If X, Y, Z are independent random variables, then X + Y, Z are independent random variables.
$X$,$Y$,$Z$ mutually independent implies $X+Y$ independent of $Z$
How does one know when to use a gerund or an infinitive? Ordered N-Tuples as Cartesian Products of Indexed Families In Introduction to Topology, Mendelson defines the Cartesian Product of an Indexed Family as follows:     DEFINITION  Let $\{X_\alpha\}_{\alpha \in I}$ be an indexed family of sets. The product  of the sets $\{X_\alpha\}_{\alpha \in I}$,    written  $\prod_{x\in I}X_\alpha$ consists of all functions $x$ with domain the indexing set $I$ having the property that for each $\alpha \in I$, $x(\alpha)\in X_\alpha$.    Halmos, in Naive Set Theory, makes the following observation about this sort of product:     Ordered triples, ordered quadruples, etc., may be defined as families whose index sets are unordered triples, quadruples, etc.   However, it is not immediately apparent to me why this is true. It makes sense if we understand the prior assertions in Halmos that $\prod_{i \in I} X_i = X_i$ and that $\prod_{i \in I} X_i = X_a \times X_b$ when $I = \{a, b\}$, but these already pose difficulty for me.  Let $I = \{0, 1\}$, let $X_0 = \{A, B\}$, and let $X_1 = \{C, D\}$. Then by Mendelson's (and Halmos's) definitions, it would seem to me that . . .  $$\prod_{i \in I} X_ i = \{\{(0, A), (1, C)\}, \{(0, A), (1, D)\}, \{(0, B), (1, C)\}, \{(0, B), (1, D)\}\}$$  This would appear to me to be the set of all functions matching the definition given, but it is not equal to $X_0 \times X_1$. Moreover, extending this to a case where the index set $I$ is a triple, we would not end up with ordered triples, but with each subset of the product containing three ordered pairs.  When Halmos posits the equalities he does, the idea seems easy to grasp, but the peculiar wording of Mendelson's definition suggests something different to me. When clicking &quot;edit&quot; I see:  Your edit will be placed in a queue until it is peer reviewed. We welcome edits that make the post easier to understand and more valuable for readers. Because community members review edits, please try to make the post substantially better than how you found it, for example, by fixing grammar or adding additional resources and hyperlinks.  Like:     Likewise, when I am not logged in, clicking &quot;improve this question&quot; or &quot;improve this answer&quot; shows:  You are not . Your edit will be placed in a queue until it is .  And when I click 'Save', I get:  Thanks for submitting an edit. It is only visible to you until it's been approved by trusted community members  Like:     Who can suggest an edit? Where do suggested edits go? Who can vote on a suggested edit? How do suggested edits get reviewed? Can I earn reputation? What about abuse or bad edits? Sometimes the Community user approves or rejects my edit. Why is that? When can I not suggest an edit? Is there a minimum change requirement for a suggested edit? How do I check my suggested edit history? I disagree with the reasons why my suggested edit was rejected. What can I do? How do I know whether there are edits waiting for review? Someone suggested an edit to my own post, and the reviewers took the wrong action. What can I do? Are tag wikis included in this scheme? What do the &quot;Improve Edit&quot; and &quot;Reject and Edit&quot; buttons do? Why are apparently valid links to rejected edits resulting in 404 errors?   For more information about editing, see &quot;&quot; in the . Select list of node titles for a view exposed filter Picking a lock typically requires a DC 15 Dexterity check (Basic rules, p. 49). Likewise for manacles (p. 50), which can alternately be escaped with a DC 20 Dexterity check or broken with a DC 20 Strength check. But what happens if you fail?  The general rules for ability checks state (p. 58):     If the total equals or exceeds the DC, the ability check is a success—the creature overcomes the challenge at hand. Otherwise, it’s a failure, which means the character or monster makes no progress toward the objective or makes progress combined with a setback determined by the DM.   As a DM, how can I handle this in practice? In our last session, the players wanted to get inside a locked building to rescue people trapped inside. Picking the lock seemed like an easy enough task for the rogue, but unluckily he failed the check, and I wasn’t prepared for the consequences. There wasn’t much time pressure and no obvious reason why he couldn’t try again, so I allowed another attempt, but it just didn’t feel right.  I’m sure that similar situations will come up again, so I’d like some advice on how to deal with checks so that they don’t just turn into a tedious exercise of rolling until you succeed. I presume that  was left out of the game for good reason, but I’m not sure what to do instead. I suspect that this was a good occasion to use “progress with a setback” but that’s a fairly new thing for D&amp;D and I’m not sure how to go about it. I’m unsure when I should use setbacks versus lack of progress, and in the latter case I’m unsure when I should allow players to try again versus requiring a different approach.  For example, suppose that a manacled player tries to pick the lock and fails the DC 15 Dexterity check. Now, the player could try to pick it again, or he could try one of the other escape methods, or he could wait for somebody else to rescue him, or he could simply succeed with some setback like a minor injury. How do I choose how to proceed? Or do I let the players choose? Inbox chops messages in the middle of an HTML character reference The messages in the inbox get chopped if they too long, and the chop can occur in the middle of an HTML character reference as can be seen in this screenshot:    The character being chopped is [ and it appears inside a code snippet, in case it makes a difference. chemformula: No auto-completion in TeXstudio? I have trouble getting auto-completion to work with chemformula in TeXstudio. I have other packages working just fine.   The .cwl-file is showing up in "Options > Configure TeXstudio > Completion", and I have it ticked. However, it doesn't work when typing.  I have contacted Clemens Niederberger (who maintains the package), however he recommended I ask here instead.  It is mainly the \ch command I would like to have auto-complete, but all commands would be ideal.  N.B. I'm running TeXStudio 2.9.4 TeXLive 2015 (both newest, at the time of writing) What's going on with "compact implies sequentially compact"? I've seen both counterexamples and proofs to &quot;compact implies sequentially compact&quot;, and I'm not sure what's going on. Apparently there are compact spaces which are not sequentially compact; quick googling and wikipedia checks will turn up examples floating around; they tend to be variants of $[0,1]^{[0,1]}$ with the product topology.  Here's a demonstration(?):  $[0,1]^{[0,1]}$ is compact by Tychonoff's theorem.  So, we demonstrate failure of sequential compactness. Choose a unique binary representation for each $x\in [0,1]$. For each $n\in\mathbb{N}$, let $f_n : [0,1]\to[0,1]$ (an element of $[0,1]^{[0,1]}$) be the function which maps each $x$ to its $n$-th place digit in its binary expansion. Let $f_{n_k}$ be a subsequence of this sequence. Let $x'\in [0,1]$ be such that the $n_{2m}$-th digit is $0$ and the $n_{2m+1}$-th digit is $1$, for all $m\in\mathbb{N}$. Then $f_{n_k} (x')$ does not converge (it alternates between $0$ and $1$), and hence $f_{n_k}$ cannot converge. Thus we have found a sequence in $[0,1]^{[0,1]}$ without any convergent subsequence and so it is not sequentially compact.  (Aside: this is apparently based off of an example in Steen's Counterexamples in Topology, according to  ) Nevertheless, there's also some proofs(?) floating around that compactness of a space implies sequential compactness, along these lines (this proof(?) is of the contrapositive):  Suppose $X$ is not sequentially compact. By definition, this means there is some sequence $(x_n)$ over $X$ with no convergent subsequence. If any $x\in X$ had for each of its neighborhood $U$ infinitely many $n$ for which $x_n \in U$, then we could define a convergent subsequence of $(x_n)$, contradicting our assumption.  (Presumably this is done by choosing for each neighborhood a sufficiently-large-indexed term in that neighborhood.) Thus, for each $x \in X$ we can select an open set $U_x$ such that $x\in U_x$ but with $x_n \in U_x$ for only finitely many $n$. The collection $\mathcal{U}=\{U_x : x\in X\}$ is clearly an open cover of $X$. If $\mathcal{U}$ had a finite subcover $\{U_1, \dots, U_k\}$ then the union $U_1 \cup \cdots \cup U_k$ would contain all of $X$ but only contain $x_n$ for finitely many $n$, which is impossible. Thus $X$ is not compact, since we have found an open cover without a finite subcover.  (Aside: this proof is essentially the same as the proof appearing in Rudin's Principles of Mathematical Analysis for Thm 2.37 that infinite subsets of compact spaces have limit points.) So, what's going on here?  It can't be that what both the counterexample and the proof are telling us is correct.  Is there some subtle (or more embarassingly for me -- glaring) flaw in the proof? I wanted to remove and reinstall the software-center package.  I removed it, but now I can't reinstall.  I tried these commands without success:  sudo apt-get update sudo apt-get install software-center   How can I reinstall software-center?    When I performed the install command this is what I got:  $ sudo apt-get install software-center Reading package lists... Done Building dependency tree        Reading state information... Done Package software-center is not available, but is referred to by another package. This may mean that the package is missing, has been obsoleted, or is only available from another source  E: Package 'software-center' has no installation candidate Suppose $u$ is an element of a ring with a right inverse. I'm trying to understand why the following are equivalent.   $u$ has at least two right inverses $u$ is a left zero divisor $u$ is not a unit   If $v$ and $w$ are distinct right inverse of $u$, then $u(v-w)=0$, but $v-w\neq 0$, so $u$ is a left zero divisor. It's also clear that if $u$ is a left zero divisor, it cannot be a unit (else I could cancel $u$ from $ub=0$ to see $b=0$).  I'm having a heck of a time seeing why $u$ is not a unit implies $u$ has at least two right inverses. I tried the contrapositive, but saw no good approach. What am I missing? With ksh I'm using read as a convenient way to separate values: $ echo 1 2 3 4 5 | read a b dump $ echo $b $a  2 1 $  But it fails in Bash: $ echo 1 2 3 4 5 | read a b dump $ echo $b $a     $  I didn't find a reason in the man page why it fails, any idea? Multiple regression or partial correlation coefficient? And relations between the two I don't even know if this question makes sense, but what is the difference between multiple regression and partial correlation (apart from the obvious differences between correlation and regression, which is not what I am aiming at)?  I want to figure out the following: I have two independent variables ($x_1$, $x_2$) and one dependent variable ($y$). Now individually the independent variables are not correlated with the dependent variable. But for a given $x_1$ $y$ decreases when $x_2$ decreases. So do I analyze that by means of multiple regression or partial correlation?  edit to hopefully improve my question: I am trying to understand the difference between multiple regression and partial correlation. So, when $y$ decreases for a given $x_1$ when $x_2$ decreases, is that due to the combined effect of $x_1$ and $x_2$ on $y$ (multiple regression) or is it due to removing the effect of $x_1$ (partial correlation)? Can a full version of Ubuntu be Installed on a USB Drive? A rearrangement of an absolutely convergent complex series is also absolutely convergent I just completed the following proof.  Is it valid?   Let $\sum_{k=1}^{\infty} a_k$ be an arbitrary convergent series that also converges absolutely.  Then $\sum_{k=1}^{\infty} a_k \in \mathbb{C}$ and $\sum_{k=1}^{\infty} |a_k| \in \mathbb{R}$. Let $\{b_k\}$ denote a rearrangement of the $\{a_k\}$.  That is, we have that $\{a_k\} = \{b_k\}$ yet $a_i = b_i$ needn't hold. We will first show that $\sum_{k=1}^{\infty} b_k$ converges absolutely (immediately implying it converges non-absolutely).  We will use the Cauchy Criterion on the partial sums of $\sum_{k=1}^{\infty} |b_k|$ to demonstrate this. First let $\epsilon &gt; 0$. Consider that since $\sum_{k=1}^{\infty} |a_k| \in \mathbb{R}$, we have that $\exists N_1 \in \mathbb{N}$ s.t. for all $N_1 &lt; m &lt; n$ we have   $$ \sum_{k=m}^n |a_k| \le \epsilon. $$ Now let $0 &lt; N_2$ be large enough s.t $\{a_1, \ldots , a_{N_1}\} \subseteq \{b_1, \ldots , b_{N_2}\}$.  Let $M &gt; max\{N_1, N_2\}$.  Let $M &lt; m' &lt; n'$.  Then since $\{b_k\} - \{b_1, \ldots , b_{m'}\} \subseteq \{a_{N_1}, a_{N_{1}+1},  a_{N_{1}+2} \ldots\}$, we will have that   $$ \sum_{k=m}^n |a_k| \le \epsilon \implies \sum_{k={m'}}^{n'} |b_k| &lt; \epsilon. $$ Then we have that   $$ \left| \sum_{k={m'}}^{n'} |b_k| \right| = \sum_{k={m'}}^{n'} |b_k| &lt; \epsilon $$  so that since $\epsilon$ was arbitrary, we have obtained that   $$ \left| \sum_{k=1}^n |b_k| - \sum_{k=1}^m |b_k|  \right| = \left| \sum_{k=m}^n |b_k|  \right| \rightarrow 0 $$  so that via the Cauchy Criterion for Convergence  we have that $\sum_{k=1}^{\infty} |b_k|$ converges to some  value in $\mathbb{R}$ as desired. Then it follows that $\sum_{k=1}^{\infty} b_k$ is absolutely convergent (and hence also just convergent).   $\square$ Over 20GB of disk space is gone? System drive, OS is Windows 7 x64 SP1 Ultimate.    Windows Explorer shows 100GB of used space.  TreeSizeFree shows 75GB of used space.  Total Commander and FAR confirm TreeSizeFree's measurements (counting hidden files).  Shadow copy is using up to 5GB with the current usage of 1.36GB.  Running FAR from Win7 PE shows the same.  Checkdisk shows no bad sectors. Disk is an OCZ SSD.  SMART monitor suggests SSD condition is perfect.    Where is the rest of space gone? Trajectory of a photon around a Schwarzschild black hole? Consider a photon coming from the infinity in a unbounded orbit to a Schwarzschild black hole (Schwarzschild radius $r_{s}$) (see  for illustration). Its impact parameter is $b$ and its distance of closest approach is $r_{0}$ with $$b^2=\frac{r_{0}^{3}}{r_{0}-r_{s}}$$.  Then its trajectory in polar coordinates is defined by :  $$\frac{d\varphi}{dr} = \frac{1}{r^2\sqrt{\frac{1}{b^2}-\left(1-\frac{r_s}{r}\right)\frac{1}{r^2}}}$$  Consequently : $$\varphi\left(r\right) = \int_{r_{0}}^{r} \frac{dp}{p^2\sqrt{\frac{1}{b^2}-\left(1-\frac{r_s}{p}\right)\frac{1}{p^2}}}$$  and one can compute the total deviation using :  $\Delta\varphi = 2\times\left(\lim_{r\to+\infty}\varphi\left(r\right)-\frac{\pi}{2}\right)$  But my question is : how can I plot/draw the trajectory using the integral expression of $\varphi\left(r\right)$ ?    Because if I compute : $$f\left(r\right) = 2\times\left(\int_{r_{0}}^{r} \frac{dp}{p^2\sqrt{\frac{1}{b^2}-\left(1-\frac{r_s}{p}\right)\frac{1}{p^2}}}-\frac{\pi}{2}\right)$$ I obtain $f\left(r_{0}\right) = -\pi$, and then $f$ increases up to zero, crosses zero, and tends to its positive value at infinity $\Delta\varphi$. It does not make sense for me and I do not understand how to compute the trajectory from that.    If I compute : $$g\left(r\right) = \int_{r_{0}}^{r} \frac{dp}{p^2\sqrt{\frac{1}{b^2}-\left(1-\frac{r_s}{p}\right)\frac{1}{p^2}}}$$ it starts from $0$, and increase up to $\frac{\pi}{2}+\frac{\Delta\varphi}{2}$.    I would like to compute the trajectory in the $\left(x, y\right)$ plane, so how to use the values of $f\left(r\right)$ or $g\left(r\right)$ to compute the function $y\left(x\right)$ ? Converting FLV to MP4 Using FFMPEG and Preserving the Quality Im trying to use ffmpeg to convert my flv files to mp4 to play them on iOS devices but the converted video has a much worse quality than the original one.  Here is the command i use:  ffmpeg -i input.flv -ar 22050 output.mp4   I would really appreciate if someone could provide me with the best settings for flv to mp4 conversion. Let's say we are given a set of positive reals, and we're told that these are the edges of a convex  $n$-gon, and we must compute it's area.   For $n = 3$ there is the famous Heron's formula:  $$A = {1\over 4}\sqrt{4a^2b^2 - (a^2 + b^2 - c^2)^2}$$  For $n = 4$ there is Brahmagupta's formula:  $$A = {1\over 4}\sqrt{(a^2 + b^2 + c^2 + d^2)^2 + 8abcd - 2(a^4 + b^4 + c^4 + d^4)}$$  Is there a generalization for arbitrary $n$? Does it even exist? How to guess a curve distribution from count data I have a sample composed by 2500 count data values. I've plotted in R the corresponding histogram and ecdf. I've run the One-Sample Kolmogorov-Smirnov test to check if the distribution is either exponential or poisson, but I got negative results.  Is it possible that the One-Sample Kolmogorov-Smirnov test, it's not the correct test to work with this count data.  What is the best way to identify a distribution for this kind of data? Rank of a Matrix Sum I have $3\times3$ matrices such that S=A+B. I know there is an inequality connecting rank of the matrices A, B and its sum S. Could you write down that here? It will be a great help for me.  Means equation  or inequality connecting rank(S), rank(A) and rank(B).