer for the occasion.
Her look was certainly one of two political occasions within the area Saturday. In Los Angeles, U.S. Sen. Bernie Sanders of Vermont, a Democratic 2020 presidential candidate, visited a mosque to commemorate the victims of the March 15 mass capturing in New Zealand, the Los Angeles Occasions reported.
Later within the day, he spoke to an estimated 12,000 individuals at a downtown Los Angeles rally.


"Jasmine ""Jazz"" Callender is at the downhill slide to wreck. as soon as a embellished murder detective, she's misplaced all of it: her former partner's been convicted of homicide, she's been cashiered out, and she's ingesting away what little self-respect she's received left. yet Jazz has a expertise for hassle, and anyone is aware it. while a mysterious, horny stranger comes trying to find her with a fateful crimson envelope in his hand, she's approximately to make the deal of her existence ... for solid or undesirable. The deal calls for her to go into right into a partnership with a stranger and examine situations that arrive in targeted crimson envelopes ... that's unusual sufficient, yet will get more bizarre as she and her new accomplice Lucia observe that they're operating for somebody with supernatural talents. and perhaps they don't seem to be at the facet of the angels anymore. "
Driven to discover the reality concerning the mysterious dying of his ladylove, the Duke of Hawkscliffe will visit any lengths to unmask a assassin. whether it skill jeopardizing his recognition by means of accomplishing a scandalous affair with London's such a lot provocative courtesan--the fascinating yet aloof Belinda Hamilton.
Bel has used her intelligence and wit to appeal the city's titled gents, whereas suffering to place the items of her lifestyles again jointly. She wishes a protector, so she accepts Hawk's invitation to develop into his mistress in identify in basic terms. He asks not anything of her physique, yet seeks her assist in snaring a similar guy who shattered her advantage. jointly they tempt the unforgiving wrath of society--until their dicy charade becomes a perilous appeal, and Bel needs to make a devastating determination which may wreck her final probability at love. . . .
From long island instances bestselling writer Julia London comes the second one novel in her irresistibly romantic determined Debutantes trilogy, within which 3 unexpectedly destitute aristocratic younger girls needs to inn to determined ability to take care of appearances. .. and locate the husbands in their desires. girl Greer Fairchild's purely desire of fending off marriage to the 1st bidder lies in traveling into the untamed Welsh geographical region looking for an inheritance she's unsure even exists -- one reportedly managed by way of Rhodrick Glendower, Earl of Radnor, often referred to as the Prince of Powys.
It truly is curious how one girl can hang-out a man… Ten years as a high-stakes jewel thief must have helped me put out of your mind Sophie Huntington Ramsey. yet regardless of how a long way I traveled, she was once by no means faraway from my techniques. Her earthy sensuality. Her wild ardour. For ten lengthy years, my reminiscence of our one forbidden evening jointly burned as scorching because the come upon itself.


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This entry was posted in Outdoor Furniture and tagged aluminum, furniture, offers, patio. Bookmark the permalink.


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Beginner's Guide Creators' New Game is About Keanu Reeves Stabbing People
By Ethan Gach on at
"Welcome player. It's a fucking video game. Get your stats ready," Absolutely: A True Crime Story begins. Then it asks you to boost one of your starting character's stats like strength, dexterity, or something called J3V. None of these stats matter: the browser game is just a random, nihilistic parody. But it's funny, mostly because your character is named Keanu Reeves.
I made a short video game with @dualryan. It is the culmination of decades of growth of video games as a medium. It is about a real criminal named Keanu Reeves.https://t.co/5hgAo48esK
— Davey Wreden (@HelloCakebread) November 28, 2017
Absolutely: A True Crime Story is made by Davey Wreden, creator of The Beginner's Guide and The Stanley Parable. Its description reads that it's "a real story, about an actual criminal named Keanu Reeves," but that's where the connections to Keanu Reeves end unless I'm missing some especially subtle deep cut references to Keanu lore. The three to five minute RPG Maker project stars a sprite that doesn't even look like Keanu. This, I suspect, was by design. Keanu is no one. He is also all of us.
At the beginning of the game, your character is accused of murder. To prove his innocence he tries recounting to the police what really happened. The truth as Keanu sees it consists mostly of selling drugs to kids and making small talk with the neighbours. Sometimes these conversations are funny, at least if you are broken inside like I am. Walking along the lone city block and occasionally stabbing people in short one-dimensional battles was as satisfyingly bleak and streamlined a deconstruction of JRPGs as I've come across.
If you thought Davey Wreden's previous games were funny, you'll probably dig the grim absurdity of Absolutely: A True Crime Story. It's not nearly as involved as those games, but it pushes some of the same buttons in only a hundred or so words of dialogue. Ryan Roth, who worked on Beginner's Guide with Wreden and also did The Yawhg's soundtrack, is the other half of the collaboration behind the game. I'm not sure if he put together the score but it's a deadly serious one, because joke games only work if you treat them with the utmost care and sincerity.
The first time I played through Absolutely: A True Crime Story, I lost. The player definitely loses this game every time, but depending on your actions Keanu can avoid his fate of getting sent to the electric chair. Why are you stabbing people in this game? Or better yet, why is an RPG Maker sprite named Keanu stabbing people? Because game designers are cruel people and those were the only options afforded to him. The turn-based JRPG battle menus consist simply of things like "Stab," "Magic Stab," or "Don't Stab Yet."
If you abhor the limits of these options and don't feel like dishing out thousands of damage points to elaborate JRPG pixel portraits of innocent bystanders, this might not be the game for you. "Destiny claims yet another victim" is what the game says after every battle, and I don't think it's talking about the NPCs you just shived. Anyone who plays this game is a victim of having played this game.
The game's epilogue concludes that "All of our dreams will die before they are realised," and the game's brevity bares that out. For a meaningless parody project, Absolutely: A True Crime Story does a great job of showing just how compelling purposelessness referentiality can be. I was genuinely a little disappointed when the the "game over" screen popped up. That's possibly the weirdest thing of all about the game: despite all the nonsense, I'm ready to find out what happens to Keanu Reeves next.
This Interactive Timeline For Netflix's Witcher Is Super Helpful
By Alistair Jones on at
Timing Might Be Everything for the Pokémon-Inspired Temtem
Destiny 2 Players are Hard at Work Trying to Crack a Mysterious New Puzzle
Kotaku East
Kotaku Australia
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Contract Law in Australian and Chinese Courts: A Comparative Case Study
November 2015–November 2019
Funding source:
Contract law provides an essential legal framework for every business transaction. However there are fundamental differences between the contract law of Australia and that of its biggest trade partner, China. This project aims to compare Australia and China's different judicial solutions to shared real-life contract problems. The findings, it is hoped, will facilitate mutual understanding and economic competiveness, produce reflections and advice on the reform of Australian contract law, assist in businesses' contract drafting, and provide Australian courts and government agencies with information critical to decision-making in the context of trade relations with China.
Professor Paul J Gollan
Faculty of Business, Economics & Law
Privacy & Terms of use | Feedback | Updated: 7 Jun 2016


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[ January 21, 2020 ] How far would you go to help a friend in need? This story is something to reflect on Featured
eRosary: Prayer platform launched to teach young people how to pray the rosary
October 18, 2019 Cha Echaluce WORLD, World News 0
Image capture from YouTube | Click To Pray App
A prayer platform called Click To Pray was launched to teach young people how to pray the rosary
The interactive, smart, and app-driven device can be worn as a bracelet
It is described as "the app that connects your prayer with the world"
You can now pray using an e-Rosary.
"Click To Pray" — an interactive, smart, and app-driven device can be worn as a bracelet — was launched recently to teach young people how to pray the rosary; calling it an app "that connects your prayer with the world".
According to the Click To Pray website, the app of the Pope's Worldwide Prayer Network invites users to pray together for his intentions which address current challenges facing humanity and the mission of the Church.
"And because 'every day is different,' the app offers you a prayer 365 days a year to help you discover each day how God calls you to live. Join us as we pray together with the Pope, united with the Heart of Christ, thousands of us throughout the world now form a truly global prayer network," the website encourages the Catholics.
Moreover, the prayer platform connects one with thousands of people worldwide who are praying every day for the challenges of humanity and for the mission of the Church as the Pope proposes in his monthly prayer intentions.
"Pray, live, and build a world which tastes of the Gospel. Give meaning to our life and put you prayer into action," it stated. "You are invited to join the Pope's call to connect your heart with others seeking to serve the mission of Jesus, a mission of compassion for the world. Together, we make each day different. Live each day with a twist."
Click To Pray also gives one three brief moments of prayer each day and let the person share his/her own prayer intentions for everyone to pray.
Here's a short but informative video that summarizes the purpose of an eRosary:
What do you think of this app?
Sources: Click To Pray, Click To Pray App, Vatican News
[Throwback] Sushmita Sen shares that special moment she felt sure she was going to be Miss Universe 1994
February 2, 2017 C Echaluce 0
"I knew something was going to happen. I didn't know what, but something." Those were the words of Miss Universe 1994 Sushmita Sen; the first Miss Universe winner from India. In a one-on-one interview with [Read More]
Rodjun-Dianne's secret to their decade-long romance: Christ at the center of relationship
October 21, 2017 Cha Echaluce 0
Actor Rodjun Cruz recently proposed to his long-time girlfriend actress and news anchor Dianne Medina The couple have been dating for more than ten years Medina admitted that keeping a relationship for over a decade has [Read More]
Comelec Chair Bautista seeks God's guidance on decision to resign or not
August 28, 2017 Sally E 0
Comelec Chairman Andres Bautista during an interview with Ted Failon on DZMM Teleradyo last Monday was adamant that he is still biding his time on whether or not to resign from his post. Chair Bautista [Read More]
How far would you go to help a friend in need? This story is something to reflect on
A story on Facebook helps us understand what true friendship means A man needs money for his sick mother and he thought his friend had forsaken him His friend helped him in a way he [Read More]