TITLE: Show that $f(x):=g(\Vert x\Vert_2)x$ has an antiderivative
QUESTION [0 upvotes]: Let be $g\in\mathcal{C}^1((0,\infty))$ and $f:\mathbb{R}^n\setminus\{0\}\to\mathbb{R}^n$ where $f(x):=g(\Vert x\Vert_2)x$.
Show that $f$ has an antiderivative.

By plain luck I found out the following:
Let be $h:\mathbb{R}^n\to\mathbb{R}$ with $h(x):=\Vert x\Vert_2$ and $G:\mathbb{R}\to\mathbb{R}$ with $G(y):=\int\limits_1^yg(t)t~dt$. Then, by chain rule we can conclude
$$
D_i(G\circ h)(x)=G'(h(x))D_ih(x)=g(\Vert x\Vert_2)\Vert x\Vert_2\frac{x_i}{\Vert x\Vert_2}=g(\Vert x\Vert_2) x_i\\\implies F:=G\circ h \text{ is an antiderivative of }f.
$$
However, this problem was stated in the context of line integrals so I am wondering if there is an appropriate curve and a line integral to derive an antiderivative.
I have already tried the easiest curve i.e. a straight line between some fixed point $a\in\mathbb{R}^n\setminus\{0\}$ and $x$ but then it might be that $0$ lies on this curve. Any attempts to create a curve that circumvents the point $0$ lead to some messy expressions.
Do you have any ideas which curve to use?

REPLY [0 votes]: You cannot use a line that pass over zero as $f$ is not defined there, however you can use any straight line from $a \mathbf{w}$ to $b \mathbf{w}$, where $0<a<b$ and $\mathbf{w}\in \{\mathbf{v}\in \mathbb{R}^n:\|\mathbf{v}\|_2=1\}$, then a parametrization of the path is $\gamma :[a,b]\to \mathbb{R}^n,\, t\mapsto t\mathbf{w}$. Then
$$
\int_{\gamma }f(\mathbf{x})\cdot \,d \mathbf{x}=\int_{a}^b\langle (f\circ \gamma )(t),\dot \gamma (t) \rangle \,d t=\int_{a}^b g(t)t\,d t=H(b)-H(a)=H(\|b\mathbf{w}\|_2)-H(\|a\mathbf{w}\|_2)$$
where $H$ is any antiderivative of the map $t\mapsto g(t)t$. So a candidate for an antiderivative of $f$ is $F:=H\circ \|\cdot\|_2$, now you only need to check that $\nabla  F(\mathbf{x})=f(\mathbf{x})$ for all $\mathbf{x}\neq 0$.∎


text:
{\bf Problem.} Functions that aren't invertible can be made invertible by restricting their domains. For example, the function $x^2$ is invertible if we restrict $x$ to the interval $[0,\infty)$, or to any subset of that interval. In that case, the inverse function is $\sqrt x$. (We could also restrict $x^2$ to the domain $(-\infty,0]$, in which case the inverse function would be $-\sqrt{x}$.)

Similarly, by restricting the domain of the function $f(x) = 2x^2-4x-5$ to an interval, we can make it invertible. What is the largest such interval that includes the point $x=0$?
{\bf Level.} Level 4
{\bf Type.} Intermediate Algebra
{\bf Solution.} Completing the square, we have $f(x)=2(x-1)^2-7$. The graph of this function is a parabola with its vertex at $x=1$. To the left of that point, $f(x)$ is decreasing; to the right, it's increasing. Thus, by restricting the domain to either $(-\infty,1]$ or $[1,\infty)$, we make $f$ invertible. The choice that includes $x=0$ is $\boxed{(-\infty,1]}$.

set_name: MATH


TITLE: Is there $\sigma_\infty$ such that $\sigma_\infty$ is a subsequence of $\sigma_n$ for all $n \in \mathbb N$?
QUESTION [1 upvotes]: I have come across this "Diagonal method" in this lecture note.

Theorem: Let $S$ be a non-empty set. For each $n\in \mathbb N$, let $\sigma_n$ be a sequence of elements of $S$. We denote by $\sigma_{n}(m)$ the $m$-th coordinate of $\sigma_n$. Assume that $\sigma_{n+1}$ is a subsequence of $\sigma_n$ for all $n\in \mathbb N$. Then there is a sequence $\sigma_\infty$ such that the restriction
$$
\sigma_\infty \restriction \{n, n+1, \ldots\}
$$
is a subsequence of $\sigma_n$.

Proof: Clearly, the sequence $\sigma_\infty$ with $\sigma_\infty (n) := \sigma_n(n)$ satisfies the requirement.
I think a stronger statement i.e.,

there is a sequence $\sigma_\infty$ such that $\sigma_\infty$ is a subsequence of $\sigma_n$ for all $n \in \mathbb N$,

could not hold in general. Could you provide me with a counter-example of above statement?

REPLY [1 votes]: Let $\sigma_n(m) = n+m$ (so $S = \mathbb N$, for example).

Claim : $\sigma_{i+1}$ is a subsequence of $\sigma_i$ for all $i
\geq 1$. Furthermore, there is no sequence $\sigma_{\infty}$ such that $\sigma_{\infty}$ is a subsequence of $\sigma_i$ for all $i \geq 1$.

Proof : One checks that $$\sigma_{i+1}(k) = (i+1)+k = i+(k+1) = \sigma_i(k+1)$$
for all $i , k \geq 1$. Therefore, $\sigma_{i+1}$ is a subsequence of $\sigma_i$, in particular if $f(i) = i+1$ then $\sigma_{i+1}(\cdot) = \sigma_{i}(f(\cdot))$.
Finally, if $\sigma_{\infty}$ is a sequence which is a subsequence of every $\sigma_i$, then consider the element $\sigma_{\infty}(1)$. It certainly belongs to $\mathbb N$, otherwise the subsequence property can't hold since the image of every $\sigma_i$ lies in $\mathbb N$. If $\sigma_{\infty}(1) = K \in \mathbb N$, then obviously it cannot be a subsequence of $\sigma_{K+1}$, whose image doesn't contain $K$. We are done.

You may have wondered why I took the natural numbers above and , perhaps, why not a simpler set (At least I wondered about this!). The answer to that is : if $S$ is finite then the desired property is in fact , true. One can show that using a fairly clever trick. More precisely :

If $S$ is finite and $\sigma_i$ is a sequence of sequences with values in $S$ with $\sigma_{i+1}$ being a subsequence of $\sigma_i$ for all $i$, then there exists $\sigma_{\infty}$ , a sequence which is a subsequence of all the $\sigma_i$.

Let's introduce a notation : Let $[a,b]$ denote the set of all natural numbers between $a$ and $b$, with both included. For example, $[3,7]= \{3,4,5,6,7\}$.
Proof : For each $s \in S$, let $$I_s = \{i \in \mathbb N :|\{k : \sigma_i(k) = s\}| = \infty\}$$ In other words, the set of all $i$ such that $s$ occurs infinitely many times in $\sigma_i$.
Since $\sigma_{i+1}$ is a subsequence of $\sigma_i$ for each $i$, it follows that if $j+1 \in I_s$ then $j \in I_s$ for all $j \in \mathbb N, s \in S$. Therefore, every $I_s$ is either of the form $[1,J]$ for some $J$, or $I_s = \mathbb N$.
Now, suppose that $I_s = \mathbb N$ for some $s \in S$. By definition, the sequence $\sigma_{\infty} \equiv s$ (constant sequence) is a subsequence of all the $\sigma_i$.
We claim that for some $s$ it must happen that $I_s = \mathbb N$. Suppose not : then, every $I_s = [1,J_s]$ for some $J_s$. Let $J = \max\{J_s : s \in S\}$, which is finite because $S$ is finite.
Look at $\sigma_{J+1}$ : this is a sequence in which none of the elements of $S$ occur infinitely many times. That's clearly impossible : $\sigma_{J+1}$ is a map from $\mathbb N$ to $S$ i.e. an infinite set to a finite set, therefore at least one element must have infinite preimage.
Therefore, $I_s = \mathbb N$ for some $s \in S$, and $\sigma_{\infty}$ as described before does the job.$\blacksquare$


text:
{\bf Problem.} Find all real solutions to
\[\frac{1}{(x - 1)(x - 2)} + \frac{1}{(x - 2)(x - 3)} + \frac{1}{(x - 3)(x - 4)} = \frac{1}{6}.\]Enter all solutions, separated by commas.
{\bf Level.} Level 3
{\bf Type.} Intermediate Algebra
{\bf Solution.} By partial fractions,
\begin{align*}
\frac{1}{(x - 1)(x - 2)} &= \frac{1}{x - 2} - \frac{1}{x - 1}, \\
\frac{1}{(x - 2)(x - 3)} &= \frac{1}{x - 3} - \frac{1}{x - 2}, \\
\frac{1}{(x - 3)(x - 4)} &= \frac{1}{x - 4} - \frac{1}{x - 3},
\end{align*}so the given equation reduces to
\[\frac{1}{x - 4} - \frac{1}{x - 1} = \frac{1}{6}.\]Multiplying both sides by $6(x - 4)(x - 1),$ we get
\[6(x - 1) - 6(x - 4) = (x - 4)(x - 1),\]which simplifies to $x^2 - 5x - 14 = 0.$  This factors as $(x - 7)(x + 2) = 0,$ so the solutions are $\boxed{7,-2}.$


TITLE: If $\phi$ is a orientation preserving linear automorphism, do we really need to choose the same orientation for the domain and the codomain?
QUESTION [1 upvotes]: In the book of Linear Algebra by Werner Greub, at pages $131-132$, it is given that

and

However, when $\phi$ is a linear automorphism, it says that $\Delta_F = \Delta_E$, and derives the following result, however, as far as I see, there is no obligation for such thing, so is this really just an assumption of the author, or do we really need to set $\Delta_F$ as $\Delta_E$ ? Note that the author is saying that "Then we have $\Delta_F = \Delta_F$", which totally confuses me.
I mean I tried with the case where $\Delta_F$ and $\Delta_E$ represents different orientations for $E$, and the result also changed with this choice, so is there anything I'm missing ? and is my results are correct ?

REPLY [1 votes]: In discussing orientation of linear transformations, one (naturally and customarily) fixes an orientation for each vector space under consideration. Consequently, if $E = F$, then $\Delta_{E} = \Delta_{F}$.


text:
\begin{document}
\parindent=0pt
\maketitle

\begin{abstract}
We present a new algorithm for the automatic one-shot generation of scattered node sets on irregular 2D and 3D domains using Poisson disk sampling coupled to novel parameter-free, high-order parametric Spherical Radial Basis Function (SBF)-based geometric modeling of irregular domain boundaries. Our algorithm also automatically modifies the scattered node sets \emph{locally} for time-varying embedded boundaries in the domain interior. We derive complexity estimates for our node generator in 2D and 3D that establish its scalability, and verify these estimates with timing experiments. We explore the influence of Poisson disk sampling parameters on both quasi-uniformity in the node sets and errors in an RBF-FD discretization of the heat equation. In all cases, our framework requires only a small number of ``seed'' nodes on domain boundaries. The entire framework exhibits $O(N)$ complexity in both 2D and 3D.
\end{abstract}
\begin{keywords}
Node generation, Geometric Modeling. Radial Basis Functions, Poisson disk sampling, Embedded Boundaries.
\end{keywords}

\begin{AMS}
  68Q25, 68U05
\end{AMS}

\input{Introduction}
\input{Overview}
\input{NodeGeneration}

\input{Discussion}
\section*{Acknowledgments}
The authors were supported by NSF DMS-1521748. VS was also supported by DMS-1160432.

\bibliographystyle{siamplain}
\bibliography{article}

\end{document}

config: arxiv


TITLE: Integral Method to find upper and lower bounds on the sum
QUESTION [0 upvotes]: I got this problem wrong on my exam and I know why I got it wrong but I am not sure how to find the right answer.
This is the question:
Consider$\sum_{k=1}^n (3+ 2/k^3)$. Use the integral method to obtain upper and lower bound on the sum. Do not split the summation into $\sum_{k=1}^n 3+\sum_{k=1}^n ( 2/k^3)$
This is the formula we had to use: Click here to see the approximation integral formula
This function is decreasing and I was able to calculate the lower bound, but I am not sure how to do the upper bound becuase this is what I get $ \int_{0}^{n}$(3+2/k^3)dx and after taking the antiderivative. I get 3k - (1/k^2) from 0 to n. so if I take the integral from 0 to n then 1/k^2 becomes undefined. I tried to change the integral from -1 to n but I got it wrong on the exam.
Could someone please tell me how to solve this? Thanks

REPLY [0 votes]: Hint. You may write
$$
\begin{align}
\sum_{k=1}^n\left(3+\frac1{k^3} \right)&=\left(3+\frac1{1^3} \right)+\sum_{k=\color{red}{2}}^n\left(3+\frac1{k^3} \right)
\\\\&\le \left(3+\frac1{1^3}\right)+\int_{\color{red}{2-1}}^n\left(3+\frac1{x^3} \right)dx
\end{align}
$$ then getting easily an upper bound.


TITLE: Counter example involving primary ideals (Atiyah-Mac Donald)
QUESTION [1 upvotes]: It is written in chapter $4$, page $51$ of Atiyah-Mac Donald's $\textit{Introduction to Commutative Agebra}$:

"a prime power $\mathfrak{p}^n$ is not necessarily primary, although its
  radical is the prime ideal $\mathfrak{p}$. For example, let $A = k[x, y, z]/(xy - z^2)$ and let $\bar{x}, \bar{y}, \bar{z}$ denote the images of $x, y, z$ respectively in $A$. Then $\mathfrak{p} = (\bar{x}, \bar{z})$ is prime
  (since $A/\mathfrak{p}\sim k[y]$, an integral domain); we have $\bar{x}\bar{y} = \bar{z}^2\in \mathfrak{p}^2$ but $\bar{x} \notin \mathfrak{p}^2$ and $\bar{y} \notin r(\mathfrak{p}^2) = \mathfrak{p}$; hence $\mathfrak{p}^2$ is not primary"

By definition, $I$ is a primary ideal $\iff \forall a,b\in A:(ab\in I \implies a\text{ or }b\in\sqrt{I})$. I don't understand why $\bar{x}\bar{y}$ is valid for Atiyah's argument, since $\bar{y}\bar{x}\in\mathfrak{p}^2$ and $\bar{x}\in \sqrt{\mathfrak{p}^2}=\mathfrak{p}$, therefore the proposition $(ab\in I \implies a\text{ or }b\in\sqrt{I})$ is satisfied. Shouldn't he find $ab\in \mathfrak{p}^2$ such that this proposition is $\textit{refuted}$? What am I missing?

REPLY [3 votes]: Your definition of primary ideal seems to be wrong. The definition is

$I$ is a primary ideal if and only if for all $a,b \in R$, we have $ab \in I \implies a \in I  \text{ or } b\in\sqrt{I}$.

Here $\bar x \bar y \in \mathfrak p^2 = I$, but $\bar x \not \in I$ and $\bar y \not \in \sqrt{I} = \mathfrak p$, showing that $I$ is not a primary ideal.

To understand what is wrong with your definition, consider the ideal $I = (x^2,xy) \triangleleft k[x,y]=R$. Its radical is $(x)$, and $I$ is not primary. However, if $ab \in I$, then $x$ divide $ab$, so that $x \mid a$ or $x \mid b$, which implies that $a \in \sqrt I$ or $b \in \sqrt I$. Therefore $I$ satisfies your definition, without being primary.
The definition of "primary ideal" seems to be not symmetric. See this question to have more details.


TITLE: A rectangle has diagonals $d$ and the acute angle between the diagonals $\alpha$, express...
QUESTION [0 upvotes]: Problem: A rectangle has diagonals with length $d$ and the acute angle between the diagonals $\alpha$. Let $\cos{\alpha}=p,$ express the length short side in terms of $p$ and $d.$
Attempt: Drawing the problem, it gives us an Isosceles with sides $d/2, \ d/2$ and let's denote the sought short side with $s:$

The law of cosines gives: $$s^2=\left(\frac{d}{2}\right)^2+\left(\frac{d}{2}\right)^2-2\cdot\frac{d}{2}\cdot\frac{d}{2}\cdot\cos{\alpha}=\frac{d^2}{2}(1-p).$$
So, $$s=\sqrt{\frac{d^2}{2}(1-p)}=\frac{d\sqrt{2}\sqrt{1-p}}{2}.$$
I cant figure out why the answer in the book is $$s=\frac{d\sqrt{2}\sqrt{1-p^2}}{2},$$
why $p^2$ instead of just $p$ ?

REPLY [1 votes]: The answer in the book is wrong. Your answer is correct.


TITLE: Is the sequence space over natural numbers $\ell^2(\mathbb{N})$ a Hilbert space?
QUESTION [1 upvotes]: In my homework for functional analysis, we had $\mathcal{H} = \ell^2(\mathbb{N})$. But the usual sequence space is defined to be a vector space over complex or real numbers. I'm not sure how $\ell^2(\mathbb{N})$ is a Hilbert space.

REPLY [1 votes]: $\ell^2(\mathbb{N})$ is the space of the (complex or real) sequences $x = (x_n)_{n\in\Bbb N}$ s.t.
$$\sum_{n\in\Bbb N}|x_n|^2 < \infty.$$
The scalar product is
$$(x,y) = \sum_{n\in\Bbb N}x_n\bar{y_n}.$$


TITLE: Markov process in interacting particle systems book
QUESTION [2 upvotes]: I have trouble understanding the setup of a Markov process in Liggett's interacting particle systems book.

Let $X$ be a compact metric space with measurable structure given by the $\sigma-$algebra of Borel sets. Let Let $D[0, \infty)$ be the set of all functions $\eta_.$ on $[0, \infty)$ with values in $X$ which are right continuous and have left limits. This is the canonical path space for a Markov process with state space $X$. A Markov process on $X$ is a collection $\{P^\eta, \eta \in X\}$ of probability
measures on $D[0, \infty)$ indexed by $X$. The expectation corresponding to $P^\eta$ will be denoted by $E^\eta$. Thus:
$$E^\eta Z = \int_{D[0,\infty)}ZdP^\eta\tag{1}$$
for any measurable function $Z$ on $D[0,\infty)$ which is integrable relative to $P^{\eta}$. let $C(X)$ be the collection of continuous functions on $X$. For $f\in C(X)$:
$$S(t)f(\eta):=E^\eta f(\eta_t) \tag{2}$$
where $S(t)$ is a linear operator on $C(X)$.
Questions:

$1)$ Let $S$ be a finite set and $X=\{0,1\}^S$ i.e. $X$ is a collection of configurations where each element in $S$ is a $0$ or $1$. In first para, does $\eta_.$ refer to a path through these configurations i.e. a path is an element of $D[0,\infty)$?

$2)$ If the above is right, then is $P^\eta$ a prob measure on paths or on individual configurations? I'm thinking on paths because in $(1)$, $Z$ is a function on $D[0,\infty)$ i.e. $Z:D[0,\infty)\to \mathbb{R}$.

$3)$ If $P^\eta$ is a prob measure on paths, I don't understand what $E^\eta f(\eta_t)$ means in $(2)$ because $\eta_t$ is the configuration at time $t$ and I don't see how $f(\eta_t)$ is a function on $D[0,\infty)$? because as defined in $(1)$, $Z$ has to be a function on $D[0,\infty)$.

REPLY [3 votes]: Concerning 1): Yes, as stated in the general setting, $\eta_.: \mathbb{R}_+ \ni t \mapsto \eta_t \in X$, so in your situation $\eta_.: t \mapsto \{0,1\}^S$, so indeed, each $\eta_t$ is a configuration.

Again, as in your general presentation, each $P^{\eta}$ is a measure on $D[0,\infty)$, so it is a probability measure on PATHS of configuration, i.e. on functions $\eta_.$ as above and not on single configurations. The parameter $\eta$ in the notation $P^{\eta}$ intuitively corresponds to the starting point of the measure $P^{\eta}$, i.e. usually $P^{\eta}$ has mass on paths $\gamma: [0,\infty) \to X$ with $\gamma_0 = \eta$.

The notation in (2) is slightly confusing, which causes your question. Precisely, for each $t \in \mathbb{R}_+$, $S(t): f \mapsto S(t)f$, where $S(t)f: \{0,1\}^S \to \mathbb{R}$ via $S(t)f(\eta) := \int_{\{0,1\}^S}f\circ\pi_tdP^{\eta},$where
$$\pi_t: D[0,\infty) \to \{0,1\}^S, \pi(\eta_.) := \eta_t$$denotes the canonical projection at time $t$. The slightly suggestive notation $f(\eta_t)$ in formula (2) should be read in this spirit: $\eta_.$ denotes a generic element in $D[0,\infty)$ and $\eta_t$ is its value at time $t$. The integral is taken over all paths, projects each path $\eta_.$ to $\eta_t$ and applies $f$ to it. Everything is rigorously well-defined.

I hope this clarifies your issues!


TITLE: Calculating the right angle vector on a line?
QUESTION [0 upvotes]: In the section, Inverse Kinetics in 2D of this article: http://freespace.virgin.net/hugo.elias/models/m_ik.htm

There is Vector R. How could I find this vector R? I know the length of the bone and its angle, how could I then calculate vector R in this example?
Thanks
Would it simply be taking the vector of the angle and returning v.y, -v.x?

REPLY [1 votes]: The base of the vector $R$ is just $(L\cos(\theta),L\sin(\theta))$ where $L$ is the length of the bone and $\theta$ is its angle, usually measured from horizontal to the left.  If the desired length of $R$ is $r$, the offset is $(-r\sin(\theta),r\cos(\theta))$, so the end point is $(L\cos(\theta)-r\sin(\theta),L\sin(\theta)+r\cos(\theta))$ .  Is this what you are after?  I don't see anything to set $r$ in the problem.


TITLE: Sufficient statistics for two-parameter binomial distribution
QUESTION [0 upvotes]: Given a family $K$ of Binomial distributions $B(\cdot|m,p)$ with two unknown parameters $m\in N$ and $0<p<1$. Find a sufficient statistics for $K$.
My attempt: The joint distribution of i.i.d random variables $X_1,\ldots, X_n$ is $f(X_1,\ldots, X_n|m,p) = {m \choose x_1}{m \choose x_2}\ldots {m \choose x_n}p^{x_1+\ldots + x_n}(1-p)^{nm-x_1-\ldots - x_n}$. But I could not find another term, besides $x_1+\ldots + x_n$ that would be a sufficient statistics for the unknown parameter $m$. The term $x_1+\ldots + x_n$ is only a sufficient statistics for $p$.
My question: Could someone please help me complete the proof above? I would sincerely appreciate any help or inputs.

REPLY [0 votes]: In case $mp\gg 1$ and $m(1-p)\gg 1$, the binomial distribution $B(m,p)$ approaches a Gaussian distribution $N(\mu=mp,\sigma^2=mp(1-p))$. Then the likelihood function
$$L(x_1,\cdots,x_n|\mu,\sigma^2)=\frac{1}{(2\pi)^{n/2}\sigma^n}\prod_{i=1}^n\exp\left(-\frac{(x_i-\mu)^2}{2\sigma^2}\right).$$
The data groups into
$$\frac{1}{2\sigma^2}\sum_{i=1}^n(x_i-\mu)^2=\frac{1}{2\sigma^2}\left(\sum_{i=1}^nx_i^2\right)-\frac{\mu}{\sigma^2}\left(\sum_{i=1}^nx_i\right)+\frac{\mu^2}{2\sigma^2}.$$
So under the Gaussian approximation, the two sufficient statistics are the sum $\sum_ix_i$ and the sum of squares $\sum_ix_i^2$, which could help determine $\mu$ and $\sigma^2$ (or equivalently $m$ and $p$).
In case $m\gg 1$ but $mp\rightarrow\mbox{const}$, the binomial distribution $B(m,p)$ approaches a Poisson distribution $\mbox{Poisson}(\mu=mp)$. Then the likelihood function
$$L(x_1,\cdots,x_n|\mu)=e^{-n\lambda}\prod_{i=1}^n\frac{\lambda^{x_i}}{x_i!}=\prod_{i=1}^n\frac{1}{x_i!}\times\lambda^{\sum_ix_i}\times e^{-n\lambda}.$$
Only one statistic, the sum $\sum_ix_i$ is needed and only one parameter $\lambda=mp$ can be determined. It is not possible to know $m$ and $p$ individually from a nearly Poisson distribution.
In case $m\gg 1$ does not hold, the binomial distribution $B(m,p)$ does not allow any approximation. Then the likelihood function
$$L(x_1,\cdots,x_n|m,p)=\frac{m!}{x_1!(m-x_1)!}\times\cdots\times\frac{m!}{x_n!(m-x_n)!}p^{\sum_ix_i}(1-p)^{mn-\sum_ix_i}.$$
The sum $\sum_ix_i$ is the sufficient statistic for $p$ if $m$ is known. If $m$ is not known, since the factorials $(m-x_1)!(m-x_2)!\cdots(m-x_n)!$ in the denomenator all factorize, there's no way to aggregate $x_1,\cdots,x_n$ to some statistic while still allowing the factorial product to be calculable. So I think the sufficient statistic is the entire data set $x_1, x_2,\ldots,x_n$.


TITLE: Sets and Quantifiers
QUESTION [1 upvotes]: I have difficulty converting sentences into 'mathematical language'.
According to my work above
1. Have I understood the problem correctly?
2. Is my answer correct?

REPLY [1 votes]: add:
$\land (\lnot [\forall x \in P , \forall t \in T , F(x,t)])$


TITLE: Dimension of the affine variety associated to $\langle zw-y^2, xy-z^3 \rangle $
QUESTION [5 upvotes]: Find the dimension of the affine variety $V(I)$, where $I=\left\langle zw-y^2,xy-z^3\right\rangle \subseteq k[x,y,z,w]$, with $k$  algebraicaly closed field.

I tried to solve the system $zw-y^2=0$, $xy-z^3=0$ and find the maximal dimension of coordinate subspaces, but I did not succeed. Could anyone help me? Thanks!

REPLY [1 votes]: A computational approach.
Set $R=k[x,y,z,w]$ and $I=\langle zw-y^2,xy-z^3\rangle$.

Show that $zw-y^2,xy-z^3$ is a Grobner basis for $I$ with respect to the grevlex order. Thus the initial ideal $\langle{\rm LT}(I)\rangle$ of $I$ is generated by $y^2,z^3$.
As it is well known (see e.g. Cox et al., Ideals, Varieties, and Algorithms, Chapter 9), if $k$ is an algebraically closed field $$\dim V(I)=\deg{}^aHP_{R/I},$$ where ${}^aHP_{R/I}$ is the affine Hilbert polynomial of $R/I$.

But ${}^aHP_{R/I}={}^aHP_{R/\langle{\rm LT}(I)\rangle}={}^aHP_{R/\sqrt{\langle{\rm LT}(I)\rangle}}$. In our case $\sqrt{\langle{\rm LT}(I)\rangle}=\langle y,z\rangle$ and thus we get $\dim V(I)=\deg{}^aHP_{R/I}=2$.


text:
\begin{document}

\begin{frontmatter}
\pretitle{Research Article}

\title{Large deviations of regression parameter estimator in
continuous-time models with sub-Gaussian noise}

\author{\inits{A.V.}\fnms{Alexander V.}~\snm{Ivanov}\ead[label=e1]{alexntuu@gmail.com}\orcid{0000-0001-5250-6781}}
\author{\inits{I.V.}\fnms{Igor V.}~\snm{Orlovskyi}\thanksref{cor1}\ead[label=e2]{orlovskyi@matan.kpi.ua}\orcid{0000-0003-0922-1611}}
\thankstext[type=corresp,id=cor1]{Corresponding author.}
\address{\institution{National Technical University of Ukraine}, <<Igor
Sikorsky Kyiv Polytechnic Institute>>,
Department of Mathematical Analysis and Probability Theory, Peremogi
avenue 37, 03056 Kiev, \cny{Ukraine}}

\markboth{A.V. Ivanov, I.V. Orlovskyi}{Large deviations of
regressionparameter estimator}

\begin{abstract}
A continuous-time regression model with a jointly strictly sub-Gaussian random
noise is considered in the paper. Upper exponential bounds for
probabilities of large deviations of the least squares
estimator for the regression parameter are obtained.
\end{abstract}

\begin{keywords}
\kwd{Continuous-time nonlinear regression}
\kwd{jointly strictly sub-Gaussian noise}
\kwd{least squares estimator}
\kwd{probabilities of large deviations}
\end{keywords}

\begin{keywords}[MSC2010]
\kwd[Primary ]{60G50}
\kwd{65B10}
\kwd{60G15}
\kwd[; Secondary ]{40A05}

\end{keywords}

\received{\sday{31} \smonth{1} \syear{2018}}

\revised{\sday{19} \smonth{4} \syear{2018}}

\accepted{\sday{22} \smonth{4} \syear{2018}}

\publishedonline{\sday{7} \smonth{5} \syear{2018}}
\end{frontmatter}

\section{Introduction}

Theory of large deviations in mathematical statistics and statistics of
stochastic processes deals with the asymptotic behaviour of tails of
distribution functions
of parametric and nonparametric statistical estimators. Concerning
parametric estimators it is necessary to refer to the monograph of
Ibragimov and Has'minskii~\cite{IbHas_SEAS} where the exponential
convergence rate of probabilities of large deviations for maximum
likelihood estimator was obtained. This result led to the appearance of
a large number of publications on the subject of large deviations of
statistical estimators.

Further we will speak about least squares estimators (l.s.e.'s) for
parameters of a nonlinear regression model. In the paper of Ivanov~\cite
{Iv_AE4DoLSEpNLRP}
a statement was proved on the power decreasing rate for
probabilities of large deviations of l.s.e.
for a scalar parameter in the nonlinear regression model with i.i.d.
observation errors having moments of finite order. Prakasa Rao~\cite
{Rao_oERCoLSEiNLRMwGE} obtained a similar result with the exponential
decreasing rate in the Gaussian nonlinear regression.

In the paper of Sieders and Dzhaparidze~\cite{SiDzha_LDR4PEaIA} a
general Theorem~2.1 on probabilities of large deviations for
$M$-estimators based on a data set of any structure was proved that
generalizes the mentioned result in \cite{IbHas_SEAS} with an
application to l.s.e. for parameters of the nonlinear regression with
pre-Gaussian and sub-Gaussian i.i.d. observation errors (Theorems 3.1
and 3.2 in \cite{SiDzha_LDR4PEaIA}). Some results in this direction are
obtained by Ivanov~\cite{Iv_AToNLR}.

The results on probabilities of large deviations
of an l.s.e. in a nonlinear regression model with correlated
observations can be found in the works of Ivanov and Leonenko~\cite
{IvLe_SAoRF}, Prakasa Rao~\cite{Rao_RoC4LSEiNLRMwDE}, Hu~\cite
{Hu_LDR4LSEiNR}, Yang and Hu~\cite{YaHu_LD4LSEiNRM}, Huang~et~al.~\cite
{Huetal_LD4LSEoNLRMBoWODE}.

Upper exponential bounds for probabilities of large deviations
of an l.s.e. for a parameter
of the nonlinear regression in discrete-time models with a jointly
strictly sub-Gaussian (j.s.s.-G.) random noise were obtained in
Ivanov~\cite{Iv_LDoRPEiMwSsGN}. In the present paper we extend some
results of~\cite{Iv_LDoRPEiMwSsGN} to continuous-time observation models.

Consider a regression model

\begin{equation}
X(t)=a(t,\,\theta)+\varepsilon(t),\quad t\ge0,
\\
\label{c_reg_m}
\end{equation}

where $a(t,\,\tau)$, $(t,\,\tau)\in\mathbb{R}_+\times\varTheta^c$, is a
continuous function, a true parameter value $\theta=(\theta_1,\ldots,\theta
_q)'$\ belongs to an open bounded convex set $\varTheta\subset\mathbb
{R}^q$ and a random noise $\varepsilon= \{\varepsilon(t),\ t\in
\mathbb{R} \}$ satisfies the following condition.

{\bf N1.} $\varepsilon$ is a mean-square and almost sure (a.s.)
continuous stochastic process defined on the probability space $
(\varOmega,\ \mathfrak{F},\ P )$, $E\varepsilon(t)=0$, $t\in\mathbb{R}$.

We shall write $\int=\int_0^T$.

\begin{defin}
Any random vector $\theta_T= (\theta_{1T},\,\ldots,\,\theta
_{qT} )'\in\varTheta^c$ having the property

\[
Q_T(\theta_T)=\underset{\tau\in\varTheta^c}
\inf{Q_T(\tau)},\: Q_T(\tau)=\int \bigl[X(t)-a(t,\,\tau)
\bigr]^2dt.
\]

is said to be the l.s.e. for an unknown parameter $\theta$, obtained by
the observations $ \{X(t),\ t\in[0,T] \}$.
\end{defin}

Under assumptions introduced above there exists at least one such
random vector $\theta_T$~\cite{Pf_oMaCoMCE}.

In the asymptotic theory of nonlinear regression in the problem of
normal approximation of the distribution
of an l.s.e., the difference $\theta_T-\theta$ is normed by diagonal
matrix~\cite{IvLe_SAoRF}

\[
d_T(\theta)=\diag \bigl(d_{iT}(\theta),\ i=\overline{1,q}
\bigr),\qquad d_{iT}^2(\theta)=\int\; \biggl(
\frac{\partial}{\partial\theta_i}a(t,\,\theta ) \biggr)^2dt.
\]

Further it is supposed that the function $a(t,\,\cdot)\in C^1(\varTheta)$
for any $t\ge0$.

The paper is organized in the following way. In Section \ref
{sec_LDiMwJSsGn} an upper exponential bound is obtained for large
deviations of $d_T(\theta) (\theta_T-\theta )$ in the
regression model \eqref{c_reg_m} with a j.s.s.-G. random noise
$\varepsilon$. In Section \ref{sec_LDitCoSJSsGN} the results of Section
\ref{sec_LDiMwJSsGn} are applied to a stationary j.s.s.-G. noise
$\varepsilon$. Section \ref{sec_2_Ex} contains examples of regression
functions $a$ and noise $\varepsilon$ satisfying the conditions of our theorems.

\section{Large deviations in models with a jointly strictly
sub-Gaussian noise}
\label{sec_LDiMwJSsGn}

\begin{defin}\label{defin_SsGRV}
A random vector $\xi= (\xi_1,\ \ldots,\ \xi_n )'\in\mathbb
{R}^n$ is called strictly sub-Gaussian (s.s.-G.) if for any $\varDelta
= (\varDelta_1,\ \ldots,\ \varDelta_n )'\in\mathbb{R}^n$

\[
E\exp \bigl\{\anglel \xi,\varDelta\angler \bigr\}\le\exp \biggl\{
\frac{1}2\anglel B\varDelta,\varDelta\angler \biggr\},
\]

where $\anglel \xi,\varDelta\angler =\sum_{i=1}^n\;\xi_i \varDelta_i$,
$B= (B(i,j) )_{i,j=1}^n$ is the covariance matrix of $\xi$,
that is $B(i,j)=E\xi_i\xi_j$, $i,j=\overline{1,n}$, $\anglel B \varDelta
,\varDelta\angler = \sum_{i,j=1}^n\;B(i,j)\varDelta_i \varDelta_j$.
\end{defin}

Note that we obtain from Definition \ref{defin_SsGRV} the definition of
an s.s.-G. random variable (r.v.) $\xi$ taking $n=1$.

\begin{defin} \label{defin_JSsGRP}
$ \{\xi(t),\ t\in\mathbb{R} \}$ is said to be jointly
strictly sub-Gaussian (j.s.s.-G.) sto\-chastic process, if for any $n\ge
1$, and any $t_1,\ \ldots,\ t_n\in\mathbb{R}$ the random vector $\xi
_n= (\xi(t_1),\ \ldots,\ \xi(t_n) )'$ is s.s.-G.
\end{defin}

These definitions and a more detailed information on sub-Gaussian
r.v.'s, vectors and stochastic processes can be found in the
monograph~\cite{BuKo_MCoRVRP} by Buldygin and Kozachenko.

Concerning the random noise $\varepsilon$ in the model \eqref{c_reg_m}
we introduce the following assumption.

{\bf N2(i)} $\varepsilon$ is a j.s.s.-G. stochastic process with the
covariance function $B(t,s)=E\varepsilon(t)\varepsilon(s)$, $t,s\in
\mathbb{R}$.

\hspace*{4mm}{\bf(ii)} For any $T>0$, $\varDelta(\cdot)\in L_2
([0,T] )$

\begin{equation}
\anglel B \varDelta,\varDelta\angler _T=\int\int\;B(t,s)\varDelta(t)
\varDelta (s)dtds\le d_0\|\varDelta\|^2_T
\label{inq_SP_By,y}
\end{equation}

for some constant $d_0>0$, $\|\varDelta\|_T= (\int\;\varDelta^2(t)dt
)^{\frac{1}2}$.

For a fixed $T$ the exact bound in \eqref{inq_SP_By,y} is

\[
\anglel B \varDelta,\varDelta\angler _T\le\|B\|_T\|
\varDelta\|^2_T,
\]

where $\|B\|_T$ is the norm of a self-adjoint positive semidefinite
operator $B$ in\break $L_2([0,T])$. Note that $\|B\|_T$ is a
nondecreasing function of $T>0$, so there exists

\[
\underset{T\to\infty}\lim\,\|B\|_T\le d_0<\infty,
\]

if \eqref{inq_SP_By,y} is fulfilled.

\begin{exm} Assume the covariance function $B(t,s)$ is such that

\[
\text{1)}\quad b_1^2=\int\limits
_0^\infty\int\limits
_0^\infty
\; B^2(t,s)dtds<\infty\qquad \text{or\quad 2)}\quad b_2=\underset{t\in
\mathbb{R}_+}\sup\; \int\limits
_0^\infty\;\bigl|B(t,s)\bigr|ds<\infty.
\]

Then using condition 1) and the Fubini theorem we get

\[

\begin{aligned} \anglel B \varDelta,\varDelta\angler _T&=
\int \biggl(\int\;B(t,s)\varDelta(s)ds \biggr)\varDelta(t) dt
\\
&\le \biggl(\int \biggl(\int\;B(t,s)\varDelta(s)ds \biggr)^2dt
\biggr)^{\frac{1}{2}} \|\varDelta\|_T\le b_1\|
\varDelta\|_T^2, \end{aligned}
\]

and we can take $d_0=b_1$. On the other hand,

\[
\anglel B \varDelta,\varDelta\angler _T\le\int\int\;\bigl|B(t,s)\bigr|
\varDelta^2(t)dtds \le b_2\|\varDelta\|_T^2,
\]

and we can take $d_0=b_2$.
\end{exm}

Let $\varDelta(t),\ t\in\mathbb{R}_+$, be a continuous function. Then
condition {\bf N1} implies the existence of the integral

\begin{equation}
I(T)=\int\;\varDelta(t)\varepsilon(t)dt, \label{def_I_T}
\end{equation}

determined for almost all paths of the process $\varepsilon(t),\ t\in
[0,T]$, as the Riemann integral. Consider partitions $r(n)$

\[
0=t_0^{(n)}<t_1^{(n)}<
\cdots<t_n^{(n)}=T
\]

of the interval $[0,T]$ such that $\max_{1\le k\le n}
(t_k^{(n)}-t_{k-1}^{(n)} )\to0$, as $n\to\infty$, and the
corresponding integral sums

\[
I_n(T)=\sum\limits
_{k=1}^n
\,u_k^{(n)}\varepsilon \bigl(t_k^{(n)}
\bigr),\qquad u_k^{(n)}=\varDelta \bigl(t_k^{(n)}
\bigr) \bigl(t_k^{(n)}-t_{k-1}^{(n)}
\bigr),\qquad k=\overline{1,n}.
\]

Then

\begin{equation}
I_n(T)\to I(T)\ \text{a.s.},\quad \text{as}\ n\to\infty. \label{I_nT_2_I_T_as}
\end{equation}

It is obvious also that

\begin{equation}
EI_n^2(T)\to EI^2(T)=\anglel B \varDelta,
\varDelta\angler _T,\quad \text{as}\ n\to \infty. \label{EI_nT^2_2_EI_T^2}
\end{equation}

\begin{lemma}\label{lema_Int_Dlta_Eps_SsG}
Under conditions {\bf N1}, {\bf N2} integral \eqref{def_I_T} is an s.s.-G.
r.v., for any $T>0$.
\end{lemma}

\begin{proof}
From Definition \ref{defin_JSsGRP} it follows that the process
$\varepsilon(t)$, $t\in[0,T]$, is j.s.s.-G., if for any $n\ge1$, and
$t_1,\ \ldots,\ t_n\in[0,T]$, $u_1,\ \ldots,\ u_n\in\mathbb{R}$,
$\lambda\in\mathbb{R}$,

\[
E\exp \Biggl\{\lambda\sum\limits
_{k=1}^n
\,u_k\varepsilon(t_k) \Biggr\}\le \exp \Biggl\{
\frac{1}2\lambda^2\sum\limits
_{i,j=1}^n
\,B(t_i,\, t_j)u_iu_j \Biggr
\}.
\]

Taking $t_k=t_k^{(n)}$, $u_k=u_k^{(n)}$, $k=\overline{1,n}$, we obtain

\[
E\exp \bigl\{\lambda I_n(T) \bigr\}\le\exp \biggl\{\frac{1}2
\lambda^2E I_n^2(T) \biggr\}.
\]

Due to \eqref{I_nT_2_I_T_as} and \eqref{EI_nT^2_2_EI_T^2} by the Fatou
lemma (see, for example,~\cite{GiSko_I2tToRP})

\begin{align*}
E\exp\bigl\{\lambda I(T)\bigr\}&=E\underset{n\to\infty}\lim\,\exp\bigl\{\lambda
I_n(T)\bigr\} \le\underset{n\to\infty}\liminf\,E\exp\bigl\{\lambda
I_n(T)\bigr\}
\\
&\le\underset{n\to\infty}\lim\,\exp \biggl\{\frac{1}2
\lambda^2E I_n^2(T) \biggr\}=\exp \biggl\{
\frac{1}2\lambda^2E I^2(T) \biggr\}.\qedhere
\end{align*}

\end{proof}

The following statement on the exponential bound
for distribution tails of integrals \eqref{def_I_T}
plays an important role in subsequent proofs.

\begin{lemma}\label{lema4P_I_T}
Under conditions {\bf N1} and {\bf N2} for any $T>0$, $x>0$,

\begin{equation}

\begin{aligned}
P \bigl\{I(T)\ge x \bigr\}&\le G_T(x),\qquad P
\bigl\{I(T)\le-x \bigr\}\le G_T(x),
\\
P \bigl\{ \bigl\llvert I(T) \bigr\rrvert\ge x \bigr\}&\le2G_T(x),
\end{aligned}
\label{inq_P_I_T}
\end{equation}

where

\[
G_T(x)=\exp \biggl\{-\frac{x^2}{2d_0\|\varDelta\|_T^2} \biggr\}.
\]

\end{lemma}

\begin{proof}
The proof is obvious (see, for example,~\cite{BuKo_MCoRVRP}). For any
$x>0$, $\lambda>0$ by the Chebyshev--Markov inequality, \eqref
{inq_SP_By,y}, and Lemma \ref{lema_Int_Dlta_Eps_SsG}

\begin{equation}
P \bigl\{I(T)\ge x \bigr\}\le\exp\{-\lambda x\}\exp \biggl\{\frac{1}2
\lambda ^2\anglel B \varDelta,\varDelta\angler _T \biggr\}
\le\exp \biggl\{\frac{1}2\lambda ^2d_0\|\varDelta
\|_T^2-\lambda x \biggr\}. \label{inq_P_I_T_d_lmba}
\end{equation}

Minimization of the right-hand side of \eqref{inq_P_I_T_d_lmba} in
$\lambda$ gives the first inequality in \eqref{inq_P_I_T}. The proof of
the second inequality is similar. The third inequality follows from the
previous ones.
\end{proof}

We need some notation to formulate conditions on the regression
function $a(t,\,\theta)$ using the approach of the paper~\cite
{SiDzha_LDR4PEaIA} (see also~\cite{Iv_AToNLR,IvLe_SAoRF}). Write

\[
U_T(\theta)=d_T(\theta) \bigl(\varTheta^c-
\theta \bigr),\qquad \varGamma_{T,\theta
,R}=U_T(\theta)\cap\{u\,:\,R\le
\|u\|\le R+1\},
\]

$u=(u_1,\ \ldots,\ u_q)'\in\mathbb{R}^q$. Denote by $G$ the family of
all functions $g=g_T(R)$, $T>0$, $R>0$, having the following properties:

1) for fixed $T$, $g_T(R)\uparrow\infty$, as $R\to\infty$;

2) for any $r>0$,

\[
R^r\exp \bigl\{-g_T(R) \bigr\}\to0,\quad \text{as}\ R,\,T\to
\infty.
\]

Let $\gamma(R)$ be polynomials of $R$ (possibly different) with
coefficients that do not depend on values $T,\ \theta,\ u,\ v$. Set also

\[

\begin{aligned} \varDelta(t,\,u)&=a\bigl(t,\,\theta+d_T^{-1}(
\theta)u\bigr)-a(t,\,\theta),\quad t\in [0,T],
\\
\varPhi_T(u,v)&= \int \bigl(\varDelta(t,\,u)-\varDelta(t,\,v)
\bigr)^2dt,\quad u,v\in U_T(\theta). \end{aligned}
\]

Assume the existence of a function $g\in G$, constants $\delta\in
(0,\frac{1}2 )$, $\varkappa>0$, $\rho\in(0,1]$ and polynomials
$\gamma(R)$ such that for sufficiently large $T,\ R$ (we will write
$T>T_0$, $R>R_0$) the following conditions are fulfilled.

{\bf R1(i)} For any $u,v\in\varGamma_{T,\theta,R}$ such that $\|u-v\|\le
\varkappa$

\begin{equation}
\varPhi_T(u,v)\le\gamma(R)\|u-v\|^{2\rho}; \label{inq_Phi_uv<=u-v_p}
\end{equation}

\hspace*{6mm}{\bf(ii)} for any $u\in\varGamma_{T,\theta,R}$ $\varPhi
_T(u,0)\le\gamma(R)$.

{\bf R2.} For any $u\in\varGamma_{T,\theta,R}$

\begin{equation}
\varPhi_T(u,0)\ge2d_0\delta^{-2}g_T(R).
\label{inq_Phi_u0_>=g_T}
\end{equation}

\begin{thm} \label{thm_PLD_sGN}
If conditions {\bf N1}, {\bf N2}, {\bf R1} and {\bf R2} are fulfilled, then there
exist constants $B_0$, $b_0>0$ such that for $T>T_0$, $R>R_0$

\begin{equation}
P \bigl\{ \bigl\llVert d_T(\theta) (\theta_T- \theta )
\bigr\rrVert \ge R \bigr\} \le B_0\exp \bigl\{-b_0g_T(R)
\bigr\}, \label{inq_gPLD_sGN}
\end{equation}

where for any $\beta>0$ the constant $B_0$ can be chosen so that

\begin{equation}
b_0\ge\frac{\rho}{\rho+q}-\beta. \label{inq4a_0dep_on_A_0}
\end{equation}

\end{thm}

\begin{proof}
Set

\[
I(T,\,u)=\int\;\varDelta(t,\,u)\varepsilon(t)dt,\qquad \zeta_T(u)=I(T,
\,u)-\frac{1}2\varPhi_T(u,0).
\]

To prove the theorem it is sufficient to check the fulfilment of
assumptions (M1) and (M2) of the Theorem 2.1 in \cite
{SiDzha_LDR4PEaIA}, reformulated in the manner similar to that used in
the proof of Theorem 3.1, ibid.:

{\bf(M1)} for any $m>0$ and $u,v\in\varGamma_{T,\theta,R}$ such that $\|
u-v\|\le\varkappa$,

\begin{equation}
E \bigl\llvert \zeta_T(u)-\zeta_T(v) \bigr\rrvert
^m\le\gamma(R)\|u-v\|^{\rho m}; \label{asum_M1_4LSE}
\end{equation}

{\bf(M2)}

\begin{equation}
P \biggl\{\zeta_T(u)-\zeta_T(0)\ge- \biggl(
\frac{1}2-\delta \biggr)\varPhi _T(u,0) \biggr\}\le\exp \bigl
\{-g_T(R) \bigr\}. \label{asum_M2_4LSE}
\end{equation}

From the first inequality in \eqref{inq_P_I_T} of Lemma \ref
{lema4P_I_T} for $\varDelta(t)=\varDelta(t,\,u)$, $x=\delta\varPhi_T(u,0)$,
condition {\bf R2}, taking into account that $\zeta_T(0)=0$ in our
particular case, we obtain

\[

\begin{aligned}
P \biggl\{\zeta_T(u)-
\zeta_T(0)\ge- \biggl(\frac{1}2-\delta \biggr)\varPhi
_T(u,0) \biggr\}
&=P \bigl\{I(T,\,u)\ge\delta\varPhi_T(u,0)
\bigr\}
\\
&\le\exp \bigl\{-\delta^2 (2d_0 )^{-1}
\varPhi_T(u,0) \bigr\}\\
&\le\exp \bigl\{-g_T(R) \bigr\},
\end{aligned}
\]

i.e. \eqref{asum_M2_4LSE} is true.

On the other hand,

\begin{align}
&E \bigl\llvert \zeta_T(u)-
\zeta_T(v) \bigr\rrvert ^m\notag\\
&\quad\le\max\bigl(1,2^{m-1}\bigr)\cdot \bigl(E \bigl\llvert I(T,\,u)-I(T,\,v)
\bigr\rrvert ^m+2^{-m} \bigl\llvert \varPhi_T(u,0)-
\varPhi_T(v,0) \bigr\rrvert ^m \bigr),

\label{1inq_E_dif_zeta^m}\\

&\bigl\llvert \varPhi_T(u,0)-
\varPhi_T(v,0) \bigr\rrvert \notag\\
&\quad\le\int\;\bigl|\varDelta(t,\,u)-\varDelta (t,
\,v)\bigr|\cdot\bigl|\varDelta(t,\,u)+\varDelta(t,\,v)\bigr|dt\notag\\
&\quad\le\varPhi_T^{\frac{1}2}(u,v)\cdot \bigl(\varPhi_T^{\frac{1}2}(u,0)+
\varPhi_T^{\frac{1}2}(v,0) \bigr)\notag\\
&\quad\le2\bigl(\gamma(R)
\bigr)^{\frac{1}2}\|u-v\|^\rho\bigl(\gamma(R)\bigr)^{\frac{1}2}\notag\\
&\quad\le\bigl(\gamma(R)+\gamma(R)\bigr)\|u-v\|^\rho=\gamma(R)\|u-v
\|^\rho\notag
\end{align}

according to {\bf R1} (polynomials $\gamma(R)$ are different in the
last two lines). Thus we obtain the bound

\begin{equation}
\bigl\llvert \varPhi_T(u,0)-\varPhi_T(v,0) \bigr\rrvert
^m\le\gamma(R)\|u-v\|^{\rho m}. \label{inq_dif_Phi^m}
\end{equation}

By the formula for the moments of nonnegative r.v. (see, for example,
\cite{Fe_I2PTaIA} and compare with \cite{Hu_LDR4LSEiNR}) and the third
inequality of Lemma \ref{lema4P_I_T} being applied to $\varDelta(t)=\varDelta
(t,\,u)-\varDelta(t,\,v)$, $t\in[0,T]$, it holds

\begin{equation}

\begin{aligned}
E \bigl\llvert I(T;u,v) \bigr\rrvert^m &=m\int\limits
_0^\infty\;x^{m-1}P \bigl\{ \bigl\llvert
I(T;u,v) \bigr\rrvert \ge x \bigr\}dx\\
&\le2m\int\limits
_0^\infty\;x^{m-1}\exp \biggl\{-
\frac{x^2}{2d_0\varPhi
_T(u,v)} \biggr\}dx \\
&=\sqrt{2\pi}m d_0^{\frac{m}2}
\varPhi_T^{\frac{m}2}(u,v)E|Z|^{m-1}, \end{aligned}

\label{inq_EI_T-uv^m}
\end{equation}

where $I(T;u,v)=I(T,\,u)-I(T,\,v)$, $Z$ is a standard Gaussian r.v.,

\begin{equation}
E|Z|^{m-1}=\pi^{-\frac{1}2}2^{\frac{m-1}2}\varGamma \biggl(
\frac{m}2 \biggr) ,\quad m>0. \label{E_StGaus^m-1}
\end{equation}

Relations \eqref{inq_EI_T-uv^m}, \eqref{E_StGaus^m-1}, and \eqref
{inq_Phi_uv<=u-v_p} lead to the bound

\begin{equation}
E \bigl\llvert I(T;u,v) \bigr\rrvert ^m\le2^{\frac{m}2}m\varGamma
\biggl(\frac{m}2 \biggr)d_0^{\frac{m}2}
\varPhi_T^{\frac{m}2}(u,v)\le\gamma(R)\|u-v\|^{\rho m}.
\label{f_inq_EI_T-uv^m}
\end{equation}

From \eqref{1inq_E_dif_zeta^m}, \eqref{inq_dif_Phi^m}, and \eqref
{f_inq_EI_T-uv^m} it follows \eqref{asum_M1_4LSE}.
\end{proof}

Suppose there exist a diagonal matrix $s_T=\diag (s_{iT},\
i=\overline{1,q} )$ with elements that do not depend on $\tau\in
\varTheta$, and constants $0<\underline{c}_i<\overline{c}_i<\infty$,
$i=\overline{1,q}$, such that uniformly in $\tau\in\varTheta$ for $T>T_0$

\begin{equation}
\underline{c}_i<s_{iT}^{-1}d_{iT}(
\tau)<\overline{c}_i,\quad i=\overline{1,q}. \label{inq4s_iT&d_iT}
\end{equation}

Then instead of the matrix $d_T(\theta)$ (at least in the framework of
the topic of this paper) it is possible to consider, without loss of
generality, the normalizing matrix $s_T$.

The next condition is more restrictive than {\bf R1} and {\bf R2},
however it is simpler due to requirement \eqref{inq4s_iT&d_iT}.

{\bf R3.} There exist numbers $0<c_0<c_1<\infty$ such that for any
$u,v\in U_T(\theta)=s_T(\varTheta^c-\theta)$ and $T>T_0$,

\begin{equation}
c_0\|u-v\|^2\le\varPhi_T(u,v)\le
c_1\|u-v\|^2. \label{inq4int_dif_H^2}
\end{equation}

It goes without saying that in the expression for $\varPhi_T(u,v)$ in
\eqref{inq4int_dif_H^2} we use the matrix $s_T^{-1}$ instead of
$d_T^{-1}(\theta)$.

A condition of the type \eqref{inq4int_dif_H^2} has been introduced in
\cite{Iv_AE4DoLSEpNLRP} and used in \cite
{Rao_oERCoLSEiNLRMwGE,SiDzha_LDR4PEaIA,Hu_LDR4LSEiNR} and other works.
The next theorem generalizes Theorem 3.2 from \cite{SiDzha_LDR4PEaIA}.

\begin{thm}\label{thm_sPLD_sGN}
Under conditions {\bf N1}, {\bf N2} and {\bf R3} there exist constants $B$,
$b>0$ such that for $T>T_0$, $R>R_0$

\begin{equation}
P \bigl\{ \bigl\llVert s_T (\theta_T- \theta ) \bigr
\rrVert \ge R \bigr\} \le B\exp \bigl\{-b R^2 \bigr\}, \label{inq_PLD_sGN}
\end{equation}

moreover for any $\beta>0$ the constant $B$ can be chosen so that

\begin{equation}
b\ge\frac{c_0} {8d_0(1+q)}-\beta. \label{inq4a_dep_on_A_0}
\end{equation}

\end{thm}

\begin{proof}
We will show that {\bf R3} implies

conditions {\bf R1} and {\bf R2}. Inequality \eqref{inq_Phi_uv<=u-v_p}
of the condition {\bf R1(i)} follows from the right-hand side of
inequality \eqref{inq4int_dif_H^2}, if we take $\rho=1$, $\gamma
(R)=c_1$. Inequality of the condition {\bf R1(ii)} follows as well from
the right-hand side of \eqref{inq4int_dif_H^2}, if we take $v=0$,
$\gamma(R)=c_1(R+1)^2$.

To check the fulfilment of condition {\bf R2} we should rewrite the
left-hand side of \eqref{inq4int_dif_H^2} for $v=0$:

\[
\varPhi_T(u,0)\ge c_0\|u\|^2
\ge2d_0\delta^{-2} \biggl(\frac{1}2\delta
^2d_0^{-1}c_0R^2
\biggr),
\]

i.e., in the inequality \eqref{inq_Phi_u0_>=g_T} one can take
$g_T(R)=\frac{1}2\delta^2d_0^{-1}c_0R^2$. In this case for the exponent
in \eqref{inq_gPLD_sGN} we have

\[
-b_0g_T(R)=- \biggl(\frac{1}2
\delta^2b_0d_0^{-1}c_0
\biggr)R^2.
\]

Now, since $\rho=1$ in \eqref{inq4a_0dep_on_A_0}, for any $\beta>0$ in
\eqref{inq_PLD_sGN} we can take

\[
b=b_\delta=\frac{1}2\delta^2b_0d_0^{-1}c_0
\ge\frac{\delta^2c_0}{
2d_0(1+q)}-\beta.
\]

By {\bf R3} and {\bf N1}, {\bf N2}, inequality \eqref{inq_Phi_u0_>=g_T} with
$g_T(R)=\frac{1}2\delta^2d_0^{-1}c_0R^2$ holds for any $\delta\in
(0,\frac{1}2 )$. We get inequality \eqref{inq4a_dep_on_A_0} as
$\delta\to\frac{1}2$.
\end{proof}

\section{Large deviations in the case of a stationary jointly strictly
sub-Gaussian noise}
\label{sec_LDitCoSJSsGN}

We impose an additional restriction on the noise process $\varepsilon$.

{\bf N3.} The stochastic process $\varepsilon$ is stationary with the
covariance function $B(t)=E\varepsilon(0)\varepsilon(t),\ t\in\mathbb
{R}$, and the bounded spectral density $f(\lambda),\ \lambda\in\mathbb{R}$:

\[
f_0=\underset{\lambda\in\mathbb{R}}\sup\;f(\lambda)<\infty.
\]

Under assumption {\bf N3} the following corollaries of the theorems
proved in Section \ref{sec_LDiMwJSsGn} are true.

\begin{cor} \label{cor_PLD_gSt_sGN}
If conditions {\bf N1}, {\bf N2(i)}, {\bf N3}, {\bf R1} and {\bf R2} are fulfilled, then
the statement of Theorem \ref{thm_PLD_sGN} is true with $d_0=2\pi f_0$.
\end{cor}

\begin{proof}
We just need to show that condition {\bf N2(ii)} is fulfilled. Indeed,
by the\break Plancherel identity,

\[
\anglel B \varDelta,\varDelta\angler _T=\int\limits
_{-\infty}^\infty
\;f(\lambda ) \biggl\llvert \int\;e^{i\lambda t}\varDelta(t) dt \biggr\rrvert
^2d\lambda\le2\pi f_0\| \varDelta\|_T^2.
\qedhere
\]

\end{proof}

\begin{cor} \label{cor_sPLD_gSt_sGN}
Under conditions {\bf N1}, {\bf N2(i)}, {\bf N3} and {\bf R3} the statement of
Theorem \ref{thm_sPLD_sGN} is true with inequality \eqref
{inq4a_dep_on_A_0} rewritten in the form

\begin{equation}
b\ge\frac{c_0} {16\pi f_0(1+q)}-\beta. \label{inq4a_dep_on_A_0_St}
\end{equation}

\end{cor}

Our next assumption is a particularization of the requirements {\bf N2}
and {\bf N3}.

{\bf N4(i).} The random noise $\varepsilon$ is of the form

\begin{equation}
\varepsilon(t)=\int\limits
_{-\infty}^t\;\psi(t-s)d\xi(s)=\int\limits
_0^\infty
\;\psi(s)d\xi(t-s), \label{int_repr_epsln}
\end{equation}

where $\xi= \{\xi(t),\ t\in\mathbb{R} \}$ is a mean-square
continuous j.s.s.-G. stochastic process with orthogonal increments,
$E\xi(t)=0$,

\[
E\bigl|\xi(t+s)-\xi(t)\bigr|^2=s,\quad t\in\mathbb{R},\ s>0;
\]

$\psi(t)=0$ as $t<0$ and

\[
\int\limits
_0^\infty\;\psi^2(t)dt<\infty.
\]

The stochastic integral in \eqref{int_repr_epsln} is understood as a
mean-square Stieltjes integral \cite{GiSko_I2tToRP}. The process $\xi$
is an integrated white noise, $\varepsilon$ can be considered as a
stationary process at the output of a physically realizable filter with
the covariance function (see ibid.)

\[
B(t)=\int\limits
_0^\infty\;\psi(t+u)\psi(u)du
\]

and the spectral density

\[
f(\lambda)= \bigl\llvert h(i\lambda) \bigr\rrvert ^2,\qquad h(i
\lambda)=(2\pi)^{-\frac{1}2}\int\limits
_0^\infty\;\psi(t)e^{-i\lambda t}dt.
\]

{\bf N4(ii).} $f_0=\sup_{\lambda\in\mathbb{R}} \llvert h(i\lambda
) \rrvert ^2<\infty$.

Obviously {\bf N4(ii)} holds, if $\int_0^\infty\, \llvert \psi
(t) \rrvert dt<\infty$.

\begin{lemma}\label{lema_Int2jssG}
If condition {\bf N4(i)} holds, then $\varepsilon$ in \eqref
{int_repr_epsln} is a j.s.s.-G. process.
\end{lemma}

\begin{proof}
Let $n\ge1$ be a fixed number and $t_1,\ \ldots,\ t_n$, $\varDelta_1,\
\ldots,\ \varDelta_n$ be arbitrary real numbers. It is necessary to prove that

\begin{equation}
E\exp \Biggl\{\sum\limits
_{k=1}^n\;\varDelta_k
\varepsilon(t_k) \Biggr\}\le \exp \Biggl\{\frac{1}2\sum
\limits
_{i,j=1}^n\;B(t_i-t_j)
\varDelta_i \varDelta _j \Biggr\}. \label{cndn4sGP4eps}
\end{equation}

Formula \eqref{int_repr_epsln} can be rewritten in the form

\[
\varepsilon(t)=\int\limits
_{-\infty}^\infty\;\psi(t-s)d\xi(s),\quad \psi (t)=0,\ \text{as}
\ t<0.
\]

Denote $\psi_k(s)=\psi (t_k-s )$, $k=\overline{1,n}$. Then

\[
\varepsilon (t_k )=\int\limits
_{-\infty}^\infty\;\psi_k(s)d
\xi (s),\quad k=\overline{1,n}.
\]

Let a sequence of simple functions

\[
\psi_k^{(m)}(s)=\sum\limits
_{l=1}^{r(k,m)}
\;c_{kl}^{(m)}\chi_{\pi
_{kl}^{(m)}}(s),\quad m\ge1,
\]

where $\pi_{kl}^{(m)}=[\alpha_{kl}^{(m)},\,\beta_{kl}^{(m)}
)$, $k=\overline{1,n}$, $l=\overline{1,r(k,m)}$, $\chi_A(s)$ is
indicator of a set $A$, approximate the function $\psi_k(s)$ in
$L_2(\mathbb{R})$:

\[
\int\limits
_{-\infty}^\infty\; \bigl\llvert \psi_k(s)-
\psi_k^{(m)}(s) \bigr\rrvert ^2ds\to0,\quad
\textnormal{as}\ m\to\infty.
\]

Then the sequences of random variables

\[
\varepsilon_k^{(m)}=\sum\limits
_{l=1}^{r(k,m)}
\; c_{kl}^{(m)} \bigl(\xi \bigl(\beta_{kl}^{(m)}
\bigr)- \xi \bigl(\alpha_{kl}^{(m)} \bigr) \bigr) = \int
\limits_{-\infty}^\infty
\;\psi_k^{(m)}(s)d\xi(s)
\]

mean-square converge to $\varepsilon(t_k)$ in $L_2(\varOmega)$:

\begin{equation}
E \bigl\llvert \varepsilon(t_k)-\varepsilon_k^{(m)}
\bigr\rrvert ^2\to0,\qquad k=\overline {1,n},\quad \text{as}\ m\to\infty.
\label{cnv_eps_m_2_eps_L2}
\end{equation}

For any fixed $m$, the random vector with coordinates $\varepsilon
_k^{(m)}$, $k=\overline{1,n}$ is s.s.-G. Indeed,

\[

\begin{aligned} \sum\limits
_{k=1}^m\,
\varDelta_k\varepsilon_k^{(m)}&=\sum
\limits
_{k=1}^m\, \varDelta_k\sum
\limits
_{l=1}^{r(k,m)}\; c_{kl}^{(m)} \bigl(
\xi \bigl(\beta _{kl}^{(m)} \bigr)- \xi \bigl(
\alpha_{kl}^{(m)} \bigr) \bigr)
\\
&=\sum\limits
_{k'=1}^{n'(m)}\,u_{k'}^{(m)}
\xi \bigl(\eta _{k'}^{(m)} \bigr), \end{aligned}
\]

where $u_{k'}^{(m)}$ are real numbers and $\eta_{k'}^{(m)}$ are
different real numbers. By condition {\bf N4(i)} the random vector with
coordinates $\xi (\eta_{k'}^{(m)} )$, $k'=\overline{1,n'(m)}$,
is s.s.-G., and therefore,

\begin{equation}

\begin{aligned} E\exp \Biggl\{\sum\limits
_{k=1}^m
\,\varDelta_k\varepsilon_k^{(m)} \Biggr\}&= E
\exp \Biggl\{\sum\limits
_{k'=1}^{n'(m)}\,u_{k'}^{(m)}
\xi \bigl(\eta _{k'}^{(m)} \bigr) \Biggr\}
\\
&\le\exp \Biggl\{\frac{1}2E \Biggl(\sum\limits
_{k'=1}^{n'(m)}
\, u_{k'}^{(m)}\xi \bigl(\eta_{k'}^{(m)}
\bigr) \Biggr)^2 \Biggr\} \\
&=\exp \Biggl\{\frac{1}2 \Biggl(\sum
\limits
_{k=1}^m\,\varDelta_k
\varepsilon_k^{(m)} \Biggr)^2 \Biggr\}.
\end{aligned}
\label{eps_k^m_is_SsG}
\end{equation}

From \eqref{cnv_eps_m_2_eps_L2} it follows that $\varepsilon
_k^{(m)}\overset{P}\to\varepsilon_{t_k}$, $k=\overline{1,n}$, as $m\to
\infty$, and thus there exists some subsequence of indexes $m^\prime\to
\infty$,
independent of $k$, such that $\varepsilon_k^{(m')}\to\varepsilon
_{t_k}$ a.s., $k=\overline{1,n}$, as $m^\prime\to\infty$.

Finally, by the Fatou lemma and \eqref{eps_k^m_is_SsG}

\[

\begin{aligned} E\exp \Biggl\{\sum\limits
_{k=1}^n
\;\varDelta_k\varepsilon(t_k) \Biggr\} &=E
\underset{m^\prime\to\infty}\lim\;\exp \Biggl\{\sum
\limits_{k=1}^n
\; \varDelta_k\varepsilon_k^{(m')} \Biggr\}
\\
&\le\underset{m^\prime\to\infty}\liminf\;E\exp \Biggl\{\sum
\limits
_{k=1}^n\;\varDelta_k\varepsilon_k^{(m')}
\Biggr\}\\
&\le \underset{m^\prime\to\infty}\lim\;\exp \Biggl\{
\frac{1}2E \Biggl(\sum\limits
_{k=1}^n\;
\varDelta_k\varepsilon_k^{(m')}
\Biggr)^2 \Biggr\}
\\
&=\exp \Biggl\{\frac{1}2E \Biggl(\sum\limits
_{k=1}^n
\;\varDelta_k\varepsilon (t_k) \Biggr)^2
\Biggr\}\\
&=\exp \Biggl\{\frac{1}2 \sum\limits
_{i,j=1}^n
\; B(t_i-t_j)\varDelta_i
\varDelta_j \Biggr\}. \end{aligned}
\]

So, we have obtained \eqref{cndn4sGP4eps}.
\end{proof}

\begin{cor} \label{cor_PLD_St_sGN}
If conditions {\bf N1}, {\bf N4(i)}, {\bf N4(ii)}, {\bf R1} and {\bf R2} are fulfilled,
then the conclusion of Theorem \ref{thm_PLD_sGN} is true with $d_0=2\pi f_0$.
\end{cor}

\begin{cor} \label{cor_sPLD_St_sGN}
If conditions {\bf N1}, {\bf N4(i)}, {\bf N4(ii)} and {\bf R3} are fulfilled, then
the conclusion of Theorem \ref{thm_sPLD_sGN} is true with a constant
$b$ satisfying \eqref{inq4a_dep_on_A_0_St}.
\end{cor}

Assume

\begin{equation}
\underset{T\to\infty} {\liminf}\;T^{-\frac{1}2}s_{iT}>0,\quad i=
\overline{1,q}. \label{liminf_T^-1/2s_iT}
\end{equation}

\begin{cor}\label{cor_LSE_Spd_of_Cnv}
Under conditions of Theorem \ref{thm_sPLD_sGN} or Corollaries \ref
{cor_sPLD_gSt_sGN}, \ref{cor_sPLD_St_sGN}, and \eqref
{liminf_T^-1/2s_iT} for any $\rho>0$, $\nu\in [0,\frac{1}2 )$,
and $T>T_0$

\begin{equation}
P \bigl\{ \bigl\llVert T^{-\frac{1}2}s_T (\theta_T-
\theta ) \bigr\rrVert \ge \rho T^{-\nu} \bigr\} \le B\exp \bigl\{-b\rho
T^{1-2\nu} \bigr\}. \label{inq_PLD4cons_rho_dlta}
\end{equation}

\end{cor}

\begin{proof}
To show \eqref{inq_PLD4cons_rho_dlta} it is sufficient to take $R=\rho
T^{\frac{1}2-\nu}$ in \eqref{inq_PLD_sGN}.
\end{proof}

For $\nu=0$ we arrive at a quite strong result on
the weak consistency of l.s.e. Similarly, in the conditions of
Corollary \ref{cor_LSE_Spd_of_Cnv} the following result on
probabilities of moderate deviations for l.s.e. holds: for any $h>0$, $T>T_0$

\[
P \bigl\{ \bigl\llVert s_T (\theta_T- \theta ) \bigr
\rrVert \ge h \ln^{\frac{1}2}T \bigr\} \le B T^{-bh^2}.
\]

Obviously, Gaussian stochastic processes $\varepsilon$ are j.s.s.-G.
ones, and all the previous results are valid for them.

\section{Two examples}
\label{sec_2_Ex}

In this section, we consider an example of a regression
function satisfying the condition {\bf R3} and an example of the
j.s.s.-G. process $\xi$ from expression \eqref{int_repr_epsln} in
condition {\bf N4(i)}.

\begin{exm} Suppose

\[
a(t,\,\tau)=\exp \bigl\{\bigl\anglel \tau,\,y(t)\bigr\angler \bigr\},
\]

where $\anglel \tau,\,y(t)\angler =\sum_{i=1}^q\,\tau_iy_i(t)$,
regressors $y(t)= (y_1(t),\,\ldots,\,y_q(t) )'$, $t\ge0$, take
values in a compact set $Y\subset\mathbb{R}^q$.

Let us assume

\begin{equation}
J_T= \biggl(T^{-1}\int\limits
_0^Ty_i(t)y_j(t)dt
\biggr)_{i,j=1}^q\ \to\ J= (J_{ij}
)_{i,j=1}^q,\quad \text{as}\ T\to\infty, \label{cnv_J_T_2_J}
\end{equation}

$J$ is a positive definite matrix. In this case the regression function
$a(t,\,\tau)$ satisfies condition {\bf R3}. Indeed, let

\[
H=\underset{y\in Y,\ \tau\in\varTheta^c}\max\,\exp \bigl\{\anglel y,
\,\tau \angler \bigr\},\qquad L=\underset{y\in Y,\ \tau\in\varTheta^c}\min
\,\exp \bigl\{ \anglel y,\,\tau\angler \bigr\}
\]

Then for any $\delta>0$ and $T>T_0$

\[
L^2 (J_{ii}-\delta )< T^{-1}
d_{iT}^2(\theta)< H^2 (J_{ii}+\delta
),\quad i=\overline{1,q},
\]

and according to \eqref{inq4s_iT&d_iT} we can take $s_T=T^{\frac{1}2}\mathbb{I}_q$
with the identity matrix $\mathbb{I}_q$ of order $q$.

For a fixed $t$

\[

\begin{aligned}
&\exp \bigl\{\bigl\anglel y(t),\theta+T^{-\frac{1}2}u
\bigr\angler \bigr\}-\exp \bigl\{ \bigl\anglel y(t),\theta+T^{-\frac{1}2}v\bigr
\angler \bigr\}
\\
&\quad =T^{-\frac{1}2}\sum\limits
_{i=1}^q
\,y_i(t)\exp \bigl\{\bigl\anglel y(t),\theta +T^{-\frac{1}2}
\bigl(u+\eta(v-u) \bigr)\bigr\angler \bigr\} (u_i-v_i ),
\ \eta\in(0,1), \end{aligned}
\]

and therefore for any $\delta>0$ and $T>T_0$

\[
\varPhi_T(u,v)\le H^2 \biggl(T^{-1}\int\;
\bigl\|y(t)\bigr\|^2dt \biggr)\|u-v\|^2< H^2 (\Tr J+
\delta )\|u-v\|^2,
\]

So we obtain the right-hand side of \eqref{inq4int_dif_H^2} with the
constant $c_1>H^2\Tr J$.

On the other hand, for a fixed $t$

\[

\begin{aligned}
& \bigl(\varDelta(t,u)-\varDelta(t,v) \bigr)^2
\\
&\quad = \bigl(\exp \bigl\{\bigl\anglel y(t),\theta+T^{-\frac{1}2}u\bigr
\angler \bigr\} -\exp \bigl\{\bigl\anglel y(t),\theta+T^{-\frac{1}2}v\bigr
\angler \bigr\} \bigr)^2
\\
&\quad =\exp \bigl\{2\bigl\anglel y(t),\theta+T^{-\frac{1}2}v\bigr\angler
\bigr\} \bigl(\exp \bigl\{\bigl\anglel y(t),T^{-\frac{1}2}(u-v)\bigr\angler
\bigr\}-1 \bigr)^2. \end{aligned}
\]

Since $ (e^x-1 )^2\ge x^2$, $x\ge0$, and $ (e^x-1
)^2\ge e^{2x}x^2$, $x<0$, it holds

\[
\bigl(\exp \bigl\{\bigl\anglel y(t),T^{-\frac{1}2}(u-v)\bigr\angler \bigr\}-1
\bigr)^2\ge L_tT^{-1}\bigl\anglel y(t),u-v\bigr
\angler ^2,
\]

with $L_t=\min \{1,\,\exp \{2\anglel y(t),T^{-\frac{1}2}(u-v)\angler  \} \}$, and

\[

\begin{aligned}
& \bigl(\varDelta(t,u)-\varDelta(t,v)\bigr)^2
\\
&\quad \ge\min \bigl\{\exp \bigl\{2\bigl\anglel y(t),\theta+T^{-\frac{1}2}v
\bigr\angler \bigr\},\,\exp \bigl\{2\bigl\anglel y(t),\theta+T^{-\frac{1}2}u
\bigr\angler \bigr\} \bigr\} T^{-1}\bigl\anglel y(t),u-v\bigr\angler
^2\\
&\quad>L^2 T^{-1}\bigl\anglel y(t),u-v\bigr\angler^2.
\end{aligned}
\]

Thus for any $\delta>0$ and $T>T_0$

\[
\varPhi_T(u,v)\ge L^2\bigl\anglel J_T(u-v),
\,u-v\bigr\angler >L^2 \bigl(\lambda_{\min
}(J)-\delta \bigr)
\|u-v\|^2,
\]

and we have obtained the left-hand side of \eqref{inq4int_dif_H^2} with
the constant $c_0<L^2\lambda_{\min}(J)$, where $\lambda_{\min}(J)$ is
the least eigenvalue of the positive definite matrix $J$.

The next fact is a reformulation of Corollary \ref{cor_sPLD_St_sGN} for
the regression function $a(t,\,\tau)$ of our example.

\begin{core}
Under conditions {\bf N1}, {\bf N4(i)}, {\bf N4(ii)} and \eqref{cnv_J_T_2_J} there
exist constants $B$, $b>0$ such that for $T>T_0$, $R>R_0$

\[
P \bigl\{ \bigl\llVert T^{\frac{1}2} (\theta_T- \theta ) \bigr
\rrVert \ge R \bigr\} \le B\exp \bigl\{-b R^2 \bigr\}.
\]

Moreover for any $\beta>0$ the constant $B$ can be chosen so that

\[
b\ge\frac{L^2\lambda_{\min}(J)} {16\pi f_0(1+q)}-\beta.
\]

\end{core}

\end{exm}

\begin{exm}
Here we offer an example of the j.s.s.-G. stochastic process $\xi$
with orthogonal increments in the formula \eqref{int_repr_epsln} using
the Ito--Nicio series (see \cite{ItoNi_otCoSoIBSVRV} and references therein).

Consider any orthonormal basis $ \{\varphi_k,\,k\ge1 \}$\ in\ $L_2 (\mathbb{R}_+ )$ and a sequence $\{Z_k,\break k\ge
1\}$ of independent $N(0,1)$ r.v.'s. Then

\[
w_0(t)=\sum\limits
_{k=1}^\infty
\,Z_k\int\limits
_0^t\,\varphi_k(u)du,\quad t
\ge0,
\]

is a standard Wiener process with covariances $Ew_0(t)w_0(s)=\min\{t,s\}
$. We need some kind of the Wiener process on the real line $\mathbb
{R}$. Let $ \{w_1(t),\,t\ge0 \}$, $ \{w_2(t),\,t\ge0
\}$ be two independent Wiener processes

of the following form:

\[
w_i(t)=\sum\limits
_{k=1}^\infty
\,Z_{ik}\int\limits
_0^t\,\varphi _k(u)du,\quad t
\ge0,\ i=1,2,
\]

and $ \{Z_{ik},\,k\ge1,\,i=1,2 \}$ be independent $N(0,1)$
r.v.'s. Then the required Wiener process on $\mathbb{R}$ can be defined
as $w(t)=w_1(t)$, $t\ge0$, and $w(t)=w_2(|t|)$, $t<0$. For any real
$t_1<t_2\le t_3<t_4$

\begin{equation}
E \bigl(w(t_2)-w(t_1) \bigr) \bigl(w(t_4)-w(t_3)
\bigr)=0, \label{indep_incr_exle}
\end{equation}

i.e. increments are orthogonal. On the other hand, for any $t>s$

\[
E \bigl(w(t)-w(s) \bigr)^2=t-s.
\]

Let $ \{\xi_{ik},\,k\ge1,\,i=1,2 \}$ be i.i.d. s.s.-G. r.v.'s
(and non-Gaussian!) with unit variance. Some examples of s.s.-G. r.v.'s
can be found in \cite{BuKo_MCoRVRP}. Thus the Bernoulli r.v. and the
r.v. uniformly distributed in $ [-\sqrt{3},\,\sqrt{3} ]$ are
s.s.-G. and have unit variances.

Let us introduce the stochastic processes

\[
\xi_i(t)=\sum\limits
_{k=1}^\infty\,
\xi_{ik}\int\limits
_0^t\,\varphi _k(u)du,\quad t
\ge0,\ i=1,2,
\]

$\xi(t)=\xi_1(t)$, $t\ge0$, and $\xi(t)=\xi_2(|t|)$, $t<0$. Then $\xi
= \{\xi(t),\,t\in\mathbb{R} \}$ is a process with orthogonal
increments and is not a Gaussian one.

However, it is a j.s.s.-G. process. To prove this statement consider
arbitrary numbers $t_1<\cdots<t_n$, where the first $m$ numbers, $0\le
m\le n$, are negative and the rest $n-m$ numbers are positive. Let
$\varDelta= (\varDelta_1,\,\ldots,\,\varDelta_n )'\in\mathbb{R}^n$ be
any vector. Then

\[

\begin{aligned} \varSigma_2&=\sum
\limits_{k=1}^N
\,\xi_{2k} \Biggl(\sum\limits
_{i=1}^m\,
\varDelta_i\int\limits
_0^{ \llvert t_i \rrvert }\,\varphi_k(u)du
\Biggr)\to\sum\limits
_{i=1}^m\,\varDelta_i
\xi_2 \bigl( \llvert t_i \rrvert \bigr)\ \text {a.s.},\quad
\text{as}\ N\to\infty,
\\
\varSigma_1&=\sum\limits
_{k=1}^N\,
\xi_{1k} \Biggl(\sum\limits
_{i=m+1}^n\,
\varDelta_i\int\limits
_0^{t_i}\,\varphi_k(u)du
\Biggr)\to\sum\limits
_{i=m+1}^n\,\varDelta_i
\xi_1 (t_i )\ \text{a.s.},\quad \text{as}\ N\to \infty.
\end{aligned}
\]

By the Fatou lemma

\begin{align*}
E\exp \Biggl\{\sum\limits
_{i=1}^n\,\varDelta_i
\xi(t_i) \Biggr\}&=E\underset {N\to\infty}\lim\,\exp \{
\varSigma_2+\varSigma_1 \}
\\
&\le\underset{N\to\infty}\liminf\,\prod\limits
_{k=1}^N
\, E\exp \Biggl\{ \xi_{2k} \Biggl(\sum\limits
_{i=1}^m
\,\varDelta_i\int\limits
_0^{ \llvert t_i \rrvert }\,\varphi_k(u)du
\Biggr) \Biggr\}\cdot
\\
&\qquad\cdot \prod\limits
_{k=1}^N\, E\exp \Biggl
\{\xi_{1k} \Biggl(\sum\limits
_{i=m+1}^n\,
\varDelta_i\int\limits
_0^{t_i}\,\varphi_k(u)du
\Biggr) \Biggr\}\le
\\
&\quad\le\underset{N\to\infty}\lim\,\exp \Biggl\{\frac{1}2 \Biggl(\sum
\limits
_{k=1}^N\, \Biggl(\sum\limits
_{i=1}^m
\,\varDelta_i\int\limits
_0^{ \llvert t_i \rrvert }\,\varphi_k(u)du
\Biggr)^2
\\
&\qquad+\sum\limits
_{k=1}^N\,
\Biggl(\sum\limits
_{i=m+1}^n\, \varDelta_i\int
\limits_0^{t_i}
\, \varphi_k(u)du \Biggr)^2 \Biggr) \Biggr\}\\

&\quad=\underset{N\to\infty}
\lim\,\exp \Biggl\{\frac{1}2 \Biggl(\sum\limits
_{i,j=1}^m
\, \Biggl(\sum\limits
_{k=1}^N\,\int\limits
_0^{ \llvert t_i \rrvert }
\,\varphi_k(u)du \int\limits
_0^{ \llvert t_j \rrvert }\,\varphi
_k(u)du \Biggr)\varDelta_i\varDelta_j
\\
&\qquad+\sum\limits
_{i,j=m+1}^n\,
\Biggl(\sum\limits
_{k=1}^N\,\int\limits
_0^{t_i}\,
\varphi_k(u)du \int\limits
_0^{t_j}\, \varphi_k(u)du
\Biggr)\varDelta_i\varDelta_j \Biggr) \Biggr\}.
\end{align*}

By Parseval's identity

\[

\begin{aligned} &\underset{N\to\infty}\lim\,\sum
\limits_{k=1}^N
\,\int\limits
_0^{ \llvert t_i \rrvert }\,\varphi_k(u)du \int\limits
_0^{ \llvert t_j \rrvert }
\,\varphi _k(u)du
\\
&\quad=\sum\limits
_{k=1}^\infty\,\int\limits
_0^\infty\,
\chi_{
[0,\, \llvert t_i \rrvert  ]}(u)\varphi_k(u)du \int\limits
_0^\infty\,
\chi_{ [0,\, \llvert t_j \rrvert  ]}(u)\varphi_k(u)du
\\
&\quad=\int\limits
_0^\infty\,\chi_{ [0,\, \llvert t_i \rrvert  ]}(u)\chi
_{ [0,\, \llvert t_j \rrvert  ]}(u)du=\min \bigl\{ \llvert t_i \rrvert ,\, \llvert
t_j \rrvert \bigr\}. \end{aligned}
\]

Similarly

\[
\underset{N\to\infty}\lim\,\sum\limits
_{k=1}^N\,\int
\limits_0^{t_i}
\, \varphi_k(u)du \int\limits
_0^{t_j}\,\varphi_k(u)du=
\min \{t_i,\, t_j \}.
\]

It means that

\[
E\exp \Biggl\{\sum\limits
_{i=1}^n\,\varDelta_i
\xi(t_i) \Biggr\}\le\exp \biggl\{ \frac{1}2\anglel B
\varDelta,\,\varDelta\angler \biggr\}
\]

with

\[
B=\lleft(
\begin{array}{cc} B_2 & 0 \\ 0 & B_1
\end{array}
 \rright),
\]

where $B_2= (\min \{ \llvert t_i \rrvert ,\,  \llvert t_j \rrvert  \}
 )_{i,j=1}^m$, $B_1= (\min \{t_i,\,t_j \}
)_{i,j=m+1}^n$, and the process $\xi$ is j.s.s.-G.
\end{exm}


TITLE: Existence of a Coend in a Monoidal Category Part II
QUESTION [1 upvotes]: I asked a very similar question but it was suggested that I rephrase and re-ask it.
Let $B$ be a monoidal category with multiplication $\Box$. Let $P$ be a category and let $T \colon P^\mathrm{op} \to B$ and $S \colon P \to B$ be functors. MacLane [CWM, p226] says that these two functors have a "tensor product"
$$ T \Box_P S = \int^{p\colon P} (Tp) \Box (Sp) .$$
It is not obvious to me that this coend exists. What are the correct assumptions for this to be true? Should it be

If $B$ is cocomplete and $P$ is small then...

or maybe

If $B$ is cocomplete and $P$ is any category...

I am very interested in the second case and I was wondering if this is known to be true or not. Note that it does not make sense to assume $B$ has all (co)limits since necessarily this would force $B$ to be a preorder. So cocomplete of course means "has all small colimits".

REPLY [1 votes]: You are maybe interested in observing that the functor tensor product
$$
T*S =\int^{p\in P} Tp\square Sp
$$
arises as a particular instance of the "nerve-realization" paradigm, namely the fact that there is an equivalence
$$
[\mathbf P^\text{op},\mathbf B](T, N_S(M))\cong {\bf B}(T*S,b)
$$
where $N_S(-)\colon b\mapsto \Big(p\mapsto \mathbf B(Sp,b)\Big)$.
In fact this is not the most general situation in which the aforementioned isomorphism holds: you can suppose $T,S$ do not share the same codomain, and instead ${\rm trg}(T)=\bf V$, a monoidal category where ${\rm trg}(S)=\mathbf B$ is enriched. If $\bf B$ is cocomplete and cotensored, namely there is a bifunctor
$$
-\boxtimes-\colon \bf V\times B\to B
$$
then you can define $T*S :=\int^p Tp\boxtimes Sp$.


TITLE: Why does the same proportion of a radioactive substance decay per time period? (half life)
QUESTION [65 upvotes]: Just wondering, if decay is random, why does the activity half every half life, as in, why does it have to reduce by the same proportion in the same time period?

REPLY [0 votes]: You may view radioactive decay as a random process. The half-life period tends to some constant value. It is because of a large number of atoms. This is not just true for half-life period but for any ratio period.
Consider an example below:
We define $T_{1/10}$ : Time after which 1/10th of isotope mass left of the radioactive substance.
Initial number of atoms N = 10 atoms
After completing first $T_{1/10}$ period number of un-decayed atoms be $X$ and it would follow binomial distribution $Binom(N,p)$.
$X \sim Binom(10,0.1)$ and P(X) the probability for $X$ atoms remaining for each of $X$ can be given in the following table.

X    P(X)
0    0.3486784401
1    0.3874204890
2    0.1937102445
3    0.0573956280
4    0.0111602610
5    0.0014880348
6    0.0001377810
7    0.0000087480
8    0.0000003645
9    0.0000000090
10   0.0000000001

Expected value (mean) for number of remaining atoms is 1. If we want to estimate how many atoms would remain with 99% confidence then we get values to be from 0 to 4 atoms i.e. $P(0 \le X \le 3)$. These values are (-100% to 300%) deviation from the mean value.
This means remaining mass can have -100% to 300% fluctuation from expected value.
With N = 1000 atoms,
Expected number of remaining atoms = 100
Possible remaining atoms from 75 to 125 with probability $P(75 \le X \le 125) = 0.9928133 $ has deviation from expected value $\pm 25\% $
With N = 10000 atoms,
Expected number of remaining atoms = 1000
Possible remaining atoms to make similar effect as above are from 920 to 1080 with probability $P(920 \le X \le 1080) = 0.9928133 $ has deviation from expected value $\pm 8\% $
With N = 100000 atoms,
Expected number of remaining atoms = 10000
Possible remaining atoms to make similar effect as above are from 9745 to 10255 with probability $P(920 \le X \le 1080) = 0.9929232 $ has deviation from expected value $\pm 2.55\% $
Here you can observe degree at which deviation is decreasing as initial number of atoms are increased. This deviation is also equivalent to deviation in expected value of mass.
At larger scale values of N are very high which are of the order of $6.022×10^{23}$ for each mol.
Hence not exactly the same proportion decays every time but the proportion value  saturates to a fixed value in the participation of very large number of radioactive atoms.


TITLE: Maxima/Minima of absolute function
QUESTION [2 upvotes]: Given $a_i=\{a_1,\dots,a_n\}$ and function
$$f(x)=\sum_{i=1}^n{|x-a_i|}^3$$
I need to find minimum value of $f(x)$. As far my understanding goes the derivative is given by:
$$f'(x) = \sum_{i=1}^n\frac{3*{(x-a_i)}^3}{{|x-a_i|}^3}$$
After this I have no clue how to solve $f'(x)=0$. Any suggestions?

REPLY [2 votes]: I have seen this problem on a programming competition and solved it as follows:

Sort $a_1\leq a_2 \leq ...\leq a_n$.
Consider your function on the following intervals: $(\infty, a_1], [a_1, a_2],...,[a_n,+\infty)$. On every such interval $f(x)$ is just a polynomial of degree at most $3$, because you can get rid of absolute values. E.g. on $[a_1,a_2]$ we have $|x-a_3|^3=(a_3-x)^3$.
For every interval minimize $f$ on it. This can be done by looking at the derivative etc.

Answer the minimum over all intervals.


TITLE: How to find Integrating factor of this equation
QUESTION [0 upvotes]: Equation
$$
    (x^2 e^{-y/x} + y^2 ) dx -xy dy =0
$$
Not Exact
Q) Further solution is possible ? means Integrating factor

REPLY [1 votes]: $$x^2e^{-y/x}+y^2=xy\frac{dy}{dx}$$
is a homogeneous differential equation, so the substitution $y=vx$ can make progress.
$$x^2e^{-v}+v^2x^2=vx^2\left(v+x\frac{dv}{dx}\right)$$
$$e^{-v}=vx\frac{dv}{dx}$$
$$\frac{dx}x=ve^vdv$$
$$\ln|x|=(v-1)e^v+C=\left(\frac yx-1\right)e^{y/x}+C$$