I am reading David Williams' "Probability with Martingales" and in this book he supposes that you are given the following information about some 'outcome' $\omega \in \Omega$: $$X_n(\omega) \ \text{s.t.} \ n\in \mathbb{N}$$ for a series of random variables $X_n:\Omega \rightarrow \mathbb{R}$. He then says, based on this information, $\sigma(X_n : n \in \mathbb{N})$ consists of 'events' $F$ such that for each point $\omega$, you can tell whether the event $F$ has occurred or not (that $\omega \in F$).  However, no matter how hard I try to wrap my head around this vague idea, I can't seem to understand it. I don't understand how the 'information' provided helps to understand whether some event has occurred or not and what this means in terms of a sigma algebra generated by random variables (which also confuses me). Is there any easier way to understand what a sigma algebra generated by these functions actually is?
I've been trying to understand $\sigma$-algebras and how it encodes information in context of filtration. While certain parts seem clear and logical, I can't say I get the whole picture.  I'll try to explain the counter-intuition I get with the classical example of : the probability space $\Omega = \{ HH, HT, TH, TT \}$ and a r.v. $X(\omega)$ equal to the number of heads.  At times $0$, $1$ and $2$ the available information is represented using $\sigma$-algebras $\mathcal{F}_0=\{\emptyset,\Omega\}$, $\mathcal{F}_1=\{\emptyset, \Omega, \{HH,HT\},\{TH,TT\}\}$ and $\mathcal{F}_2=\{\emptyset, \Omega,\{HH,HT\},\{TH,TT\},\{HH\},\{HT\},\{TH\},\{TT\}\}$.  One can notice that $X(\omega)$ is not measurable with respect to $\mathcal{F}_0$ and $\mathcal{F}_1$, because $X^{-1}((\frac{3}{2}; +\infty))=\{HH\}$. To me it is quite surprising: intuitively $X$ makes perfect sense at all times. In particular it has an expected value at time $0$, which I interpret as that the probability and value of all outcomes $\{\omega\}$ can be computed. How do you think of a non-measurable function?  Here's another way of expressing the same confusion. The most natural choice of $\sigma$-algebra in a finite discrete case is $\mathcal{F}=2^\Omega$, and it is implicitly used in all elementary probability problems. However, this choice of $\mathcal{F}$ does not reflect the fact that some information is known or unknown, conditional probability does. Does it mean that the statement "$\sigma$-algebra is known information" make sense only in conditioning? Why is it convenient then?
Let $W^{1,p}(U)$ be the Sobolev space. Suppose that $U$ is connected bounded domain in $\mathbb{R}^n$ and $u \in W^{1,p}(U)$ satisfies $Du=0$ a.e. in $U$. How can I prove that $u$ is constant a.e. in $U$? If $|\lbrace g \in G: \pi (g)=g^{-1} \rbrace|>\frac{3|G|}{4}$, then $G$ is an abelian group. Assume that $\pi$ is an automorphism of a finite group $G$. Let $S$ denote the set $\lbrace g \in G: \pi (g)=g^{-1} \rbrace$. Show that if $|S|&gt;\frac{3|G|}{4}$, then $G$ is an abelian group.    Anyone has any idea on how to solve this ? I have no idea how to start. Why don't self-closing script elements work? What is the reason browsers do not correctly recognize:  &lt;script src="foobar.js" /&gt; &lt;!-- self-closing script element --&gt;   Only this is recognized:  &lt;script src="foobar.js"&gt;&lt;/script&gt;   Does this break the concept of XHTML support?  Note: This statement is correct at least for all IE (6-8 beta 2). My interest in combinatorics was recently sparked when I read about the many things that the Catalan numbers count, as found by Richard Stanley. I picked up a copy of Brualdi's Combinatorics, and while browsing the section on counting sequences I found a nice little puzzle that has definitely puzzled me.     Let $m$ and $n$ be nonnegative integers with $n\geq m$. There are $m+n$ people in line to get into a theater for which admission if $50$ cents. Of the $m+n$ people, $n$ have a $50$-cent piece and $m$ have a $\$ 1$ dollar bill. The box offices opens with an empty cash register. Show that the number of ways the people can line up so that change is available when needed is   $$ \frac{n-m+1}{n+1}\binom{m+n}{m}. $$     I first noted that the first person to enter must be one of the $n$ with a half-dollar. Now the register has a half-dollar change. The second person can be either a person with a half-dollar or a dollar. In the first case, the register will now have two half-dollars, in the second case, the register will now have one dollar bill. So it seems to me that when one of the $n$ people with a half-dollar enters, the number of half-dollars in the register increases by $1$, and when one of the $m$ people with a bill enters, the number of half-dollars decreases by $1$ but the number of bills increases by $1$.   I tried to model this by looking at paths in $\mathbb{Z}^2$. The $x$-axis is like the number of half-dollars, and the $y$-axis is the number of bills. You start at $(0,0)$, and you can take steps forward $(1,0)$ or backwards diagonally $(-1,1)$ corresponding to who enters, but you must always stay in the first quadrant of the plane without crossing over the axes. The goal is to make $m+n$ moves, and I figured maybe the number of such paths is counted by $\frac{n-m+1}{n+1}\binom{m+n}{m}$, but I'm not sure how to show this. I don't know if this observation simplifies the problem at all, as I don't know how to finish up. I'd be happy to see how this problem is done, thank you. Solvability of Poisson's equation for $p$-forms directly from Hodge decomposition If $M$ is a closed manifold, it's well known that $\Delta\phi=\alpha$, where $\alpha$ is a $p$-form, has a solution if and only if $\alpha$ is orthogonal to the space of harmonic $p$-forms. This is proved on page 224 of Warner, but the proof depends heavily on his approach to Hodge theory. There are other methods that avoid his main technical lemma 6.6. So, can one prove this fact using only the "standard" facts of the theory?  For functions it is easy, for one has $$\alpha=\delta \psi+h.$$ But $(\alpha,h)=0$, so $\alpha=\delta\psi$. Then decompose $\psi$ to get $\alpha=\delta d\zeta$ for some function $\zeta$. But on functions, $\Delta=\delta d$. But for $p$-forms in general one has $$\alpha=d\xi+\delta \psi+h,$$ since $\Omega^{p-1}$ is nontrivial. Using the decompositions again, one has $$\alpha=d\delta \rho+\delta d\sigma,$$ where $\rho,\sigma$ are $p$-forms. I don't see how $\rho$ and $\sigma$ are related, but it looks close. How to install Python and wxPython offline? I could not install my modem driver so I have to update and install applications offline. For this I want to use keryx but to install keryx, Python and wxPython have to be installed first. So how do I install Python and wxPython offiline on Ubuntu 13.04?  Note : I don't have internet connection in Ubuntu. So I'll have to download the required file from my Windows 7. If a function returns no value, with a valid return type, is it okay to for the compiler to return garbage? Where do I find system sounds? After installing gm-notify I see the option of hearing a sound every time a new e-mail arrives in my Gmail inbox, but I don't know where Ubuntu (10.4) stores the system sounds to assign one. Any ideas? Thanks. How to manage a redirect request after a jQuery Ajax call I'm using $.post() to call a servlet using Ajax and then using the resulting HTML fragment to replace a div element in the user's current page. However, if the session times out, the server sends a redirect directive to send the user to the login page. In this case, jQuery is replacing the div element with the contents of the login page, forcing the user's eyes to witness a rare scene indeed.   How can I manage a redirect directive from an Ajax call with jQuery 1.2.6? Exporting video to .mov in Blender 2.79 In the past, Blender (at least the Mac version) has had the ability to encode video files as .mov, the format used by Quicktime Player. However, with Blender 2.79 this feature seems to have vanished, for both the Internal and Cycles render engines. The release log happens to have the entry "Video encoding settings have been simplified, along with the addition of a Constant Rate Factor (CRF) mode" which I suppose could be the cause.  Does anyone know what happened—whether it is the aformentioned reason or something else—and how to export videos as Quicktime .mov in Blender 2.79? Convert HTML entities to Unicode and vice versa How to urlencode data for curl command? I am trying to write a bash script for testing that takes a parameter and sends it through curl to web site.  I need to url encode the value to make sure that special characters are processed properly.  What is the best way to do this?    Here is my basic script so far:  #!/bin/bash host=${1:?'bad host'} value=$2 shift shift curl -v -d "param=${value}" http://${host}/somepath $@ ACPI error when trying to make dual boot with Ubuntu 18.04 and Windows 10 What does it mean if a professor does not answer your email in time? I recently joined a research group as a postdoc. I wrote the professor an email to ask quick but important questions including asking him when when he wanted me to start. It's been over a week and I haven't received get an answer. Other emails I've sent previously have also gone unanswered for long periods as well.   With that background, my question is: do professors sometimes (intentionally or otherwise) not respond to emails in a timely manner? Flash drive .docx disappeared files centering column headers in siunitx table column Why is log-of-sum-of-exponentials $f(x)=\log\left(\sum_{i=1}^n e^ {x_i}\right)$ a convex function for $x \in\mathbb R^n$? How to prove $f(x)=\log\left(\displaystyle\sum_{i=1}^n e^{x_i}\right)$ is a convex function?  EDIT1: for above function $f(x)$, following inequalities hold:  $$\max\{x_1,x_2,\ldots,x_n\}\leqslant f(x)\leqslant\max\{x_1,x_2,\ldots,x_n\}+\log n$$  and I have tried proving its convexity via definition of a convex function with above inequalities, but that didn't work.  EDIT2: I have posted my answers below. How to reduce AWS EBS root volume size? Where does Google Chrome for Linux store user specific data? How to use WPF Background Worker