text:
The proofs of both \Cref{thm:main2} $(i)$ and \Cref{thm:main} $(i)$, that treat the case $\beta>0$, are based on distributional limits for the empirical $U$-quantile processes $\mathbb{U}_n^{-1}\coloneqq\sqrt{n}\big(U_n^{-1}\allowdisplaybreaks -U^{-1}\big) $ and $\mathbb{V}_m^{-1}\coloneqq\sqrt{m} \big(V_m^{-1}-V^{-1} \big)$ in $\ell^\infty[\beta,1-\beta]$. These limits are derived in Section B.2 of \citet{supplement} using the Hadamard differentiability of the inversion functional $\phi_{inv}:F\mapsto F^{-1}$ regarded as a map from the set of restricted distribution functions into the space $\ell^\infty[\beta,1-\beta]$ \cite[Lemma 3.9.23]{vaartWeakConvergenceEmpirical1996}. Once the distributional limits of $\mathbb{U}_n^{-1}$ and $\mathbb{V}_m^{-1}$ are established, both \Cref{thm:main2} $(i)$ and \Cref{thm:main} $(i)$ follow by standard arguments (see Section B.2 and Section B.4 of \cite{supplement}). Therefore, we focus in the following on the proofs of \Cref{thm:main2} $(ii)$, \Cref{thm:finite sample bound for p=2} and \Cref{thm:main} $(ii)$, which are concerned with the case $\beta=0$. As the proof of \Cref{thm:finite sample bound for p=2} is essential for the derivation of the second part of \Cref{thm:main2}, we prove it first.
\subsection{Proof of Theorem \ref*{thm:finite sample bound for p=2}}\label{subsec:proof of the finite sample boud}
 The key idea is to exploit the dependency structure of the samples $\left\lbrace \dX\left(X_i,X_j \right) \right\rbrace_{1\leq i <j \leq n} $ and $\left\lbrace \dY\left(Y_k,Y_l \right) \right\rbrace_{1\leq k <l \leq m}$. Let us consider the set $\left\lbrace \dX\left(X_i,X_j \right) \right\rbrace_{1\leq i <j \leq n} $. Since $X_1,\dots,X_n$ are independent, it follows that the random variable $\dX(X_i,X_j)$ is independent of $\dX(X_{i'},X_{j'})$, whenever $i,j,i',j'$ are pairwise different. This allows us to divide the sample into relatively large groups of independent random variables. This idea is represented in \Cref{fig:idearepresentation}. It highlights a possibility to divide the set $\left\lbrace \dX\left(X_i,X_j \right) \right\rbrace_{1\leq i <j \leq 6} $ into five sets of three independent distances.
\begin{figure}\centering
	\begin{subfigure}[c]{0.2\textwidth}
		\centering
		\includegraphics[width = \textwidth,height=\textwidth]{Graphics/full.png}
	\end{subfigure}
\hspace*{0.5cm}
	\begin{subfigure}[c]{0.2\textwidth}
		\centering
		\includegraphics[width = \textwidth, height=\textwidth]{Graphics/fullonedist2.png}
	\end{subfigure}
\hspace*{0.5cm}
	\begin{subfigure}[c]{0.2\textwidth}
		\centering
		\includegraphics[width = \textwidth, height=\textwidth]{Graphics/fulltwodist2.png}
	\end{subfigure}\vspace*{2mm}\\
	\begin{subfigure}[c]{0.2\textwidth}
		\centering
	\includegraphics[width = \textwidth,height=\textwidth]{Graphics/secondfull.png}
\end{subfigure}
\hspace*{0.5cm}
\begin{subfigure}[c]{0.2\textwidth}\centering
	\includegraphics[width = \textwidth, height=\textwidth]{Graphics/secondonedist2.png}
\end{subfigure}
\hspace*{0.5cm}
\begin{subfigure}[c]{0.2\textwidth}\centering
	\includegraphics[width = \textwidth, height=\textwidth]{Graphics/secondtwodist2.png}
\end{subfigure}
\caption{\textbf{Partitioning the Distances:} Illustration how to partition the set $\left\lbrace \dX\left(X_i,X_j \right) \right\rbrace_{1\leq i <j \leq 6} $ successively into five set of independent distances of size three. Top row: Left: All pairwise distances between six points. Middle: All distances (blue) that are independent of one chosen distance (red). Right: All distances that are independent of two chosen, independent distances (same color code, right) are shown. Bottom row: The same selection process for the set, where the independent distances displayed in the top right plot were removed.}\label{fig:idearepresentation}
\end{figure}
Generally, one can prove the following (see Section B.3 of the supplementary material \cite{supplement}).

\begin{lemma}\label{lem:Ustatsamplesplitup}
	Let $n\geq 3$ and let $X_1,\dots,X_n\iid\muX$. If $n$ is even, there exists a partition $\left\lbrace \Pi^n_k\right\rbrace_{1\leq k\leq n-1}$ of $\left\lbrace(i,j) \right\rbrace_{1\leq i<j\leq n}$ such that $|\Pi^n_k|=n/2$ for each $k$ and such that the random variables in the set $\left\lbrace \dX(X_i,X_j)\right\rbrace_{(i,j)\in\Pi^n_k}$ are independent, $1\leq k\leq n/2$. If $n$ is odd, there exists a partition $\left\lbrace \Pi^n_k\right\rbrace_{1\leq k\leq n}$ of $\left\lbrace(i,j) \right\rbrace_{1\leq i<j\leq n}$ such that $|\Pi^n_k|=(n-1)/2$ for each $k$ and such that the random variables in the set $\left\lbrace \dX(X_i,X_j)\right\rbrace_{(i,j)\in\Pi^n_k}$ are independent, $1\leq k\leq n/2$.
\end{lemma}
The idea of the proof of \Cref{thm:finite sample bound for p=2} is now to write the problem at hand as a certain assignment problem and then to restrict the assignments to assignments between groups of independent distances.

	\noindent We observe that for $\beta\in[0,1/2)$
	\[\Eargs{ \dodstatnn }\leq\Eargs{\int\limits_{0}^{1}\!|U_n^{-1}(t)-V_m^{-1}(t)|^2 \, dt }=\Eargs{\Wtwo{\muUn}{\muVm}},\]
	where $\muUn$ and $\muVm$ are the empirical measures corresponding to $U_n$ and $V_m$, i.e., for $A\in\borelsetsR$
	\[\muUn(A)=\frac{2}{n(n-1)}\sum_{1 \leq i <j\leq n}\indifunc{\dX(X_i,X_j)\in A}\]
	and
	\[\muVm(A)=\frac{2}{m(m-1)}\sum_{1 \leq k <l\leq m}\indifunc{\dY(Y_k,Y_l)\in A}.\]
	Since it holds $\muU=\mu^V$ by assumption, we obtain by the triangle inequality
	\[\Eargs{\Wtwo{\muUn}{\muVm}}\leq2\left( \Eargs{\Wtwo{\muUn}{\muU}}+\Eargs{\Wtwo{\muVm}{\mu^V}}\right). \]
	Thus, it remains to show
\begin{equation}\label{eq:aimfinitesamplebound}
	\Eargs{\Wtwo{\muUn}{\muU}}\leq\frac{4}{n+1}J_2\left(\muU\right).
\end{equation}
In order to demonstrate \eqref{eq:aimfinitesamplebound}, we will use \Cref{lem:Ustatsamplesplitup}. Hence, it is notationally convenient to distinguish the cases $n$ even and $n$ odd, although the proof is essentially the same in both settings. In the following, we will therefore restrict ourselves to $n$ odd.

\noindent The first step to prove \eqref{eq:aimfinitesamplebound} is to realize (cf. \citet[Sec. 4]{bobkov2016one}), that
		\begin{equation}\label{eq:firstU-quantileLpnorm estimate}
\Eargs{\Wtwo{\muUn}{\muU}}\leq\Eargs{\Wtwo{\muUn}{\nuUn}},
\end{equation}
where $\nuUn$ denotes an independent copy of $\muUn$, i.e.,
\[\nuUn(A)=\frac{2}{n(n-1)}\sum_{1 \leq i <j\leq n}\indifunc{\dX(X_i',X_j')\in A},\]
for $X_1',\dots,X_n'\iid \muX$. By \Cref{lem:Ustatsamplesplitup}, there exists a partition $\Pi^n_1,\dots,\Pi^n_{n}$ of the index set $\left\lbrace (i,j)\right\rbrace_{1\leq i<j\leq n}$ with $|\Pi^n_k|=(n-1)/2$, $1\leq k\leq n$, such that the random variables in the sets $\{\dX(X_i,X_j)\}_{(i,j)\in\Pi^n_k}$ and the ones in the sets $\{ \dX(X_i',X_j')\}_{(i,j)\in\Pi^n_k}$ are independent. Let $\left\lbrace d_{(i)}^{\Pi^n_k,X}\right\rbrace_{1\leq i\leq(n-1)/2}$ stand for the ordered sample of $\lbrace \dX(X_i,X_j )\rbrace_{(i,j)\in\Pi^n_k}$ and the let $\Big\lbrace d_{(i)}^{\Pi^n_k,X'}\Big\rbrace_{1\leq i\leq(n-1)/2}$ stand for the one of $\left\lbrace \dX\left(X_i',X_j' \right)\right\rbrace_{(i,j)\in\Pi^n_k}$, $1\leq k\leq n$. The application of Corollary B.12 with this partition yields that
\begin{align*}
\Eargs{\Wtwo{\muUn}{\nu^U_n}}\leq&\frac{2}{n(n-1)}\Eargs{\sum_{k=1}^{n}\sum_{i=1} ^{(n-1)/2}\left| d^{\Pi^n_k,X}_{(i)}-d^{\Pi^n_k,X'}_{(i)}\right|^2}.
\end{align*}
Furthermore, we realize that, as $X_1,\dots,X_n,X_1',\dots,X_n'$ are independent, identically distributed, it holds for $1\leq k,l\leq n$ that
\begin{equation*}
\sum_{i=1} ^{(n-1)/2}\left| d^{\Pi^n_k,X}_{(i)}-d^{\Pi^n_k,X'}_{(i)}\right|^2\overset{D}{=}	\sum_{i=1} ^{(n-1)/2}\left| d^{\Pi^n_l,X}_{(i)}-d^{\Pi^n_l,X'}_{(i)}\right|^2.
\end{equation*}
Consequently, we have
\begin{align*}
\Eargs{\sum_{k=1}^{n}\sum_{i=1} ^{(n-1)/2}\left| d^{\Pi^n_k,X}_{(i)}-d^{\Pi^n_k,X'}_{(i)}\right|^2}=n\Eargs{\sum_{i=1} ^{(n-1)/2}\left| d^{\Pi^n_1,X}_{(i)}-d^{\Pi^n_1,X'}_{(i)}\right|^2}.
\end{align*}
We come to the final step of this proof. Let for any $A\in\borelsetsR$
\[\mu_n^*(A)=\frac{2}{n-1}\sum_{i=1} ^{(n-1)/2}\indifunc{d_{(i)}^{\Pi^n_1,X}\in A}\]
and let $\nu_n^*(A)$ be defined analogously.

Then, Theorem 4.3 of \citet{bobkov2016one} implies that
\[\Eargs{\Wtwo{\mu_n^*}{\nu_n^*}}=\frac{2}{n-1}\Eargs{\sum_{i=1} ^{(n-1)/2}\left| d^{\Pi^n_1,X}_{(i)}-d^{\Pi^n_1,X'}_{(i)}\right|^2}.\]
By construction, the samples $\left\lbrace d_{i}^{\Pi^n_1,X}\right\rbrace_{1\leq i\leq (n-1)/2}$ and $\left\lbrace d_{i}^{\Pi^n_1,X'}\right\rbrace_{1\leq i\leq (n-1)/2}$ consist of independent random variables and are independent of each other. Furthermore, we have $\Eargs{\mu_n^*}=\Eargs{\nu_n^*}=\muU$. Since $J_2\left(\mu^U\right)<\infty$ by assumption, it follows by Theorem 5.1 of \citet{bobkov2016one} that
\begin{align*}
\frac{2}{n-1}\Eargs{\sum_{i=1} ^{(n-1)/2}\left| d^{\Pi^n_1,X}_{(i)}-d^{\Pi^n_1,X'}_{(i)}\right|^2}=\Eargs{\Wtwo{\mu_n^*}{\nu_n^*}}\leq \frac{4}{n+1}J_2\left(\mu^U\right).
\end{align*}
This yields \eqref{eq:aimfinitesamplebound} and thus concludes the proof. \hfill\qed

\subsection{Proof of Theorem \ref*{thm:main2} \texorpdfstring{$(ii)$}{(ii)}}

\noindent For notational convenience we restrict ourselves from now on to the case $n=m$. However, the same strategy of proof also gives the general case $n\neq m$ (for some additional details on this issue see \cite[Sec. B.2.2]{supplement}). We first demonstrate that $\Big\{\XiUVn (\beta)\,|\,\beta\in[0,1/2]\Big\}_{n\in\mathbb{N}}\subset\left(C[0,1/2],||\cdot||_\infty\right),$ where
		\begin{align*}
		\XiUVn(\beta)=\frac{n}{2}\int_{\beta}^{1-\beta}\!\left( U_n^{-1}(t)-V_n^{-1}(t)\right)^2 \, dt,
	\end{align*}
	is tight. Under \Cref{cond:secondcondition}, which implies \Cref{cond:firstcondition} for $\beta\in (0,1)$, we already have by \Cref{thm:main2} $(i)$ that $\XiUVn(\beta)\rightsquigarrow\Xi(\beta)$ for all $\beta\in(0,1/2)$. In order to prove tightness of the sequence $\left\{\XiUVn\right\}_{n\in\mathbb{N}}\subset C[0,1/2]$ we process the subsequent steps:
	\begin{enumerate}
		\item Show that the sequence of real valued random variables $\left\lbrace \XiUVn(0)\right\rbrace_{n\in\N}$ is tight;
		\item Control the following expectations for small $\beta$
		\[	\Eargs{\int_{0}^{\beta}\! \left(U_n^{-1}(t)-V_n^{-1}(t)\right)^2\,dt } \text{  and  } \Eargs{\int_{1-\beta}^{1}\! \left(U_n^{-1}(t)-V_n^{-1}(t)\right)^2\,dt }.\]
	\end{enumerate}
While the first step of the above strategy directly follows by a combination of Lemma B.6 and \Cref{thm:finite sample bound for p=2}, we have to work for the second. Using a technically somewhat more involved variation of the partitioning idea of the proof of \Cref{thm:finite sample bound for p=2}, we can demonstrate the subsequent bounds.
\begin{lemma}\label{lemma:integrated difference for small beta}
	Suppose \Cref{cond:setting} and \Cref{cond:secondcondition} are met. Let $\muU=\mu^V$, let $n\geq 100$, let $0\leq\beta=\beta_n<1/6$ and let $n\beta > 8$. Then, it follows that
	\begin{align}\label{eq:important estimate1}
	&\Eargs{\int_{0}^{\beta}\! \left(U_n^{-1}(t)-V_n^{-1}(t)\right)^2\,dt } \leq \frac{2C_1}{n-1}\left( 4\beta\left( 1+2\frac{\sqrt{\log\left( n\right)}}{\sqrt{n }}\right)\right)^{2\gamma_1+2}+o\left(n^{-1}\right)
	\end{align}
	as well as
	\begin{align}\label{eq:important estimate2}
	&\Eargs{\int_{1-\beta}^{1}\! \left(U_n^{-1}(t)-V_n^{-1}(t)\right)^2\,dt } \leq \frac{2C_2}{n-1}\left( 4\beta\left( 1+2\frac{\sqrt{\log\left( n\right)}}{\sqrt{n }}\right)\right)^{2\gamma_2+2} +o\left(n^{-1}\right),
	\end{align}
	where $C_1$ and $C_2$ denote finite constants that are independent of $\beta$.
\end{lemma}
Based on this, we prove the subsequent technical lemma.
\begin{lemma}\label{lemma:Stetigkeitsmodul}
	Under \Cref{cond:secondcondition}, it holds for all $\epsilon>0$ that
	\[\lim_{\delta\to0^+}\limsup_{n\to\infty}\prob\left(\omega(\XiUVn,\delta)>\epsilon)\right)=0,\]
	where $\omega(\cdot,\cdot)$ is defined for $f:[0,1/2]\to\R$ and $0<\delta\leq 1/2$ as $\omega(f,\delta)=\sup_{|s-t|\leq \delta}|f(s)-f(t)|$.
\end{lemma}
\begin{proof}
	Let $0<\delta<1/20$ and let $0\leq s, t\leq1/2$. We have that
	\begin{align}
	&\prob\left( \omega(\XiUVn,\delta)>\epsilon\right)= \prob\left( \sup_{|s-t|\leq \delta}\left|\XiUVn(s)-\XiUVn(t) \right| >\epsilon\right)\nonumber\\
	 = &\prob\left( \sup_{\substack{|s-t|\leq \delta, \\ t< 2\delta}}\left|\XiUVn(s)-\XiUVn(t) \right| >\epsilon \text{ or }  \sup_{\substack{|s-t|\leq \delta,\\ t\geq 2\delta}}\left|\XiUVn(s)-\XiUVn(t) \right| >\epsilon\right)\nonumber\\
	\leq& \prob\left( \sup_{\substack{|s-t|\leq \delta, \\ t< 2\delta}}\left|\XiUVn(s)-\XiUVn(t) \right| >{\epsilon} \!\right)\!+\!\prob\left(\sup_{\substack{|s-t|\leq \delta, \\ t\geq 2\delta}}\left|\XiUVn(s)-\XiUVn(t) \right| >{\epsilon}\!\right)\nonumber\\
	=& \text{ \textbf{I}}+\text{\textbf{II}}\label{eq:twosummands}.
	\end{align}
	In the following, we consider both summands separately.\vspace*{2mm}

	\noindent\textit{First Summand:} Since $\XiUVn(\beta)$ is monotonically decreasing in $\beta$,
	\begin{align*}
	\text{ \textbf{I}}\leq& 	\prob\left( \XiUVn(0)-\XiUVn(3\delta) >{\epsilon} \right)\\
	=&	\prob\left( \int_{0}^{3\delta}\!|\UV(s)|^2\diff s+ \int_{1-3\delta}^{1}\!|\UV(s)|^2\diff  s >{\epsilon} \right),
	\end{align*}
	where $\UV=\sqrt{\frac{n}{2}}\left(\qu-\qv\right)$. Further, we obtain that
	\begin{align*}
	&\prob\left( \int_{0}^{3\delta}\!|\UV(s)|^2\diff s+ \int_{1-3\delta}^{1}\!|\UV(s)|^2\diff  s >{\epsilon} \right)\\

	\leq &\,\frac{2}{\epsilon} \left(\Eargs{ \int_{0}^{3\delta}\!|\UV(s)|^2\diff s}+   \Eargs{ \int_{1-3\delta}^{1}\!|\UV(s)|^2\diff  s  }\right).
	\end{align*}

	As $3\delta<1/6$, we can conclude with \Cref{lemma:integrated difference for small beta} that
	\begin{align*}
	&\limsup_{n\to\infty}\frac{2}{\epsilon}\mathbb{E}\left[ \int_{0}^{3\delta}\!\left|\UV(s)\right|^2\diff s\right]\\

	\leq& \limsup_{n\to\infty} \frac{n}{\epsilon}\left( \frac{2C_1}{n-1}\left( 6\delta\left( 1+2\frac{\sqrt{\log\left( n\right)}}{\sqrt{n }}\right)\right)^{2\gamma_1+2} +o\left(n^{-1}\right)\right)
	\leq C_2\delta^{2\gamma_1+2}.
	\end{align*}
	Here, $C_1$ and $C_2$ denote finite constants that are independent of $\delta$. Similarly,
	\[\limsup_{n\to\infty}\Eargs{ \int_{1-3\delta}^{1}\!|\UV(s)|^2\diff  s  }\leq C_3\delta^{2\gamma_2+2},\]
	where $C_3$ denotes a finite constant independent of $\delta$. Since we have by assumption that $\gamma_1,\gamma_2>-1$, i.e., $2\gamma_1+2>0$ and $2\gamma_2+2>0$, it follows that
	\[\lim_{\delta\to0+}\limsup_{n\to\infty}\prob\left( \sup_{\substack{|s-t|\leq \delta, \\ t< 2\delta}}\left|\XiUVn(s)-\XiUVn(t) \right| >{\epsilon} \right)=0.\]
	\textit{Second Summand:} In order to handle \textbf{II} in \eqref{eq:twosummands}, we want to make use of the fact that for $\delta >0$ the process $\UV=\sqrt{\frac{n}{2}}(U_n^{-1}-V_n^{-1})$ converges to a Gaussian process in $\ell^\infty[\delta,1-\delta]$ given \Cref*{cond:secondcondition} (see Lemma B.7 of \citet{supplement}). To this end, we verify that the function
	\begin{align*}&\Upsilon:\ell^\infty[\delta,1-\delta]\times \ell^\infty[\delta,1-\delta] \to\R,\\&(f,g)\mapsto \sup_{\substack{|s-t|\leq \delta, \\ t\geq 2\delta}}\left|\int_{s}^{1-s}\!f^2(x) \diff x -\int_{t}^{1-t}\! g^2(x)\diff x \right| \end{align*}
	is continuous.

	\noindent Let in the following $||\cdot||_\infty$ denote the norm of $\ell^\infty[\delta,1-\delta]$. Let $\left((f_n,g_n)\right)_{n\in\N}\subset \ell^\infty[\delta,1-\delta]\times \ell^\infty[\delta,1-\delta]$ be a sequence such that $(f_n,g_n)\to(f,g)$ with respect to the product norm $||(f,g)||\coloneqq||f||_\infty+||g||_\infty$. Then, the inverse triangle inequality yields
	\begin{align*}
	&\left| \Upsilon(f_n,g_n)-\Upsilon(f,g)\right|\\
	\leq& \sup_{\substack{|s-t|\leq \delta, \\ t\geq 2\delta}} \Bigg|\int_{s}^{1-s}\! f_n^2 (x)-f^2(x)\diff x -\int_{t}^{1-t}\! g_n^2(x)-g^2(x) \diff x \Bigg|\\

	\leq&  \sup_{\substack{|s-t|\leq \delta, \\ t\geq 2\delta}} \left|\int_{s}^{1-s}\! \left\|  f_n +f\right\|_\infty \left\|  f_n -f\right\|_\infty \diff x +\int_{t}^{1-t}\! \left\| g +g_n\right\| \left\| g-g_n\right\|_\infty  \diff x \right|\\
	\leq& (1-\delta)\max\left\lbrace  \left\|f_n\right\|_\infty +\left\|f\right\|_\infty,~ \left\|g_n\right\|_\infty +\left\|g\right\|_\infty \right\rbrace \left\|\left( f-f_n, g-g_n\right)\right\|\overset{n\to\infty}{\to}0.
	\end{align*}
	Thus, we have shown $\lim_{n\to\infty}\Upsilon(f_n,g_n)=\Upsilon(f,g)$, i.e., that $\Upsilon$ is sequentially continuous. Hence, a combination of Lemma B.7 and the Continuous Mapping Theorem \citep[Thm. 1.3.6]{vaartWeakConvergenceEmpirical1996} yields that
	\[\Upsilon\left( \UV,\UV\right) \rightsquigarrow \Upsilon\left( \G,\G\right),\]
	where $\G$ denotes the centered Gaussian process defined in Theorem 2.5. Furthermore, Lemma B.8 shows that $\XiUVn(\beta)$ is measurable for $\beta\in[\delta,1-\delta]$ and $n\in\N$. As $\XiUVn$ is continuous in $\beta$ this induces the measurability of $\Upsilon\left( \UV,\UV\right)$ for $n\in\N$. Thus, we find that
	\[\Upsilon\left( \UV,\UV\right) \Rightarrow\Upsilon\left( \G,\G\right).\]
	Let $A=[\epsilon,\infty)\subset\R$. Then, the set $A$ is closed and $(\epsilon,\infty)\subset A$. Hence, an application of the Portmanteau-Theorem \citep[Thm. 2.1]{BillingsleyConvergenceProbabilityMeasures2013} yields that
	\begin{align*}
	\limsup_{n\to\infty} \prob\left(\!\sup_{\substack{|s-t|\leq \delta, \\ t\geq 2\delta}}\left|\XiUVn(s)-\XiUVn(t) \right| >{\epsilon}\!\right)&\leq\limsup_{n\to\infty}\prob\left( \Upsilon\left( \UV,\!\UV\right) \!\! \in\! A\right)\\&\leq\prob\left( \Upsilon\left( \G,\G\right)  \geq{\epsilon}\right).
	\end{align*}
	Next, we remark that
	\[\sup_{\substack{|s-t|\leq \delta, \\ t\geq 2\delta}}\left|\XiUVn(s)-\XiUVn(t) \right|=\sup_{\substack{|s-t|\leq \delta, \\ s\leq t, t\geq 2\delta}}\left|\XiUVn(s)-\XiUVn(t) \right|.\]
	Hence, we can assume for the treatment of this summand that $s\leq t$. With this, we obtain that
	\begin{align*}
	\limsup_{n\to\infty} \,&\prob\left(\sup_{\substack{|s-t|\leq \delta, \\ t\geq 2\delta}}\left|\XiUVn(s)-\XiUVn(t) \right| >{\epsilon}\right)\\
	\leq&  \prob\left(\sup_{\substack{|s-t|\leq \delta, \\ t\geq 2\delta}}\left|\int_{s}^{1-s}\!\G^2(x) \diff x -\int_{t}^{1-t}\! \G^2(x)\diff x \right|\geq {\epsilon}\right)\\

	\leq&\prob\left(\sup_{\substack{|s-t|\leq \delta, \\ t\geq 2\delta}}\int_{s}^{t}\!\G^2(x) \diff x \geq \frac{\epsilon}{2}\right)+\prob\left(\sup_{\substack{|s-t|\leq \delta, \\ t\geq 2\delta}}\int_{1-t}^{1-s}\!\G^2(x) \diff x \geq\frac{\epsilon}{2}\right).
	\end{align*}
	In the following, we focus on the first term. As $0<\delta<1/20$ and $ t\leq1/2$, it holds
	\begin{align*}
	&\prob\left(\sup_{\substack{|s-t|\leq \delta, \\ t\geq 2\delta}}\int_{s}^{t}\!\G^2(x) \diff x \geq \frac{\epsilon}{2}\right)

	\!\leq\!  \prob\left(\sup_{\substack{|s-t|\leq \delta, \\ t\geq 2\delta}} \left( \sup_{x\in[\delta,1-\delta]}\G^2(x) \right)(t-s) \geq \frac{\epsilon}{2}\right)\\
	=& \prob\left( \sup_{x\in[\delta,1-\delta]}\G^2(x)  \geq \frac{\epsilon}{\delta}\right)
	\leq\sqrt{\frac{\delta}{\epsilon}}\Eargs{\sup_{x\in[\delta,1-\delta]}\left| \G(x)\right| }.
	\end{align*}
	By Lemma D.11 the Gaussian process $\G$ is continuous on $[\delta,1-\delta]$ under the assumptions made, i.e., almost surely bounded on $[\delta,1-\delta]$. Thus, Theorem 2.1.1 of \citet{AdlerRandomFieldsGeometry2007} ensures that $\Eargs{\sup_{x\in[\delta,1-\delta]}\left| \G(x)\right| }<\infty$. Hence, we find that
	\begin{align*}
	\lim_{\delta\to0+} \prob\left(\sup_{\substack{|s-t|\leq \delta, \\ t\geq 2\delta}}\int_{s}^{t}\!\G^2(x) \diff x \geq \frac{\epsilon}{2}\right)\leq \lim_{\delta\to0+}\sqrt{\frac{\delta}{\epsilon}}\Eargs{\sup_{x\in[\delta,1-\delta]}\left| \G(x)\right| }=0.
	\end{align*}
Analogously,
	\[\lim_{\delta\to0+}\prob\left(\sup_{\substack{|s-t|\leq \delta, \\ t\geq 2\delta}}\int_{1-t}^{1-s}\!\G^2(x) \diff x >\frac{\epsilon}{2}\right)=0.\]
	This concludes the treatment of the second summand in \eqref{eq:twosummands}.\vspace*{2mm}

	\noindent Combining the results for \textbf{I} and \textbf{II}, we find that
	\begin{align*}&\lim_{\delta\to0+}\limsup_{n\to\infty}\prob\left( \sup_{|s-t|\leq \delta}\left|\XiUVn(s)-\XiUVn(t) \right| >\epsilon\right)

	=0.\end{align*}
	Thus, we have proven \Cref{lemma:Stetigkeitsmodul}.
\end{proof}
Now, we obtain the tightness of the sequence $\left\{\XiUVn\right\}_{n\in\mathbb{N}}$ in $C[0,1/2]$ as a simple consequence of the above results.
\begin{cor}\label{cor:tightness}
	Under \Cref*{cond:secondcondition}, the sequence $\left\{\XiUVn\right\}_{n\in\mathbb{N}}$ is tight in $\left( C[0,1/2],||\cdot||_\infty\right)$.
\end{cor}
\begin{proof}
	By Theorem 7.3 in \citet{BillingsleyConvergenceProbabilityMeasures2013} (and a rescaling argument) it is sufficient to prove that the sequence $\left\{\XiUVn(0)\right\}_{n\in\mathbb{N}}$ is tight in $\R$ and that\[\lim_{\delta\to0^+}\limsup_{n\to\infty}\prob\left(\omega(\XiUVn,\delta))>\epsilon)\right)=0.\] We have already noted that $\left\{\XiUVn(0)\right\}_{n\in\mathbb{N}}$ is tight (by \Cref{thm:finite sample bound for p=2}, which is applicable due to Lemma B.6) and thus \Cref{lemma:Stetigkeitsmodul} yields \Cref{cor:tightness}.
\end{proof}
We conclude the proof of \Cref{thm:main2} $(ii)$ by using the Skorohod Representation Theorem \cite[Thm. 6.7]{BillingsleyConvergenceProbabilityMeasures2013} to verify (see Section B.2.2) that the tightness of $\left\{\XiUVn \right\}_{n\in\mathbb{N}}$ induces that
	\[\frac{nm}{n+m}\int_{0}^{1}\!\left( U_n^{-1}(t)-V_m^{-1}(t)\right)^2 \, dt\rightsquigarrow\int_{0}^{1}\!(\G(t))^2 \, dt.\]

\subsection{Proof of Theorem \ref*{thm:main} \texorpdfstring{$(ii)$}{(ii)}}\label{subsec:proof of thm:main2}

The most important step of the proof of \Cref{thm:main} $(i)$ (see \cite[Sec. B.4]{supplement}) is to derive the limit distributions of
 \begin{align}\label{eq:limit1}
 \int_{\beta}^{1-\beta}\!\zeta(t)\mathbb{U}_n^{-1}(t) \, dt  \text{ and } \int_{\beta}^{1-\beta}\!\zeta(t)\mathbb{V}_m^{-1}(t) \, dt
 \end{align}
 under \Cref{cond:firstcondition}, where $\zeta(t)\coloneqq U^{-1}(t)-V^{-1}(t)$.

 Under \Cref{cond:firstcondition} these can be derived via the distributional limits for the empirical $U$-quantile processes $\mathbb{U}_n^{-1}\coloneqq\sqrt{n}\big(U_n^{-1}\allowdisplaybreaks -U^{-1}\big) $ and $\mathbb{V}_m^{-1}\coloneqq\sqrt{m} \big(V_m^{-1}-V^{-1} \big)$ in $\ell^\infty[\beta,1-\beta]$. However, as already argued, we cannot expect $\ell^\infty(0,1)$-convergence of $\mathbb{U}_n^{-1}$ and $\mathbb{V}_m^{-1}$ under \Cref{cond:secondcondition}. Reconsidering \eqref{eq:limit1}, we realize that $\ell^1(0,1)$-convergence of $\mathbb{U}_n^{-1}$ and $\mathbb{V}_m^{-1}$ is sufficient to derive the corresponding limiting distributions. Convergence in $\ell^1(0,1)$ is much weaker than convergence in $\ell^2(0,1)$ or $\ell^\infty(0,1)$. Indeed, it turns out that this convergence can quickly be verified, since $\phi_{inv}$ is Hadamard differentiable in the present setting as a map from $\mathbb{D}_2\subset D[a,b]\to \ell^1(0,1)$. Using ideas from \citet{kaji2017switching} we can show the following.
 \begin{lemma}\label{lem:Hadamardfiff2}
 	Let $F$ have compact support on $[a,b]$ and let $F$ be continuously differentiable on its support with derivative $f$ that is strictly positive on $(a,b)$ (Possibly, $f(a)=0$ and/or $f(b))=0$). Then the inversion functional $\phi_{inv}:F\mapsto F^{-1}$ as a map $\mathbb{D}_2\subset D[a,b]\to \ell^1(0,1)$ is Hadamard-differentiable at $F$ tangentially to $C[a,b]$ with derivative $\alpha\mapsto-(\alpha/f)\circ F^{-1}$.
 \end{lemma}
 \begin{proof}
 	Let $h_n\to h$ uniformly in $D[a,b]$, where $h$ is continuous, $t_n\to 0$ and $F+t_nh_n\in\mathbb{D}_2$ for all $n\in\N$. Let $\HadPhi(\alpha)=-(\alpha/f)\circ F^{-1}$. We have to demonstrate that
 	\[\left\| \frac{\phi_{inv}(F+t_nh_n)-\phi_{inv}(F)}{t_n}-\HadPhi(h)\right\|_{\ell^1(0,1)} \to 0,\]
 	as $n\to \infty$. We realize that for every $\epsilon>0$ there exist $a_\epsilon,b_\epsilon\in[a,b]$ such that $\max\lbrace F(a_\epsilon),1-F(b_\epsilon)\rbrace<\epsilon $ and $f$ is strictly positive on $[a_\epsilon,b_\epsilon]$. It follows that
 	\begin{align*}
 	&\left\| \frac{\phi_{inv}(F+t_nh_n)-\phi_{inv}(F)}{t_n}-\HadPhi(h)\right\|_{\ell^1(0,1)}\\\leq& \int_{F(a_\epsilon)+\epsilon}^{F(b_\epsilon)-\epsilon}\!\left| \frac{\phi_{inv}(F+t_nh_n)-\phi_{inv}(F)}{t_n}-\HadPhi(h)\right|(s)\,ds\\+&\int_{0}^{2\epsilon}\!\left| \frac{\phi_{inv}(F+t_nh_n)-\phi_{inv}(F)}{t_n}-\HadPhi(h)\right|(s)\,ds\\ +&\int_{1-2\epsilon}^{1}\!\left| \frac{\phi_{inv}(F+t_nh_n)-\phi_{inv}(F)}{t_n}-\HadPhi(h)\right|(s)\,ds.
 	\end{align*}
 Next, we treat the summands separately. The claim follows once we have shown that the first summand vanishes for all $\epsilon>0$ as $n\to\infty$ and the other two summands become arbitrarily small for $\epsilon$ small and $n\to\infty$.

 \noindent\textit{First summand:} We start with the first summand. Since its requirements are fulfilled for all $\epsilon>0$, we have by Lemma 3.9.23 of \citet{vaartWeakConvergenceEmpirical1996} that
 	\begin{align}\label{eq:li-Hadamard first summand}
 	\sup_{s\in [F(a_\epsilon)+\epsilon,F(b_\epsilon)-\epsilon]}\left| \frac{\phi_{inv}(F+t_nh_n)-\phi_{inv}(F)}{t_n}-\HadPhi(h)\right|(s)\to 0,
 	\end{align}
 	as $n\to\infty$. Thus, the same holds for
 	\[\int_{F(a_\epsilon)+\epsilon}^{F(b_\epsilon)-\epsilon}\!\left| \frac{\phi_{inv}(F+t_nh_n)-\phi_{inv}(F)}{t_n}-\HadPhi(h)\right|(s)\,ds,\]
 	which is bounded by \eqref{eq:li-Hadamard first summand}.

 	\noindent\textit{Second summand:} We have
 	\begin{align}
 	&\int_{0}^{2\epsilon}\!\left| \frac{\phi_{inv}(F+t_nh_n)-\phi_{inv}(F)}{t_n}-\HadPhi(h)\right|(s)\,ds\nonumber\\
 	\leq &\int_{0}^{2\epsilon}\!\left| \frac{\phi_{inv}(F+t_nh_n)-\phi_{inv}(F)}{t_n}\right|(s)\,ds +\int_{0}^{2\epsilon}\!\left|\HadPhi(h)\right|(s)\,ds.\label{eq:nikolaus}
 	\end{align}
 	In the following, we consider both terms separately. For the first term, we find that
 	\begin{align*}
 	&\int_{0}^{2\epsilon}\!\left| \frac{\phi_{inv}(F+t_nh_n)-\phi_{inv}(F)}{t_n}\right|(s)\,ds\\=&\frac{1}{|t_n|}\int_{0}^{2\epsilon}\!\left| (F+t_nh_n)^{-1}(s)-F^{-1}(s)\right| \,ds.
 	\end{align*}
 	Next, we realize that for $G\in\mathbb{D}_2\subset D[a,b]$ and $s\in(0,1)$, we have that
 	\begin{align*}G^{-1}(s)=-\int_a^{b}\!\indifunc{ s\leq G(x)}\,dx+b.\end{align*}
 	Since $F\in \mathbb{D}_2$ and $F+t_nh_n\in\mathbb{D}_2$ for all $n\in\N$, this yields that
 	\begin{align*}
 	&\frac{1}{|t_n|}\int_{0}^{2\epsilon}\!\left| (F+t_nh_n)^{-1}(s)-F^{-1}(s)\right| \,ds\\

 	\leq&\frac{1}{|t_n|} \int_a^{b}\!\int_{0}^{2\epsilon}\!\left|\indifunc{s\leq (F+t_nh_n)(x)}-\indifunc{s\leq F(x)}\right|\,ds\,dx,
 	\end{align*}
 	where we applied the Theorem of Tonelli/Fubini \cite[Thm. 18.3]{billingsley2008probability} in the last step. Let in the following $F_{t_nh_n}=F+t_nh_n$. Then, we obtain that

 	\begin{align*}&\int_{0}^{2\epsilon}\!\!\left|\indifunc{s\leq F_{t_nh_n}(x)}-\indifunc{s\leq F(x)}\right|\,ds\\\leq&
 	\begin{cases}
 	\left|F(x)-F_{t_nh_n}(x) \right| , &\mathrm{if}~ \min\left\lbrace F(x),F_{t_nh_n}(x)\right\rbrace \leq 2\epsilon.\\
 	0, & \mathrm{else}.
 	\end{cases}\end{align*}
 	Let now $x\geq F^{-1}\left(2\epsilon+\left\| h_nt_n\right\|_\infty \right)$, then it follows that
 	\[F(x)\geq2\epsilon+\left\| h_nt_n\right\|_\infty\geq2\epsilon\] { as well as } \[F_{t_nh_n}(x)\geq 2\epsilon+\left\| h_nt_n\right\|_\infty+t_nh_n(x)\geq 2\epsilon.\]
 	Combining these findings, we obtain that
 	\begin{align*}
 	&\frac{1}{|t_n|}\int_{0}^{2\epsilon}\!\left| (F+t_nh_n)^{-1}(s)-F^{-1}(s)\right| \,ds\\\leq&\frac{1}{|t_n|}\int_{a}^{F^{-1}\left(2\epsilon+\left\| h_nt_n\right\|_\infty \right)}\!\left|(F+t_nh_n)(x)-F(x)\right| \,dx\\
 	\leq &\left\|h_n-h \right\|_{\ell^1(a,b)}+ \int_{a}^{F^{-1}\left(2\epsilon+\left\| h_nt_n\right\|_\infty \right)}\!\left|h(x)\right| \,dx.
 	\end{align*}
 	We realize that the first term goes to zero as $n\to\infty$ by construction. Further, since $t_n\to0$ for $n\to\infty$, we obtain that $F^{-1}\left(2\epsilon+\left\| h_nt_n\right\|_\infty \right)\to F^{-1}(2\epsilon) $. Thus, for $n\to\infty$ the second term can be made arbitrarily small by the choice of $\epsilon$.

 	\noindent For the second term in \eqref{eq:nikolaus}, we obtain by a change of variables that
 	\begin{align*}
 	\int_{0}^{2\epsilon}\!\left|\HadPhi(h)\right|(s)\,ds=\int_{0}^{2\epsilon}\!\frac{h\left( F^{-1}(u)\right) }{f\left( F^{-1}(u)\right)}\,du=\int_{a}^{F^{-1}(2\epsilon)}\!h (u) \,du.
 	\end{align*}
 	Thus, this term will be arbitrarily small for $\epsilon$ small.

 	\noindent\textit{Third summand:} The third summand can be treated with the same arguments as the second.

 \end{proof}
With the above lemma established it is straight forward to prove that
\begin{equation*}\mathbb{U}_n^{-1}\rightsquigarrow\mathbb{G}_1 \text{ in } \ell^1(0,1) \text{ and }
 \mathbb{V}_m^{-1}\rightsquigarrow\mathbb{G}_2 \text{ in } \ell^1(0,1).\end{equation*}
where $\mathbb{G}_1$ and $\mathbb{G}_2$ are centered, independent, continuous Gaussian processes with covariance structures
\begin{align*}
&\Cov\left(\mathbb{G}_1(t),\mathbb{G}_1(t') \right)= \frac{4}{(u\circ U^{-1}(t))(u\circ U^{-1}(t'))}\Gamma_{\dX}(U^{-1}(t),U^{-1}(t'))
\end{align*}
and
\begin{align*}
&\Cov\left(\mathbb{G}_2(t),\mathbb{G}_2(t') \right)= \frac{4}{(v\circ V^{-1}(t))(v\circ V^{-1}(t'))}\Gamma_{\dY}(V^{-1}(t),V^{-1}(t')),
\end{align*}
where $u=U'$ and $v=V'$. This allows us to essentially repeat the arguments of the proof of \Cref{thm:main} $(i)$. For the missing details we refer to \cite[Sec. B.4]{supplement}.


TITLE: Non-linear differential equation involving a function and its first and second derivatives
QUESTION [1 upvotes]: I was wondering if anyone could get me started on solving the following:
$$(u')^2-u\cdot u''=0$$
I have tried letting $v=u'$, but I don't seem to make progress with such a substitution.

REPLY [1 votes]: You can also observe that if you divide by $u'^2$ then it's a derivative
$$(u')^2-u\cdot u''=0 \implies \frac {(u')^2-u\cdot u''}{u'²}=0$$
$$ \left (\frac {u}{u'} \right )'=0$$
Integrate :
$$u=Ku' \implies \frac {u'}{u}=C \implies (\ln u)'=C$$
$$ \ln u = Cx+K \implies u(x)=Ke^{Cx}$$


TITLE: Another recursive definition question
QUESTION [0 upvotes]: http://vvcap.net/db/vPo_GbDDwvkN3_ASoW6g.htp
I have tried multiples definitions but can't seem to get the correct one
Subfive(0)= 5
Subfive(x)= 250-subfive(x-1)-5
subfive(1)=250-5-5 which is obviously wrong...
subfive(x)= 250-subfive(x-1)
subfive(0) = 5
subfive(1)= 250-subfive(0)
subfive(2)=250-subfive(1)
again, wrong..

REPLY [0 votes]: I think what you want to look at is what the first term is, and what the rate of change is. In this case, the first term is 250, and the rate of change is -5. Therefore, the previous function is five numbers higher than the current one. Taking the first term, the complete recursive function would be the following:
f(0) = 250, f(x) = f(x-1) - 5
If you want the explicit, this is what it would be:
f(x) = 250 - 5x
Always remember to figure out the first term, and the rate of change when finding a recursive function. This will make it much easier for you to figure out function like this.


text:
\section{Problem Setup}
\label{sec:setup}

\begin{figure*}[t]
\centering

    \begin{subfigure}[ ]{
        \centering
        \includegraphics[width=0.24\textwidth]{ggm.pdf}
        \label{subfig:toy:ggm}
    }
    \end{subfigure}
    \begin{subfigure}[ ]{
        \centering
        \includegraphics[width=0.175\textwidth]{ggmMatrix.pdf}
        \label{subfig:toy:ggmMatrix}
    }
    \end{subfigure}
    \begin{subfigure}[ ]{
        \centering
        \includegraphics[width=0.24\textwidth]{lvggm.pdf}\label{subfig:toy:lvggm}
    }
    \end{subfigure}
    \begin{subfigure}[ ]{
        \centering
        \includegraphics[width=0.20\textwidth]{lvggmMatrix.pdf}    \label{subfig:toy:lvggmMatrix}
    }
    \end{subfigure}
\caption{Illustrations of a sparse Gaussian graphical model (GGM) (left) and a latent variable Gaussian graphical model (LVGGM) (right).  (A) Example of a sparse GGM with only observed variables, (B) Sparsity pattern of example sparse GGM's precision matrix, (C) Example of a LVGGM with both observed and latent variables, (D) Sparsity pattern of example LVGGM's precision matrix.}
\end{figure*}

In this section, we review Gaussian graphical models and formulate the problem of latent variable Gaussian graphical model estimation via a regularized maximum likelihood optimization.

\subsection{Gaussian Graphical Models}
Consider a $p$-dimensional random vector $\mbf{x}$ associated with an undirected graph $\mcal{G} = (V_G,E_G)$, where $V_G$ is a set of nodes corresponding to elements of $\mbf{x}$ and $E_G$ is a set of edges connecting nodes (including self-edges for each node).
Then $\mbf{x}$ follows a graphical model distribution if it satisfies the Markov property with respect to $\mcal{G}$: for any pair of nonadjacent nodes in $\mcal{G}$, the corresponding pair of variables in $\mbf{x}$ are conditionally independent given the remaining variables, \emph{i.e.,} $x_i \perp x_j \ | \ \mbf{x}_{\bs i, j}$, for all $(i, j) \notin E_G$.

If $\mbf{x}$ follows a multivariate Gaussian distribution, the corresponding graphical model is called a Gaussian graphical model (GGM).  We assume without loss of generality that $\mbf{x}$ has zero mean.
The Markov property in GGM is manifested in the sparsity pattern of the inverse covariance matrix $\mbf{J}$:
\ALGNN{
\mbf{J}_{i,j} = 0 \text{ for all }   i \neq j, (i,j) \notin E.

}
An example of this property for sparse GGM is shown in Figure~\ref{subfig:toy:ggm} and~\ref{subfig:toy:ggmMatrix}.

The precision matrix parameterization arises in many statistical inference problems for Gaussian distributions, in areas such as belief propagation~\citep{malioutov2006walk},
linear prediction,
portfolio selection in financial data~\citep{ledoit2003improved},
and anomaly detection~\citep{chen2011robust}.
Estimation of the precision matrix in GGM is the first step in these inference problems.

\subsection{Latent Variable Gaussian Graphical Models}
\label{sec:LVGGMsetup}

Unfortunately, due to the presence of global factors that destroy sparsity, real-world observations often do not conform exactly to a sparse GGM~\citep{choi2010gaussian,choi2011learning}.  By introducing latent variables (denoted as a $r$-dimensional random vector $\mbf{x}_L$) to capture global factors, we can generalize the GGM.
Specifically, we construct a model that is conditionally a GGM, {\em i.e.,} one that has a sparse precision matrix given knowledge of latent variables, ${\bf x}_L$.

Defining the $p$ observed variables as $\mbf{x}_O$, we assume the joint distribution of the $(p+r)$-dimensional concatenated random vector $\mbf{x} = (\mbf{x}_{O}, \mbf{x}_{L})$ follows a Gaussian distribution with covariance matrix $\mbf{\Omega}$ and precision matrix $\mbf{J} = \mbf{\Omega}^{-1}$.   An example of this structure can be seen in Figure~\ref{subfig:toy:lvggm} and~\ref{subfig:toy:lvggmMatrix}.
Marginalizing over the latent variables $\mbf{x}_L$, the distribution of the observed variables $\mbf{x}_O$ remains Gaussian with observed covariance matrix, $\mbf{\Sigma} = \mbf{\Omega}_{O, O}$.  The observed precision matrix $\Theta \in \mathbb{R}^{p \times p}$ satisfies:
\ALGNN{
\Theta & = \mbf{\Sigma}^{-1} = \underbrace{\mbf{J}_{O,O}}_{\mbf{S}} \underbrace{- \mbf{J}_{O,L} \mbf{J}_{L,L}^{-1} \mbf{J}_{L,O}}_{\mbf{L}},
\label{eq:theta}
}
where we have defined $\mbf{S} := \mbf{J}_{O,O}$ and $\mbf{L} := - \mbf{J}_{O,L} \mbf{J}_{L,L}^{-1} \mbf{J}_{L,O}$. Thus, the marginal precision matrix can be written as $\Theta = \mbf{S} + \mbf{L}$, the sum of a sparse and a low-rank matrix.
Similar to standard GGM, we parameterize the marginal distribution through the precision matrix.   We refer to this model as the latent variable GGM, or LVGGM.

The LVGGM is a hierarchical model that generalizes the (sparse) GGM.  Note that $\mbf{S}^{-1} = \mbf{J}_{O,O}^{-1} = \mbf{\Omega}_{O, O} - \mbf{\Omega}_{O, L} \mbf{\Omega}_{L,L}^{-1} \mbf{\Omega}_{L, O}$ is the covariance matrix of the conditional distribution of the observed variables.  The matrix is not generally sparse, even though $\mbf{S}$ is assumed to be sparse.  We will also assume that the number of latent variables is much smaller than the number of observed variables, \emph{i.e.,} $r \ll p$.  We place no sparsity restrictions on the dependencies between the observed and latent variables -- the submatrices $\mbf{J}_{O,L}$ and $\mbf{J}_{L,O}$ could be dense. As a result, the $p \times p$ matrix $\mbf{L} = -\mbf{J}_{O,L} \mbf{J}_{L,L}^{-1} \mbf{J}_{L,O}$ is low-rank and potentially dense. The sparse plus low-rank structure of the marginal precision matrix $\Theta$ is the key property of the precision matrix that will be exploited for model estimation.

The structural assumptions on the precision matrix of the LVGGM can be further motivated and validated on real-world recommender system data and stock return data.  Due to the space limits, we defer these two motivating examples to Section~\ref{sec:motivatingExp} in the Appendix.

\subsection{Effective Rank of Covariance Matrix}
\label{sec:effectiveRank}

We introduce the \emph{effective rank} of a matrix, which will be useful to derived high-dimensional error bounds.  The effective rank of a matrix $\mbf{\Sigma}$ is defined as~\citep{vershynin2010introduction}:
\ALGNN{
\reff(\mbf{\Sigma}) := {\text{tr}(\mbf{\Sigma})}/{\| \mbf{\Sigma} \|_2}.
}

The effective rank can be considered a measure of the concentration level of the spectrum of $\mbf{\Sigma}$. As we will show in Section~\ref{sec:effrankexp}, in many situations the effective rank of the covariance matrix corresponding to a LVGGM is much smaller than $p$. Under this condition, our theoretical results in the sequel provide a tight Frobenius norm estimation error bound, which is significantly improved upon the error bound derived without the effective rank assumption.

\subsection{Regularized ML Estimation of LVGGM}
\label{sec:regML}

Available are $n$ samples $x_1, x_2, \ldots, x_n$ from a LVGGM model $\mbf{x}_O$, concatenated into a data matrix $\mbf{X} \in \mathbb{R}^{p \times n}$.  The negative log-likelihood function is
\ALGNN{
\mcal{L}(\Theta; \mbf{X}) = \langle \Sighat, \Theta \rangle - \log \det (\Theta),
\label{eq:llh}
}
where $\Sighat := \frac{1}{n} \mbf{X}^T \mbf{X}$ is the sample covariance matrix. The regularized ML estimate minimizes the objective function $\mcal{L}(\Theta; \mbf{X}) + \la \mcal{R}(\Theta)$, where the regularization parameter $\lambda > 0$, and the regularization function $\mcal{R} ( \Theta )$ is designed to enforce the sparse plus low-rank structure on $\Theta$.

Similar to~\citet{chandrasekaran2012latent}, we consider the following regularized ML estimation problem:
\ALGNN{
\begin{split}
\min_{\mbf{S}, \mbf{L}} \ & \ \mathcal{L}(\mbf{S} + \mbf{L}; \mbf{X}) + \lambda \| \mbf{S} \|_1 + \mu \| \mbf{L} \|_* \\%\mathcal{R}(\mbf{S}, \mbf{L}), \\
{\text{s.t.}} \ & \ - \mbf{L} \succeq \mbf{0}, \ \mbf{S} + \mbf{L} \succeq \mbf{0},
\end{split} \label{eq:MestLVGGM}
}
where the corresponding regularization function is the sum of two regularizers:
$\mathcal{R}(\Theta) = \| \mbf{S} \|_1 + \frac{\mu}{\lambda} \| \mbf{L} \|_*$,
each of which has been shown to promote sparse (low-rank) structure in $\mbf{S}$ ($\mbf{L}$, respectively)~\citep{negahban2012unified}. Constants $\la, \mu >0$ are regularization parameters corresponding to the two functions, respectively.  The LVGGM estimator is defined as a solution to the above convex optimization problem~\eqref{eq:MestLVGGM}. Efficient convex solver, such as~\citet{ma2012alternating}, can be used to solve.

config: arxiv


TITLE: Superconductivity resistance
QUESTION [4 upvotes]: I'm confused about the resistance of superconductors. It is known that the resistance at the the critical temperature goes exactly to zero (the resistance is not very small, its really zero, right?). But in a superconducting state, not all electrons form Cooper pairs right? so why then is the resistance exactly equal to zero?
Greetings

REPLY [6 votes]: It is absolutely true that the resistance of a superconductor goes to zero at at absolute zero, but just below the critical temperature, most of the conduction electrons are still in the normal state.  The current density has two parts:  J=...A+...E, the coefficients being proportional to the numbers of electrons in the paired and normal states.  The DC resistance vanishes because it is possible to have current without any electric field, but there can still be some AC resistance.


TITLE: Need help understanding why formula for prime number is correct
QUESTION [0 upvotes]: I was reading a Mathematical paper from the '40s, and came across the following formula:
$$q = \theta\sqrt{\frac{n}{\nu}}$$
Where $q$ is a prime number, $\nu$ is a given integer, and $\theta \in (1, 1+\vartheta)$, where $\vartheta$ is a given real. $n$ is an integer.
For the argument in the paper to work, I believe this formula may only fail for finitely many $n$. The value of $\theta$ may of course depend on $n$. The only reference why this should be the case, is that: "According to the law of prime numbers, for sufficiently large $n$ such a prime $q$ exists." I was not able to find any such law, and this is were I need help.
Why should this be the case?
Link to the original paper, which is in Russian: http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=sm&paperid=5985&option_lang=eng

REPLY [1 votes]: It seems the full claim is:

Let $\vartheta >0$ and $\nu\in\Bbb N$. Then for almost all $n\in\Bbb N$, there exists $\theta\in(1,1+\vartheta)$ such that $\theta\sqrt{\frac n\nu}$ is prime.

or equivalently

Let $\vartheta >0$ and $\nu\in\Bbb N$. Then for almost all $n$, there exists  a prime $q$ such that $$\sqrt{\frac n\nu}<q<(1+\vartheta)\sqrt{\frac n\nu}.$$

For this it is sufficient to show

Let $\vartheta>0$. Then for all $x\gg0$,
$$ \pi((1+\vartheta) x)>\pi(x).$$

What the Prime Number Theorem gives us is: For $\alpha <1<\beta$ of our choice, we have
$$ \frac{\alpha x}{\ln x}<\pi(x)<\frac{\beta x}{\ln x}$$
for all $x\gg 0$. So we can certainly achieve $\frac \beta\alpha < 1+\vartheta$.
Then
$$\begin{align} \pi(x+\vartheta x)-\pi(x)
&>\frac{\alpha x(1+\vartheta)}{\ln x+\ln(1+\vartheta)}-\frac{\beta x}{\ln x}\\
&>\frac{\alpha x(1+\vartheta)}{\ln x+\vartheta}-\frac{\beta x}{\ln x}\\
&=\frac{\alpha x(1+\vartheta)\ln x - \beta x\ln x - \beta\vartheta x}{(\ln x+\vartheta)\ln x}\\
&=\frac{x}{(\ln x+\vartheta)\ln x}\cdot\bigl(\overbrace{(\alpha(1+\vartheta)-\beta)}^{>0} \ln x-\beta\vartheta\bigr ).
\end{align}$$
This is positive for $x\gg0$.


TITLE: Proving Jensen's discrete inequality holds with equality if and only if x1=x2=x3=...=x1 (given a twice differentiable function)
QUESTION [0 upvotes]: Edit. This is the last question, my monster. I kind if have some work into it now and forgot to mention the function was twice differentiable

Well I'm not sure if the image worked but it was just the question written down. The professor advised we "lift" some of the proofs online, as some of these are not meant to be solved creatively as an undergraduate.
This is one such problem.  I can prove if f is concave then the inequality holds, but showing it only holds
with equality when all inputs are equal is not going well.
I have 1 other question like this this and a couple that I actually have some more usable ideas to go off of.
This is for a final but I assure you we are allowed any resources we want. The style of teaching seems suprising but he seems to know what he's doing. He didn't even bother to mention we make sure we completely understand the proof but I suppose it goes without saying.
I have hours and hours of unusable rabbit holes but no starting idea worth posting on this one yet. Will add to thread if I bring one into existence.

REPLY [3 votes]: Hint: Consider what happens if you look at the proof of the generalization:
$$ f \left( \sum_{i=1}^n \lambda_i x_i \right) \ge \sum_{i=1}^n \lambda_i f(x_i) $$
given that each $\lambda_i > 0$ and $\lambda_1 + \cdots + \lambda_n = 1$.  Then in the inductive step, you consider:
$$ f \left( \sum_{i=1}^n \lambda_i x_i \right) = f \left( (1-\lambda_n) \sum_{i=1}^{n-1} \frac{\lambda_i}{1-\lambda_n} x_i + \lambda_n x_n \right) \\ \ge (1 - \lambda_n) f \left( \sum_{i=1}^{n-1} \frac{\lambda_i}{1-\lambda_n} x_i \right) + \lambda_n f(x_n) \\ \ge (1-\lambda_n) \left( \sum_{i=1}^{n-1} \frac{\lambda_i}{1-\lambda_n} f(x_i)\right) + \lambda_n f(x_n) \\ = \sum_{i=1}^n \lambda_i f(x_i).$$
Now, if you have equality between the two ends, then each inequality in this chain must be equality.  Also, the hypothesis that $f$ is strictly concave pretty much directly gives that for $n=2$, equality holds if and only if $x_1 = x_2$.  Now, if you use induction, can you see how to complete the proof that equality holds in the generalized inequality if and only if $x_1 = \cdots = x_n$?  And then, how to specialize this result to get what you wanted?


TITLE: Critique of this subgroup proof?
QUESTION [5 upvotes]: This is an exercise out of a Dover book on abstract algebra. I was hoping I could get some feedback on this proof, as I'm still fairly new to this level of proof writing. Thanks! Here's the exercise:
Let $G$ be an Abelian group whose operation is multiplication, let $H$ be a subgroup of $G$, and let $K = \{x \in G:$ for some positive integer $ n$, $x^n \in H \}$. Prove that $K$ is a subgroup of $G$.
My proof:
To show that $K$ is a subgroup of $G$, it is sufficient to show that $K$ is a subset of $G$ and that $K$ is closed under multiplication and inverses. The first condition is trivially satisfied, because $K$ consists only of elements of $G$. We now show that $K$ is closed under multiplication.
Let $x,y \in K$. We know $x,y \in G$ and $x^n,y^m \in H$ for some positive integers $m,n$. Additionally, $x^m,y^n \in H$. Since $x,y \in G$ and $G$ is closed under multiplication, $x \cdot y \in G$. Similarly, $x^n \cdot x^m \cdot y^n \cdot y^m \in H$. Now $(x \cdot y)^{n+m} = \underbrace{(x \cdot y) \cdot (x \cdot y) \cdots (x\cdot y)}_\text{n+m}$. Since $G$ is Abelian, we can rearrange this as $\underbrace{x \cdot x \cdots x}_\text{n+m} \cdot \underbrace{y \cdot y \cdots y}_\text{n+m} \\ = \underbrace{x \cdot x \cdots x}_\text{n} \cdot \underbrace{x \cdot x \cdots x}_\text{m} \cdot \underbrace{y \cdot y \cdots y}_\text{n} \cdot \underbrace{y \cdot y \cdots y}_\text{m} = x^n \cdot x^m \cdot y^n \cdot y^m \in H$. Therefore, since $x \cdot y \in G$, $n+m$ is a positive integer, and $(x \cdot y)^{n+m} \in H$, $x \cdot y \in K$. Therefore, $K$ is closed with respect to multiplication.
Now we prove that $K$ is closed with respect to inverses. Let $x \in K$. Therefore $x \in G$ and $x^n \in H$ for some positive integer $n$. Since $x \in G$ and $G$ is closed to inverses, $\frac{1}{x} \in G$. Similarly, $\frac{1}{x^n} \in H$. $(\frac{1}{x})^n = \frac{1}{x^n} \in H$. Therefore, $\frac{1}{x} \in K$, and $K$ is closed with respect to inverses. Thus, $K$ is a subgroup of $G$.

REPLY [1 votes]: There is no reason $x^m$ or $y^n$ should be in $H$. So you need to change the proof that $xy\in K$.
Also you may prove directly that if $x,y\in K$ then $xy^{-1}\in K$ which can prove the same subgroup property in one step.


TITLE: question on poisson random measure
QUESTION [0 upvotes]: Assume $(S,\mathcal{S},\eta)$ is an arbitrary $\sigma$-finite measure space. Let $N:\mathcal S\rightarrow\{0,1,2,\ldots\}\cup\{\infty\}$ in a way such that $\{N(A):A\in\mathcal S\}$ are random variables defined on the probability space $(\Omega,\mathcal{F},\mathbb{P})$. Then $N$ is called a Poisson random measure on  $(S,\mathcal{S},\eta)$ if
(i) for mutually disjoint $A_1,\ldots,A_n$ in $\mathcal{S}$, the variables $N(A_i),\ldots,N(A_n)$ are independent,
(ii) for each $A\in\mathcal{S}$, $N(A)$ is Poisson variable distributed with parameter $\eta(A)$.
(iii) $\mathbb{P}$-almost surely, $N$ is a measure.
In the context of a Levy process, I can see that $\eta$ corresponds to the product measure of the jump measure in a Levy process and Lebesgue measure on $t$, and ($S$, $\mathcal{S}$) is the measure space generated by the Borel $\sigma$-algebra on $\mathbb{R}\times[0,\infty)$.
now here is a colloary to a proof such measure exists
Suppose that $N$ is a Poisson random measure on $(S,\mathcal{S}, \eta)$., the the support of $N$ is $\mathbb{P}$ almost surely countable.
I don't quite understand what is the support of $N$ here means and would appreciate an answer.

REPLY [0 votes]: To say that the support of $N$ is countable means that there exists a countable random set $T$ in $\mathcal S$ such that, $P$-almost surely,  $N(S\setminus T)=0$ and $N(\{t\})=1$ for every $t$ in $T$.
Thus, $N=\sum\limits_{t\in T}\delta_t$.
Equivalently, for every (deterministic) $A$ in $\mathcal S$, one asks that $N(A)(\omega)=\# (A\cap T(\omega))$ for $P$-almost every $\omega$ in $\Omega$.


TITLE: Sum of squared arctangents
QUESTION [9 upvotes]: I discovered a curious identity
$$\small{2939\arctan^2 2+450\arctan^2 8+84\arctan^2 13+330\arctan^2 18+147\arctan^2 38=\\1250\arctan^2 3+252\arctan^2 4+360\arctan^2 5+870\arctan^2 7+210\arctan^2 21+210\arctan^2 47.}\tag{$\diamond$}$$
Is there a simpler identity of this kind? All coefficients must be positive integers and all arguments of squared arctangents must be distinct integers $\ge2$.
Is there a systematic way to find all identities of this kind? Is there an infinite number of them?

REPLY [9 votes]: What does "simpler" mean? Fewer terms? Smaller Coefficients? Smaller arguments?
The identity you wrote down is easy to verify using the fact that
$$4 \arctan(n) = \log(1 + n i)^2 + \log(1 - n i )^2 - 2 \log(1 + n i) \log(1 - ni).$$
In order to find identities, you want to consider a collection of $n$ such that the Gaussian integers $1 \pm n i$ all have a fixed collection of Gaussian prime divisors, or equivalently $n^2 + 1$ has a small number of prime factors. For example, all the arguments in your formula have $n^2 + 1$ divisible only by the primes $2$, $5$, $13$, $17$. The complete collection of integers with this property is
$$1, 2, 3, 4, 5, 7, 8, 13, 18, 21, 38, 47, 57, 239, 268.$$
Ignoring the term $1$ (I'm not sure why you want to ignore this), the space of values $\arctan(n)^2$ for the $14$ values of $n$ generates a vector space over $\mathbf{Q}$ of dimension at most $10$, and so any $11$ terms are linearly dependent over the field $\mathbf{Q}$.
To prove the dimension is at leat $10$ would require various algebraic independence results for logarithms which are probably not known, the point being that one has quadratic terms in logarithms which are beyond the usual transcendence results.
Suppose you look for $n$ so that $n^2 + 1$ is only divisible by the primes $2$, $5$, and $13$. Then the only such $n$ are
$$1, 2, 3, 5, 7, 8, 18, 57, 239.$$
Ignoring $1$, there are $8$ numbers, which generate a vector space of dimension at most $7$, and so any $8$ terms give a linear relation. The one with smallest coefficients is probably given by taking all $9$ terms, namely
$$21 \arctan(5)^2 + 34 \arctan(7)^2 + 5 \arctan(8)^2 = 75 \arctan(2)^2 + \arctan(3)^2 + 5 \arctan(18)^2 + 5 \arctan(57)^2 + \arctan(239)^2.$$
Of course, the appearance of $239$ will not be a huge surprise to arctan fans.
So there are certainly identities with smaller coefficients and a smaller number of terms. There are probably not identies where the exponents are smaller than $47$, however, because there is not enough cancellation in terms of the factors of $n^2 + 1$.
As to whether there are infinitely many identities, that's a slightly sticky problem. If you fix a set of prime numbers $S$ (necessarily in this case of the form $2$ or $1 \pmod 4$), then there are only finitely many integers $n$ such that $n^2 + 1$ is only divisible by primes in $S$ (a non-trivial but standard number theory exercise). However, it's quite unclear exactly how many there should be. For example, if you include the next prime $29$ as well, then you seem to get
$$2, 3, 4, 5, 7, 8, 12, 13, 17, 18, 21, 38, 41, 47, 57, 70, 99, 157, 239, 268, 307$$
There are new relations here, the simplest one found by LLL turning out to have all coefficients under $10$, namely
$$\sum a_i \log(b_i)^2 = 0,$$ where $(a_i)$ and $(b_i)$ are given by
$$\{-9, -8, 1, -3, -8, -1, 2, 3, -4, 1, -2, 3, -3, -1, 7, 7, -7, 7, 7, 0, 0\},$$
$$\{2, 3, 4, 5, 7, 8, 12, 13, 17, 18, 21, 38, 41, 47, 57, 70, 99, 157, 239, 268, 307\}.$$
respectively.
I suspect that there are probably infinitely many more relations but that proving it will depend on subtle (and probably open) number theoretical problems concerning the number of integers of the form $n^2 + 1$ with restricted prime factors.


TITLE: Find the domain and range of the following functions.
QUESTION [2 upvotes]: a. $f(x)=x^5+\sqrt{x}$
b. $g(y)=\frac{1}{y-2}$
c. $h(z)=\sqrt{z^2-2z-3}$

I know the solutions of all three, but in order to show the proper format of work on paper, what should I do? I'm not sure if my explanations are good enough. Here is my work and solutions for all three:

a. Domain: $x\geqslant0$; Range: $f(x)\geqslant0$.
My reasoning: In order for $f(x)$ to be real, $x$ must be greater or equal to $0$, since only the square root of a non-negative real number gives a real solution. Therefore, the smallest possible $f(x)$ is $0$.
b. Domain: $y\neq2$; Range: $g(y)\neq0$.
My reasoning: A fraction whose denominator equals to $0$ cannot possibly exist. When $y$ is $2$, the denominator is $0$; thus, $y\neq2$. The only value $g(y)$ cannot be is $0$ because the numerator of the function is $1$; the numerator must equal to $0$ in order for $g(y)$ to be $0$. This is not possible, so $g(y)\neq0$.
c. Domain: $z\geqslant3 \cup z\leqslant-1$; Range: $h(z)\geqslant0$.
My reasoning: In order for $h(z)$ to be real, the value inside the square root sign must be greater or equal to $0$, since only the square root of a non-negative real number gives a real solution. Only $z\geqslant3$ and $z\leqslant-1$ make this happen. Therefore, the smallest possible $h(z)$ is $0$.

REPLY [1 votes]: Your explanations all convey the right idea and would likely be more than enough for a high school class. But since you solicited feedback, I'll provide some.
A. You wrote:

$x$  must be greater or equal to $0$, since only the square root of a non-negative real number gives a real solution. Therefore, the smallest
  possible $f(x)$ is $0$.

This argument is not quite complete: you have not explained the "therefore." You are right that $\sqrt{x}$ is defined only when $x\geq0$ (at least in the setting of real numbers). But you need to explain why this fact implies $x^5+\sqrt{x}\geq0$. Of course, this extra step is very easy to fill in, but as it stands, it is missing from your explanation. Consider, instead, what you would have written if the function had been $g(x)=-x^5+\sqrt{x}$. In this case, you could write the same first sentence that you wrote for $f(x)=x^5+\sqrt{x}$, but it does not follow that $g(x)\geq0$. This shows that you have left something out.
B. You wrote:

A fraction whose denominator equals to $0$ cannot possibly exist.

A mathematician would not speak of "existence" here. What you mean to say is that the expression $(y-2)^{-1}$ is undefined precisely when $y=2$. It is not that no such fraction exists (consider, for example, the fraction $1/0$); it is that such a fraction is meaningless (in this context, at least). A fraction is a string of symbols, not a number. We use (some) fractions to refer to numbers, but the fraction is not the same thing as the number, just as the word cat is not the same thing as a  cat. We often get away with confusing a fraction with the number it represents, but strictly speaking, that way of talking doesn't make sense.


TITLE: Compute Pointwise Multiplication by hand
QUESTION [0 upvotes]: How could we compute the pointwise multiplication of two vectors?
In example:
A = [1 2 3]'

What is the pointwise multiplication of A.^2? (like computational programs using)
In short: what steps should be take to convert:
 A = [1 2 3]' to A^2 = [1 4 9]'

REPLY [0 votes]: $A = \{a_1, a_2, a_3\}$
$A\cdot A= A^2 = \{a_1\cdot a_1, a_2\cdot a_2, a_3\cdot a_3\}$
In your example:
$A.A = A^2 = \{1^2,2^2,3^2\} = \{1,4,9\}$


TITLE: Proof for All rows/cols for A has same sum $s$ $\implies$ $s$ is an eigenvalue?
QUESTION [0 upvotes]: To prove for All rows/cols for A has same sum $s$ $\implies$ $s$ is an eigenvalue
So in order for $s$ to be an eigenvalue, it has to satisfy Ax=$\lambda$x where x$\ne 0$. But how do you simplify A to find $\lambda$?
Someone please help?

REPLY [1 votes]: Hint:
Find $Au$ for $u=(1,1,1,1,....,1)^T$ ( all components$= 1$).


TITLE: Two continuous functions on closed interval and smallest element
QUESTION [0 upvotes]: Assume $f,g\colon [0,1]\to [0,1]$ are continuous functions and $$ g(0)<f(0) \ \ \text{and}\ \ g(1)>f(1)$$Is it true that there exists smallest $t\in(0,1)$ such that $$ g(t)\geq  f(t)$$
I know continuous will have max and min since it defined on closed interval. But I know If we could ensure such smallest element.

REPLY [1 votes]: Let $h(x) = f(x) - g(x)$, $h$ is continuous.
$$h(0)= f(0) -g(0) \gt 0$$
$$h(1)=f(1)-g(1) \lt 0$$
By the intermediate value theorem there is some $x_0$ in $(0,1)$ such that $h(x_0)=0$.
Also, since $h(x)=0$ is a level curve then its preimage is closed, since it is also bounded in $(0,1)$ then it is compact; so by the minimum value theorem there is such a least element.