I have an upcoming statistics exam and I'm studying it on my own. I was recommended Hogg's Introduction to Mathematical Statistics, but I didn't find it helpful. I just want a book which covers basic mathematical statistics and statistical inference: stochastic convergence, estimation of parameters, testing hypothesis, etc. all at an elementary level (suitable for a math undergraduate), with good introduction to the required background for such concepts. Can you provide me a good source?
To give you some background, I have a grasp on the basics of statistics and probability theory and even remember touching Bayes theorem at the university data mining course. But being a few years away from the university made my math got extremely rusty (so much for last-minute pre-exam cramming). While I remember various random basic concepts, there are a lot of gaps in my understanding of them.  What would be a good material (a book, a site, or otherwise equally accessible medium) to revise the fundamentals and go beyond basics? I'd like a book that can be actually read as a book (most statistics books are really dry and are close to being reference material, rather than a book).
Reference to a figure uses the section's number What are attention mechanisms exactly? Attention mechanisms have been used in various Deep Learning papers in the last few years. Ilya Sutskever, head of research at Open AI, has enthusiastically praised them:    Eugenio Culurciello at Purdue University has claimed that RNNs and LSTMs should be abandoned in favor of purely attention-based neural networks:    This seems an exaggeration, but it's undeniable that purely attention-based models have done quite well in sequence modeling tasks: we all know about the aptly named paper from Google,   However, what exactly are attention-based models? I've yet to find a clear explanation of such models. Suppose I want to forecast the new values of a multivariate time series, given its historical values. It's quite clear how to do that with an RNN having LSTM cells. How would I do the same with an attention-based model? Space before chapters and contents I am using LaTeX to write my dissertation, and I have realised that on the table of contents and on every chapter page, LaTeX leaves a lot of empty space.  What I mean is that lets say in the page where chapter 1 begins, from the top of the page until the chapter title there is much more vertical space than on any other page. The same goes with chapter 2 etc... and for contents page as well.   Can anybody help me to remove that space? Is there any mistake in my code or something like that? Is this a default of LaTeX?  the preamble is  \documentclass[12pt,twoside]{report} \pagestyle{plain} \usepackage{a4paper} \usepackage{setspace} \doublespacing  \usepackage{amssymb} \usepackage{amsmath}    I've tried the methods below, but none of them work properly. Group With an Endomorphism That is "Almost" Abelian is Abelian. Suppose a finite group has the property that for every $x, y$, it follows that   \begin{equation*} (xy)^3 = x^3 y^3. \end{equation*}  How do you prove that it is abelian?    Edit: I recall that the correct exercise needed in addition that the order of the group is not divisible by 3. Example of a ring without Invariant Basis Property I have noticed that while asking questions and scrolling to the bottom of the page, the right sidebar containing recent questions covers the footer partially. Also, a part of it goes beyond the browser's boundary. It may not be a big problem but it still doesn't look good.    This also happens on the  page I'm having a problem rendering the environment map. I downloaded a sample project that has the same settings as my project but it renders without a problem. I tried using both a .hdr and .png image. Attached are screenshots with node settings. First is the incorrect render and 2nd image is the project the renders correctly. How do I go about fixing this? I've read in several places that one motivation for category theory was to be able to give precise meaning to statements like, "finite dimensional vector spaces are canonically isomorphic to their double duals; they are isomorphic to their duals as well, but not canonically."  I've finally sat down to work through this, and -  Okay, yes, it is easy to see that the "canonical isomorphism" from $V$ to $V^{**}$ is a functor that has a natural isomorphism (in the sense of category theory) to the identity functor.  Also, I see that there is no way that the functor $V\mapsto V^*$ could have a natural isomorphism to the identity functor, because it is contravariant whereas the identity functor is covariant. My question amounts to:     Is contravariance the whole problem?   To elaborate:   I was initially disappointed by the realization that the definition of natural isomorphism doesn't apply to a pair of functors one of which is covariant and the other contravariant, because I was hoping that the lack of a canonical isomorphism $V\rightarrow V^*$ would feel more like a theorem as opposed to an artifact of the inapplicability of a definition.  Then I tried to create a definition of a natural transformation from a covariant functor $F:\mathscr{A}\rightarrow\mathscr{B}$ to a contravariant functor $G:\mathscr{A}\rightarrow\mathscr{B}$. It seems to me that this definition should be that all objects $A\in\mathscr{A}$ get a morphism $m_A:F(A)\rightarrow G(A)$ such that for all morphisms $f:A\rightarrow A'$ of $\mathscr{A}$, the following diagram (in $\mathscr{B}$) commutes:  $$\require{AMScd}\begin{CD} F(A) @&gt;m_A&gt;&gt; G(A)\\ @VF(f)VV @AAG(f)A\\ F(A') @&gt;&gt;m_{A'}&gt; G(A') \end{CD}$$  This is much more stringent a demand on the $m_A$ than the typical definition of a natural transformation.  Indeed, it is asking that $m_A=G(f)\circ m_{A'}\circ F(f)$, regardless of how $f$ or $A'$ may vary. Taking $\mathscr{A}=\mathscr{B}=\text{f.d.Vec}_k$, $F$ the identity functor and $G$ the dualizing functor, it is clear that this definition can never be satisfied unless $m_V$ is the zero map for all $V\in\text{f.d.Vec}_k$ (because take $f$ to be the zero map). In particular, it cannot be satisfied if $m_V$ is required to be an isomorphism.     Is this the right way to understand (categorically) why there is no natural isomorphism $V\rightarrow V^*$?   As an aside, are there any interesting cases of some kind of analog (the above definition or another) of natural transformations from covariant to contravariant functors?  Note: I have read a number of math.SE answers regarding why $V^*$ is not naturally isomorphic to $V$. None that I have found are addressed to what I'm asking here, which is about how categories make the question and answer precise. ( was closest.) Hence my question here. Show that $(\mathbb{Q}^*,\cdot)$ and $(\mathbb{R}^*,\cdot)$ aren't cyclic I'm reading a book about abstract algebra, but I'm having trouble solving this excercise: "Show that $(\mathbb{Q}^*,\cdot)$ and $(\mathbb{R}^*,\cdot)$ aren't cyclic"  Where $(\mathbb{Q}^*,\cdot)$ is the group of nonzero rational numbers under multiplication and $(\mathbb{R}^*,\cdot)$ is the group of nonzero real numbers under multiplication.    Here is my attempt for the first.  Suppose $(\mathbb{Q}^*,\cdot)$ is cyclic, then $\mathbb{Q}^*=\langle\frac{p}{q}\rangle=\{(\frac{p}{q})^n,n\in\mathbb{Z}\}$, where $p$ and $q$ are coprime.  $\frac{2p}{q}$ is also in $\mathbb{Q}^*$ so it must be equal to $(\frac{p}{q})^n$ for some $n\in\mathbb{Z}$.  To solve $\frac{2p}{q}=(\frac{p}{q})^n$, I take a logarithm of both sides and end up with $1+\log_\frac{p}{q}(2)=n$, since $n$ is an integer $\log_\frac{p}{q}(2)$ must be an integer too, but it is possible only when $\frac{p}{q}=2^{\frac{1}{k}}, k\in\mathbb{N}$, (i.e. $\frac{p}{q}$ is a k-th root of $2$), but $k$ must be $1$ for $2^\frac{1}{k}$ to be rational so $\frac{p}{q}=2$ contradicting the hypothesis of $p$ and $q$ being coprime.    However I don't know whether this is a proper proof and the same reasoning cannot be applied to $\mathbb{R}^*$, I'd like you to just give me an hint towards a proof, without telling me the whole proof, if possible. image of polynomial map is not an algebraic set I am doing an exercise about algebraic geometry, where the exercise  tell us to provide example of the image of polynomial map $f:\mathbb{C}^{m}\rightarrow \mathbb{C}^{n}$ that is not an algebraic set. But I am thinking that the image of every polynomial is $\mathbb{C}$ thus every polynomial map is surjective. What have I misunderstand? Do trees really get a large share of their mass from the carbon in the air? I remember hearing that trees and other plants actually obtain a large amount of their mass from the carbon floating in the air, not the ground beneath them. Does the makeup of air actually contain enough carbon to support this theory, and is a tree's surface area actually large enough to obtain the amount of carbon it needs directly from the air? Greiner in his book "Field Quantization" page 173, eq.(7.11) did this calculation:     ${\mathcal L}^\prime=-\frac{1}{2}\partial_\mu A_\nu\partial^\mu A^\nu+\frac{1}{2}\partial_\mu A_\nu\partial^\nu A^\mu-\frac{1}{2}\partial_\mu A^\mu\partial_\nu A^\nu $   $\space\space\space\space=-\frac{1}{2}\partial_\mu A_\nu\partial^\mu A^\nu+\frac{1}{2}\partial_\mu[A_\nu(\partial^\nu A^\mu)-(\partial_\nu A^\nu) A^\mu]$        The last term is a four-divergence which has no influence on the field equations. Thus the dynamics of the electromagnetic field (in the Lorentz gauge) can be described by the simple Lagrangian        ${\mathcal L}^{\prime\prime}=-\frac{1}{2}\partial_\mu A_\nu\partial^\mu A^\nu$     Yes, if it is a four-divergence of a vector whose 0-component doesn't contain time derivatives of the field, indeed according to the variational principle this four-divergence will not influence the field equation.  And actually I calculated the time-derivative dependence of 0-component of $[A_\nu(\partial^\nu A^\mu)-(\partial_\nu A^\nu) A^\mu]$, in which only $[A_0(\partial^0 A^0)-(\partial_0 A^0) A^0]$ could possibly contain time-derivative, which vanishes fortunately, so whatever the general case it doesn't matter in this present case.  But how can he seem to claim that it holds for a general four-divergence term,The last term is a four-divergence which has no influence on the field equations?  EDIT: I only assumed the boundary condition to be $A^\mu=0$ at spatial infinity, not at time infinity. And the variation of the action $S = \int_{t_1}^{t_2}L \, dt$ is due to the variation of fields which vanish at time, $\delta A^\mu(\mathbf x,t_1)=\delta A^\mu(\mathbf x,t_2)=0$, not having the knowledge of $\delta \dot A^\mu(\mathbf x,t_1)$ and $\delta \dot A^\mu(\mathbf x,t_2)$, which don't vanish generally, so the four-divergence term will in general contribute to the action, $$\delta S_j=\delta \int_{t_1}^{t_2}dt\int d^3\mathbf x \partial_\mu j(A(x),\nabla A(x),\dot A(x))^\mu =\delta \int_{t_1}^{t_2}dt\int d^3\mathbf x \dot j^0 =\int d^3\mathbf x [\delta j(\mathbf x, t_2)^0-\delta j(\mathbf x, t_1)^0]$$ which does not vanish in general! Grammatical function of "at best" idiom state that "at best" is an idiom. But, what is the grammatical function of "at best" (for example, in the below sentences?)        Their response to the proposal was, at best, cool.   The government's response seems to have been at best confused and at   worst dishonest.   If he drops the course now, at best he’ll get an incomplete, and he could fail.      Thanks Install 18.04 server without network connection My QGIS project has two layers, one layer is retrieved by Google layer plugin which fetches the static Google map image whose CRS is +proj=merc +lon_0=0 +lat_ts=0 +x_0=0 +y_0=0 +a=6378137 +b=6378137 +units=m +no_defs and another vector layer is a postgis layer whose CRS is +proj=longlat +ellps=WGS84 +datum=WGS84 +no_defs.  I digitalized some polygon features from Google Layers and added it to the vector layer. However, QGIS seems do not make the automatic CRS transformation. How can I change its projection?   Thanks "Pandora's Box" metaphors The majority of definitions give the same meaning - "Pandora's box" is a synonym for "a source of extensive but unforeseen troubles or problems."  Are there any other metaphors with the same meaning? Travelling with two different passports Is it possible to remove a terminal command? For example, if there was a command call lookup (not an alias), how would I go about removing it? XeTeX: What happened to \mathcal and \mathscr? Recently I switched from pdflatex to xelatex to be able to use OpenType fonts. I like to use variables highlighted using \mathcal for specific variables, while I used to use \mathscr-typeset variables for matrices. But when I compile with XeLaTeX they look exaclty the same.  Old code  \documentclass{standalone} \usepackage{mathrsfs} \begin{document} $\mathcal{C} = \mathscr{C}$ \end{document}   Output        New code  \documentclass{standalone} \usepackage{unicode-math} \begin{document} $\mathcal{C} = \mathscr{C}$ \end{document}   Output        Broken code  I tried to get back the glyphs from mathrsfs, but it failed this way.  \documentclass{standalone} \usepackage{unicode-math} \usepackage{mathrsfs} \begin{document} $\mathcal{C} = \mathscr{C}$ \end{document}   Output        Question: How can I get back the behaviour? In principle I'd like to have the old \mathcal letters from the pdflatex-version of Computer/Latin Modern.    Alternate solutions (Addendum 2016/05/11) dd aplied to dev USB before connect corrupt /dev/sdX and dont work I have a USB where I'm coping an ISO of 5 Mb aprox with a small OS with its own grub. If I connect the device and check it sudo fdisk -l I get:  ... Device     Boot Start   End Sectors  Size Id Type /dev/sdb1  *        1  9551    9551  4.7M cd unknown   If I copy with dd the ISO sudo dd if=my_os.iso of=/dev/sdb   14852+0 records in 14852+0 records out 7604224 bytes (7.6 MB) copied, 0.538487 s, 14.1 MB/s   It works perfectly. An boot from BIOS in other PC.  The problem is, sometime I forget to connect the USB after update the BIOS, so I do ff WITHOUT the USB connected,  the output is:  14852+0 records in 14852+0 records out 7604224 bytes (7.6 MB) copied, 0.00987684 s, 770 MB/s   Now, the /dev/sdb register is corrupted, the copy speed (770 MB/s) have no sense, and cannot user again dd until reboot.  I have tried to repair it with ddrescue and it doesn't work  sudo ddrescue /dev/zero /dev/sdb conv=noerror,sync   My question is, what is happening exactly and if is possible, how to repair /dev/sdb to avoid reboot and dd start working again?  Thanks.