TITLE: If $f$ is strictly convex prove that $f(x + f'(x)) \geq f(x)$ for every $x$.
QUESTION [0 upvotes]: Let $f$(maps from $R$ to $R$) be twice differentiable function and strictly convex. Prove that for every $x$ from $R$ it is true that $f(x + f'(x))\geq f(x)$.
Let's suppose otherwise i.e. let $c$ be such that $f(c + f'(c))<f(c)$. We have that $f$ is strictly convex if and only if $f'$ is strictly increasing. Now let $f$ reach it's minimum in $l$ i.e. $f'(l)=0$. Since $f'$ is strictly increasing it is also strictly increasing on $(l,\infty)$. Now since $f'(x)>0$ for every $x$ from $(l,\infty)$ it follows that for every $x,y$ from $(l,\infty)$, where $x<y$ we have that $f(x)<f(y)$. So if $c$ is from $(l,\infty)$ we have that $c + f'(c) > c$ but $f(c + f'(c))<f(c)$, that is contradiction. We do almost same process from $(-\infty,l)$. If $f$ doesn't have a minimum it means that $f'(x)$ is either $f'(x)>0$ or $f'(x)<0$ for whole domain. Let $f'(x)>0$ for whole domain. This means that $f$ is strictly increasing on whole domain  i.e. if $x<y$ then $f(x)<f(y)$. Since $c + f'(c) > c$ and we have that $f(c + f'(c))<f(c)$ we reach a contradiction. We do almost same process for $f'(x)<0$. Hence it follows that for every $x$ from $R$ it is true that  $f(x + f'(x))\geq f(x)$.

REPLY [4 votes]: Yeah, I don't think there is any mistake :)
But you can can show this  more nicely/rigorously by use of Lagrange mean value theorem .
Indeed,
Suppose $f'(x) >0$ for some $x$ then,
Applying LMVT to interval , $[x,x+f'(x)]$, we have,
$$ f(x+f'(x))-f(x) =  f'(a)f'(x) $$
Now , since $a \in (x,x+f'(x)$, $f'(a)>f'(x)>0$ Thus,
$f(x+f'(x)) > f(x) $.
Do the same, assuming $f'(x) <0$ for some $x$.
And when $f'(x)=0$, this relation is obviously true (equality case)


TITLE: find the probability that any student who repeats one year and find the probability that he / she has repeating the second year;
QUESTION [1 upvotes]: Only one of the three years is repeat in the faculty (year $I$ or year $II$ or year $III$). The probability that the student will repeat the first year is $0.2$, the second year $0.5$, and the third year $0.25$. The probability of a student graduating a faculty while repeating the first year is $0.4$, graduating a faculty while repeating a second year is $0.5$, and graduating a faculty while repeating a third year is $0.9$.
$a)$ find the probability that any student who repeats one year (whether $I$, $II$ or $III$) will graduate from faculty
$b)$ if we know that the student has graduated from the faculty, find the probability that he / she has repeating the second year;
My attemp is: Write the set $A={\text {graduated the faculty}}$, and $B={\text {repeat the faculty}}$
$I^c,II^c,III^c,$ student repeat the first, second or third year in the faculty
$$P(I^c)=0,25$$
$$P(II^c)=0,5$$
$$P(III^c)=0,25$$
and
$$P(A|I^c)=0.4$$
$$P(A|II^c)=0.5$$
$$P(A|III^c)=0.9$$
I didnt know how to continue. Please help me. Thanks for your help. Thanky very much

REPLY [0 votes]: Here are some hints.
Start from first principles.  $$\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}$$
In part a), A is the event that the student repeats a year, and $B$ is the event that a student graduates.  For $A$, we have $$A=I^c\cup II^c\cup III^c$$ and we are given that $I^c, II^c, III^c$ are mutually exclusive.  Therefore,
$$\Pr(A\cap B) = \Pr(I^c\cap B)+\Pr(II^c\cap B)+\Pr(III^c\cap B)\tag1$$
Can you compute this probability?
For the denominator, we just have to add the probability that the student graduates without repeating any year to the numerator we computing in $(1)$.
What is that probability?
In part b), we want the probability that the student repeated the second year, given that he has graduated.  Since we know the probability that the student graduates, given that he repeats the second year, we can use Bayes' rule.
EDIT
In reply to OP's comment.
To compute $(1)$ we use $$\Pr(X\cap Y) = \Pr(X)\Pr(Y|X).$$  Let me just comment on why this definition makes intuitive sense.  In order that $X$ and $Y$ both occur, we first need $X$ to occur, and then, knowing that $X$ has occurred, we need $Y$ to occur.  ("First" and "then" aren't intended to imply that $X$ precedes $Y$ in time.)  So, to compute $\Pr(I^c\cap B)$ we have a  choice of computing
$$\Pr(I^c)\Pr(B|I^c)\tag2$$ or
$$\Pr(B)\Pr(I^c|B)\tag3$$
We are given the quantities in $(2)$, but not those in $(3)$, so naturally, we choose the first alternative.
$$\Pr(I^c\cap B)=\Pr(I^c)\Pr(B|I^c)=.2\cdot.4$$
In the same way, we get $$
\Pr(A\cap B) = .2\cdot.4+.5\cdot.5+.25\cdot.95=.5675$$
To finish part a), we need to know $\Pr(B).$  We've just computed the probability that a student who repeats a year graduates, so we need to add in the probability that a student graduates without repeating any year.  The probability that a student does not repeat a year is $$1-.2-.5-.25=.05$$
so that $\Pr(B)=.5675+.05=.6575+.05=.6175$ and
$$\Pr(A|B)=\frac{.5675}{.6175}\approx\boxed{.9190}$$


TITLE: Any algebraic number with modulus $1$ is root of a polynomial with positive coefficients
QUESTION [4 upvotes]: Given a complex number on the unit circle $e^{i\theta}\neq 1$ that is the root of some polynomial in $\mathbb Z[x]$, can we always construct a polynomial $p(x)$ with positive integer coefficients such that $e^{i\theta}$ is a root of $p(x)$?

REPLY [7 votes]: Nope, a counter-example are the complex numbers $\mu_{\pm} =-\frac{1 \pm i\sqrt{\varphi}}{\varphi}$.
It is easy to check $|\mu_{\pm}| = 1$ and $\mu_{\pm}$ are the roots of the irreducible polynomial $$q(x) = x^4 -2x^3 -2x^2 -2x + 1$$
This means $\mu_{\pm}$ are algebraic integers.
On the other hand, $q(x)$ have positive real roots $\nu_{\pm} = \varphi \pm \sqrt{\varphi}$. If any $\mu_{\pm}$ is root of $p(x) \in \mathbb{Z}[x]$, then $q(x) | p(x) \implies p(\nu_{\pm}) = 0$. This implies coefficients of $p(x)$ cannot be all positive.


TITLE: Flawed proof that real numbers are countable
QUESTION [1 upvotes]: I know there is something wrong with this argument, I am just not sure where.
Consider constructing the real numbers are equivalence classes of cauchy sequences of rationals. So in some sense the real numbers are a subset of equivalence classes of cauchy sequences of rationals so lets just consider those. In fact lets just consider all sequences of rationals.
A sequence of rationals has countably infinite many terms. Each term has countably infinite possibilities (any rational numbers). So the number of sequences is really just the same amount as
$$\bigotimes_{i=1}^\infty\mathbb{Q} $$
which would be countable.
Where is the flaw here?

REPLY [3 votes]: While the union of countably many countable sets is countable and the product of finitely many countable sets is countable, the product of countably many countable sets is not countable (unless all but finitely many of those sets are singletons, or at least one of those sets is empty).
So you are right that $$\mathbb{R}\equiv\prod_{i\in\mathbb{N}}\mathbb{Q}$$ ("$\mathbb{Q}^\mathbb{N}$" would also be appropriate notation here, but the symbol "$\otimes$" refers to something quite different). However, your error is that this does not imply that $\mathbb{R}$ is countable.

As a quick aside:
Let's say we have a set $X$ with some distinguished element $x_0$. We can consider the subset of $X^\mathbb{N}$ consisting of all sequences of elements of $X$ which are "mostly $x_0$," that is, such that all but finitely many elements of the sequence are $x_0$. This describes a countable set, and is basically analogous to how the countable set of reals with terminating (= eventually $0$) decimal expansions sits inside the uncountable set $\mathbb{R}$.
This sort of operation comes up a lot in abstract algebra; the relevant term is "direct sum" (as opposed to product), and this is denoted by "$\bigoplus$."


TITLE: 2-dimensional uniform distribution on a given set.
QUESTION [1 upvotes]: If $X,Y$ are uniformely  distributed on a quadrilateral with coordinates{$(0,0),(2,0),(1,1),(3,1)$}.
I want to find their marginal densities and $P(X>1|Y=\frac{1}{2}$).
I get that the joint pdf is  area$^{-1}$ multiplied by the  indicator over the domain of each variable and the double integral of this is equal to $1$.
In this shape that is given, the area is $2$ so our constant is $\frac{1}{2}$.
But I cant figure out how should I define the domain for the variables, should I make a piece wise function?
Also, is there a way to figure out the marginals just by looking at the shape of the given set?
Any help is greatly appreciated!

REPLY [1 votes]: Yes, there is a way to figure out the marginals just by looking at the shape of the set. The pdf of $X$ is a distance between lower and upper boundaries of quadrilateral, divided by the area of quadrilateral. That is
$$
f_X(x)=\frac12\times \begin{cases} x,  & 0\leq x <1,\cr
1, & 1\leq x < 2, \cr
1-(x-2), & 2\leq x \leq 3, \cr
0, & x\not\in[0,\,3].\end{cases}
$$
The pdf of $Y$ is a distance between left and right boundaries of quadrilateral, divided by area of quadrilateral:
$$
f_Y(y)=\frac12\times \begin{cases} 2,  & 0\leq y \leq 1,\cr
0, & y\not\in[0,\,1].\end{cases}
$$
This is just a consequence of definition of pdf and geometric probability. Say, pdf $f_X(x)$ is a function s.t. for any interval $(a,b)$
$$
\mathbb P(X\in (a,b)) = \int_a^b f_X(x)dx. \tag{1}
$$
We can restrict ourselves with $(a,b)\subset [1,\,3]$ only since outside this interval the pdf is zero. And this is the probability that the point which is chosen randomly in quadrilateral has $x$-coordinate in $(a,b)$. This is exactly ratio of two areas: area of the part of quadrilateral over $(a,b)$ and the total area which is $2$.
And the area of the part of quadrilateral over $(a,b)$ is the integral over $(a,b)$ from distance between upper boundary $u(x)$ and lower boundary $l(x)$. So,
$$
\mathbb P(X\in (a,b)) = \frac{\displaystyle\int_a^b (u(x)-l(x))\, dx}{2} \tag{2}
$$
Comparing (1) and (2) get $f_X(x)=\frac{u(x)-l(x)}{2}$.


TITLE: Least squares coordinate-wise smoothness
QUESTION [1 upvotes]: What is coordinate-wise smoothness
constant $L_i$ for least squares objective:
$$f(\mathbf{x}):=\frac{1}{2}\|A \mathbf{x}-\mathbf{b}\|^{2}$$
We define coordinate-wise smoothness as:
$$f\left(\mathbf{x}+\gamma \mathbf{e}_{i}\right) \leq f(\mathbf{x})+\gamma \nabla_{i} f(\mathbf{x})+\frac{L_i}{2} \gamma^{2} \quad \forall \mathbf{x} \in \mathbb{R}^{d}, \forall \gamma \in \mathbb{R}$$
A is a $m \times n$ matrix, $A=\left[\mathbf{a}_{1}, \ldots \mathbf{a}_{n}\right] \text { with columns } \mathbf{a}_{i}$
where $e_i$ is a standard basis vector, namely e.g: $e_i \in R^d$ is the vector where ith coordinate is 1 and all other are 0.
I seem to fail to get any of the possible answers. I plugged in the objective to the definition of coordinate-wise smoothness, my steps:
$$ ||A(x+\gamma e_i)-b|| = ||Ax + \gamma A e_i - b||$$
$$ ||Ax + \gamma A e_i - b|| = ||Ax-b + \gamma a_i||$$
by triangle inequality:
$$ ||Ax-b + \gamma a_i|| \leq ||Ax-b|| + \gamma||a_i||$$
$$ ||Ax-b + \gamma a_i||^2 \leq ||Ax-b||^2 + \gamma^2||a_i||^2 + 2||a_i||||Ax-b||$$
$$ \frac{1}{2}||Ax-b + \gamma a_i||^2 \leq \frac{1}{2}||Ax-b||^2 + \frac{\gamma^2}{2}||a_i||^2 + ||a_i||||Ax-b||$$
so it seems like it could be $||a_i||^2$ but the last term does not match ith coordinate of the gradient in my opinion so the first definition doesn't hold. On the other hand I found in other textbook that general smoothness constant for least squares is $\lambda_{max}(A^TA)$. Any guidance how to proceed or hint of what I am overlooking would be greatly appreciated, thanks!

REPLY [2 votes]: The smoothness parameter w.r.t. the $l_2$ norm is $\lambda_{\max}(A^T A)$.
The reason is that $f$ is L-smooth w.r.t. the $l_p$ norm if and only if $\|\nabla^2 f(\mathbf{x})\|_{p,q}\leq L$, where $p,q \geq 1$ and $\frac{1}{p}+\frac{1}{q}=1$. The proof of this Theorem is not very complicated and you can easily find it online.
Since in your case $\nabla^2 f(\mathbf{x}) = A^TA$ the result follows immediately. If you wish to look at it "component-wise", then simply look at the $i$-th column of A. Then an individual component $L_i = \lambda_{\max}(A_i^T A_i)$, and clearly $\forall i, \;L_i\leq L$.


TITLE: Confusion About Operators
QUESTION [2 upvotes]: Hello I am currently studying introductory QM and am confused about bases and operators. If I have an operator $\hat{Q}$, does this represent a change of basis matrix? In other words, does $\hat{Q} | \psi \rangle$ for any state $ | \psi \rangle$ simply represent the expansion of $|\psi \rangle$ in the eigenbasis of $\hat{Q}$?

REPLY [5 votes]: No, an arbitrary operator does not represent a change of basis. And even those that can be used to perform changes of basis should not always be interpreted as such.
A "change of basis" in a Hilbert space is usually meant to be a change from one orthonormal basis to another. The operators that map orthonormal systems to orthonormal systems are precisely the unitary operators. But to speak of a change of basis for a unitary operator you have to simultaneously transform all states by $\lvert\psi\rangle\mapsto U \lvert \psi \rangle$ and all operators by $O\mapsto UOU^\dagger$. Just applying a unitary operator to a state is not a "change of basis", it is transforming the state.
As an example, all symmetry operators are unitary operators. So when you rotate a state, you apply a unitary operator to it. The "proper" physical interpretation here is not that you changed you basis, it is that you rotated the state. What pointed into the $x$-direction before now points into the $y$-direction, and unless you applied the rotation to all states and operators, you may not take the passive view and declared that you just switched the basis labels $x$ and $y$.


TITLE: Example for: $Cov(X,Y)=0$ but $Cov(h(X),Y)\neq0$ for some function $h(.)$.
QUESTION [0 upvotes]: Can you help me find a simple example for this particular situation?
$X$ and $Y$ are random variables such that $Cov(X,Y)=0$ but $Cov(h(X),Y)\neq0$
for some function $h(.)$.
Thank you

REPLY [0 votes]: I was just checking Graham's answer.It's a really nice simple example, but I found different values:
Assume $X$ takes value $-1$ with probability $1/3,$ $0$ with probability
$1/3$ and 1 with probability $1/3$. Therefore
\begin{align*}
E\left(X\right)=-1\times\frac{1}{3}+0\times\frac{1}{3}+1\times\frac{1}{3}=0,
\end{align*}
\begin{align*}
E\left(X^{2}\right) & =\left(-1\right)^{2}\times\frac{1}{3}+0\times\frac{1}{3}+\frac{1}{3}\\
 & =\frac{2}{3},
\end{align*}
\begin{align*}
E\left(X^{3}\right) & =\left(-1\right)^{3}\times\frac{1}{3}+0\times\frac{1}{3}+\frac{1}{3}\\
 & =0
\end{align*}
and
\begin{align*}
E\left(X^{4}\right) & =\left(-1\right)^{4}\times\frac{1}{3}+0\times\frac{1}{3}+\frac{1}{3}\\
 & =\frac{2}{3}.
\end{align*}
Let $Y=X^{2}$. Then
\begin{align*}
Cov(X,Y) & =Cov\left(X,X^{2}\right)\\
 & =E\left(X^{3}\right)-E\left(X\right)E\left(X^{2}\right)\\
 & =0
\end{align*}
and
\begin{align*}
Cov(X^{2},Y) & =Cov\left(X^{2},X^{2}\right)\\
 & =E\left(X^{4}\right)-\left(E\left(X^{2}\right)\right)^{2}\\
 & =\frac{2}{3}-\left(\frac{2}{3}\right)^{2}\\
 & =\frac{2}{3}\left(1-\frac{2}{3}\right)\\
 & =\frac{2}{9}.
\end{align*}


TITLE: Does Cayley's formula count isomorphic trees?
QUESTION [2 upvotes]: Cayley's formula tells us that on $n$ vertices, there exist $n^{n-2}$ trees.
Now I am wondering if this formula considers each set of isomorphic trees as 1 or if it counts every one of them.

REPLY [4 votes]: First, note that there is no known closed form expression for the number of unlabeled trees up to isomorphism.
Cayley's Formula on the other hand gives us the number of labeled trees up to isomorphism. Note that in labeled case, for example the following graphs are not considered isomorphic although if they were unlabeled, we would consider them isomorphic. But if you look at their labeling carefully, they are all different.


TITLE: Endomorphisms of the ring $(\mathbb{Z}/p\mathbb{Z})^n$
QUESTION [2 upvotes]: Describe the set of Endormorphisms of the ring $ K = (\mathbb{Z}/p\mathbb{Z})^n$ where $p$ is prime.

I think my main difficulty is viewing $K$ as ring versus viewing it as a $\mathbb{Z}/p\mathbb{Z}$-vector space. Where viewing it as a vector space wouldn't be right because we don't multiply vectors. So $n \times n$ matrices wouldn't be right, right?

REPLY [2 votes]: The set of ring-endomorphisms of a commutative ring is just a monoid in general.
Let $X = \{1,\ldots, n\}$ and $k = \mathbb{F}_p$. Then there is an isomorphism of rings $\psi : K \to Fun(X,k)$ given by $\psi(v)(i) = v_i$ for all $i \in X$.
Now $Y \mapsto Fun(Y,k)$ is a contravariant functor from sets to commutative rings. This means that there is a monoid anti-homomorphism
$$\varphi : End(X) \to End(Fun(X,k))$$
given by
$$\varphi(\alpha)(f) = f \circ \alpha.$$
We will show that $\varphi$ is a bijection. It will follow that $End(K)$ is anti-isomorphic to the monoid of all functions $X\to X$ under composition.
Note that $Fun(X,k)$ is a $k$-vector space, with basis given by the delta functions $\{\delta_x : x\in X\}$ which are defined by $\delta_x(y) = 1$ if $y = x$ and $\delta_x(y) = 0$ otherwise.
If $\alpha, \beta : X\to X$ are distinct functions then $\alpha(x) \neq \beta(x)$ for some $x \in X$, but then $\delta_{\alpha(x)}\circ \alpha \neq \delta_{\alpha(x)} \circ \beta$ because the first takes value $1$ at $x$ and the second takes value $0$ at $x$. So $\varphi(\alpha)\neq \varphi(\beta)$ and $\varphi$ is injective.
If $\theta : Fun(X,k) \to Fun(X,k)$ is a ring homomorphism then $\theta(\delta_x)$ is an idempotent for all $x\in K$ so $\theta(\delta_x) = \delta_{S(x)}$ for some subset $S(x)$ of $X$. Since $\{\delta_x : x \in X\}$ forms a complete set of orthogonal idempotents in $Fun(X,k)$, their image under $\theta$, namely $\{\delta_{S(x)} : x\in K\}$, must also form a complete set of orthogonal idempotents. This means that $S(x) \cap S(y) = \emptyset$ if $x\neq y$ and $\cup_{x\in X}S(x) = X$. In other words, $\{S(x) : x \in X\}$ is a partition of $X$ into at most $n$ parts (some $S(x)$ could well be empty). Let $g : X \to X$ be defined by $g(y) = x$ if $y \in S(x)$. This is well-defined since $\{S(x) : x \in X\}$ is a partition, and then $S(x) = g^{-1}(\{x\})$ for all $x \in X$ by construction. Now, using the Kronecker delta we compute
$$[\varphi(g)(\delta_x)](y) = [\delta_x \circ g](y) = \delta_{x,g(y)} = \delta_{y \in g^{-1}(\{x\})} = \delta_{g^{-1}(x)}(y) = \delta_{S(x)}(y) = \theta(\delta_x)(y)$$
for all $x,y\in X$. Now, the subring of constant functions in $Fun(X,k)$ is the prime subfield of $Fun(X,k)$ because $k = \mathbb{F}_p$. So the restriction of $\theta$ to $k$ is the identity map. Hence $\theta$ is a $k$-linear, and therefore is completely determined by its values on the basis $\{\delta_x : x\in X\}$. Hence $\theta = \varphi(g)$ and $\varphi$ is surjective.
This can be viewed as a consequence of the equivalence of categories between the category of commutative rings and the opposite of the category of affine schemes.


TITLE: When is the spin of an electron changing?
QUESTION [1 upvotes]: An electron can have a half spin up or down. The up spin can become a down spin to lose his weak charge. But when are electrons changing their spin?

REPLY [0 votes]: If initially you had the spin projection in a certin (eigen)state and then you apply a spin-dependent interaction, the spin wave function becomes a superposition of up and down eigenstates. Measurements will find the up-state with a probability $P_{\text{up}}$ and the down-state with the probability $P_{\text{down}}=1-P_{\text{up}}$, the mean value being between up and down.


text:
\section{Elliptical confidence sets}
\label{sect:ellip}

 Confidence sets of an elliptical shape are one of the major type of confidence regions in statistical theory and applications. They are commonly constructed for parameters of (generalized) linear regression models, in ANOVA methods, and in various parametric models where a multivariate statistic is asymptotically normal. As for example, in the case of the score function considered in Section \ref{sect:score}. See, for instance, \cite{Friendly2013elliptical} for an overview of statistical problems and applications involving elliptical confidence regions.

In this section we construct confidence regions for an expected value of i.i.d. random vectors $\{X_{i}\}_{i=1}^{n}$,  using the bootstrap-based quantiles.
Since the considered set-up is rather general, the provided results can be used for various applications, where one is interested in estimating a mean of an observed sample in a nonasymptotic and multivariate setting. See, for example, \cite{ArlotBlanch2010}, where the authors constructed nonasymptotic confidence bounds in $\ell_{r}$-norm for the mean value of high dimensional random vectors and considered a number of important practical applications.

Let $W \in \R^{d\times d}$ be a p.d. symmetric matrix. $W$ is supposed to be known, it defines the quadratic form of an elliptical confidence set:
$B_{W}(x_{0},r)\define \{x \in \R^{d}:  (x-x_{0})^{T}W(x-x_{0})\leq r\},$
for $x_{0}\in \R^{d}, r \geq 0$.
There are various ways of how one can interpret and use $W$ in statistical models. For example, $W$ can serve for weighting an impact of residuals in linear regression models in the presence of errors' heteroscedasticity (cf. weighted least squares estimation); for  regularized least squares estimators in the linear regression model (for example, ridge regression), $W$ denotes a regularized covariance matrix of the LSE; see \citep*{Friendly2013elliptical} for further examples.

In Proposition \ref{prop:ellip2} below we construct an elliptical confidence region for $\E X_{1}$ based on the bootstrap approximation established in  Section \ref{section:ES_bootstr}.  Let $ \Xbar^{\ast}\define n^{-1}\sum_{j=1}^{n}\Xs_{j}$ for the i.i.d. bootstrap sample $\{\Xs_{j}\}_{j=1}^{n}$ generated from the empirical distribution of $\{X_{i}\}_{i=1}^{n}$. Let  also
$$q^{\ast}_{\alpha}\define \inf\{t>0: (1-\alpha) \leq \Pb( n^{1/2}\|W^{1/2} \Xbar^{\ast} \| \leq t)\}$$ denote $(1-\alpha)$-quantile of the bootstrap statistic $n^{1/2}\|W^{1/2} \Xbar^{\ast} \|$ for arbitrary $\alpha \in (0,1)$.
We assume that  coordinates of vectors $\{W^{1/2}(X_{i}-\E X_{i})\}_{i=1}^{n}$ are sub-Gaussian with variance factor $\sigma_{W}^{2}>0$ (i.e. condition \eqref{condit:expm1} is fulfilled).
Let also $d,n$ be such that

$ \sigma_{W}^{2}(d/\sqrt{n})C_{1\ast}
< \lambda_{\min}(W^{1/2}\Sigma W^{1/2})$ (for $C_{1\ast}$ defined in \eqref{def:tast_Cast}). Theorem \ref{theorem:bootst} implies the following statement

\begin{proposition}
\label{prop:ellip2}
If the conditions above are fulfilled, it holds
\begin{align*}
\bigl|\P\bigl( n^{1/2}\|W^{1/2} (\Xbar - \E X_{1})\|  \leq q^{\ast}_{\alpha} \bigr) - (1-\alpha)\bigr| &\leq \delta_{W}.
\end{align*}
$\delta_{W}$ is analogous to $\delta_{\BCl}$, where we take $\Sigma \define W^{1/2} \Sigma W^{1/2}$, $\sigma^{2}\define \sigma_{W}^{2}$, etc. A detailed definition of $\delta_{W}$ is given in \eqref{def:deltaW}, see also Remark \ref{remark:delta} for the discussion about its dependence on $d$ and n.
\end{proposition}

config: arxiv


TITLE: On relationships involving $\int_0^\infty \rho(u)e^{-u^2}du$ and $\int_0^\infty \rho'(u)e^{-u^2}du$, being $\rho(u)$ the Dickman function
QUESTION [0 upvotes]: In this Wikipedia is defined the so-called Dickman–de Bruijn function. If we assume that $$I:=\int_0^\infty \rho(u)f(u)du$$ is convergent, for example you can think in the function $f(u)=e^{-u^2}$, and we assume that $$J:=\int_0^\infty \rho'(u)f(u)du$$ is also convergent, I've curisosity if there is a way for which one can dilucidate some relationship involving $I$ and $J$.
I don't know if previous exercise was in the literature. In order to establish a clear question, I consider next particular case

Question. Let $\rho(u)$ the Dickman function. If we define $$I:=\int_0^\infty \rho(u)e^{-u^2}du$$ and $$J:=\int_0^\infty \rho'(u)e^{-u^2}du,$$  how can we find a concise relationship between $I$ and $J$? I am asking about identities, or well inequalities involving these real numbers $I$ and $J$. Thanks in advance.

REPLY [1 votes]: In a concise way, the Dickman function can be defined as the continuous function $\rho:\mathbb{R}^+\to (0,1)$ which equals $1$ on the interval $(0,1)$ and fulfills the differential equation $\rho'(u) = -\frac{1}{u}\rho(u-1)$. In particular $I<\int_{0}^{+\infty}e^{-u^2}\,du=\tfrac{1}{2}\sqrt{\pi}$ is trivial and
$$\begin{eqnarray*} I &=& \int_{0}^{1}e^{-u^2}\,du +\int_{1}^{+\infty}\rho(u)e^{-u^2}\,du\\&=&\sqrt{\pi}\,\text{Erf}(1)+\tfrac{\sqrt{\pi}}{2}\int_{1}^{+\infty}\rho(u-1)\frac{\text{Erf}(u)}{u}\,du\end{eqnarray*} $$
can be deduced from integration parts and leads to better bounds.
$J$ can be managed in a similar way,
$$ J=\int_{1}^{+\infty}\rho'(u)e^{-u^2}\,du = -\int_{0}^{+\infty}\rho(u)\frac{e^{-(u+1)^2}}{(u+1)}\,du $$
leads to $J\in\left(-\frac{1}{9},0\right)$ and better bounds can be deduced by applying integration by parts and recalling the differential equation fulfilled by $\rho$.