TITLE: Anomalies in Global Symmetries (Srednicki ch 76)
QUESTION [2 upvotes]: In chapter 76 of Srednicki's QFT book, he defines $C^{\mu\nu\rho}(p,q,r)$ via (76.21)
\begin{equation}
(2\pi)^{4}\delta^{4}(p+q+r)C^{\mu\nu\rho}(p,q,r)\equiv \int d^{4}xd^{4}yd^{4}z e^{-i(px+qy+rz)}\langle 0|\textrm{T}j^{\mu}(x)j^{\nu}(y)j^{\rho}_{\textrm{A}}(z)|0\rangle
\end{equation}
and rewrites (76.15)
\begin{equation}
\langle p,q|j^{\rho}_{\textrm{A}}(z)|0\rangle = (ig)^{2}\epsilon_{\mu}\epsilon'_{\nu} \int d^{4}xd^{4}y e^{-i(px+qy)}\langle 0|\textrm{T}j^{\mu}(x)j^{\nu}(y)j^{\rho}_{\textrm{A}}(z)|0\rangle
\end{equation}
as (76.22)
\begin{equation}
\langle p,q|j^{\rho}_{\textrm{A}}(z)|0\rangle = -g^{2}\epsilon_{\mu}\epsilon'_{\nu} C^{\mu\nu\rho}(p,q,r)e^{irz}|_{r=-p-q}
\end{equation}
but I don't understand this transformation.
When I multiply (76.21) by $e^{irz}|_{r=-p-q}$ on both sides, I get
\begin{equation}
(2\pi)^{4}\delta^{4}(p+q+r)C^{\mu\nu\rho}(p,q,r)e^{irz}|_{r=-p-q}= \int d^{4}xd^{4}yd^{4}z e^{-i(px+qy)}e^{-i(p+q+r)z}\langle 0|\textrm{T}j^{\mu}(x)j^{\nu}(y)j^{\rho}_{\textrm{A}}(z)|0\rangle
\end{equation}
It seems though that the right hand side cannot be written  as
\begin{equation}
(2\pi)^{4}\delta^{4}(p+q+r) \int d^{4}xd^{4}y e^{-i(px+qy)}\langle 0|\textrm{T}j^{\mu}(x)j^{\nu}(y)j^{\rho}_{\textrm{A}}(z)|0\rangle
\end{equation}
because there is a $z$ dependence in $j^{\rho}_{\textrm{A}}(z)$ and thus (76.22) doesn't hold. Am I wrong?

REPLY [1 votes]: Prof. Srednicki multiplies the first line that you wrote,
$(2\pi)^{4}\delta^{4}(p+q+r)C^{\mu\nu\rho}(p,q,r)\equiv \int d^{4}xd^{4}yd^{4}w e^{-i(px+qy+rw)}\langle 0|\textrm{T}j^{\mu}(x)j^{\nu}(y)j^{\rho}_{\textrm{A}}(w)|0\rangle$,
by $e^{i r z}$ and integrates on $r$.
This is explicitly $$\int d^4r e^{i r z} (2\pi)^{4}\delta^{4}(p+q+r)C^{\mu\nu\rho}(p,q,r)= \int d^4r \int d^{4}xd^{4}yd^{4}w e^{-i(px+qy)}e^{ir(w-z)}\langle 0|\textrm{T}j^{\mu}(x)j^{\nu}(y)j^{\rho}_{\textrm{A}}(w)|0\rangle$$
The left side is easily evaluated with the delta function. The right side, if one integrates on $r$ first, enjoys a factor of $(2\pi)^4 \delta(w-z)$.
Thus we have
$(2\pi)^{4}e^{i r z}C^{\mu\nu\rho}(p,q,r)|_{p+q+r=0} = (2\pi)^{4}\int d^{4}xd^{4}yd^{4}w e^{-i(px+qy)}\delta(w-z)\langle 0|\textrm{T}j^{\mu}(x)j^{\nu}(y)j^{\rho}_{\textrm{A}}(w)|0\rangle .$
Dividing both sides by $(2 \pi)^4$ and integrating over $w$ nets
$e^{i r z}C^{\mu\nu\rho}(p,q,r)|_{p+q+r=0} = \int d^{4}xd^{4}y e^{-i(px+qy)}\langle 0|\textrm{T}j^{\mu}(x)j^{\nu}(y)j^{\rho}_{\textrm{A}}(z)|0\rangle .$
Using your second line gives us the desired result.


TITLE: How to show $|\sin(x+iy)|^2=\sin^2x+\sinh^2y$
QUESTION [6 upvotes]: How would I show that $|\sin(x+iy)|^2=\sin^2x+\sinh^2y$?
Im not sure how to begin, does it involve using $\sinh z=\frac{e^{z}-e^{-z}}{2}$ and $\sin z=\frac{e^{iz}-e^{-iz}}{2i}$?

REPLY [1 votes]: \begin{align}
\sin(z)^2 &= (\sin x \cos (iy))^2 +(\cos x \sin(iy))^2 \\
&=\sin^2x \cosh^2y+\cos^2x \sinh^2y \\
&= \sin^2x (1+\sinh^2y )+(1-\sin^2x ) \sinh^2y \\
&=\sin^2 x+\sin^2 x \sinh^2y+ \sinh^2y-\sin^2x \sinh^2y \\
&=\sin^2x+\sinh^2y
\end{align}


TITLE: Mock Theta Functions
QUESTION [4 upvotes]: I am studying about Mock modular forms and Mock theta functions. I wonder how Zwegers connected mock theta functions with Harmonic Maass Forms? I mean, what was the philosophy/idea of Mock Theta function that motivated him to make a connection with the holomorphic projection of weight 1/2 harmonic Maass forms?
Any reference or help will be highly appreciated.

REPLY [8 votes]: In addition to Zagier's excellent Bourbaki seminar I would also recommend some notes by Ken Ono that include both a summary of the history of mock theta functions and mock modular forms and a survey of applications. They are available at http://swc.math.arizona.edu/aws/2013/2013OnoNotes.pdf.
You might also want to take a look at Brunier and Funke, "On Two Geometric Theta Lifts," Duke Math. Journal 125 (2004).
Zwegers is on MO but I don't think he is a regular participant. Perhaps someone who knows him will see this and ask him to describe his motivation to you directly.


TITLE: Help finding integral of chord length divided by Cauchy PDF
QUESTION [0 upvotes]: I've tried many different ways to find this integral but can't find it. Is anyone able to help me by finding the integral for this? If it's easier to find the definite integral, I only need to integrate it from $a-r$ to $a+r$. Also, all variables will always be positive, real integers. If it helps, this equation is based on the chord length of a circle divided by the Cauchy-Lorentz distribution function.
$\int \frac{\sqrt{r^2-(a-x)^2}}{(x-x_0)^2+\gamma^2}\,dx$

REPLY [1 votes]: Using Mathematica (and lots of simplification) I get the following:
$$\frac{\pi  \sqrt{-a^2+\sqrt{\left(-(a-x_0)^2+\gamma ^2+r^2\right)^2+4 \gamma ^2 (a-x_0)^2}+2 a x_0+\gamma ^2+r^2-x_0^2}}{\sqrt{2} \gamma }-\pi$$
For $a=3$, $r=5$, $x_0=1$, and $\gamma=2$, I get
$$\frac{1}{4} \left(\sqrt{2 \left(\sqrt{689}+25\right)}-4\right) \pi \simeq 4.80987$$
If I use numerical integration, I get the same value:
parms = {a -> 3, r -> 5, x0 -> 1, γ -> 2};
NIntegrate[Sqrt[r^2 - (a - x)^2]/((x - x0)^2 + γ^2) /. parms,
  {x, a - r, a + r} /. parms]
(* 4.80987 *)


TITLE: Hausdorff iff convergent filterbase has a unique point of convergence. Is this correct?
QUESTION [3 upvotes]: Suppose $(X, \mathcal{T})$ is a Hausdorff space. Consider a filterbase $\mathcal{B}$ in $X$ such that $\mathcal{B}$ converges to $a\in X$.We show that, if $b \in X$ and $a\neq b$, then $\mathcal{B}$ cannot converge to $b$. Since, $a\neq b$ and $X$ is Hausdorff, there exist open neighbourhoods $U(a) \in \mathcal{T}$ and $V(b) \in \mathcal{T}$ such that $U(a) \cap V(b) =\emptyset$.  Since $\mathcal{B}$ converges to $a$, there exists a filterbasis element $B_j\in \mathcal{B}$ for every open neighbourhood $U(a)$ of $a$ such that $B_j \subset U(a)$. Consequently, $B_j \cap V(b) =\emptyset$. Therefore, $\mathcal{B}$ cannot converge to $b$. [\textit{If it did, there would exist $B_k\in \mathcal{B}$ such that $B_k\subset V(b)$, as a consequence of which $B_j\cap V(b) \neq \emptyset$}].\ Conversely, suppose $X$ is not Hausdorff. We shall construct a filterbase $\mathcal{B}$ that converges to more than one point in $X$. Consider $a$ and $b$ in $X$, since $X$ is not Hausdorff $U(a) \cap V(b) \neq \emptyset$ for all open neighbourhoods $U$ of $a$ and $V$ of $b$. \ Define $\mathcal{B}= \{B_i \mid B_i=U_i \cap V_i; i \in I\}$ where $U$ and $V$ are open neighbourhoods of two distinct points $x$ and $y$ respectively. Clearly, $B_i \neq \emptyset$ for all $B_i \in \mathcal{B}$ and since we can choose $U=X$ for any $i, j \in I$ we can find $k \in I$ such that $B_k \subset B_i \cap B_j = X \cap V_i(y) \cap V_j(y)$. Thus, $\mathcal{B}$ is a filterbase. Every $B_i \in \mathcal{B}$ is such that $B_i = U_i(x) \cap V_i(y)$ and so for every $U(x) and V(y)$ we have a $B_i \in \mathcal{B}$ such that $B_i \subset U(x)$ and $B_i \subset V(y)$. Therefore, $\mathcal{B}$ converges to both $x$ and $y$ in $X$.

REPLY [2 votes]: There are a few problems, mostly minor.

Since, $a\neq b$ and $X$ is Hausdorff, there exist open neighbourhoods $U(a) \in \mathcal{T}$ and $V(b) \in \mathcal{T}$ such that $U(a) \cap V(b) =\emptyset$.

Although your notation $U(a)$ suggests that $U(a)$ is an open nbhd of $a$, and similarly for $V(b)$ and $b$, you never actually said so, and you really should. And since you’re already distinguishing these as $U$ and $V$, there’s no real need for the extra notational complication of $U(a)$ and $V(b)$.

Since $a\ne b$ and $X$ is Hausdorff, there are $U,B\in\mathcal{T}$ such that $a\in U$, $b\in V$ and $U\cap V=\varnothing$.

The rest of this direction of the proof is okay, though you could stand to make the contradiction in the parenthetical If it did remark at the end a bit more explicit:

... as a consequence of which $B_j\cap V\ne\varnothing$, since $B_j\cap B_k\ne\varnothing$ by the definition of a filterbase.

And you could organize it a bit more tightly without changing the idea at all:

Since $\mathcal{B}$ converges to $a$, there is a filterbasis element $B_a\in\mathcal{B}$ such that $B_a\subseteq U$. If $\mathcal{B}$ also converged to $b$, there would be a $B_b\in\mathcal{B}$ such that $B_b\subseteq V$. But then $B_a\cap B_b\subseteq U\cap V=\varnothing$, contradicting the definition of a filterbase. Thus, $\mathcal{B}$ cannot converge to $B$.

In the other direction there’s a mistake near the beginning. You have:

Consider $a$ and $b$ in $X$, since $X$ is not Hausdorff $U(a) \cap V(b) \neq \emptyset$ for all open neighbourhoods $U$ of $a$ and $V$ of $b$.

First, you can’t let $a$ and $b$ be arbitrary points of $X$: there might be only two points in $X$ that can’t separated by disjoint open sets, as is the case, for instance, with the line with two origins. You must specify that you’re starting with points $a$ and $b$ that can’t separated by disjoint open sets. Finally, as a minor point, $U(a)$ and $U$ are not the same thing, so you can’t switch back and forth between them as if they were the same.

Since $X$ is not Hausdorff, there are distinct points $a,b\in X$ such that $U\cap V\ne\varnothing$ whenever $U$ is an open nbhd of $a$ and $V$ is an open nbhd of $b$.

Then you say:

Define $\mathcal{B}= \{B_i \mid B_i=U_i \cap V_i; i \in I\}$ where $U$ and $V$ are open neighbourhoods of two distinct points $x$ and $y$ respectively.

Where did $x$ and $y$ come from? You’ve not mentioned them before: they come out of nowhere. You should be talking about the specific two points $a$ and $b$ that cannot be separated by disjoint open sets, not about some new points $x$ and $y$. You also haven’t defined the index set $I$. You want $\mathcal{B}$ to be the set of all intersections of an open nbhd of $a$ with an open nbhd of $b$, so just say so:

Let $\mathcal{B}=\{U\cap V:U\text{ is an open nbhd of }a\text{ and }b\text{ is an open nbhd of }b\}$.

If you want to get rid of some of the words, you can do so like this:

Let $\mathcal{N}(a)=\{U\in\mathcal{T}:a\in U\}$, the family of open nbhds of $a$, and let $\mathcal{N}(b)=\{U\in\mathcal{T}:b\in U\}$. Let $\mathcal{B}=\{U\cap V:U\in\mathcal{N}(a)\text{ and }V\in\mathcal{N}(b)\}$.

Don’t bother to index $\mathcal{B}$: there’s no need to do so, and the only really convenient index set is $\mathcal{N}(a)\times\mathcal{N}(b)$, which gives you moderately ugly indices. With or without indices, however, you’re quite correct that it’s clear that each $B\in\mathcal{B}$ is non-empty. The next step, though, has problems. You wrote:

Clearly, $B_i \neq \emptyset$ for all $B_i \in \mathcal{B}$ and since we can choose $U=X$ for any $i, j \in I$ we can find $k \in I$ such that $B_k \subset B_i \cap B_j = X \cap V_i(y) \cap V_j(y)$.

You have to prove that for any $B,B\,'\in\mathcal{B}$ there is a $B\,''\in\mathcal{B}$ such that $B\,''\subseteq B\cap B\,'$. This means that you can’t specify what open nbhds of $a$ and $b$ are used in forming $B$ and $B\,'$.

Now let $B,B\,'\in\mathcal{B}$ be arbitrary. Then there are open nbhds $U,U'$ of $a$ and $V,V\,'$ of $b$ such that $B=U\cap V$ and $B\,'=U'\cap V\,'$. Clearly $U\cap U'$ is an open nbhd of $a$, and $V\cap V\,'$ is an open nbhd of $b$, so by definition $(U\cap U')\cap(V\cap V\,')\in\mathcal{B}$. But $$(U\cap U')\cap(V\cap V\,')=(U\cap V)\cap(U'\cap V\,')=B\cap B\,'\;,$$ so $B\cap B\,'\in\mathcal{B}$.

And this does show that $\mathcal{B}$ is a filterbase. (Indeed, it’s a rather nice one, since it’s actually closed under taking finite intersections.)
In addition to any changes needed to match the adjustments so far, your conclusion could be stated a bit more clearly:

Now let $U$ be any open nbhd of $a$. Clearly $X$ is an open neighborhood of $b$, so $U=U\cap X\in\mathcal{B}$. Thus, each open nbhd of $a$ contains (in fact is) an element of $\mathscr{B}$, and therefore $\mathcal{B}$ converges to $a$. Similarly, if $V$ is any open nbhd of $b$, then $V=X\cap V\in\mathcal{B}$, since $X$ is an open nbhd of $a$, and $\mathcal{B}$ converges to $b$ as well.


TITLE: Does the action and Lagrangian have identical symmetries and conserved quantities?
QUESTION [14 upvotes]: From the book Introduction to Classical Mechanics With Problems and Solutions by David Morin, page 236 states:

Noether's Theorem: For each symmetry of the Lagrangian, there is a conserved quantity.

Whereas the Wikipedia page states:

Noether's (first) theorem states that any differentiable symmetry of the action of a physical system has a corresponding conservation law

Does the action and Lagrangian have identical symmetries and conserved quantities?

REPLY [1 votes]: OP, I am new to stackexchange (but a physics veteran) so I am not yet allowed to comment on the post itself, but you should know that the one you chose as the correct answer is only valid for one dimensional curves, and even there it is valid only for a special definition of symmetry that allows for boundary terms (called "quasi-symmetries," as QMechanic points out).  I'm saying this because if you study anything more general than that, two-dimensional planes, QFT, string theory, for example, this result will not hold and the Lagrange symmetries may not be the same as the action symmetries because the measure (e.g. $\int d^4x$) can break the symmetry of $\mathcal L$.


text:
\begin{document}

\title{\vspace{-3cm}  {\LARGE A fiducial approach to nonparametric deconvolution problem: discrete case}}

\author{
Yifan Cui\thanks{Department of Statistics and Data Science, National University of Singapore},
Jan Hannig\thanks{Department of Statistics and Operations Research, University North Carolina at Chapel Hill}}
\date{}

\maketitle
\vspace{-2.2cm}
\begin{abstract}
Fiducial inference, as generalized by \cite{hannig2016generalized}, is applied to nonparametric $g$-modeling \citep{10.1093/biomet/asv068} in the discrete case. We propose a computationally efficient algorithm to sample from the fiducial distribution, and use the generated samples to construct point estimates and confidence intervals.
We study the theoretical properties of the fiducial distribution and perform extensive simulations in various scenarios. The proposed approach yields good statistical performance
in terms of the mean squared error of point estimators and the coverage of confidence intervals.
Furthermore, we apply the proposed fiducial method to estimate the probability of each satellite site being malignant using gastric adenocarcinoma data with 844 patients \citep{10.1093/biomet/asv068}.
\end{abstract}

\noindent {\bf keywords}
Fiducial inference, Empirical Bayes,  Confidence intervals, Nonparametric deconvolution

\section{Introduction}\label{sec:intro}

\cite{Efron2014TwoMS,10.1093/biomet/asv068,narasimhan2016g} studied the following important problem:
An unknown distribution function $F(\theta)$ yields unobservable realizations $\Theta_1$, $\Theta_2$, \ldots , $\Theta_n$, and each $\Theta_i$ produces an observable value $X_i$ according to a known probability mechanism.
The goal is to estimate the unknown distribution function from the observed data. In this paper, we aim to provide a generalized fiducial solution to the same problem in the case where $X_i$ given $\Theta_i$ follows a discrete distribution.
Following \cite{10.1093/biomet/asv068}, we use the terminology {\em deconvolution} here because the marginal distribution of $X_i$ admits the following form (see Equation~(6) of \cite{10.1093/biomet/asv068} and Equation~(7) of \cite{narasimhan2016g}):
\begin{align}
\int g_i(x_i|\theta_i) dF_i(\theta_i), \label{eq:lk}
\end{align}
where $g_i$ is the probability density for $X_i$ given $\Theta_i$.

\cite{10.1093/biomet/asv068} proposed an empirical Bayes deconvolution approach to estimating the distribution of $\Theta$ from an observed sample $\{X_i, i=1,\ldots,n\}$, where the only requirement is a known specification of the distribution for $X_i$ given $\Theta_i$.
The empirical Bayes deconvolution since developed has seen tremendous success in many scientific applications including causal inference \citep{lee2019}, single-cell analysis \citep{wang2018gene}, cancer studies \citep{gholami2015number,shen2019randomized}, clinical trials \citep{shen2018using,shen2019towards} and many other fields \citep{dulek2018empirical}.
Moreover, for a classic Bayesian data analysis, as noted in \cite[p19]{ross2018dirichletprocess} and \cite{gelman2013bayesian},
a single distribution prior may sometimes be unsuitable; hence, the prior choice is dubious.
Efron's empirical Bayes deconvolution could be one of the alternatives because the obtained estimator of the distribution of  $\Theta$ can be used as a prior distribution to produce posterior approximations \citep{narasimhan2016g}.

Fiducial inference can be traced back to R. A. Fisher \citep{Fisher1930,Fisher1933}  who introduced the concept as a potential replacement of the Bayesian posterior distribution.
\cite{Hannig2009,hannig2016generalized} showed that fiducial distributions can be related to empirical Bayes methods, which are widely used in large-scale parallel inference problems \citep{efron2012large,efron2019,efron2019rejoinder}. \cite{Efron1998} pointed out that objective Bayes theories also have connections with fiducial inference.
Other fiducial related approaches include the Dempster-Shafer theory \citep{Dempster2008, EdlefsenLiuDempster2009,MartinZhangLiu2010,HannigXie2012}, inferential models \citep{martin2013inferential,MartinLiu2013b,martin2015inferential,martin2015marginal},
 confidence distributions \citep{SchwederHjort2002,XieSinghStrawderman2011, XieSingh2013,XieEtAl2013, schweder2016confidence, HjortSchweder2018,shen2019jasa}, and higher-order likelihood expansions and implied data-dependent priors \citep{Fraser2004,Fraser2011}.

We propose a novel fiducial approach to modeling the distribution function $F$ of $\Theta$ nonparametrically.
In particular, we propose a computationally efficient algorithm to sample from the {\em generalized fiducial distribution} (GFD) (i.e., a distribution on a set of of distribution functions), and use the generated samples to construct statistical procedures.
The pointwise median of the GFD is used as a point estimate, and appropriate quantiles of the GFD evaluated at a given point
provide pointwise confidence intervals.
We also study the theoretical properties of the fiducial distribution. Extensive simulations in various scenarios show that the proposed fiducial approach is a good alternative to existing methods, including Efron's $g$-modeling.
We apply the proposed fiducial approach to intestinal surgery data to estimate the probability of each satellite site being malignant for a patient. The resulting fiducial estimate of distribution function
reflects the observed patterns of raw data.

The remainder of this paper is organized as follows. In Section~\ref{sec:method}, we present the mathematical framework for the fiducial approach to the nonparametric deconvolution problem. In Section~\ref{sec:theory}, we establish an asymptotic theory that verifies the frequentist validity of the proposed fiducial approach. Extensive simulation studies are presented in Section~\ref{sec:simulation}. We illustrate our
method using intestinal surgery data in Section~\ref{sec:realdata}. The article
concludes with a discussion of future work in Section~\ref{sec:discussion}. Some technical results and additional simulations are
provided in the Appendix and Supplementary Material.

\section{Methodology}\label{sec:method}

\subsection{Data generating equation}\label{s:DGE}
In this section, we first explain the definition of a generalized fiducial distribution and then demonstrate how to apply it to the deconvolution problem. We start by expressing the relationship between the data $X_i$ and the parameter $\Theta_i$ using
\begin{equation}\label{eq:DGE}
X_i=G_i^{-1}(U_i,\Theta_i),\quad \Theta_i = F^{-1}(W_i),\quad i=1,\ldots n,
\end{equation}
where $U_i,W_i$ are independent and identically distributed (i.i.d.) $\tunif(0,1)$, $G_i(\cdot,\theta_i)$ are known distribution functions of discrete random variables supported on integers, $G_i$ are defined and non-increasing in $\theta_i\in \mathcal S \subset \mathbbm R$ for all $i$, and $F$ is the unknown distribution function with support in the interval $\mathcal S$.
We are interested in estimating the unknown distribution function $F(\theta)$.

Recall that $F^{-1}(w)=\inf\{\theta: F(\theta)\geq w\}$ \cite[p54]{CasellaBerger2002}, and $F^{-1}(w)=\theta$ if and only if $F(\theta)\geq w>F(\theta-\epsilon)$ for all $\epsilon>0$. We denote $G_i^{*}(x_i,u_i)=\sup\{\theta:G_i(x_i,\theta)\geq u_i\}$ with the usual understanding that $\sup \emptyset$ is smaller than all elements of $\mathcal S$. If $G_i(\cdot,\theta_i)$ is continuous in $\theta_i$, $G_i^{*}(x_i,u_i)$ is the solution (in $\theta_i$) to the equation $G_i(x_i,\theta_i)= u_i$. By Lemma~\ref{lemma:equiv} in Section~\ref{thm:0} of the Appendix, $x_i=G_i^{-1}(u_i,\theta_i)$ if and only if $\theta_i\in (G_i^{*}(x_i-1,u_i),G_i^{*}(x_i,u_i)]$.

We combine $G_i^{*}(x_i-1,u_i)< \theta_i \leq G_i^{*}(x_i,u_i)$ and $F(\theta_i -\epsilon) <w_i\leq F(\theta_i)$ for all $\epsilon$. Consequently, the inverse of the data generating equation \eqref{eq:DGE} is
\begin{equation}
Q_{\bx}(\bu,\bw) =
\{F\, :\ \, F(G_i^*(x_i-1,u_i))< w_i\leq F(G_i^*(x_i,u_i)), i=1,\ldots,n\}.  \label{eq:constraint}
\end{equation}
Note that $Q_{\bx}(\bu,\bw)$ is a set of cumulative distribution functions (CDFs). By Lemma~\ref{lemma:equiv2} in Section~\ref{thm:0} of the Appendix,
$Q_{\bx}(\bu,\bw)\neq\emptyset$ if and only if $\bu, \bw$ satisfy:
\begin{equation}\label{eq:constraint2}
\mbox{whenever $G_i^*(x_i,u_i)\leq G_j^*(x_j-1,u_j)$ then $w_i < w_j$.}
\end{equation}

A generalized fiducial distribution is obtained by inverting the data generating equation. \cite{hannig2016generalized} proposed a general definition of GFD. However, to simplify the presentation, we use an earlier, less general version in \cite{Hannig2009}. These two definitions are equivalent for the models considered here.
Suppose $(\bU^\star,\bW^\star)$ are uniformly distributed on the set $\{(\bu^\star,\bw^\star): Q_{\bx}(\bu^\star,\bw^\star)\neq\emptyset\}.$ A GFD is then the distribution of any element in the random set
$
 \overline{Q_{\bx}(\bU^\star,\bW^\star)},
$
where the closure is in the weak topology on the space of probability measures on $\mathcal S$, and the element is selected so that it is measurable. Hence, the selected element is a random distribution function on $\mathcal S$.
Given the observed data $(x_1,\ldots,x_n)$, we define random functions $F^U$ and $F^L$ as follows: for each $\theta\in\mathcal S$ and $(\bU^\star,\bW^\star)$,
\[
F^U(\theta) \equiv \min\{W^\star_i, \mbox{ for all $i$ such that } \theta<G_i^*(x_i-1, U^\star_i)\},
\]
and
\[
F^L(\theta) \equiv \max\{W^\star_i, \mbox{ for all $i$ such that } \theta\geq G_i^*(x_i, U^\star_i)\},
\]
where $\min\emptyset =1$ and $\max\emptyset =0$. These functions are clearly non-decreasing and right continuous.

Note that if $Q_{\bx}(\bU^\star,\bW^\star)\neq\emptyset$, Portmanteau's theorem~\citep{Billingsley1999} and  \eqref{eq:constraint} imply that a distribution function $F\in \overline{Q_{\bx}(\bU^\star,\bW^\star)}$ if and only if  $F^L(\theta)\leq F(\theta)\leq F^U(\theta)$ for all $\theta\in\mathcal S$.
Thus the functions $F^U$ and $F^L$  will be called the {\em upper} and {\em lower fiducial bounds}, respectively. We generate realizations of $F^U$ and $F^L$ using a novel Gibbs sampler in the next section.

\subsection{Gibbs Sampling and GFD based inference}
We need to generate $(\bU^\star,\bW^\star)$ from the standard uniform distribution
on a set described by Equation~\eqref{eq:constraint2}, which is achieved using a Gibbs sampler.
For each fixed $i$, we denote random vectors with the $i$-th observation removed by $(\bU^\star_{[-i]},\bW^\star_{[-i]})$. If  $(\bU^\star,\bW^\star)$ satisfy the constraint \eqref{eq:constraint2}, so do $(\bU^\star_{[-i]},\bW^\star_{[-i]})$. The proposed Gibbs sampler is based on the conditional distribution of
\begin{equation}\label{eq:Gibbs}
 (U^\star_i,W^\star_i) \mid \bU^\star_{[-i]},\bW^\star_{[-i]},
 \end{equation}
which is a bivariate uniform distribution on a set $A$, where $A$ is a disjoint union of small rectangles. The beginnings and ends of the rectangles' bases are the neighboring points in the set $\bigcup_{j\neq i}\{G_i^*(x_i,U^\star_j), G_i^*(x_i-1,U^\star_j)\}$, while the corresponding location and height on the vertical axis are determined by \eqref{eq:constraint2}. The details are described in Algorithm~\ref{alg:gibbs}, and
a visualization of the rectangles is shown in Figure~\ref{fig:4} in the Supplementary Material.
Each marginal conditional distribution is supported on the entire $\mathcal S$; therefore, we expect the proposed Gibbs sampler to mix well.

The proposed Gibbs sampler requires starting points,
and we consider two potential initializations.
The first one starts with randomly generated $(\bU^\star,\bW^\star)$ from independent $\tunif(0,1)$ and reorders $\bW^\star$ so that the constraint~\eqref{eq:constraint2} is satisfied.
The second starting value is consistent with having a deterministic $\Theta$, e.g., $p=\frac{\sum_{i=1}^n x_i}{\sum_{i=1}^n m_i}$ for binomial data $X_i \sim \tbin(m_i,p)$ and $\lambda=\frac{\sum_{i=1}^n x_i}{n}$ for Poisson data $X_i \sim \tpoi(\lambda)$. As these two starting points are very different, they can be used to monitor convergence. To streamline our presentation, in Sections~\ref{sec:simulation} and \ref{sec:realdata}, we present the numerical results using the first initialization.

\begin{algorithm}
\SetAlgoLined
\caption{Pseudo algorithm for the fiducial Gibbs sampler} \label{alg:gibbs}
\KwIn{Dataset, e.g., $(m_i,x_i)$, $i=1,\ldots, n$ for binomial data,\\
      ~~~~~~~~~~ starting vectors $\bu, \bw$ of length $n$,\\
      ~~~~~~~~~~ $n_{\text{mcmc}}$ , $n_{\text{burn}}$, and vector $\btheta_{\text{grid}}$ of length $n_{\text{grid}}$.}
\For{$i = 1$ \textbf{to} $n$}{
$\btheta_{L}[i]=G_i^{*}(x_i-1,\bu[i]), \btheta_{U}[i]=G_i^{*}(x_i,\bu[i])$\;
}
\ShowLn Run Gibbs Sampler using the initial values $\bu,\bw,\btheta_L, \btheta_U$\;
\For{$j = 1$ \textbf{to} $n_{\text{burn}}+n_{\text{mcmc}}$}
{
\For{$i = 1$ \textbf{to} $n$}
{
\ShowLn $\bu^0=\bu[-i],\bw^0=\bw[-i],\btheta^0_{L}=\btheta_{L}[-i], \btheta^0_{U}=\btheta_{U}[-i]$\;
\ShowLn $\bu^{\text{pre}}_{L}=G_i(x_i,\btheta^0_{L})$, $\bu^{\text{pre}}_{U}=G_i(x_i-1,\btheta^0_{U})$\;
\ShowLn Sort $\bu^{\text{pre}}=(\bu^{\text{pre}}_{L}, \bu^{\text{pre}}_{U},0,1) $, denoted as $\bu^{\text{sort}}$\;
\ShowLn Sort $(\bw^0,1(n-1),1,0)$ according to the order of $\bu^{\text{pre}}$ as $\bw^\star_U$,\\~~where $1(n-1)$ is a vector with elements 1 of length $n-1$\;
\ShowLn Sort $(0(n-1),\bw^0,1,0)$ according to the order of $\bu^{\text{pre}}$ as $\bw^\star_L$,\\~~where $0(n-1)$ is a vector with elements 0 of length $n-1$\;
\ShowLn $\bw^{\text{pre}}_{U}=\text{cummin}(\bw^\star_U)$, $\bw^{\text{pre}}_{L}=\text{rev}(\text{cummax}(\text{rev}(\bw^\star_L)))$\;
\ShowLn Take the component-wise difference of $\bu^{\text{sort}}$, denoted as $\bu^{\text{diff}}$\;
\For{$k = 1$ \textbf{to} $2n-1$}{
\ShowLn $\bw^{\text{diff}}[k]= \bw^{\text{pre}}_{U}[k]- \bw^{\text{pre}}_{L}[k+1]$\;
}
\ShowLn Sample $i^\star \in \{1,\ldots,2n-1\}$ with probability $\propto \bu^{\text{diff}}\cdot \bw^{\text{diff}}$\;
\ShowLn Sample $a$ and $b$ from independent \tunif(0,1), and set \\~~~$u=\bu^{\text{sort}}[i^\star]+\bu^{\text{diff}}[i^\star]\cdot a,
        w=\bw^{\text{pre}}_U[i^\star]-\bw^{\text{diff}}[i^\star]\cdot b$,\\

~~~$\theta_{L}=G_i^{*}(x_i-1,u), \theta_{U}=G_i^{*}(x_i,u)$,\\

~~~$\bu[i]=u,\bw[i]=w,\btheta_{L}[i]=\theta_{L}, \btheta_{U}[i]=\theta_{U}$\;
}
Generate $n$ i.i.d. \tunif(0,1) and sort them according to the order of $\bw$, denoted by $\bw^*$. Replace $\bw$ by $\bw^*$, i.e., $\bw=\bw^*$\;
}
\ShowLn Evaluate the upper and lower bounds on a grid of values $\btheta_{\text{grid}}$ for each MCMC sample after burn-in, indexed by $l$\;
\For{$j = 1$ \textbf{to} $n_{\text{grid}}$}{
$F_l^{L}(\btheta_{\text{grid}}[j]) = \max( \bw[\btheta_{U}\leq \btheta_{\text{grid}}[j] ])$\;
$F_l^{U}(\btheta_{\text{grid}}[j])=  \min( \bw[\btheta_{L}\geq \btheta_{\text{grid}}[j] ])$\;
}
\Return  The fiducial samples $F_l^{U}$, $F_l^{L}$ evaluated on $\btheta_{\text{grid}}$.
\label{algorithm}
\end{algorithm}

From Algorithm~\ref{alg:gibbs}, we output two distribution functions required for the proposed mixture and conservative confidence intervals.
In the remainder of this paper, we denote the Monte Carlo realizations of the lower and upper fiducial bounds by $F_l^L$ and $F_l^U$, respectively, where $l=1,\ldots, n_{\text{mcmc}}$, and $n_{\text{mcmc}}$ is the number of fiducial samples.

We propose using the median of the $2n_{\text{mcmc}}$ samples $\{F_l^L(\theta), F_l^U(\theta), l=1,\ldots, n_{\text{mcmc}}\}$ as a point estimator of the distribution function $F(\theta)$.
We construct two types of pointwise confidence intervals,  conservative and mixture, using appropriate quantiles of fiducial samples \citep{Hannig2009}.
In particular, the $95\%$ conservative confidence interval is formed by taking the empirical 0.025 quantile of $\{F_l^{L}(\theta),l=1,\ldots, n_{\text{mcmc}}\}$ as the lower limit and the empirical 0.975 quantile of $\{F_l^{U}(\theta),l=1,\ldots, n_{\text{mcmc}}\}$ as the upper limit.
The lower and upper limits of the $95\%$ mixture confidence interval are formed by taking the empirical 0.025 and 0.975 quantiles of $\{F_l^L(\theta), F_l^U(\theta), l=1,\ldots, n_{\text{mcmc}}\}$, respectively.

\begin{remark}
Algorithm~\ref{alg:gibbs} is general to any discrete distribution of $X_i$.
For example, if $X_i$ follows a binomial distribution, $G_i$ is the CDF of the binomial distribution with number of trials $m_i$ and $G_i^{*}(x_i,u_i)$ is the $(1-u_i)$ quantile of $\tbeta(x_i+1,m_i-x_i)$;
if $X_i$ follows a Poisson distribution, $G_i$ is its CDF, and $G_i^{*}(x_i,u_i)$ is the $(1-u_i)$ quantile of $\tgamma(x_i+1,1)$.
\end{remark}

\subsection{Further illustration with a simulated dataset}\label{sec:2.3}
To streamline our presentation, we consider the binomial case as our running example hereinafter, i.e., the observed data are $(m_i,x_i),$ and $X_i \sim \tbin(m_i,P_i)$, $i=1,\ldots, n$, where $P_i\in\mathcal S= [0,1]$ plays the role of $\Theta_i$. We also provide the details of the proposed approach and some examples for the Poisson data in the Supplementary Material.

 We present a toy example to demonstrate the proposed fiducial approach. Suppose that $F$ follows the beta distribution $\tbeta(5,5)$. The number of trials $m_i=20$, $i=1,\ldots,n$. The sample size of the simulated binomial data is $n=20$. The fiducial estimates are based on 10000 iterations after 1000 burn-in times.

Figure~\ref{fig:1} presents the last MCMC sample of the lower fiducial bound $F_l^{L}(p)$ (blue line) and upper fiducial bound $F_l^{U}(p)$ (red line) for the two starting points, respectively.
As the fiducial distribution reflects the uncertainty, we do not expect every single fiducial curve to be close to the true CDF (black line).
Furthermore, Figure~\ref{fig:2} presents the mixture (blue line for lower limit; red line for upper limit) and conservative (cyan line for lower limit; magenta line for upper limit) confidence intervals (CIs) with two starting points, respectively, computed from the MCMC sample. We also plot the point estimates of the proposed approach alongside Efron's $g$-modeling. The brown curve represents the fiducial point estimate $\widehat F(p)$. The dashed curve is the point estimate of $F(p)$ for Efron's $g$-modeling without bias correction. Efron's confidence intervals with bias correction look almost the same as those without correction; thus, we omit them in the figures.
As can be seen, the proposed fiducial point estimator and confidence intervals capture the shape of the true CDF fairly well.

\begin{figure}[H]
\begin{center}
\includegraphics[width=6.5cm]{par0.pdf}
\includegraphics[width=6.5cm]{par1.pdf}
\label{fig:3}
\end{center}
\caption{The last MCMC sample from generalized fiducial distribution. The blue curve is a realization of the lower fiducial bound $F^{L}(p)$ and the red curve is a realization of the upper fiducial bound $F^{U}(p)$. The black curve is the true $F(p)$. Each panel represents a different starting value.}
\label{fig:1}
\end{figure}

\begin{figure}[H]
\begin{center}
\includegraphics[width=6.5cm]{ci0.pdf}
\includegraphics[width=6.5cm]{ci1.pdf}
\end{center}
\caption{Point estimates and 95\% CIs for $F(p)$ given a fixed simulated dataset. Each panel represents an interval computed  from a realization of the MCMC chain initiated with different starting values. The orange curve is the fiducial point estimate $\widehat F(p)$.
The dashed curve is the point estimate of $F(p)$ for Efron's $g$-modeling. The black curve is the true $F(p)$. The blue and red curves are the lower and upper limits of the mixture CIs, respectively.
The cyan and magenta curves are the lower and upper limits of the conservative CIs, respectively.}
\label{fig:2}
\end{figure}

\section{Theoretical results}\label{sec:theory}
Recall that the GFD is a data-dependent distribution defined for every fixed
dataset $\bx$. It can be made into a random measure in the same way as one defines the
usual conditional distribution, i.e., by plugging random variables $\bX$ into the observed dataset.
In this section, we study the asymptotic behavior of this random measure for a binomial distribution $X_i \sim \tbin(m_i,P_i)$ when the rate of $m_i$ is much faster than $n$, i.e., the following assumption holds.
\begin{assumption}
$\lim_{n \rightarrow \infty} n^4(\log n)^{1+\epsilon}/ (\min_{i=1,\ldots,n} m_i) = 0$ for any $\epsilon>0$.
\label{as:2}
\end{assumption}

We provide a central limit theorem for $F^L(p)$. The proof of Theorem~\ref{main} is deferred to Section~\ref{thm:3.1} of the Appendix. A similar result holds for $F^U(p)$.
\begin{theorem}\label{main}
Suppose the true CDF is absolutely continuous with a bounded density.
Based on Assumption~\ref{as:2},
\begin{equation}
n^{1/2}\{ F^L(\cdot)-\widehat F_n(\cdot)  \} \rightarrow B_F(\cdot),
\label{eq:gaussian}
\end{equation}
 in distribution on the Skorokhod space $\mathcal D[0,1]$ in probability, where
 \begin{equation}\label{eq:OracleECDF}
\widehat F_n(s) \equiv \frac{1}{n}\sum_{i=1}^n I[P_i\leq s]
\end{equation}
is the oracle empirical CDF constructed based on unobserved $P_i$ used to generate the observed $X_i,\ i=1,\ldots,n$, and $B_F(\cdot)$ is a mean zero Gaussian process with covariance $\operatorname{cov}(B_F(s),B_F(t))=F(t\wedge s)-F(t)F(s)$.
\end{theorem}

Note that the stochastic process on the left-hand side of \eqref{eq:gaussian} is naturally in $\mathcal D[0,1]$, i.e., the space of functions on $[0,1]$ that are right continuous and have left limits. Distances on $\mathcal D[0,1]$ are measured using  Skorokhod's metric, which makes it a Polish space \citep{Billingsley1999}. To understand the mode of convergence used here, note that there are two sources of randomness. One is from the fiducial distribution derived based on each fixed dataset. The other is the usual randomness of the data. Thus, \eqref{eq:gaussian} can be interpreted as
\begin{equation}\label{eq:explanation123}
 \rho\left(n^{1/2}\{ F^L(\cdot)-\widehat F_n(\cdot)\},B_F(\cdot)\right) \overset{pr}{\to} 0,
\end{equation}
where $\rho$ is any metric metrizing weak convergence of probability measures on the Polish space $\mathcal D[0,1]$, e.g., L\'evy-Prokhorov or Dudley metric \citep{Shorack2017}. The distribution of $n^{1/2}\{ F^L(\cdot)-\widehat F_n(\cdot)\}$ in the argument of $\rho$ is a fiducial distribution, i.e., induced by the randomness of $(\bU^\star,\bW^\star)$ with the data $X_i$ and $P_i$ being fixed. Consequently, the left-hand side of \eqref{eq:explanation123} is a function of $X_i$ and $P_i$, and the convergence in probability is based on the distribution of the data.

Theorem~\ref{main} establishes a Bernstein-von Mises theorem for the fiducial distribution under Assumption~\ref{as:2}, which implies that the confidence intervals described in Section~\ref{sec:method} have asymptotically correct coverage. Moreover, Theorem~\ref{main} provides a sufficient condition for the $n^{1/2}$-estimability of the binomial probability parameter's distribution function.
Although Assumption~\ref{as:2} is pretty stringent, it does not seem
likely that one can establish a unified asymptotic property of the proposed fiducial approach under a general scheme.
This is best seen by the fact that in the binomial case, $m_i$ are uniformly bounded, and there is not enough information in the data to consistently estimate the underlying distribution function of $P$.

Interestingly, looking at the fiducial solution reveals an interesting connection to a different statistical problem \citep{cui2021}. Recall that quantities $P_i,\ i=1,\ldots,n$ are only known to be inside random intervals $(G_i^{*}(x_i-1,U_i^\star), G_i^{*}(x_i,U_i^\star)]$. Therefore, the statistical problem we study here is similar to non-parametric estimation under Turnbull's general censoring scheme \citep{efron1967two,turnbull1976empirical}, in which case there is no unified theory of the nonparametric maximum likelihood estimator, but properties are investigated under various special and challenging cases, such as the $n^{1/2}$ convergence for right-censored data \citep{breslow1974large} and the $n^{1/3}$ convergence for current status data \citep{groeneboom1992information}.

\begin{remark}\label{newremark}
Note that $$pr(F \in Q_x(U^*,W^*))\propto  \prod_{i=1}^{n} \left[\int_0^1  {{m}\choose{x_i}} p^{x_i} (1-p)^{m-x_i} dF(p) \right] $$ is proportional to the nonparametric likelihood function. Some implications of this observation are provided in Section
~\ref{sec:moments} of the Supplementary Material.
\end{remark}

\section{Numerical experiments}\label{sec:simulation}

We performed simulation studies to compare the frequentist properties of the proposed fiducial confidence intervals with Efron's $g$-modeling \citep{10.1093/biomet/asv068,deconvolveR},  the nonparametric bootstrap, and a nonparametric Bayesian approach \citep{dirichletprocess}.
For each scenario, we first generated $p_i, i=1,\ldots, n$ from the distribution function $F$. Then we drew $X_i$ from the binomial distribution Bin($m_i,p_i$), where $m_i$ is described in Section~\ref{subsec:simulation}. The simulations were replicated 500 times for each scenario.

For the proposed method and other existing methods, we chose the grid $p=0.01, 0.02, \ldots, 0.99$ following \cite{narasimhan2016g}.
The fiducial estimates were based on 2000 iterations of the Gibbs sampler after 500 burn-in times.
Efron's $g$-modeling was implemented using the R package \texttt{deconvolveR} \citep{deconvolveR}. We used default values for the degree of the splines, i.e., 5.
We considered the regularization strategy with
the default value $c_0=1$.
For the nonparametric bootstrap, we first obtained the maximum likelihood estimates $\hat p_i=x_i/m_i$. We then constructed the empirical CDF as the point estimator, and used $B=1000$ bootstrap samples of $\hat p_i$ to construct confidence intervals. We also considered a fully Bayesian approach, i.e., the Dirichlet process mixture of beta binomial distributions that provides more flexibility than a beta binomial model \cite[p19]{ross2018dirichletprocess}. The default values for the prior parameters in the R package \texttt{dirichletprocess} \citep{dirichletprocess} were used. The Bayesian estimates were based on 2000 MCMC samples after 500 burn-in times.

\subsection{Simulation settings}\label{subsec:simulation}

We start with the following two scenarios with fixed $m_i$'s.
\begin{scenario} \normalfont
We consider the same setting as in  Section~\ref{sec:2.3}.
Let $F$ follow $\tbeta(5,5)$, and the number of trials $m_i=20$, $i=1,\ldots,n$. The sample size $n$ of the simulated binomial data is set to 50.
\end{scenario}

\begin{scenario} \normalfont
Let $F$ follow a mixture of beta distributions $0.5\tbeta(10,30)+0.5\tbeta (30,10)$, and the number of trials be $m_i=20$, $i=1,\ldots,n$. The sample size $n=50$.
\end{scenario}

Next, we consider three complex settings from \cite{10.2307/41724531}.

\begin{scenario} \normalfont(Beta density)
Let $F$ follow $\tbeta(8,8)$, and $m_i$'s be integers sampled uniformly between 100 and 200. The sample size $n=100$.
\end{scenario}

\begin{scenario} \normalfont (Multimodal distribution)
We consider a mixture of beta distributions $0.5\tbeta(60,10)+0.5\tbeta(10,60)$ for $F$. The sample size $n=100$, and $m_i=100$, $i=1,\ldots,n$.
\end{scenario}

\begin{scenario} \normalfont (Truncated exponential)
Let $F$ be $\texp(8)$ truncated at 1. The sample size $n = 200$ and $m_i = 100$ for $i = 1, \ldots,n.$
\end{scenario}

We also consider the above five settings for $n= 1000$. The results are reported in the Supplementary Material.

\subsection{Numerical results}

In this section, we first compare the mean squared error (MSE) of different methods of $F(p)$ for the five scenarios.
The numerical results for $p = 0.15, 0.25, 0.5, 0.75, 0.85$ for each scenario are presented in Table~\ref{table:mse}.
We see that the MSEs of the proposed fiducial point estimates are as good as and sometimes smaller than those of the competing methods.

\begin{table}[!h]
\centering
\scalebox{1.1}{
\begin{tabular}{ccccccc}
  \hline
 Scenario & $p$ & F & g & bc & BP & BA \\
  \hline
      \multirow{2}{*}{1} & 0.15 & 5 &  40  & 39  & 17  &  1\\
& 0.25 &  20 &  49 &  48 &  69  & 10\\
& 0.50 & 51 &  47  & 48 &  71 &  78\\
&0.75 & 18  & 22  & 21  & 16  & 10\\
& 0.85 & 5  & 13 &  12   & 4   & 1\\
   \hline
    \multirow{2}{*}{2} & 0.15 &  41 & 149 & 149 & 145 &   9  \\
&0.25 & 39 &  17 &  18  & 66 & 101\\
&0.50 & 49 &  32  & 33  & 54 &  53\\
&0.75 & 37 &  18 &  19  & 50 &  88\\
&0.95 & 46 &  86 &  84  & 33  & 12\\
   \hline
   \multirow{2}{*}{3}  & 0.15 &   0.14  & 8.45 & 8.16 &  0.15 &  0.04\\
& 0.25  & 2.57 & 14.89 & 14.41 & 2.87 & 1.39\\
& 0.50 & 22.85 &  26.94 & 27.04 &  24.68 &  21.92\\
&0.75  &  2.54 & 5.03 & 4.81 & 2.80 & 1.34\\
&0.85   & 0.16  & 1.46 & 1.38  & 0.15 &   0.05\\
   \hline
 \multirow{2}{*}{4}  & 0.15 &  22 &  49  & 49 &  23 &  27\\
& 0.25 & 27 &  34 &  33 &  26 &  25\\
&0.50 & 25  & 24  & 25  & 26  & 26\\
& 0.75 & 26 &  36  & 35  & 27  & 25\\
& 0.85 & 20 &  52 &  52 &  26  & 25\\
   \hline
  \multirow{2}{*}{5}  &   0.15  & 11  & 10  & 10 &  11 &  10* \\
     & 0.25 & 6 &  10 &  10  &  6   & 6* \\
     & 0.50 & 1   & 5  &  4  &  1   & 1* \\
     & 0.75 &  0.13 & 1.84 & 1.66 &  0.12 &  0.10* \\
     & 0.85 & 0.05  & 0.76 & 0.68 &  0.05  &  0.04* \\
   \hline
\end{tabular}
}
\caption{MSE ($\times 10^{-4}$) of point estimates for $F(p)$ of each scenario.
``F'' denotes the fiducial point estimates;
``g'' denotes Efron's $g$-modeling without bias correction;  ``bc'' denotes Efron's $g$-modeling with bias correction;
``BP'' denotes the bootstrap method; ``BA'' denotes the Bayesian method.
*The Bayesian results for Scenario 5 are reported based on 489 replications as 11 runs failed because of an error in the R package \texttt{dirichletprocess}.}
\label{table:mse}
\end{table}

\begin{table}[!h]
\centering
\scalebox{1.1}{
\begin{tabular}{cccccccc}
  \hline
 Scenario & $p$ & M & C & g & bc & BP & BA \\
  \hline
      \multirow{2}{*}{1} &0.15 & 99 & 100  & 25  & 27 &  74 &  88\\
&0.25 & 99 & 100  & 80 &  82 &  65  & 83\\
&0.50 & 100 & 100  & 94 &  94 &  91  & 84\\
&0.75 & 100  & 100  & 96  & 96  & 92  & 83\\
&0.85 & 100 & 100 &  90 &  91  & 54  & 89\\
   \hline
    \multirow{2}{*}{2}  & 0.15 &  98  & 99  &  1  &  1 &  27  & 97\\
&0.25 & 100 & 100  & 98 &  98  & 90   & 86\\
&0.50& 98  & 98  & 96 &  96 &  97  & 45\\
&0.75 &  100 & 100 &  96 &  96  & 85  & 87\\
& 0.85 &  97  & 99  & 31 &  31 &  79  & 97\\
   \hline
 \multirow{2}{*}{3}   & 0.15&   100  & 100  &  4  &  5  & 13 &  28\\
& 0.25 &    99 &  99  & 69  & 71  & 86  & 28\\
& 0.50 & 99 &  99  & 89 &  89 &  96  & 27\\
&0.75 &  98  & 99 &  97 &  97  & 87 &  32\\
& 0.85 &  99 & 100  & 97  & 98  & 12  & 31\\
   \hline
 \multirow{2}{*}{4}   & 0.15&   99 & 100  & 48  & 49  & 95  & 59\\
& 0.25 &  95  & 97 &  89 &  90  & 93 &  12 \\
& 0.50 & 95 &  96  & 93  & 93&   95  &  6\\
&0.75 & 95 &  95  & 89  & 89 &  93  &  9\\
& 0.85 &   99 &  99  & 75  & 75   &91  & 59\\
   \hline
  \multirow{2}{*}{5}  &   0.15 &  99  & 99  & 95  & 95  & 93 &  18* \\
     & 0.25 & 97  & 98 &  95  & 95  & 93  & 21* \\
     & 0.50 & 98 &  98 &  98  & 98 &  89  & 20* \\
     & 0.75 & 98  & 99 & 100 & 100  & 36  & 20* \\
     & 0.85 & 99 & 100 & 100 & 100 &  17 &  11* \\
   \hline
\end{tabular}
}
\caption{Coverage (in percent) of 95\% CIs for $F(p)$ of each scenario. ``M'' denotes mixture GFD confidence intervals;  ``C'' denotes conservative GFD confidence intervals;
``g'' denotes Efron's $g$-modeling without bias correction;  ``bc'' denotes Efron's $g$-modeling with bias correction;
``BP'' denotes the bootstrap method; ``BA'' denotes the Bayesian method.
*The Bayesian results for Scenario 5 are reported based on 489 replications as 11 runs failed because of an error in the R package \texttt{dirichletprocess}.}
\label{table:c}
\end{table}

\begin{table}[!h]
\centering
\scalebox{1.1}{
\begin{tabular}{cccccccc}
  \hline
 Scenario & $p$ & M & C & g & bc & BP & BA \\
  \hline
      \multirow{2}{*}{1} & 0.15 &  123 & 139 & 101 & 101 &  84 &  28\\
&  0.25 &  220 & 241 & 171 & 171 & 172 &  95\\
& 0.50 & 426 & 465 & 246  &246 & 272  & 279\\
&0.75 & 217 & 239 & 154  &154  &128 &  89\\
& 0.85 & 121 & 137 &  82 &  81 &  42 &  28\\
   \hline
    \multirow{2}{*}{2} &0.15 & 243 & 266  & 66&   66 & 185 & 101 \\
& 0.25 & 357 & 390 & 162 & 162 & 251 & 311\\
&0.50 & 305  & 330 & 206  &206  &275  & 90\\
& 0.75 & 352 & 385 & 166 & 166& 226 & 301\\
& 0.85 &245 & 268 & 143 & 143 & 133  & 107\\
   \hline
   \multirow{2}{*}{3}  & 0.15 &  35  & 42  & 46  & 46  &  4  &  2\\
& 0.25 &  77 &  84  & 81  & 81  & 51  & 10\\
& 0.50 & 243  & 259 & 168 & 168  & 195 &  33\\
& 0.75  & 78 &  86 &  65  & 65  & 51  & 10\\
 & 0.85 & 35 &  42  & 27  & 26  &  4   & 2\\
   \hline
 \multirow{2}{*}{4}  &0.15&  240 & 259 & 125 & 125 & 179 & 101\\
& 0.25 & 201 & 213 & 191 & 191  &195 &  16\\
&0.50&  193 & 202 & 188  &188 & 196  &  0.02\\
&0.75& 201 & 213 & 198 & 198  &195  & 16\\
&0.85& 240 & 259 & 178 & 178 & 173 &  98\\
   \hline
  \multirow{2}{*}{5}  &   0.15 & 160 & 171 & 125 & 125 & 126 &  18* \\
     & 0.25 & 115 & 124 & 114 & 114  & 94 &  13* \\
     & 0.50 & 45 &  50  & 74 &  73 &  35  &  6*\\
     & 0.75 &  20 &  24  & 37 &  36 &   7 &  1*\\
     & 0.85 & 17 &  20  & 23  & 22  &  3 &   1* \\
   \hline
\end{tabular}
}
\caption{Mean length ($\times 10^{-3}$) of 95\% CIs for $F(p)$ of each scenario.  ``M'' denotes mixture GFD confidence intervals;  ``C'' denotes conservative GFD confidence intervals;
``g'' denotes Efron's $g$-modeling without bias correction;  ``bc'' denotes Efron's $g$-modeling with bias correction;
``BP'' denotes the bootstrap method; ``BA'' denotes the Bayesian method.
*The Bayesian results for Scenario 5 are reported based on 489 replications as 11 runs failed because of an error in the R package \texttt{dirichletprocess}.}
\label{table:l}
\end{table}

Next, we present the coverage and average length of confidence intervals for the various methods in Tables~\ref{table:c}-\ref{table:l}.
Table~\ref{table:c} summarizes the coverage of 95\% confidence intervals of various methods, and Table~\ref{table:l} summarizes the average length of these confidence intervals. ``M'' denotes mixture GFD confidence intervals;  ``C'' denotes conservative GFD confidence intervals;
``g'' denotes Efron's $g$-modeling without bias correction;  ``bc'' denotes Efron's $g$-modeling with bias correction;
``BP'' denotes the bootstrap method; ``BA'' denotes the Bayesian method.

We see that the GFD confidence intervals maintain or exceed the nominal coverage everywhere,
whereas other methods often have coverage problems.
In particular, the Efron's confidence intervals and the nonparametric bootstrap have substantial coverage problems close to the boundary, while the Bayesian method consistently underestimates the uncertainty, resulting in credible intervals that are too narrow.

It is not surprising that the GFD confidence intervals are often longer than those of the other methods, as the GFD approach aims to provide a conservative way to quantify uncertainty.
As expected, the mean length of the mixture GFD confidence intervals is slightly shorter than the conservative GFD confidence intervals.
A potential reason for the fiducial approach outperforming Efron's $g$-modeling in terms of coverage is that Efron's $g$-modeling relies on an exponential family parametric model, and the proposed fiducial approach is nonparametric. Therefore, the proposed method is expected to perform better when the true model does not belong to an exponential family.

\section{Intestinal surgery data}\label{sec:realdata}
In this section, we consider an intestinal surgery study on gastric adenocarcinoma involving $n = 844$ cancer patients \citep{gholami2015number}.
Resection of the primary tumor with appropriate dissection of the surrounding lymph nodes is the foundation of curative care.
In addition to the primary tumor, surgeons also removed the satellite nodes for later testing.
Efron's deconvolution was used to estimate the prior distribution of the probability of one satellite being malignant in this study \citep{gholami2015number}.

The dataset consists of pairs $(m_i,X_i), i=1,\ldots,n$, where $m_i$ is the number of satellites removed, and $X_i$ is the number of satellites found to be
malignant. The $m_i$ varies from 1 to 69. Among all cases, 322 have $X_i = 0$. For the rest, $X_i /m_i$ has an approximate $\tunif(0,1)$ \citep{10.1093/biomet/asv068}. We are interested in estimating the distribution function of the probability of one satellite being malignant.
Following the model proposed in \cite{10.1093/biomet/asv068}, we assume a binomial model, i.e., $X_i \sim \tbin(m_i,P_i)$, where $P_i$ is the $i$-th patient's probability of any one satellite being malignant.

We compared the proposed mixture and conservative GFD confidence intervals to Efron's with and without bias correction, the bootstrap method, and the fully Bayesian approach. For all methods, we used the grid $[0.01, 0.02,$ $\ldots, 0.99]$ for the discretization of $p$. The fiducial and Bayesian estimates were based on 10000 iterations after 1000 burn-in times.
Other tuning parameters for each method were chosen in the same manner as in Section~\ref{sec:simulation}.

\begin{figure}[!h]
\begin{center}
\includegraphics[width=11cm]{surg.pdf}
\end{center}
\caption{GFD vs. $g$-modeling.
Estimated CDF (dashed line) and 95\% CIs for $F(p)$  of GFD and Efron's $g$-modeling. The red and orange curves are mixture and conservative confidence intervals, respectively, and the blue curve is Efron's confidence intervals without bias correction. Efron's confidence intervals with bias correction look almost the same as those without correction; thus, we omit the bias-corrected ones in the figure.
}
\label{pic:real}
\end{figure}

The overall shapes of the bootstrap and Bayesian confidence intervals are similar to the fiducial ones but much narrower.
We provide bootstrap and Bayesian point estimates and 95\% confidence intervals in the Supplementary Material.
Figure~\ref{pic:real} shows the point estimates and 95\% confidence intervals of the distribution function $F$ for the proposed GFD approach and Efron's $g$-modeling.
Overall, the GFD confidence intervals are more conservative. The GFD confidence intervals cover Efron's almost everywhere.

For the proposed fiducial approach, there is a large mode for the upper fiducial confidence interval near  $p= 0$, which reflects the fact that approximately $38\%$ of the $X_i$'s are 0 in the surgery data. However, the Bayesian method and Efron's $g$-modeling seem to quantify the uncertainty of this proportion to be lower. One exception is the nonparametric bootstrap, which gives the estimation of point mass at zero $0.38$ with 95\% confidence intervals $(0.35,0.41)$. We note that this might be overestimated, as $X_i=0$ may correspond to a non-zero probability $p$ especially when $m_i$ is small.

Moreover, the generalized fiducial confidence intervals provide a unimodal density, whereas Efron's gives a bimodal density.
Thus, we believe that the fiducial, bootstrap, and nonparametric Bayesian answers are more in line with Efron's observation that $X_i /m_i$ has an approximate $\tunif(0,1)$ for those $X_i \neq 0$ in the surgery data \citep{10.1093/biomet/asv068}.

\section{Discussion}\label{sec:discussion}
In this paper, we proposed a prior-free approach to the nonparametric deconvolution problem, and obtained valid point estimates and confidence intervals.
This was accomplished through a novel algorithm to sample from the GFD.
The median of the GFD was used as the point estimate, and appropriate quantiles of the GFD evaluated at a given $p$
provided pointwise confidence intervals.
We also studied the theoretical properties of the fiducial distribution. Extensive simulations show that the proposed fiducial approach is a good alternative to existing methods, including Efron's $g$-modeling.
We applied the proposed fiducial approach to intestinal surgery data to estimate the probability that each satellite site is malignant for patients.

We conclude by listing some open research problems:
\begin{enumerate}
\item
The proposed fiducial method seems to be a powerful nonparametric approach.
It would be interesting to implement it inside other statistical procedures, such as tree or random forest models, to include covariates \citep{wu2019uncertainty}.

\item This paper focuses on discrete data. The proposed approach can be extended to continuous data, such as $\tnorm(\Theta,1)$, where $\Theta$ follows a distribution function $F$.
    This part is currently under investigation.

\item As we can see from the simulations, the GFD approach is sometimes over-conservative. It might be possible to consider a
different choice of fiducial samples, such as log-interpolation \citep{cuihannig2019} or monotonic spline interpolation \citep{Taraldsen2019discuss,cuihannig2019rejoinder}.

\item The asymptotic distribution results in Section~\ref{sec:theory} only hold under Assumption~\ref{as:2}.
To investigate a non-$n^{1/2}$ rate of convergence should provide a fruitful avenue for future research.

\item It should be possible to use the GFD in conjunction with various functional norms to construct simultaneous confidence bands \citep{cuihannig2019,nair1984confidence, martin2019discuss}.
\end{enumerate}

\section{Acknowledgments}
 The authors are thankful to Prof. Hari Iyer for helpful discussions.

 Yifan Cui's research was supported in part by the NUS Grant R-155-000-229-133, the NIH Grant R01CA222147, and R01AI127271.
 Jan Hannig's research was supported in part by the NSF Grant DMS-1512945, IIS-1633074, and DMS-1916115.

\newpage

\begin{center}
{\LARGE \textbf{Appendix}}
\end{center}

\appendix
\section{Lemmas \label{thm:0}}
Recall that the observed data points $x_i$ are integers.
\begin{lemma}
$x_i=G_i^{-1}(u_i,\theta_i)$ if and only if $G_i^{*}(x_i-1,u_i) < \theta_i \leq G_i^{*}(x_i,u_i)$.
\label{lemma:equiv}
\end{lemma}

\begin{proof}

Recall the definition of $G_i^{-1}(u_i,\theta_i)= \inf\{x_i: G_i(x_i,\theta_i)\geq u_i\}$. So $x_i=G_i^{-1}(u_i,\theta_i)$ if and only if $G_i(x_i,\theta_i)\geq u_i >G_i(x_i-\epsilon,\theta_i)$ for all $\epsilon>0$. Then by the definition of $G_i^{*}$, it is further equivalent to $G_i^{*}(x_i-1,u_i) < \theta_i \leq G_i^{*}(x_i,u_i)$.
\end{proof}

\begin{lemma}
$Q_{\bx}(\bu,\bw)\neq\emptyset$ if and only if $\bu, \bw$ satisfy: whenever $G_i^*(x_i,u_i)\leq G_j^*(x_j-1,u_j)$ then $w_i< w_j$.
\label{lemma:equiv2}
\end{lemma}

\begin{proof}
Sufficiency: If $Q_{\bx}(\bu,\bw)\neq\emptyset$ holds, and $G_i^*(x_i,u_i)\leq G_j^*(x_j-1,u_j)$, then we know that $ w_i \leq F(G_i^*(x_i,u_i))\leq F(G_j^*(x_j-1,u_j))< w_j $.

Necessity: We prove it by contradiction. If $Q_{\bx}(\bu,\bw)$ is empty, then there must exist indices $i$ and $j$ such that, $(G_j^*(x_j-1,u_j),G_j^*(x_j,u_j)]$ is strictly larger than $(G_i^*(x_i-1,u_i),G_i^*(x_i,u_i)]$ but $w_i \geq w_j$. This contradicts with whenever $G_i^*(x_i,u_i)\leq G_j^*(x_j-1,u_j)$ then $w_i< w_j$.
\end{proof}

\section{Proof of Theorem~\ref{main} \label{thm:3.1}}
\begin{proof}
Recall that the data generating equation is
\begin{equation}\label{eq:DGEinApp}
X_i= G_i^{-1}(U_i,P_i),
\quad
P_i=F^{-1}(W_i),
\end{equation}
where $G_i$ is the CDF of binomial distribution.
Define the oracle fiducial distribution based on unobserved $P_1,\ldots,P_n$ as
\begin{equation}\label{eq:OracleFidL}
\widetilde F(s)=\sum_{i=0}^n I[
P_{(i)} \leq s< P_{(i+1)}]  W^*_{(i)},
\end{equation}
where $P_{(0)} \equiv 0$, $P_{(n+1)} \equiv 1$, $W^*_{(i)}$ are uniform order statistics, and $W^*_{(0)} \equiv 0$.
Note that \eqref{eq:OracleFidL} is the lower fiducial distribution $F^L$ in \cite{cuihannig2019} when there is no censoring.
By Corollary
~\ref{corollary} of the Supplementary Material, we have
\begin{align*}
n^{1/2}\{ \widetilde F(\cdot)-\widehat F_n(\cdot)  \} \rightarrow \{1-F(\cdot)\} B(\gamma(\cdot)),
\end{align*}
in distribution on the Skorokhod space $\mathcal D[0,1]$ in probability, where $B$ is a Brownian motion, $\gamma(t)=\int_0^t \frac{f(s)}{[1-F(s)]^2} ds= \frac{F(t)}{1-F(t)}$, and $\widehat F_n$ is the oracle empirical distribution function defined in \eqref{eq:OracleECDF}.
We also define
\begin{align*}
\widetilde F^L(s)\equiv \sum_{i=0}^n I[G^U_{(i)}\leq s < G^U_{(i+1)}] W^*_{(i)},
\end{align*}
and
\begin{equation*}
\widetilde F^U(s)\equiv \sum_{i=0}^n I[G^L_{(i)}\leq s < G^L_{(i+1)}] W^*_{(i+1)},
\end{equation*}
where $G^U_{(1)},\ldots,G^U_{(n)}$ are order statistics of $\{G_i^*(x_i,U_i^\star),i=1,\ldots,n\}$,  $G^L_{(1)},\ldots,G^L_{(n)}$ are order statistics of $\{G_i^*(x_i-1,U_i^\star),i=1,\ldots,n\}$, $G^U_{(0)} \equiv 0$, $G^U_{(n+1)} \equiv 1$, $G^L_{(0)} \equiv 0$, $G^L_{(n+1)} \equiv 1$, and  $W^*_{(n+1)} \equiv 1$. Note that $\widetilde F^L$ and $\widetilde F^U$ can be regarded as the lower and upper bounds of $F^L$ and $F^U$, respectively.

In order to obtain
\begin{align*}
n^{1/2} \{F^L(\cdot) -\widehat F_n(\cdot)\} \rightarrow \{1-F(\cdot)\} B(\gamma(\cdot)),
\end{align*}
in distribution in probability, it is enough to show that
\begin{align*}
\sup_s n^{1/2} |\widetilde F(s)- F^L(s) | \rightarrow 0,
\end{align*}
in probability, which would be implied by
\begin{align}\label{eq:conv}
\sup_s n^{1/2} \{\widetilde F^U(s)- \widetilde F^L(s) \} \rightarrow 0,
\end{align}
in probability. In order to show Equation~\eqref{eq:conv}, one essentially needs to show that for any $\epsilon>0$,
\begin{align*}
pr(\sup_s n^{1/2} \{\widetilde F^U(s)- \widetilde F^L(s) \}>\epsilon ) \rightarrow 0.
\end{align*}
By
Lemma~\ref{lemma:nointersect},
given in the Supplementary Material, there is no intersection between $\{(G_i^*(x_i-1,U_i^\star),G_i^*(x_i,U_i^\star)],i=1,\ldots,n\}$ with a large probability converging to 1.

Thus, we have that
\begin{align*}
 pr(\sup_s n^{1/2} \{\widetilde F^U(s)- \widetilde F^L(s) \}>\epsilon )
\leq & \sum_{i=0}^n pr(W^*_{(i+1)}-W^*_{(i)} > \frac{\epsilon}{n^{1/2}})  \\
= & (n+1) \times pr(Beta(1,n)>\frac{\epsilon}{n^{1/2}} )\\
= & (n+1) \times (1-\frac{\epsilon}{n^{1/2}})^n
\rightarrow  0.
\end{align*}

Therefore, we have
\begin{align*}
n^{1/2} \{F^L(\cdot) -\widehat F_n(\cdot)\} \rightarrow \{1-F(\cdot)\} B(\gamma(\cdot)),
\end{align*}
 in distribution in probability. Note that for any $t<s$,
$$\operatorname{cov}[\{1-F(s)\}B(\gamma(s)),\{1-F(t)\} B(\gamma(t))]=\gamma(t)\{1-F(s)\}\{1-F(t)\}=F(t)\{1-F(s)\},$$ which completes the proof.
\end{proof}

\newpage

\begin{center}
{\LARGE \textbf{Supplementary Material}}
\end{center}

\section{Lemma~\ref{lemma:nointersect0}  and its proof}
\begin{lemma}\label{lemma:nointersect0}
Assume the conditions of Theorem~3.1. Suppose that $X_i \sim
\tbin(m_i
, P_i)$, where $P_i$ are unobserved i.i.d. random variables with a distribution function $F$.
We have that \begin{align}\label{eq:xm}
\min_{i,j\in\{0,\ldots,n\}} \left \{\frac{X_i}{m_i}-\frac{X_j}{m_j}\right\} \succeq O\left(\frac{1}{n^2(\log n)^{\sqrt{\epsilon/2}}}\right),
\end{align}
for any $\epsilon>0$, with a large probability converging to 1.
\end{lemma}

\begin{proof}
We first show that the unobserved $P_i=F^{-1}(W_i)$ are well-separated.
Straightforward calculation with uniform order statistics shows that
\[pr\left(\min_{i\in\{0,\ldots,n\}} \left \{W_{(i+1)}-W_{(i)}\right\} >\frac{t}{n(n+1)}\right)\geq \left(1-\frac{t}{n}\right)^{n},\]
for any $t>0$, where $W_{(0)} \equiv 0$ and $W_{(n+1)} \equiv 1$. Therefore,
\begin{align*}
\min_{i\in\{0,\ldots,n\}} \left \{P_{(i+1)}-P_{(i)}\right\} \succeq O\left(\frac{1}{n^2(\log n)^{\sqrt{\epsilon/2}}}\right),
\end{align*}
with a large probability converging to 1, where $P_{(0)} \equiv 0$ and $P_{(n+1)} \equiv 1$.
Furthermore, by the following Bernstein inequality for binomial $X \sim \tbin(m,P)$,
\begin{align*}
pr(|X - m P|\geq t) \leq 2\exp \left\{-\frac{t^2}{2[mP(1-P)+t/3]}\right\},
\end{align*}
where $t>0$. Taking $t\sim O\left(\frac{m}{n^2(\log n)^{\sqrt{\epsilon/2}}}\right)$,
we have that
\begin{align*}
\min_{i,j\in\{0,\ldots,n\}} \left \{\frac{X_i}{m_i}-\frac{X_j}{m_j}\right\} \succeq O\left(\frac{1}{n^2(\log n)^{\sqrt{\epsilon/2}}}\right),
\end{align*}
with a large probability converging to 1 because $\lim_{n \rightarrow \infty} n^4(\log n)^{1+\epsilon}/ m_i = 0$ for any $m=m_i$ given by Assumption~1.
\end{proof}

\section{Lemma~\ref{lemma:nointersect}  and its proof}
\begin{lemma}\label{lemma:nointersect}
Given data that satisfy Equation~\eqref{eq:xm} of Lemma~\ref{lemma:nointersect0}, if $(\bU^\star,\bW^\star)$ are uniformly distributed on the set $\{(\bu^\star,\bw^\star): Q_{\bx}(\bu^\star,\bw^\star)\neq\emptyset\},$ we have that
\begin{equation}\label{eq:lemma2}
pr\{(G_i^*(x_i-1,U_i^\star), G_i^*(x_i,U_i^\star)]\cap (G_j^*(x_j-1,U_j^\star), G_j^*(x_j,U_j^\star)]\neq\emptyset \mbox{ ~for some~ $i\neq j$}\}\to 0.
\end{equation}
\end{lemma}

\begin{proof}
Let $\widetilde U_i$ be i.i.d. $U(0,1)$ and
denote
$L_i=G_i^*(x_i-1,\widetilde U_i) \sim \tbeta(x_i,m_i-x_i+1)$, and $R_i= G_i^*(x_i,\widetilde U_i)\sim \tbeta(x_i+1,m_i-x_i)$.
Recall that the proposed Algorithm~1 can be regarded as an importance sampling in the following way: if there is one $k$-intersections (i.e., $(k+1)$ intervals share one common area) in $\{(L_i,R_i],i=1,\ldots,n\}$, the corresponding $W^*$ has $(k+1)!$ possible permutations. So we need to show $\sum_{k=1}^{n-1} (k+1)!\times q_k \rightarrow 0$ as $n \to 0$, where $q_k$ is defined as the probability of $\{(L_i,R_i],i=1,\ldots,n\}$ having one or more $k$-intersections, and $W^*_{i_1}<\ldots<W^*_{i_{k+1}}$ (say $i_1,\ldots,i_{k+1}$ are indices of intervals that intersect).

We start with one-intersection between the $i$-th and $j$-th intervals, i.e., $(L_i,R_i)\cap (L_j,R_j) \neq \emptyset$, which is equivalent to $L_1\leq R_2$ and $L_2\leq R_1$.
Without loss of generality, we focus on $L_1\leq R_2$ as $$pr(A_1\cap A_2)\leq \min\{pr(A_1),pr(A_2)\}.$$
For a random variable $Y\sim \tbeta(\alpha,\beta)$, let $\mu=E[Y]=\frac{\alpha}{\alpha+\beta}$.
By Theorem~2.1 of \cite{marchal2017}, $Y$ is sub-Gaussian with the variance proxy parameter
\begin{align*}
\Sigma\equiv  \frac{1}{4(\alpha+\beta+1)}.
\end{align*}
Hence, by the definition of a sub-Gaussian random variable, for any $c,t \in \mathbbm{R}$,
\begin{align*}

    pr\left(Y- \mu\geq t\right)\leq \exp\left\{-\frac{t^2}{2\Sigma}\right\}.
\end{align*}

Note that
\begin{align}\label{eq:p}
pr(L_1\leq R_2) = pr(R_2-L_1\geq 0),
\end{align}
where $L_1$ is sub-Gaussian with mean $x_1/(m_1+1)$ and $\Sigma_1=1/(m_1+2)$, and $R_2$ is sub-Gaussian with mean $(x_2+1)/(m_2+1)$ and $\Sigma_2=1/(m_2+2)$.
Therefore, Equation~\eqref{eq:p} equals $pr(Z\geq x_1/(m_1+1)-(x_2+1)/(m_2+1))$, where $Z$ is a sub-Gaussian with mean $0$ and $\Sigma_Z=1/(m_1+2)+1/(m_2+2)$.

Then we further have that Equation~\eqref{eq:p} equals
\begin{align*}
& pr\{Z\geq x_1/(m_1+1)-(x_2+1)/(m_2+1)\}\\
\leq  & \exp\left\{ - \frac{T^2}{2\Sigma_Z}  \right\}\\
\leq & \exp\left\{ - \tilde m T^2    \right\},
\end{align*}
where $\tilde m = \min_{i=1,\ldots,n} m_i$, and $T= x_1/(m_1+1)-(x_2+1)/(m_2+1)$.

Thus, the probability of a single intersection $2!\times q_1$ is bounded by
\begin{align*}
2! \times {n \choose 2} \times  \exp\left[ - \tilde m T^2    \right]
\preceq \exp\left\{c_1 \log n - c_2 (\log n)^{1+\epsilon/2}\right\},
\end{align*}
for any $\epsilon>0$, and some constants $c_1$ and $c_2$. We note that the coefficient $ {n \choose 2}$ refers to the number of possible pairs.

Next, we consider the case of $k$-intersections. We only need to consider two intervals corresponding to the two farthest $P_i$ among $k$ intervals.
 Thus, the probability of existing $k$-intersections $(k+1)!\times q_k$ is bounded by
\begin{align*}
 &(k+1)! \times {n \choose k} \times  \exp\left\{ - \tilde m \left[O\left(\frac{k}{n^2(\log n)^{\sqrt{\epsilon/2}}} \right)  \right]^2    \right\} \\
 \preceq & \exp\{ c_1(k\log n + k\log k) - c_2 k^2 (\log n)^{1+\epsilon/2} \},
\end{align*}
for any $\epsilon>0$, and some constants $c_1$ and $c_2$.
Because $\sum_{k=1}^{n-1} (k+1)!\times q_k \rightarrow 0$, we
conclude \eqref{eq:lemma2}.

\end{proof}

\section{Corollary~\ref{corollary} and its proof}
We present a Bernstein-von Mises theorem for fiducial distribution associated with the empirical distribution function. This result can be viewed either as a special special case (without censoring) of \cite{cuihannig2019}, or as a particular case of the exchangeably weighted bootstrap in \cite{praestgaard1993exchangeably}.

\begin{corollary}\label{corollary}
Assume the conditions of Theorem~3.1. We have
\[
n^{1/2}\{ \widetilde F(\cdot)-\widehat F_n(\cdot)  \} \rightarrow \{1-F(\cdot)\} B(\gamma(\cdot)),
\]
in distribution on the Skorokhod space $\mathcal D[0,1]$ in probability, where $B$ is a Brownian motion, $\gamma(t)=\int_0^t \frac{f(s)}{[1-F(s)]^2} ds= \frac{F(t)}{1-F(t)}$, $\widehat F_n$ is defined in Equation~(7) of Section~3.1, and $\widetilde F$ is defined in Equation~(10) of Appendix~B.
\end{corollary}
\begin{proof}
By Theorem~2 of \cite{cuihannig2019}, we need to check their Assumptions 1-3. Their Assumption~1 satisfies with their $\pi(p)= 1-F(p)$; their Assumption~2 satisfies as we assume true CDF is absolutely continuous; their Assumption~3 satisfies as $$\int_0^p \frac{g_n(s)}{\sum_{i=1}^n I(P_i\geq s)} d [\sum_{i=1}^n I(P_i\leq s)] \rightarrow \int_0^p \frac{f(s)}{[1-F(s)]^2} ds,$$ for any $p$ such that $1-F(p)>0$ and any sequence of functions $g_n \rightarrow \frac{1}{1-F}$ uniformly.
\end{proof}

\section{Remark on binomial and Poisson data}\label{sec:moments}

In the following two theorems for binomial and Poisson data, respectively, we show implications of the fact that $pr(F \in Q_x(U^*,W^*))$ is proportional to the nonparametric likelihood function. In particular, maximizing the scaled fiducial probability in its limit provides the underlying true CDF.

\begin{theorem}\label{main2}
Suppose $\Theta_i \equiv P_i$, and $X_i \mid P_i$ follows $\tbin(m,P_i)$,
maximizing $\lim_{n \rightarrow \infty} [c_n \times pr(F\in Q_{\bx}(\bU^\star,\bW^\star))]^{1/n}$ leads to a CDF  matching the first $m$-moments of the true $F(p)$, where $c_n$ is the normalizing constant.
\end{theorem}
\begin{theorem}\label{remark:2}
Suppose $\Theta_i \equiv \Lambda_i$, and $X_i \mid \Lambda_i$ follows $\tpoi(\Lambda_i)$,
maximizing $\lim_{n \rightarrow \infty} [c_n \times pr(F\in Q_{\bx}(\bU^\star,\bW^\star))]^{1/n}$ leads to the true $F(\lambda)$ almost surely, where $c_n$ is the normalizing constant.
\end{theorem}

\begin{proof}[Proof of Theorem~\ref{main2}]
Recall the data generating equation in \eqref{eq:DGEinApp},
where $G_i$ is the CDF of binomial distribution.
The fiducial probability is
\begin{align}
&pr(F\in Q_{\bx}(\bU^\star,\bW^\star))\nonumber \\  \propto & \prod_{i=1}^{n} \left[\int_0^1  {{m}\choose{x_i}} p^{x_i} (1-p)^{m-x_i} dF(p) \right] \nonumber \\
 =& \exp\left\{  \sum_{i=1}^{n}  \log \left( E_F \left[{{m}\choose{x_i}} p^{x_i} (1-p)^{m-x_i} \right]  \right)  \right\} \nonumber \\
 = &\exp\bigg \{n_{m} \log(E_F[P^m]) + n_{m-1} \log\left(E_F\left[ {{m}\choose{m-1}} P^{m-1}(1-P)\right] \right) + \cdots  \nonumber\\
& + n_{0} \log(E_F[(1-P)^m]) \bigg \}, \nonumber
\end{align}
where $n_{k}$ is the number of samples with $X_i=k$, and $E_F$ refers to the expectation with respect to $F$ that is evaluated.
Thus, as $n$ goes to infinity,
\begin{align}
& [c_n \times pr(F\in Q_{\bx}(\bU^\star,\bW^\star))]^{1/n} \nonumber \\
\rightarrow &\exp\bigg \{ E[P^m] \log(E_F[P^m]) + E\left[ {{m}\choose{m-1}} P^{m-1}(1-P)\right] \log\left(E_F\left[ {{m}\choose{m-1}} P^{m-1}(1-P)\right] \right)\nonumber\\&+ \cdots   + E[(1-P)^m] \log(E_F[(1-P)^m]) \bigg \},\label{eq:loglikelihood3}
\end{align}
where $c_n$ is the normalizing constant, and $E$ refers to the expectation with respect to the true distribution function that generates data.
 By the method of Lagrange multipliers, \begin{align*}
 H(x) = \sum_i y_i \log x_i~~\text{subject to}~~\sum_i x_i=1,\sum_i y_i=1,
 \end{align*}
 is maximized with respect to $x$ by setting $x_k=y_k$, $k\in \mathbbm{R}^+$.
Maximizing Equation~\eqref{eq:loglikelihood3} gives
\begin{align*}
E[P^m]&=E_F[P^m], \\
E[P^{m-1}(1-P)]&=E_F[P^{m-1}(1-P)], \\
& \ldots \\
E[(1-P)^m]&=E_F[(1-P)^m].
\end{align*}
 The above equations are restrictions on the first $m$-moments of $P$, which completes the proof.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{remark:2}]
Recall the data generating equation is
\begin{align*}
X_i= G_i^{-1}(U_i,\Lambda_i),
\quad
\Lambda_i=F^{-1}(W_i),
\end{align*}
where $G_i$ is the CDF of Poisson distribution.
The fiducial probability is
\begin{align}
& pr(F\in Q_{\bx}(\bU^\star,\bW^\star)) \nonumber \\
\propto & \prod_{i=1}^n \int_0^\infty   \frac{ \lambda^{x_i} \exp\{-\lambda \}}{x_i!} dF(\lambda) \nonumber \\
=& \left\{\frac{E_F[\exp(-\Lambda)]}{0!}\right\}^{n_0}\times \left\{\frac{ E_F[\Lambda \exp(-\Lambda)]}{1!} \right \}^{n_1} \cdots \times \left\{\frac{ E_F[\Lambda^k \exp(-\Lambda)]}{k!} \right \}^{n_k}\times \cdots \nonumber\\
= & \exp\left \{n \left [ \frac{n_0}{n} \log \frac{E_F[\exp(-\Lambda)]}{0!}  +\frac{n_1}{n} \log \frac{E_F[\Lambda\exp(-\Lambda)]}{1!} + \cdots + \frac{n_k}{n} \log \frac{E_F[\Lambda^k\exp(-\Lambda)]}{k!} +\cdots \right]  \right\}, \nonumber
\end{align}
where $n_k$ is the count of $X_i=k$, and $E_F$ refers to the expectation with respect to $F$ that is evaluated.
 Thus, as $n$ goes to infinity,
\begin{align}
& [c_n \times pr(F\in Q_{\bx}(\bU^\star,\bW^\star))]^{1/n} \nonumber \\
 \rightarrow & \exp\bigg \{ \bigg [ \frac{E[\exp(-\Lambda)]}{0!} \log \frac{E_F[\exp(-\Lambda)]}{0!}  + \frac{E[\Lambda \exp(-\Lambda)]}{1!} \log \frac{E_F[\Lambda\exp(-\Lambda)]}{1!} + \cdots \nonumber \\ & +  \frac{E[\Lambda^k \exp(-\Lambda)]}{k!} \log \frac{E_F[\Lambda^k\exp(-\Lambda)]}{k!} +\cdots \bigg]  \bigg\},
\label{eq:loglikelihood2}
\end{align}
where $c_n$ is the normalizing constant, and $E$ refers to the expectation with respect to the true distribution function that generates data.
 By the method of Lagrange multipliers, maximizing Equation~\eqref{eq:loglikelihood2} gives
\begin{align*}
\frac{E[\Lambda^k \exp(-\Lambda)]}{k!} =\frac{E_F[\Lambda^k \exp(-\Lambda)]}{k!}, ~ k \in \mathbbm{R}^+.
\end{align*}
 The above equations are essentially restrictions on all derivatives of the Laplace transform at 1.
The property of the Laplace transform being analytic in the region of absolute convergence implies
the uniqueness of a distribution,
which completes the proof.
\end{proof}

\section{Rectangles of Algorithm~1}
\begin{figure}[H]
\begin{center}
\includegraphics[width=5.5cm]{rec1.pdf}
\includegraphics[width=5.5cm]{rec2.pdf}
\end{center}
\caption{A visualization of rectangles in Algorithm~1 for the final step of the Gibbs sampler ($j=n_{burn}+n_{mcmc}, i=n$) for the toy example of Section~2.3. The last $(U_n,W_n)$ was generated from the gray area, with the rectangles described in Section~2.2 indicated by black lines. Just like in Figure~1, each panel represents a different starting value. \label{fig:4}}
\end{figure}

\section{Additional simulation results with $n=1000$}

In this section, we present the results for both point estimates and 95\% CIs for $n=1000$. The simulations were replicated 500 times for each scenario. We again observe a consistent pattern that the proposed methods are comparable to and sometimes better than the competing methods.

\begin{table}[H]
\centering
\scalebox{1.1}{
\begin{tabular}{ccccccc}
  \hline
 Scenario & $p$ & F & g & bc & BP & BA \\
  \hline
      \multirow{2}{*}{1} & 0.15 & 0.50 & 1.79 & 1.74 & 10.65  & 0.04\\
& 0.25 & 1.49 & 1.46  &1.44 &52.03 &  0.65\\
& 0.50 &3.92  & 9.54  &9.52 &27.67  & 2.94\\
&0.75 & 1.32 &  0.61  & 0.62 & 5.48 &  0.67\\
& 0.85 & 0.48 &  0.17  & 0.16 & 1.30   & 0.04\\
   \hline
    \multirow{2}{*}{2} & 0.15 &  5  & 10 &   9  &122  &  2* \\
&0.25 & 4 &  8  &  8  & 19 &  14*\\
&0.50 &  3  &  3  &  3  &  3  &  4*\\
&0.75 & 4  &  3  &  3  & 17  & 16*\\
&0.95 & 4   & 1 &   1 &  18  &  2*\\
   \hline
   \multirow{2}{*}{3}  & 0.15 & 0.02 &  0.33  & 0.30   & 0.02  &  0.003\\
& 0.25  & 0.28 & 1.27  &1.21 &  0.65 &  0.10\\
& 0.50 & 3.02 & 6.98 & 6.95 & 2.81 & 2.21\\
&0.75  & 0.29 &  0.11 &  0.11 &  0.54 &  0.12\\
&0.85   & 0.02  & 0.02  &  0.01  &  0.02  &  0.002 \\
   \hline
 \multirow{2}{*}{4} & 0.15 & 3  &  5   & 5  &  3  &  2\\
& 0.25 &  3   & 2  &  2  &  3  &  2\\
&0.50 & 2  &  2  &  2 &   2  &  2\\
& 0.75 & 3  &  2  &  2  &  5  &  2\\
& 0.85 & 3 &  34  & 34 &   8   & 2\\
   \hline
  \multirow{2}{*}{5}  &   0.15 &  2  &  3  &  3   & 2  &  2*\\
     & 0.25 & 1  &  1  &  1  &  1  &  1* \\
     & 0.50 &0.20  &  0.26  &  0.25  &  0.20  &  0.17* \\
     & 0.75 & 0.02  & 0.11  & 0.10  &  0.02  &  0.02* \\
     & 0.85 & 0.01  &  0.04  &  0.04  &  0.01  &  0.01* \\
   \hline
\end{tabular}
}
\caption{MSE ($\times 10^{-4}$) of point estimates for $F(p)$ of each scenario.
``F'' denotes the fiducial point estimates;
``g'' denotes Efron's $g$-modeling without bias correction;  ``bc'' denotes Efron's $g$-modeling with bias correction;
``BP'' denotes the bootstrap method; ``BA'' denotes the Bayesian method.
*The Bayesian results for Scenarios 2, 5 are reported based on 495, 476 replications as 5, 24 runs failed because of an error in the R package \texttt{dirichletprocess}.}

\end{table}

\begin{table}[H]
\centering
\scalebox{1.1}{
\begin{tabular}{cccccccc}
  \hline
 Scenario & $p$ & M & C & g & bc & BP & BA \\
  \hline
      \multirow{2}{*}{1} &0.15 & 98 &  99 &  28  & 31  &  0  & 93\\
&0.25 & 100 & 100 &  91 &  92 &   0  & 83\\
&0.50 &  100 & 100 &  84 &  84&   10  & 70\\
&0.75 & 100 & 100 &  98  & 98  & 17 &  80\\
&0.85 & 98 &  99  & 98 &  98  & 10 &  88\\
   \hline
 \multirow{2}{*}{2}   & 0.15&   93 &  98 &  10  & 12  &  0 &  89*\\
& 0.25 &  100 & 100  & 82  & 82 &  17 &  87*\\
& 0.50 & 97  & 98 &  93  & 93  & 91  & 65*\\
&0.75 &  100  & 100  & 97  & 97  & 15 &  85*\\
& 0.85 & 95  & 97 &  98 &  98  &  0 &  89*\\
   \hline
 \multirow{2}{*}{3}   & 0.15&  99 &  99 &  74  & 80 &  72 &  31\\
& 0.25 &   98 &  99  & 68 &  71  & 77  & 29\\
& 0.50 & 99 & 100  & 71  & 71  & 94  & 20\\
&0.75 & 99  & 99  &100 &100  & 82  & 28\\
& 0.85 & 99 &  99 & 100  &100  & 69  & 28\\
   \hline
 \multirow{2}{*}{4}   & 0.15&  100 & 100 &  75 &  76  & 89  & 49\\
& 0.25 &  95  & 95  & 96 &  96 &  91  & 13 \\
& 0.50 & 96  & 96 &  96 &  96 &  96 &   3\\
&0.75 &94  & 96 &  96 &  96  & 85 &  12\\
& 0.85 & 100 & 100 &   2  &  2  & 55  & 53\\
   \hline
  \multirow{2}{*}{5}  &   0.15 &  99 &  100 &  87 &  87  & 93 &  22* \\
     & 0.25 & 99  & 99 &  96 &  96 &  96 &  21* \\
     & 0.50 & 98  & 98  & 98 &  98  & 94  & 29* \\
     & 0.75 & 99 & 100  & 99  & 99  & 90 &  29* \\
     & 0.85 & 99 &  99  & 100 & 100  & 56  & 26* \\
   \hline
\end{tabular}
}
\caption{Coverage (in percent) of 95\% CIs for $F(p)$ of each scenario. ``M'' denotes mixture GFD confidence intervals;  ``C'' denotes conservative GFD confidence intervals;
``g'' denotes Efron's $g$-modeling without bias correction;  ``bc'' denotes Efron's $g$-modeling with bias correction;
``BP'' denotes the bootstrap method; ``BA'' denotes the Bayesian method.
*The Bayesian results for Scenarios 2, 5 are reported based on 495, 476 replications as 5, 24 runs failed because of an error in the R package \texttt{dirichletprocess}.}

\end{table}

\begin{table}[H]
\centering
\scalebox{1.1}{
\begin{tabular}{cccccccc}
  \hline
 Scenario & $p$ & M & C & g & bc & BP & BA \\
  \hline
      \multirow{2}{*}{1} & 0.15 & 27  & 29  & 23 &  23 &  23 &   8\\
&  0.25 &  64  & 69 &  39 &  39 &  40 &  26\\
& 0.50 & 147 & 156  & 86 &  86   &62 &  47\\
&0.75 & 64  & 69 &  35   &35  & 32  & 26\\
& 0.85 & 26 &  29   &15  & 15  & 16 &  8\\
   \hline
    \multirow{2}{*}{2} &0.15 & 69  & 75 &  43 &  43 &  43 &  45*\\
& 0.25 & 132 & 142 &  74 &  74  & 57 & 120*\\
&0.50 & 72 &  76 &  65  & 65 &  62  & 35*\\
& 0.75 & 132 & 142  & 72 &  72 &  51&  121*\\
& 0.85 & 70 &  75  & 35 &  35 &  32  & 45*\\
   \hline
   \multirow{2}{*}{3}  & 0.15 &  7  &  7  & 12 &  12  &  3 &   0.49\\
& 0.25 &  25 &  27  & 24  & 24  & 19  &  3\\
& 0.50 & 92  & 96  & 60 &  60 &  62  &  8\\
& 0.75  &  25  & 27  & 17 &  17  & 18  &  3\\
 & 0.85 & 7  &  7  &  4 &   4  &  3  &  0.49\\
   \hline
 \multirow{2}{*}{4}  &0.15&  99 & 105 &  56  & 56 &  57  & 24\\
& 0.25 & 65 & 67 &  62  & 62  & 62  &  6\\
&0.50&  62 &  63  & 62  & 62 &  62 &   0.02 \\
&0.75&  65 &  67 &  61  & 61  & 62  &  5\\
&0.85& 99  & 105   & 57 &  57  & 55  & 24\\
   \hline
  \multirow{2}{*}{5}  &   0.15 & 81 &  86 &  52  & 52 &  56  &  8* \\
     & 0.25 & 57 &  60 &  44 &  44  & 43  &  6* \\
     & 0.50 & 21 &  22&   20 &  20  & 17  &  3*\\
     & 0.75 &  7 &   8 &  11 &  11  &  5  &  1* \\
     & 0.85 &  5  &  6 &   6  &  6   & 2   & 1* \\
   \hline
\end{tabular}
}
\caption{Mean length ($\times 10^{-3}$) of 95\% CIs for $F(p)$ of each scenario.  ``M'' denotes mixture GFD confidence intervals;  ``C'' denotes conservative GFD confidence intervals;
``g'' denotes Efron's $g$-modeling without bias correction;  ``bc'' denotes Efron's $g$-modeling with bias correction;
``BP'' denotes the bootstrap method; ``BA'' denotes the Bayesian method.
*The Bayesian results for Scenarios 2, 5 are reported based on 495, 476 replications as 5, 24 runs failed because of an error in the R package \texttt{dirichletprocess}.}

\end{table}

\section{Trace plots of the proposed Gibbs sampler for intestinal surgery data}
In this section, we present trace plots of the proposed Gibbs sampler for intestinal surgery data.
As can be seen from these figures, the fiducial MCMC samples have good variability and mix well.

\begin{figure}[H]
\begin{center}
\includegraphics[width=12cm]{mean.pdf}
\includegraphics[width=12cm]{variance.pdf}

\end{center}
\caption{Trace plots of mean and variance of $P\sim [F^L(p)+F^U(p)]/2$ for intestinal surgery data, respectively.}
\end{figure}

\section{Additional plots for intestinal surgery data}
In Figure~\ref{pic:real2}, we plot the point estimates and 95\% confidence intervals of $F(p)$ for the Bayesian method and nonparametric bootstrap, respectively.

\begin{figure}[H]
\begin{center}
\includegraphics[width=5.5cm]{surg3.pdf}
\includegraphics[width=5.5cm]{surg4.pdf}
\end{center}
\caption{Estimated CDF (dashed line) and 95\% CIs for $F(p)$. Left panel: fiducial vs. Bayesian.
Right panel: fiducial vs. nonparametric bootstrap. The red and orange curves are the mixture and conservative confidence intervals, respectively.
The brown and black curves are the Bayesian and bootstrap confidence intervals, respectively.
}
\label{pic:real2}
\end{figure}

\bibliographystyle{asa}
\bibliography{fiducial}

\end{document}

config: arxiv


TITLE: u-substitution for a definite integral
QUESTION [2 upvotes]: I want to solve the following integral:
$$\int_{-2}^2 \sqrt{4- x^2} dx$$
I am thinking of doing a u-substitution of the whole term inside the root. If I do this, I need to change the limits of the integral. Doing so, the new limits would be equal (0=0). Doesn't this mean that the value of the integral would be 0?

REPLY [2 votes]: The integral can be found with the substitution $x=\sin \theta$.
If we let $u=4-x^2$. Then $du=-2xdx$.
Note that $x=\sqrt{4-u}$ if $x\ge 0$ and $x=-\sqrt{4-u}$ if $x<0$.
So, $\displaystyle dx=\frac{du}{-2\sqrt{4-u}}$ if $x\ge 0$ and $\displaystyle dx=\frac{du}{2\sqrt{4-u}}$ if $x\le 0$.
So we actually do not have $\displaystyle \int_0^0 \frac{\sqrt{u}du}{-2\sqrt{4-u}}$.
\begin{align*}
\int_{-2}^2\sqrt{4-x^2}dx&=\int_{-2}^0\sqrt{4-x^2}dx+\int_0^2\sqrt{4-x^2}dx\\
&=\int_0^4 \frac{\sqrt{u}du}{2\sqrt{4-u}}+\int_4^0 \frac{\sqrt{u}du}{-2\sqrt{4-u}}\\
&=\int_0^4 \frac{\sqrt{u}du}{2\sqrt{4-u}}+\int_0^4 \frac{\sqrt{u}du}{2\sqrt{4-u}}\\
&=\int_0^4 \frac{\sqrt{u}du}{\sqrt{4-u}}
\end{align*}
This does not help us to find the integral. But at least the integral is not $0$.


TITLE: How to find the centroid of the intersecting region between three circles of differing diameters
QUESTION [9 upvotes]: This question is a follow-up to this question I asked earlier which deals with finding the midpoint of the intersecting region of two circles of differing diameters. Using the parametric equation of a line as suggested in the accepted answer, it works perfectly. Now I want to take it a step further. In the image below, you'll see I have three circles of differing diameters which all intersect at various points.
I would like to find the centroid of the region at which all three circles overlap. The only information available is the coordinates of the circles' centers and their respective diameters. It doesn't appear to be as straight-forward as the last question, though. The intersection of the circles does not always occur on the line between the centers. I imagine one would need to average the X and Y coordinates of the points of intersection to find it, but I'm not sure how to find the points of intersection in this scenario.
Any thoughts or guidance in the right direction is appreciated.

REPLY [3 votes]: Get the coordinates of the three intersection points between pairs of circles by solving
$$\begin{cases}(x-x_1)^2+(y-y_1)^2=r_1^2\\(x-x_2)^2+(y-y_2)^2=r_2^2\end{cases}$$
(subtract the equations to get a linear one, express $y$ in terms of $x$ ang plug in the first equation to get a quadratic equation in $x$.)
The area of the triangle is given by half the cross-product between two sides. The centroid is at the arithmetic mean of the coordinates.
For each circular segment, you know the radius and the chord length and you easily derive the aperture angle ($2\arcsin l/2r$). Hence the area, which is the difference between a sector and a triangle ($r^2(\theta-\sin\theta)/2$). The centroid is located on the symmetry axis at a distance from the arc center equal to
$$\frac{4r\sin^3\frac\theta2}{3(\theta-\sin\theta)}.$$
See http://mathworld.wolfram.com/CircularSegment.html.
For each circle, find the translation and rotation that brings the segment in the reference position (as in the link). The translation just cancels the center coordinates. The rotation angle is found from the slope of the chord.
Compute the coordinates of the centroids of the segments, counter-rotate them and translate the centers back. This will give you the absolute coordinates of the centroids.
The global centroid is the weighted average of the centroids, where the weights are the areas
$$\bar x=\frac{\bar x_{t}A_t+\bar x_{s_1}A_{s_1}+\bar x_{s_2}A_{s_2}+\bar x_{s_3}A_{s_3}}{A_t+A_{s_1}+A_{s_2}+A_{s_3}}$$
and similar for $y$.

A complete analytical formula is possible but will probably be pretty large.


TITLE: Show that $Cov(c_{11}X_1+c_{12}X_2+\dots,+c_{1p}X_p,c_{21}X_1+c_{22}X_2,\dots c_{2p}X_p)$
QUESTION [1 upvotes]: Show that
  $$Cov(c_{11}X_1+c_{12}X_2+\dots,+c_{1p}X_p,c_{21}X_1+c_{22}X_2,\dots
 c_{2p}X_p)=c_1'\Sigma_Xc_2$$ where $c_1'=[c_{11},\dots,c_{1p}]$ and
   $c_2'=[c_{21},\dots,c_{2p}]$

We have that
$$Cov(c_{11}X_1+c_{12}X_2+\dots,+c_{1p}X_p,c_{21}X_1+c_{22}X_2,\dots
 c_{2p}X_p)=Cov(\sum_{i=1}^pc_{1i}X_i,\sum_{i=1}^pc_{2i}X_i)$$
I'm not sure but I think that last equality is
$$Cov(\sum_{i=1}^pc_{1i}X_i,\sum_{i=1}^pc_{2i}X_i)=\sum_{i=1}^p\sum_{j=1}^pc_{1i}c_{2i}Cov(X_i,X_j)\textbf{(*)}$$
and
$$c'_1\Sigma_X=\begin{bmatrix}c_{11}&\dots&c_{1p}\end{bmatrix}\times\begin{bmatrix}\sigma_{11}&\dots&\sigma_{1p}\\\sigma_{21}&\dots&\sigma_{2p}\\\dots&\dots&\dots\\\sigma_{p1}&\dots&\sigma_{pp}\end{bmatrix}$$
$$=\begin{bmatrix}\sum_{i=1}^p c_{1i}\sigma_{i1}&\dots,& \sum_{i=1}^p c_{1i}\sigma_{ip}\end{bmatrix}$$
then
$$c_1'\Sigma_X c_2=\begin{bmatrix}\sum_{i=1}^p c_{1i}\sigma_{i1}&\dots,& \sum_{i=1}^p c_{1i}\sigma_{ip}\end{bmatrix}\times \begin{bmatrix}c_{21}\\\dots\\c_{2p}\end{bmatrix}$$
$$=\sum_{i=1}^p c_{1i}c_{21}\sigma_{i1}+\dots +\sum_{i=1}^p c_{
1i}c_{2p}\sigma_{ip}$$
but I'm not sure if it is equal to (*).
It make any sense ?

REPLY [1 votes]: $Cov \left( \sum \limits_{i=1}^{p}{c_{1i}X_{i}},\sum\limits_{i=1}^{p}{c_{2i}X_{i}} \right)$
$=E\left[\left(\sum \limits_{i=1}^{p}{c_{1i}X_{i}}\right).\left(\sum \limits_{i=1}^{p}{c_{2i}X_{i}}\right)\right]-E\left[\sum \limits_{i=1}^{p}{c_{1i}X_{i}}\right].E\left[\sum \limits_{i=1}^{p}{c_{2i}X_{i}}\right]$, (by definition of covariance)
$=E[\sum \limits_{i=1}^{p}\sum \limits_{j=1}^{p}{c_{1i}.c_{2j}X_{i}X_{j}}]-\left(\sum \limits_{i=1}^{p}{c_{2i}}E[X_{i}]\right).\left(\sum \limits_{i=1}^{p}{c_{2i}}E[X_{i}]\right)$, by linearity fo expectaion
$=\sum \limits_{i=1}^{p}\sum \limits_{j=1}^{p}{c_{1i}c_{2j}E[X_{i}X_{j}}]-\sum \limits_{i=1}^{p}\sum \limits_{j=1}^{p}{c_{1i}c_{2j}}E[X_{i}]E[X_{j}]$, by linearity fo expectaion
$=\sum \limits_{i=1}^{p}\sum \limits_{j=1}^{p}{c_{1i}c_{2j}(E[X_{i}X_{j}}]-E[X_{i}]E[X_{j}])$
=$  \sum \limits _{i=1}^{p} {c_{1i}.c_{21}.(E[X_iX_1]-E[X_i]E[X_1])} + \sum \limits _{i=1}^{p} {c_{1i}.c_{22}.(E[X_iX_2]-E[X_i]E[X_2])} + \ldots +
\sum \limits _{i=1}^{p} {c_{1i}.c_{2p}.(E[X_iX_p]-E[X_i]E[X_p])}$
(expanding the 2nd sum)
$=\begin{bmatrix}
  \sum \limits _{i=1}^{p} {c_{1i}.(E[X_iX_1]-E[X_i]E[X_1])} & \sum \limits _{i=1}^{p} {c_{1i}.(E[X_iX_2]-E[X_i]E[X_2])} & \ldots & \sum \limits _{i=1}^{p} {c_{1i}.(E[X_iX_p]-E[X_i]E[X_p])}  \end{bmatrix}.
\begin{bmatrix}
   c_{21} \\
   c_{22} \\
   \ldots \\
    c_{2p}
\end{bmatrix}$ (in matrix form)
$=\begin{bmatrix}
   c_{11} \\
   c_{12} \\
   \ldots \\
    c_{1p}
  \end{bmatrix}^T.
  \begin{bmatrix}
   E[X_1^2]-E[X_1]^2 & E[X_1X_2]-E[X_1]E[X_2] & \ldots & E[X_1X_p]-E[X_1]E[X_p]\\
    E[X_2X_1]-E[X_2]E[X_1] & E[X_2^2]-E[X_2]^2 & \ldots & E[X_2X_p]-E[X_2]E[X_p]\\
    \ldots & \ldots & \ldots & \ldots \\
    E[X_pX_1]-E[X_p]E[X_1] & E[X_pX_2]-E[X_pX_2] & \ldots & E[X_p^2]-E[X_p]^2  \end{bmatrix}.
\begin{bmatrix}
   c_{21} \\
   c_{22} \\
   \ldots \\
    c_{2p}
  \end{bmatrix}$
=$c_1^{\prime}.\Sigma_X.c_2$,
where $\Sigma_X(i,j)=E[X_{i}X_{j}]-E[X_{i}]E[X_{j}],\;\forall{i},\forall{j}$


TITLE: Partial differential equation - integral surface
QUESTION [1 upvotes]: I have to find the integral surface of the PDE
$$x(y^2+z)p - y(x^2+z)q = (x^2-y^2)z$$
containing the straight line $x+y=0, z=1$.
From the auxiliary equations, I can obtain $xyz=C_{1}$ for some constant $C_{1}$. To proceed further, I need to find another relation involving $x,y,z$ or at least one of them. However, I am having difficulty finding this. Would someone be able to help me determine the other relation?

REPLY [1 votes]: $x(y^2+z)p - y(x^2+z)q = (x^2-y^2)z$ is ambiguous because the symbols are not defined. I suppose that the PDE is :
$$x(y^2+z)\frac{\partial z}{\partial x} - y(x^2+z)\frac{\partial z}{\partial y} = (x^2-y^2)z(x,y)\tag 1$$
The Charpit-Lagrange characteristic ODEs are :
$$\frac{dx}{x(y^2+z)}=\frac{dy}{-y(x^2+z)}=\frac{dz}{(x^2-y^2)z}$$
Search for a first characteristic equation :
$$\frac{ydx+xdy}{y(x(y^2+z))+x(-y(x^2+z))}=\frac{dz}{(x^2-y^2)z}=\frac{d(xy)}{xy(y^2-x^2)}$$
$$\frac{dz}{z}=-\frac{d(xy)}{xy}$$
$$xyz=c_1$$
Seach for a second characteristic equation :
$$\frac{xdx+ydy}{x(x(y^2+z))+y(-y(x^2+z))}=\frac{dz}{(x^2-y^2)z}=\frac{d(x^2+y^2)/2}{(x^2-y^2)z}$$
$$dz=d(x^2+y^2)/2$$
$$z-\frac12 (x^2+y^2)=c_2$$
The general solution of the PDE on implicit form $c_2=F(c_1)$ is :
$$z-\frac12 (x^2+y^2)=F(xyz)\tag 2$$
$F$ is an arbitrary function (to be determined according to the boundary condition).
Condition : $z=1$ on $y=-x$
$$1-\frac12 (x^2+(-x)^2)=F(x(-x))$$
$$F(-x^2)=1-x^2$$
Let $X=-x^2$
$$F(X)=1+X$$
Now the function $F$ is determined. We put it into the general solution Eq.$(2)$ where $X=xyz$
$$z-\frac12 (x^2+y^2)=1+xyz$$
$$z(x,y)=\frac{1+\frac12 (x^2+y^2)}{1-xy}$$


TITLE: Subspace of Set of n by n Matrices Proof
QUESTION [0 upvotes]: I am wondering whether this proof of a set being a vector space is valid. I am allowed to assume that $M_{n,n}$ is a vector space over the field $\mathbb{R}$, and basic properties of matrices.
let A be a fixed matrix $\in M_{n,n}$. Show that $S = \{X\in M_{n,n}: AXA = X\}$ is a vector space.
$\underline{\text{Proof}}$
We assume that $M_{n,n}$ is a vector space over the field $\mathbb{R}$, and the basic properties of matrices.
We know that $S \subset M_{n,n}$ by definition of $S$, and so it suffices to show that $S$ is a subspace of $M_{n,n}$.
Now, consider the equation
\begin{align}
A\mathbf{0}A &= (A\mathbf{0})A \\
&= \mathbf{0}A \\
&= \mathbf{0}
\end{align} by associativity of matrix multiplication and properties of the n by n zero matrix.
Hence, $\mathbf{0} \in S$. Therefore $S$ is non-empty.
Now, let $X,Y \in S$. This implies that
\begin{align}
AXA &= X \\
AYA &= Y
\end{align}
Hence, $$X + Y = AXA + AYA = A(XA + YA) = A(X + Y)A$$ This implies that $X + Y \in S$. Therefore $S$ is closed under matrix addition.
Now, let $\alpha \in \mathbb{R}$ and $X \in S$. This implies that
\begin{align}
AXA &= X \\
\Rightarrow \alpha(AXA) &= \alpha X \\
\Leftrightarrow A(\alpha X)A &= \alpha X
\end{align}
Hence, $\alpha X \in S$. Therefore $S$ is closed under scalar multiplication.
Hence, by the Subspace Theorem, S is a subspace of $M_{n,n}$. Therefore $S$ is a vector space. This completes the proof.
I am not so sure about this step:
$\alpha(AXA) = \alpha X \\
\Leftrightarrow A(\alpha X)A = \alpha X$
Would this be considered a basic property of matrices that can be assumed?
Edit: My question has been resolved.

REPLY [0 votes]: Yes$$\alpha(AXA) =
 A(\alpha X)A$$
Therefore $$\alpha(AXA) = \alpha X \\
\Leftrightarrow A(\alpha X)A = \alpha X$$
Is correct.