TITLE: What about a surface determines its color?
QUESTION [6 upvotes]: Light falls on a surface. Some wavelengths get absorbed. The other are reflected. The reflected ones are the colors that we perceive to be of the surface.
What is the property that determines, what wavelengths are reflected and what are absorbed? Is it electronic configuration of the object on which the light falls?
If yes, then if we know the electronic configuration of a surface can we make a model, which will predict the color it will show?

REPLY [0 votes]: Paul G. Hewitt has a great non-mathy description of this in his book Conceptual Physics.
Strike a tuning fork and it vibrates at a characteristic frequency, its Natural Frequency. The tuning fork might put out other frequencies but they are dampened faster than the target frequency. Dampening sucks energy out of vibration and becomes thermal if it doesn't become sound.
You can think of many things as being made up of tiny tuning forks. So hitting a metal sounds different from hitting wood.
Thes stiffness of a spring can determine the frequency of its vibrations. Again it has a natural frequency.
Like these examples, you can think of atoms and molecules as tiny optical tuning forks that re-emit some of the light and absorb the rest. Some light passes through.
The color you see depends on various features of the atomic and molecular interactions. For example, how strongly bounded an electron is to its parent atom is roughly analogous to the stiffness of a spring.
Vibrations in general can often be approximated through the analysis of Simple Harmonic Motion.
This is why Planck was able to analyze the interaction of Black Body Radiation with matter by assuming the radiation interacting with tiny "Harmonic Oscillators" in the surface of the black body.


TITLE: Eigenvectors 2x2
QUESTION [2 upvotes]: I have matrix $ \begin{bmatrix}
 1&0 \\
 4&1
\end{bmatrix} $ with repeated eigenvalue of 1. How do I find the eigenvectors? All I seem to get is
$ \begin{bmatrix}
0\\
0
\end{bmatrix} $

REPLY [3 votes]: Suppose $(x_1,x_2)^{T}$ is an eigenvector, then
$$\left[\begin{array}{cc}1&0\\4&1\end{array}\right]\left[\begin{array}{c}x_1\\x_2\end{array}\right]=\left[\begin{array}{c}x_1\\x_2\end{array}\right]$$
The system of equation reduces to $$\begin{eqnarray*}x_1&=&x_1\\4x_1+x_2&=&x_2\end{eqnarray*}$$
From the second equation we get $x_1=0$ and $x_2$ can be any real. Thus the eigenvectors corresponding to $1$ will look like $(0,x)^{T}$.


TITLE: show that $\mathbb{E}[\sup_{0 \leq s \leq t} M^4_s] = \frac43 \mathbb{E}[M_t \sup_{0 \leq s \leq t} M^3_s] $
QUESTION [0 upvotes]: Show that $\mathbb{E}[\sup_{0 \leq s \leq t} M^4_s] = \frac43 \mathbb{E}[M_t \sup_{0 \leq s \leq t} M^3_s] $,
  where $M = (M_t)_{t \geq 0}$ is a non-negative, continuous martingale starting from $0$.

I have already shown that :
$a \mathbb{P}(\sup_{s \in [0, t]} M_s \geq a) = \mathbb{E}[M_t \mathbb{1}_{\{ \sup_{s \in [0, t]} M_s \geq a \}} ] \,\,\, \forall (a, t) \in \mathbb{R}^2_{+} - \{0,0\}$
It's supposed to help me prove the equality in the title.
So I integrated both sides :
\begin{align} & \frac12\int_{0}^{\infty} 2a \mathbb{P}(\sup_{s \in [0, t]} M_s \geq a) da =\int_{0}^{\infty} \mathbb{E}[M_t \mathbb{1}_{\{ \sup_{s \in [0, t]} M_s \geq a \}} ]  da \\
\iff & \frac12 \mathbb{E}[(\sup_{s \in [0, t]} M_s)^2] = \int_{0}^{\infty} \mathbb{E}[M_t \mathbb{1}_{\{ \sup_{s \in [0, t]} M_s \geq a \}} ]  da \\
\iff &  \frac12 \mathbb{E}[\sup_{s \in [0, t]} M_s^2] = \int_{0}^{\infty} \mathbb{E}[M_t \mathbb{1}_{\{ \sup_{s \in [0, t]} M_s \geq a \}} ]  da \\ \end{align}
Since the property of being a martingale is weakened down to being just a sub-martingale when squaring the martingale then we can't replace $M$ by $M^2$ and thus I don't how else to make the to the fourth power term appear, I also have no idea how to deal with the RHS.
Any help will be greatly appreciated, thanks !

REPLY [2 votes]: If $M_t$ is a nonnegative continuous martingale starting from $0$ then $M_t\equiv 0$ a.s. for all $t$ and thus both sides are $0$.


TITLE: Prove of equality of function and composition
QUESTION [1 upvotes]: $V, W$ are finite dimensional $\mathbb{K}$-Vectorspaces. $\,\,f: V \rightarrow W$ is an Isomorphism. $\,\,q_v \in \text{Quad}(V), \,\,q_w \in \text{Quad}(W)$.
It is known, that:
$$
q_v(v_1 + v_2) - q_v(v_1) - q_v(v_2) = (q_w \circ f)(v_1+v_2) - (q_w \circ f)(v_1) - (q_w \circ f)(v_2)
$$
for every $v_1, v_2 \in V$. How can one prove that $q_v = q_w \circ f$? My first guess was setting $v_1$ or $v_2$ equal to zero, but then the equation collapses into $0 = 0$.

REPLY [1 votes]: Try setting $v_2 = v_1 = v$. Then your equality says:
$$
\begin {align}
q_V(2v) - 2q_V(v) &= q_W(f(2v)) - 2q_W(f(v)) \\
2q_V(v) &= 2q_W(f(v))
\end {align}
$$


TITLE: Uncertainty Principle for Compound Objects Interpretation
QUESTION [1 upvotes]: The most common misstatement of the HUP seems to be that it's a constraint on knowledge of a system or particle, rather than a description of the behavior of the system itself.  E.g. an electron is completely described by the wavefunction in a hydrogen orbital, rather than having a definite position within the orbital that we discover by measurement.
While this seems intuitively easy to grasp when considering electrons and individual particles, I struggle with how to conceptualize this "completely described by the wavefunction" existence for compound objects like protons, nuclei, and molecules that need to maintain strict geometric relations to one another.
Put another way:
I can visualize that the electron IS the wavefunction describing the electron field in the double slit experiment, which is why interference effects are observed.
For a compound object like a molecule that still displays interference effects, I'm finding it difficult to reconcile the uncertainty principle that limits localization of these tightly bound atoms, and yet allows for interference effects.
How does the HUP allow for the tightly geometrically confined configurations of atoms within molecules, and simultaneously for large compound objects like molecules to be de-localized to the extent they can exhibit interference effects in a double slit setup?

REPLY [1 votes]: Suppose you take two weights and put them between two walls, and attach all of them with springs. So it's wall, spring, weight, spring, weight, spring, wall. This system will have two oscillation modes, and these two oscillation modes can be excited independently. If we keep it simple and say that the weights have the same mass, and all the springs have the same spring constant, then the weights could oscillate back and forth in unison (so the distance between them stays the same), or they could oscillate towards and away from each other (so the position of their center of mass stays constant). Every oscillation of the system can be written as a linear combination of these two oscillations. The set of possible oscillations is a vector space, and those two oscillations make up a basis for the space. Thus, we can analyze the system as being an "atom" of two weights, and the atom has an excitation amplitude, and there's an independent excitation amplitude within the "atom"; the two amplitudes are independent and can be analyzed separately.
Similarly, when you have a hydrogen atom, there will be a space of quantum states the system can be in. We can choose a basis for this system in which all the excitations within the atom (that is, the electron orbitals) are elementary vectors. We can then treat the entire system as being a state of the entire atom combined with a state within the atom; those two types of states can be separated into two linearly independent spaces. When we have an entire molecule, the number of internal states increases, but the basic principle remains. I'll continue discussing a hydrogen atom, rather a molecule, since the analysis is simpler
Since a hydrogen atom can be split into external states and internal states, it's quite possible for the position uncertainty to be quite high for the external state, while quite low for the internal state. That is, it's quite possible that we have very little idea where the proton is, and very little idea where the electron is, but still have a very good idea what the distance between them is. Even though the proton is a "blur", and the electron is a "blur", these are not distinct blurs. These are highly correlated blurs; measuring the position of the proton highly localizes the position of the electron, and vice versa. The two particles are entangled; there isn't a position for one particle that's independent of the position of the other. Neither particle has its own wavefunction; the two-particle system has an overall wavefunction, and that wavefunction has a "where is the atom" part and a "where are the two particles in relation to each other" part.
Note that if you had a free electron and a free proton, the two particles would make up a system that would still have an overall wavefunction, but this overall wavefunction could be split into a "electron part" and a "proton part" that would be largely distinct. It's once they form an atom that the wavefunction can  no longer be approximated by treating it as a simple decomposition into two wavefunctions that each deal with only one particle.


TITLE: Stokes' theorem for a flat surface
QUESTION [0 upvotes]: Let $\vec{F}=\langle x^2,y^2,-z\rangle$. I want calculate the line integral of $F$ on curve $C$ which is a closed loop consisting of the line segments that form the triangle with vertices $(0,0,0), (0,2,0),(0,0,2)$.
Since $C$ is a simple, closed, smooth curve with positive orientation that encloses an oriented smooth surface, the Stokes' Theorem gives $$\oint_C\vec{F}\cdot d\vec{r}=\iint_S\text{curl}\vec{F}\cdot d\vec{S}$$ but $$\text{curl}\vec{F}=\vec{0}$$ since it's a conservative vector field (there exists its potential function). Thus $$\oint_C\vec{F}\cdot d\vec{r}=\iint_S\vec{0}\cdot d\vec{S}=0.$$ Is it correct?

REPLY [1 votes]: Yes, everything is fine, well-done.


TITLE: Type of discontinuities in Dirichlet function
QUESTION [1 upvotes]: I notice three types of discontinuities "removable", "jump", "infinite" defined here Classification of Discontinuities involve limit. Then I am puzzled that what is the type of discontinuities in the Dirichlet function, in which all rational numbers are mapped to 1 and irrational numbers are mapped to 0? They seem not to fall in any of these three categories? Thank you.

REPLY [2 votes]: The classification you link to defines an "infinite discontinuity" to mean the case where "one or both of the limits $L^{-}$ and $L^{+}$ does not exist or is infinite".
That's precisely what happens with the Dirichlet function everywhere.


text:
\begin{document}

\title[Inequalities in noncommutative probability spaces]{Inequalities for sums of random variables in noncommutative probability spaces}
\author{Ghadir Sadeghi}
\address{$^1$ Department of Mathematics and Computer
Sciences, Hakim Sabzevari University, P.O. Box 397, Sabzevar, Iran}
\email{ghadir54@gmail.com, g.sadeghi@hsu.ac.ir}

\author{Mohammad Sal Moslehian}
\address{$^2$ Department of Pure Mathematics, Center of Excellence in
Analysis on Algebraic Structures (CEAAS), Ferdowsi University of
Mashhad, P. O. Box 1159, Mashhad 91775, Iran}
\email{moslehian@um.ac.ir, moslehian@member.ams.org}

\subjclass[2010]{Primary 46L53, 47A30; Secondary 60B20,60B11.}
\keywords{Noncommutative probability space, von Neumann algebra, noncommutative Rosenthal inequality, noncommutative Bennett inequality, noncommutative Hoeffding inequality.}

\begin{abstract}
In this paper, we establish an extension of a noncommutative Bennett inequality with a parameter $1\leq r\leq2$ and use it together with some noncommutative techniques to establish a Rosenthal inequality. We also present a noncommutative Hoeffding inequality as follows: Let $(\mathfrak{M}, \tau)$ be a noncommutative probability space, $\mathfrak{N}$ be a von Neumann subalgebra of $\mathfrak{M}$ with the corresponding conditional expectation $\mathcal{E}_{\mathfrak{N}}$ and let subalgebras $\mathfrak{N}\subseteq\mathfrak{A}_j\subseteq\mathfrak{M}\,\,(j=1, \cdots, n)$ be successively independent over $\mathfrak{N}$. Let $x_j\in\mathfrak{A}_j$ be self-adjoint such that $a_j\leq x_j\leq b_j$ for some real numbers $a_j<b_j$ and $\mathcal{E}_{\mathfrak{N}}(x_j)=\mu$ for some $\mu\geq 0$ and all $1\leq j\leq n$. Then for any $t>o$ it holds that
\begin{eqnarray*}
{\rm Prob}\left(\left|\sum_{j=1}^n x_j-n\mu\right|\geq t\right)\leq 2 \exp\left\{\frac{-2t^2}{\sum_{j=1}^n(b_j-a_j)^2}\right\}.
\end{eqnarray*}
\end{abstract}

\maketitle

\section{Introduction}

By a noncommutative probability space $(\mathfrak{M}, \tau)$ we mean a von Neumann algebra $\mathfrak{M}$ on a Hilbert
space $\mathfrak{H}$ with unit element $1$ equipped with a faithful normal finite trace $\tau$ such that $\tau(1)=1$. The modules $|x|$ of $x\in \mathfrak{M}$ is defined by continuous functional calculus as $|x|=(x^*x)^{1/2}$.

For each self-adjoint operator $x\in \mathfrak{M}$, there exists a unique spectral measure $E$ as a $\sigma$-additive mapping with respect to the strong operator topology from the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$ of $\mathbb{R}$ into the set of all orthogonal projections such that for every Borel function $f: \sigma(x)\to \mathbb{C}$ the operator $f(x)$ is defined by $f(x)=\int f(\lambda)dE(\lambda)$, in particular, $x=\int \lambda dE(\lambda)$ and $\chi_B(x)=\int_BdE(\lambda)=E(B)$. In addition,
\begin{eqnarray}\label{tc}
\tau\left(\chi_{[t,\infty)}\left(\left|x\right|\right)\right)
=\tau\left(\chi_{[t,\infty)}\left(x\right)\right)+\tau\left(\chi_{[t,\infty)}\left(-x\right)\right)\,.
\end{eqnarray}
Further, if $x \geq 0$ and $t>0$, then $\chi_{[t,\infty)}(x)t \leq x$. Hence we get the inequality
\begin{eqnarray*}
\tau(\chi_{[t,\infty)}(x))\leq t^{-1}\tau(x)\,,
\end{eqnarray*}
which is called the Markov inequality in the literature. For a self-adjoint element $x\in \mathfrak{M}$, it follows from the Markov inequality that $$\tau(\chi_{[t,\infty)}(x))=\tau(\chi_{[e^t,\infty)}(e^x))\leq e^{-t}\tau(e^x),$$
fromwhere we reach the exponential Chebyshev inequality as follows:
\begin{eqnarray}\label{GH1}
\tau(\chi_{[t,\infty)}(x))\leq e^{-t}\tau(e^x)\,.
\end{eqnarray}
As in the commutative context, we use the notation ${\rm Prob}(x\geq t):=\tau(\chi_{[t,\infty)}(x))$.

For any Borel set $A\subseteq\mathbb{R}$, we define $\nu(A)=\tau(E(A))$. Then $\nu$ is a scalar-valued spectral measure for $x$ and $\nu(\mathbb{R})=1$. In addition,
\begin{align}\label{MOS1}
\tau(f(x))=\tau\left(\int f(\lambda)dE(\lambda)\right)=\int f(\lambda)d\tau\left(E(\lambda)\right)=\int f(\lambda)d\nu(\lambda)\,.
\end{align}
By the measurable functional calculus \cite{CO} there is a $*$-homomorphism $\pi:L^{\infty}(\nu)\to\mathfrak{M}$ depending on $x$ such that $\pi(f)=f(x)$ for all $f\in L^{\infty}(\nu)$.

For $1\leq p<\infty$, $L^p(\mathfrak{M},\tau)$ is defined as the completion of $\mathfrak{M}$ with respect to the norm $\|x\|_p=\tau(|x|^p)^{\frac{1}{p}}$. Important special cases of these noncommutative spaces are the usual $L^p$-spaces and the Schatten $p$-classes. For further information we refer the reader to \cite{Y, P, MTS, SAD}.
Let $x\in\mathfrak{M}$ be positive. For $p\geq 1$ and positive $x\in\mathfrak{M}$, from \eqref{MOS1}, we have
\begin{eqnarray*}
\|x\|_p^p&=& \tau(x^p)=\int_0^\infty \lambda ^pd\nu(\lambda)=\int_0^{\infty} pt^{p-1}\nu([t,\infty))dt\\&=&\int_0^{\infty} pt^{p-1}\tau(\chi_{[t,\infty)}(x))dt.
\end{eqnarray*}

Let $\mathcal{P}$ be the lattice of projections of $\mathfrak{M}$. Set $p^{\perp}=1-p$ for $p\in \mathcal{P}$. Given a family of projections $(p_{\lambda})_{\lambda\in\Lambda}\subseteq\mathcal{P}$, we denote by $\vee_{\lambda\in\Lambda}p_{\lambda}$ (resp., $\wedge_{\lambda\in\Lambda}p_{\lambda}$) the projection from $\mathfrak{H}$ onto the closed subspace generated by $p_{\lambda}(\mathfrak{H})$ (resp. onto the subspace $\cap_{\lambda\in\Lambda}p(\mathfrak{H})$). Consequently, $\left(\vee_{\lambda\in\Lambda}p_{\lambda}\right)^{\perp}=\wedge_{\lambda\in\Lambda}p_{\lambda}^{\perp}$. Two projections $p$ and $q$ are said to be equivalent if there exists a partial isometry $u\in\mathfrak{M}$ such that $u^*u=p$ and $uu^*=q$. In this case, we
write $p\sim q$. If $p$ is equivalent to a projection $q_1 \leq q$, we write $p\prec q$. We need the following elementary properties of projections (see \cite{N}).

\begin{lemma}\label{NE}
Let $p$ and $q$ be two projection of $\mathfrak{M}$. Then\\
(i) $p\vee q-q\sim p-p\wedge q$.\\
(ii) If $p\wedge q=0$ then $p\prec q^{\perp}$\\
(iii) If $p$ and $q$ are equivalent projections in $\mathfrak{M}$ then $\tau(p)=\tau(q).$\\
(iv) If $(p_{\lambda})_{\lambda\in\Lambda}$ is a family of projections in $\mathfrak{M}$ then $\tau\left(\vee_{\lambda\in\Lambda}p_{\lambda}\right)\leq\sum_{\lambda\in\Lambda}\tau(p_{\lambda})$.
\end{lemma}

Let $\mathfrak{N}$ be a von Neumann subalgebra of $\mathfrak{M}$. Then there exists a map $\mathcal{E}_{\mathfrak{N}}:\mathfrak{M}\to\mathfrak{N}$ satisfying the following properties:\\
(i) $\mathcal{E}_{\mathfrak{N}}$ is a normal contraction positive map projecting $\mathfrak{M}$ onto $\mathfrak{N}$;\\
(ii) $\mathcal{E}_{\mathfrak{N}}(axb) =a\mathcal{E}_{\mathfrak{N}}(x)b$ for any $x\in\mathfrak{M}$ and $a, b\in\mathfrak{N}$;\\
(iii) $\tau\circ\mathcal{E}_{\mathfrak{N}}=\tau$.\\
Moreover, $\mathcal{E}_{\mathfrak{N}}$ is the unique map satisfying (i) and (ii). The map $\mathcal{E}_{\mathfrak{N}}$ is called the conditional expectation of $\mathfrak{M}$ with respect to $\mathfrak{N}$. We say that two subalgebras $\mathfrak{N}\subseteq \mathfrak{A}, \mathfrak{B}\subseteq\mathfrak{M}$ are independent over $\mathfrak{N}$ if $\mathcal{E}_{\mathfrak{N}}(xy)=\mathcal{E}_{\mathfrak{N}}(x)\mathcal{E}_{\mathfrak{N}}(y)$ for all $x\in\mathfrak{A}, y\in\mathfrak{B}$. In particular, two random variable $X$ and $Y$ of a commutative von Neumann algebra $L^\infty(\mu)$ in which $\mu$ is a probability measure are independent if the algebras they generate are independent over the complex field $\mathbb{C}$. A sequence of subalgebras
$\mathfrak{N}\subseteq\mathfrak{A}_1,\ldots,\mathfrak{A}_n\subseteq\mathfrak{M}$ is
called successively independent over $\mathfrak{N}$ if $\mathfrak{A}_{k+1}$ is independent of the algebra $\mathfrak{M}(k)$ generated by $\mathfrak{A}_1,\ldots,\mathfrak{A}_k$. For further information the reader is refereed to \cite{Xu1, Xu2}.

In 1970, Rosenthal \cite{ROS} presented an inequality to describe isomorphic types of some subspaces in $L_p$-spaces. It indeed gives a bound for the $p$-norm of independent mean $0$ random variables. More precisely, it says that for any $p\geq 2$ there exists a constant $c(p)$ such that for any $n\in \mathbb{N}$ and any independent mean $0$ random variables $f_1, \cdots, f_n$ it holds that
\begin{equation}\label{MOS0}
{\mathbb E}\Biggl|\sum_{k=1}^n
f_k\Biggr|^p \le c(p) \Biggl( \Biggl( \sum_{k=1}^n {\mathbb E}|f_k|^2
\Biggr)^{p/2}+ \sum_{k=1}^n
{\mathbb E}|f_k|^p \Biggr)\,.
\end{equation}\
Burkholder \cite{BUR} generalized Rosenthal's inequality in the context of martingales. Since then this inequality has been generalized and applied by many mathematicians; see, e.g., \cite{JSZ, JX} and references therein. Recently, Junge and Zeng \cite{JZ} extended the Bennett and Bernstein inequalities to the noncommutative setting and derive a version of the Rosenthal inequality from Bernstein's inequality by using the properties of Gamma function. In probability theory, the Bennett inequality (Bernstein inequality, resp.) gives an upper bound on the probability that the sum of independent random variables deviates from its expected value (deviates from its mean, resp.) by more than a fixed amount, see \cite{BEN}.

In this paper, we establish an extension of the noncommutative Bennett inequality due to Junge and Zeng \cite{JZ} and use it together with some noncommutative techniques to prove the Rosenthal inequality with a parameter $1\leq r\leq2$. We also prove a noncommutative Hoeffding inequality. The Hoeffding inequality \cite{HOE} gives a probability bound for the deviation between the average of $n$ independent bounded random variables and its mean (see Corollary \ref{cor1}). There have been several generalizations and applications of this significant inequality, see \cite{1, 3}.

\section{Bennet inequality}

We provide an improved version of the noncommutative Bennett inequality based on the arguments of \cite[Theorem 01]{JZ}.
\begin{theorem} (Noncommutative Bennett inequality)\label{B}
Let $\mathfrak{N}\subseteq\mathfrak{A}_j\subseteq\mathfrak{M}$ be successively independent over $\mathfrak{N}$ and $x_j\in\mathfrak{A}_j$ be self-adjoint and $1\leq r\leq2$ such that
\begin{itemize}
\item $\mathcal{E}_{\mathfrak{N}}(x_j)\leq0$,

\item $\mathcal{E}_{\mathfrak{N}}(|x_j|^r)\leq b_j^r$,

\item $\|x_j\|\leq M$,

\end{itemize}
for some $M>0$ and all $1\leq j\leq n$. Then for each $t\geq0$,
\begin{align}\label{MOS2}
{\rm Prob}\left(\sum_{j=1}^nx_j\geq t\right)=\tau\left(\chi_{[t,\infty)}\left(\sum_{j=1}^nx_j\right)\right)\leq \exp\left(\frac{-b}{M^r}\Phi\left(\frac{tM^{r-1}}{b}\right)\right),
\end{align}
where $\Phi(\alpha)=(1+\alpha)\log(1+\alpha)-\alpha$ and $b=\sum_{j=1}^nb_j^r$.
\end{theorem}
\begin{proof}
Let $\lambda\geq0$. We have
\begin{eqnarray*}
\mathcal{E}_{\mathfrak{N}}\left(e^{\lambda x_n}\right)&=&\mathcal{E}_{\mathfrak{N}}\left(\sum_{k=0}^{\infty}\frac{(\lambda x_n)^k}{k!}\right)
=\sum_{k=0}^{\infty}\frac{\lambda^k}{k!}\mathcal{E}_{\mathfrak{N}}(x_n^k)\\
&=&1+\mathcal{E}_{\mathfrak{N}}(x_n)+\sum_{k=2}^{\infty}\frac{\lambda^k}{k!}\mathcal{E}_{\mathfrak{N}}(x_n^k)\\
&\leq&1+\sum_{k=2}^{\infty}\frac{\lambda^k}{k!}\mathcal{E}_{\mathfrak{N}}(x_n^k)\qquad\qquad\qquad\quad ({\rm by~} \mathcal{E}_{\mathfrak{N}}(x_n)\leq 0)\\
&\leq&1+\sum_{k=2}^{\infty}\frac{\lambda^k}{k!}\mathcal{E}_{\mathfrak{N}}(|x_n|^k)\qquad\qquad\qquad ({\rm by~} x_n^k\leq|x_n|^k)\\
&=&1+\sum_{k=2}^{\infty}\frac{\lambda^k}{k!}\mathcal{E}_{\mathfrak{N}}(|x_n|^r|x_n|^{k-r})\\
&\leq&1+\sum_{k=2}^{\infty}\frac{\lambda^k}{k!}\|x_n\|^{k-r}\mathcal{E}_{\mathfrak{N}}(|x_n|^r)\\
&&\qquad\qquad\qquad\qquad({\rm by~} |x_n|^r|x_n|^{k-r}\leq \|x_n\|^{k-r}|x_n|^r)\\
&\leq&1+\sum_{k=2}^{\infty}\frac{\lambda^k}{k!}\|x_n\|^{k-r}b_n^r\\
&=&1+\frac{b_n^r}{\|x_n\|^{r}}(e^{\lambda \|x_n\|}-1-\lambda \|x_n\|)\\
&\leq&\exp\left(\frac{b_n^r}{\|x_n\|^{r}}(e^{\lambda \|x_n\|}-1-\lambda \|x_n\|)\right).
\end{eqnarray*}
Note that the function $f(s):=\exp\left(\frac{e^{\lambda s}-1-\lambda s}{s^r}\right)$ is increasing for $s> 0$. It follows that
\begin{eqnarray}\label{GH2}
\mathcal{E}_{\mathfrak{N}}\left(e^{\lambda x_n}\right)\leq \exp\left(\frac{b_n^r}{M^{r}}(e^{\lambda M}-1-\lambda M)\right)\,.
\end{eqnarray}

It follows from \eqref{GH1} that
\begin{align}\label{GH3}
\tau\left(\chi_{[t,\infty)}\left(\sum_{j=1}^nx_j\right)\right)&=\tau\left(\chi_{[\lambda t,\infty)}\left(\sum_{j=1}^n\lambda x_j\right)\right)\nonumber\\
&\leq \exp(-\lambda t)\tau\left(e^{\sum_{j=1}^n\lambda x_j}\right).
\end{align}
Recall that the Golden--Thompson inequality \cite{RUS} stating that for all self-adjoint elements $z_1, z_2 \in \mathfrak{M}$,
\begin{eqnarray}\label{T1}
\tau(e^{z_1+z_2})\leq \tau(e^{z_1} e^{z_2})\,.
\end{eqnarray}
Hence
\begin{eqnarray*}
\tau\left(e^{\sum_{j=1}^n\lambda x_j}\right)&\leq&\tau\left(e^{\sum_{j=1}^{n-1}\lambda x_j}e^{\lambda x_n}\right) \qquad\qquad\qquad\qquad\qquad\quad({\rm by~ \eqref{T1}})\\
&=&\tau\left(\mathcal{E}_{\mathfrak{N}}\left(e^{\sum_{j=1}^{n-1}\lambda x_j}e^{\lambda x_n}\right)\right)\\
&=&\tau\left(\mathcal{E}_{\mathfrak{N}}\left(e^{\sum_{j=1}^{n-1}\lambda x_j}\right)\mathcal{E}_{\mathfrak{N}}\left(e^{\lambda x_n}\right)\right)\\
&\leq& \exp\left(\frac{b_n^r}{M^{r}}(e^{\lambda M}-1-\lambda M)\right)\tau\left(\mathcal{E}_{\mathfrak{N}}\left(e^{\sum_{j=1}^{n-1}\lambda x_j}\right)\right)\\
&&\qquad\qquad\qquad\qquad\qquad\quad({\rm by~ \eqref{GH2} ~and~traciality~of~\tau})\\
&\leq&\exp\left(\frac{\sum_{j=1}^nb_j^r}{M^{r}}(e^{\lambda M}-1-\lambda M)\right)\\
&&\qquad \qquad \qquad \qquad \qquad \qquad {\rm (by~iterating~} n-2 {\rm ~times)}
\end{eqnarray*}
We infer from the latter inequality together with \eqref{GH3} that
\begin{eqnarray*}
\tau\left(\chi_{[t,\infty)}\left(\sum_{j=1}^nx_j\right)\right)\leq\exp\left(-\lambda t+\frac{\sum_{j=1}^nb_j^r}{M^{r}}(e^{\lambda M}-1-\lambda M)\right).
\end{eqnarray*}
By basic calculus method we find the minimizing value
\begin{eqnarray*}
\lambda=\frac{1}{M}\log\left(1+\frac{tM^{r-1}}{\sum_{j=1}^nb_j^r}\right),
\end{eqnarray*}
which yields \eqref{MOS2}.
\end{proof}

Since both inequalities $\Phi(t)\geq \frac{t^2}{2+2t/3}$ and $\Phi(t)\geq\frac{t}{2}{\rm arcsinh}(\frac{t}{2})$ are valid for all $t\geq0$, so one can get the Bernstein and Prohorov inequalities from the Bennett's inequality as follows.
\begin{corollary}
Under the same hypothesis of Theorem \ref{B},
\begin{eqnarray}\label{BER}
\tau\left(\chi_{[t,\infty)}\left(\sum_{j=1}^nx_j\right)\right)\leq \exp\left(-\frac{t^2M^{r-2}}{2b+(2/3)tM^{r-1}}\right)
\end{eqnarray}
and
\begin{eqnarray*}
\tau\left(\chi_{[t,\infty)}\left(\sum_{j=1}^nx_j\right)\right)\leq \exp\left(-\frac{t}{2M}{\rm arcsinh}\left(\frac{tM}{2b}\right)\right)
\end{eqnarray*}
\end{corollary}
We can immediately deduce the following commutative Bernstein's Inequality.
\begin{corollary}
Let $X_1, \cdots, X_n$ be independent Bernoulli random variables taking values $1$ and $-1$ with probability $1/2$. Then
$${\rm Prob}\left(\left|\frac{1}{n}\sum_{i=1}^nX_i\right|\geq \varepsilon\right) \leq 2e^{\frac{-n\varepsilon^2}{2+(2\varepsilon/3)}}\,.$$
\end{corollary}

\section{Hoeffding inequality}

In this section we provide a noncommutative version of Hoeffding's inequality and present some consequences.

\begin{theorem} (Noncommutative Hoeffding inequality)\label{H}
Let $\mathfrak{N}\subseteq\mathfrak{A}_j\subseteq\mathfrak{M}$ be successively independent over $\mathfrak{N}$ and let $x_j\in\mathfrak{A}_j$ be self-adjoint such that $a_j\leq x_j\leq b_j$ for some real numbers $a_j<b_j$ and $\mathcal{E}_{\mathfrak{N}}(x_j)=\mu$ for some $\mu\geq 0$ and all $1\leq j\leq n$. Then
\begin{eqnarray}\label{MOS222}
{\rm Prob}\left(\left|S_n-n\mu\right|\geq t\right)\leq 2 \exp\left\{\frac{-2t^2}{\sum_{j=1}^n(b_j-a_j)^2}\right\}.
\end{eqnarray}
for any $t>0$, where $S_n=\sum_{j=1}^n x_j$.
\end{theorem}
\begin{proof}
First we show that if $x\in\mathfrak{A}_j$ is self-adjoint such that $a \leq x\leq b$ and $\mathcal{E}_{\mathfrak{N}}(x)=0$, then
\begin{eqnarray}\label{H1}
\mathcal{E}_{\mathfrak{N}}(e^{sx})\leq \exp\left\{\frac{s^2(b-a)^2}{8}\right\}
\end{eqnarray}
for any $s>0$.\\
Let $s>0$. Note that $t\mapsto e^{ts}$ is convex, therefore for any $a\leq \alpha\leq b$,
\begin{eqnarray*}
e^{s\alpha}\leq e^{sb}\frac{\alpha-a}{b-a}+e^{sa}\frac{b-\alpha}{b-a}\,.
\end{eqnarray*}
By the functional calculus we have
\begin{eqnarray*}
e^{sx}\leq e^{sb}\frac{x-a}{b-a}+e^{sa}\frac{b-x}{b-a}.
\end{eqnarray*}
Since $\mathcal{E}_{\mathfrak{N}}$ is a positive map and $\mathcal{E}_{\mathfrak{N}}(x)=0$, we reach
\begin{eqnarray*}
\mathcal{E}_{\mathfrak{N}}(e^{sx})\leq \frac{-a}{b-a}e^{sb}+\frac{b}{b-a}e^{sa}=e^{h(\alpha)},
\end{eqnarray*}
where $\alpha=s(b-a)$, $h(\alpha)=-\gamma \alpha+\log(1-\gamma+\gamma e^{\alpha})$ and $\gamma=-a/(b-a)$. In addition, $h(0)=h'(0)=0$ and $h''(\alpha)\leq \frac{1}{4}$ for all $\alpha>0$. By Taylor's Theorem there exists a real number $\xi\in(0,\alpha)$ such that
$$h(\alpha)=h(0)+\alpha h'(0)+\frac{\alpha^2}{2}h''(\xi)\leq\frac{\alpha^2}{8}=\frac{s^2(b-a)^2}{8}\,.$$
Hence
\begin{eqnarray*}
\mathcal{E}_{\mathfrak{N}}(e^{sx})\leq \exp\left\{\frac{s^2(b-a)^2}{8}\right\}\,.
\end{eqnarray*}

Second, for arbitrary value of $\mathcal{E}_{\mathfrak{N}}(x)$, setting $y:=x-\mu$ we get $a-\mu \leq y\leq b-\mu$ and $\mathcal{E}_{\mathfrak{N}}(y)=0$. Employing \eqref{H1} we reach
\begin{eqnarray}\label{H2}
\mathcal{E}_{\mathfrak{N}}\left(e^{s(x-\mu)}\right) \leq \exp\left\{\frac{s^2(b-a)^2}{8}\right\}\,.
\end{eqnarray}

Next, by the same reasoning as in the proof of Theorem \ref{B}, we get from \eqref{H2} that
\begin{eqnarray*}
\tau\left(e^{\lambda\sum_{j=1}^n(x_j-\mu)}\right)&\leq& \tau\left(\mathcal{E}_{\mathfrak{N}}\left(e^{\lambda\sum_{j=1}^{n-1}(x_j-\mu)}\right)\mathcal{E}_{\mathfrak{N}}\left(e^{\lambda (x_n-\mu)}\right)\right)\\
&\leq& e^{\frac{\lambda ^2(b_n-a_n)^2}{8}}\tau\left(\mathcal{E}_{\mathfrak{N}}\left(e^{\lambda\sum_{j=1}^{n-1}(x_j-\mu)}\right)\right)\leq \cdots\\
&\leq& \exp\left\{\frac{\lambda^2\sum_{j=1}^n(b_j-a_j)^2}{8}\right\}\,.
\end{eqnarray*}
Therefore \eqref{tc} and the exponential Chebyshev inequality \eqref{GH1} yield that
\begin{eqnarray*}
{\rm Prob}(\left|S_n-n\mu\right|\geq t)&=&2{\rm Prob}(S_n- n\mu\geq t)\\
&\leq& 2e^{-\lambda t}\exp\left\{\frac{\lambda^2\sum_{j=1}^n(b_j-a_j)^2}{8}\right\}.
\end{eqnarray*}
This is minimized when $\lambda=\frac{4t}{\sum_{j=1}^n(b_j-a_j)^2}$. Thus
\begin{eqnarray*}
{\rm Prob}(\left|S_n-n\mu\right|\leq 2\exp\left\{\frac{-2t^2}{\sum_{j=1}^n(b_j-a_j)^2}\right\}\,,
\end{eqnarray*}
which is the desired inequality.
\end{proof}

\begin{remark}
Under the same hypothesis of Theorem \ref{H} we can show that the bound in the Hoeffding inequality \eqref{MOS222} is sharper than that in the Bernstein inequality \eqref{BER} for $r=2$. Let us assume that We may assume that $\mathcal{E}_{\mathfrak{N}}(x_j)=0$ and $-1 \leq x_j \leq 1$. By the functional calculus $ |x_j|\leq 1$ so
$\mathcal{E}_{\mathfrak{N}}(|x_j|^2)\leq 1$ and $\|x_j\|\leq 1$. So $b=n$ and $M=1$ in the notation of Theorem \ref{B}. Then Heoffding inequality gives rise to
\begin{eqnarray*}
{\rm Prob}\left(\sum_{j=1}^nx_j\geq t\right) \leq \exp\left\{\frac{-t^2}{2n}\right\}\,.
\end{eqnarray*}
and from Bernstein inequality we have
\begin{eqnarray*}
{\rm Prob}\left(\sum_{j=1}^nx_j\geq t\right) \leq \exp\left\{\frac{-t^2}{2n+(2t/3)}\right\}\,.
\end{eqnarray*}
\end{remark}

The next result is the classical (commutative) version of the Hoeffding inequality.
\begin{corollary}[Hoeffding's Inequality \cite{HOE}]\label{cor1}
Let $a \leq X_1, \cdots, X_n\leq b$ be independent random variables with the expectation $\mathbb{E}(X_i)=\mu\,\,(i=1, \cdots, n)$. If $\overline{X}_n=\left(\sum_{i=1}^nX_i\right)/n$, then
$${\rm Prob}(\left|\overline{X}_n-\mu\right|\geq t) \leq 2e^{-2nt^2/(b-a)^2}\,.$$
\end{corollary}
In the special case, we immediately get
\begin{corollary}\label{cor1}
Let $0 \leq X_1, \cdots, X_n\leq 1$ be independent random variables with common mean $\mu$. Then with probability at least $1-\varepsilon$,
$$\left|\frac{1}{n}\sum_{i=1}^nX_i-\mu\right| \leq \sqrt{\frac{\log(2/\varepsilon)}{2n}}\,.$$
\end{corollary}

\section{Rosenthal inequality}

In this section we intend to prove a noncommutative Rosenthal inequality by using our noncommutative Bennet inequality. Our argument seems to be simpler than that of \cite{HU} for the case of usual random variables. We should notify that there is a refinement of it in the literature in which various approach is used, see \cite[Theorem 0.4]{JZ} and \cite{DIR}.
\begin{theorem}
Let $1\leq r\leq2\leq p<\infty, \mathfrak{N}\subseteq\mathfrak{A}_j\subseteq\mathfrak{M}$ be successively independent over $\mathfrak{N}$ and $x_j\in\mathfrak{A}_j$ be self-adjoint such that $\mathcal{E}_{\mathfrak{N}}(x_j)=0$. Then there exists a constant $C(p, r)$ such that
\begin{eqnarray*}
\left\|\sum_{j=1}^nx_j\right\|_p^p\leq C(p, r)\left\{\sum_{j=1}^n\|x_j\|_p^p+\left(\sum_{j=1}^n\|x_j^r\|\right)^{\frac{p}{r}}\right\}.
\end{eqnarray*}
\end{theorem}
\begin{proof}
We use the noncommutative Bennett inequality, but for this end, we replace $\Phi(\alpha)$ by $\alpha\log(1+\alpha)-\alpha$, which is clearly smaller
than $\Phi(\alpha)$ for any $\alpha\geq0$.

Let us fix a number $s\geq0$ and consider $y_j=x_j\chi_{(-\infty,s]}(x_j)\in \mathfrak{A}_j$. It follows from $y_j\leq x_j$ and the positivity of $\mathcal{E}_{\mathfrak{N}}$ that $\mathcal{E}_{\mathfrak{N}}(y_j)\leq\mathcal{E}_{\mathfrak{N}}(x_j)=0$. In addition, $\mathcal{E}_{\mathfrak{N}}$ is norm decreasing, so
\begin{eqnarray*}
\sum_{j=1}^n\mathcal{E}_{\mathfrak{N}}(|x_j|^r)\leq \sum_{j=1}^n\|x_j^r\|:=B\quad {\rm and} \quad b:=\sum_{j=1}^n\|y_j^r\|\geq
\sum_{j=1}^n\mathcal{E}_{\mathfrak{N}}(|y_j|^r).
\end{eqnarray*}
Further, $b \leq B$ since $\|y_j\|\leq \|x_j\|\leq M$, where $M:=\max_{1\leq j\leq n}\|x_j\|$.
It follows from the noncommutative Bennett inequality \eqref{MOS2} that
\begin{align}\label{T}
\tau\left(\chi_{[t,\infty)}\sum_{j=1}^ny_j\right)&\leq \exp\left\{\frac{-b}{M^r}\Phi\left(\frac{tM^{r-1}}{b}\right)\right\}\\\nonumber
&\leq \exp\left\{\frac{-b}{M^r}\left[\frac{tM^{r-1}}{b}\log\left(1+\frac{tM^{r-1}}{b}\right)-\frac{tM^{r-1}}{b}\right]\right\}\\\nonumber
&\leq\exp\left\{-\frac{t}{M}\left(\log\left(1+\frac{tM^{r-1}}{B}\right)-1\right)\right\}
\end{align}
for all $t>0$. We have
\begin{eqnarray}\label{P}
\chi_{[t,\infty)}\left(\sum_{j=1}^nx_j\right)\prec \chi_{[t,\infty)}\left(\sum_{j=1}^ny_j\right)\vee\left(\vee_{j=1}^n\chi_{[s,\infty)}(x_j)\right)
\end{eqnarray}
for all $t>0$. To show this we have to prove that
\begin{eqnarray*}
\chi_{[t,\infty)}\left(\sum_{j=1}^nx_j\right)\wedge\left(\chi_{[0,t)}\left(\sum_{j=1}^ny_j\right)\wedge
\left(\wedge_{j=1}^n\chi_{(-\infty,s)}(x_j)\right)\right)=0.
\end{eqnarray*}
Let $\xi\in\chi_{[t,\infty)}\left(\sum_{j=1}^nx_j\right)(\mathfrak{H})\bigcap\left(\chi_{[0,t)}\left(\sum_{j=1}^ny_j\right)(\mathfrak{H})\bigcap
\left(\cap_{j=1}^n\chi_{(-\infty,s)}(x_j)(\mathfrak{H})\right)\right)$. Then
$$\left\langle\left(\sum_{j=1}^nx_j\right)\xi,\xi\right\rangle\geq t$$ and
\begin{eqnarray*}
t>\left\langle\left(\sum_{j=1}^ny_j\right)\xi,\xi\right\rangle&=&\left\langle\left(\sum_{j=1}^nx_j\chi_{(-\infty,s)}(x_j)\right)\xi,\xi\right\rangle=\left\langle\left(\sum_{j=1}^nx_j\right)\xi,\xi\right\rangle\geq t
\end{eqnarray*}
since $\xi\in\cap_{j=1}^n\chi_{(-\infty,s)}(x_j)(\mathfrak{H})$.
Therefore
\begin{eqnarray*}
\chi_{[t,\infty)}\left(\sum_{j=1}^nx_j\right)\wedge\left(\chi_{[0,t)}\left(\sum_{j=1}^ny_j\right)\wedge
\left(\wedge_{j=1}^n\chi_{(-\infty,s)}(x_j)\right)\right)=0.
\end{eqnarray*}
We deduce from Lemma \ref{NE}(ii) that
\begin{eqnarray*}
\chi_{[t,\infty)}\left(\sum_{j=1}^nx_j\right)\prec\left(\chi_{[0,t)}\left(\sum_{j=1}^ny_j\right)\wedge
\left(\wedge_{j=1}^n\chi_{(-\infty,s)}(x_j)\right)\right)^{\perp}.
\end{eqnarray*}
and this gives us inequality \eqref{P}.\\
Using \eqref{P}, we get
\begin{eqnarray*}
\tau\left(\chi_{[t,\infty)}\left(\sum_{j=1}^nx_j\right)\right)
&\leq& \tau\left(\chi_{[t,\infty)}\left(\sum_{j=1}^ny_j\right)\right)+\tau\left(\vee_{j=1}^n\chi_{[s,\infty)}(x_j)\right)\\
&&\qquad\qquad\qquad\qquad ({\rm by~ Lemma~ \ref{NE}(ii-iii)})\\
&\leq&\exp\left\{-\frac{t}{M}\left(\log\left(1+\frac{tM^{r-1}}{B}\right)-1\right)\right\}\\
&&\quad+\sum_{j=1}^n\tau\left(\chi_{[s,\infty)}(x_j)\right)\\
&&\qquad\qquad\qquad \quad ({\rm by~ Lemma~ \ref{NE}(iv)}, ~\eqref{T})
\end{eqnarray*}
for any $t>0$. An easy investigation shows that the latter inequality holds if we replace $M$ by any number $L$ with $L\geq M$. In addition, it holds for all $0<L \leq M$ since the function
 $$f(\alpha)=\exp\left\{-\frac{t}{\alpha}\left(\log\left(1+\frac{t\alpha^{r-1}}{B}\right)-1\right)\right\}$$
is decreasing for any $\alpha>0$. Thus
 \begin{align}\label{IN1}
\tau\left(\chi_{[t,\infty)}\left(\sum_{j=1}^nx_j\right)\right)
\leq \exp\left\{-\frac{t}{L}\left(\log\left(1+\frac{tL^{r-1}}{B}\right)-1\right)\right\}+\sum_{j=1}^n\tau\left(\chi_{[s,\infty)}(x_j)\right)
\end{align}
for all $t>0, s>0, L>0$.

Next, we deal with the modulus of $\sum_{j=1}^nx_j$. Inequality \eqref{IN1} together with \eqref{tc} imply that
\begin{eqnarray*}
&&\hspace{-1cm}\tau\left(\chi_{[t,\infty)}\left(\left|\sum_{j=1}^nx_j\right|\right)\right)\\
&\leq&2\exp\left\{-\frac{t}{L}\left(\log\left(1+\frac{tL^{r-1}}{B}\right)-1\right)\right\}+ \sum_{j=1}^n\tau\left(\chi_{[s,\infty)}(|x_j|)\right)\,.
\end{eqnarray*}
Now, by putting first $L=s$ and $s=\frac{t}{\gamma}$, where $\gamma>0$ we obtain that
\begin{eqnarray*}
&&\hspace{-1cm}\tau\left(\chi_{[t,\infty)}\left(\left|\sum_{j=1}^nx_j\right|\right)\right)\\
&\leq&
2\exp\left\{-\gamma\left(\log\left(1+\frac{t^r}{\gamma^{r-1} B}\right)-1\right)\right\}+ \sum_{j=1}^n\tau\left(\chi_{[t,\infty)}\left(\gamma|x_j|\right)\right).
\end{eqnarray*}
For $p\geq2$, we have
\begin{align}\label{IN2}
&\hspace{-1cm}\left\|\sum_{j=1}^nx_j\right\|_p^p\nonumber\\
&=\int_0^{\infty}pt^{p-1}\tau\left(\chi_{[t,\infty)}\left(\left|\sum_{j=1}^nx_j\right|\right)\right)dt\nonumber\\
&\leq \sum_{j=1}^n\int_0^{\infty}pt^{p-1}\tau\left(\chi_{[t,\infty)}(\gamma\left|x_j\right|)\right)dt\nonumber\\
&\quad+
2p\int_0^{\infty}t^{p-1}\exp\left\{-\gamma\left(\log\left(1+\frac{t^r}{\gamma^{r-1}B}\right)-1\right)\right\}dt\nonumber\\
&\leq \gamma^p\sum_{j=1}^n\|x_j\|_p^p+\frac{2p}{r}(\gamma^{r-1} B)^{\frac{p}{r}}e^{\gamma}\int_0^{\infty}\beta^{\frac{p-r}{r}}(1+\beta)^{-\gamma}d\beta\,,
\end{align}
where we use the change of variable $\beta=\frac{t^r}{\gamma^{r-1} B}$.
Next let $\gamma$ be such that the last integral of the above inequality is convergent, i.e. let us choose $\gamma>\frac{p}{r}$. With this choice
of $\gamma$ inequality \eqref{IN2} implies the Rosenthal inequality with
\begin{eqnarray*}
C(p, r)=\max\left\{ \gamma^p, \frac{2p}{r}\gamma^{\frac{p(r-1)}{r}}e^{\gamma}\int_0^{\infty}\beta^{\frac{p-r}{r}}(1+\beta)^{-\gamma}d\beta\right\}.
\end{eqnarray*}
\end{proof}

In the commutative setting, we have
\begin{corollary}
Let $1\leq r\leq2\leq p<\infty$ and $X_1, \cdots, X_n$ be independent real random variables with expected values $\mathbb{E}(X_j)=0\,\,(j=1, \cdots, n)$. Then
\begin{eqnarray*}
\mathbb{E}\left(\left|\sum_{j=1}^nX_j\right|^p\right)\leq C(p)\left\{\sum_{j=1}^n\mathbb{E}\left(|X_j|^p\right)+\left(\sum_{j=1}^n\mathbb{E}\left(|X_j|^r\right)\right)^{p/r}\right\},
\end{eqnarray*}
where $C(p)=\displaystyle{\min_{\gamma>p/2}}\max\left\{ \gamma^p, \frac{2p}{r}\gamma^{\frac{p(r-1)}{r}}e^{\gamma}\int_0^{\infty}\beta^{\frac{p-r}{r}}(1+\beta)^{-\gamma}d\beta\right\}$.
\end{corollary}

\bigskip

\textbf{Acknowledgment.} We thank S. Dirksen for pointing out the reference \cite{DIR} to our attention.

\bigskip

config: arxiv


TITLE: What is the probability that Tom's finding $22$ of the end of the process?
QUESTION [2 upvotes]: Tom's calculator , when $3$ is being pressed

$3$ with probability $\frac {1}{6}$
$4$ with probability $\frac {1}{3}$
$6$ with probability $\frac {1}{2}$

Perceives them as seen above.
Tom will do the process with this calculator that only $3$ is broken
$$23 -\biggr(\frac{12}{3}\biggr)$$
According to this, What is the probability that Tom's finding $22$ of the end of the process?

REPLY [0 votes]: If I did not misunderstood your question, there are $3$ method that will get $22$.
$23-1, 24-2, 26-4$
So, we start of with fraction part of the question. You will never get $1.
\frac {12}{6}=2$ which satisfy $24-2$ and $\frac {12}{3}=4$, Which satisfy $26-4$. $2$ ways to obtain $22$.
Since there are $3\times3$ possible events.
Chances of obtaining through $\frac {12}{6}$ is $(\frac{1}{3}\times \frac{1}{2})$
Chances of obtaining through $\frac{12}{3}$ is $(\frac{1}{2}\times \frac{1}{6})$
sums up to be $0.25$, $25$% Chance of getting.
Hope I did not mistaken your question.
Good luck!
Notice: Edited after Ross Millikan mentioned what I did wrong.