Using a variable both as fixed and random effect?
Using the same variable as a fixed and random effect in Mixed Effect Models
Why doesn't 'find' show this file? If someone could claim me as a dependent, but they will not, am I required to check the "Someone can claim me..." box on my 1040? Let's say that an adult is living with his parents.  He meets all of the  of his parents.  Now let's say that he wants to file a tax return and claim the , as he did not receive any of the stimulus payments in 2020.  His parents decide not to claim him as a dependent on their taxes.  Can he leave the &quot;Someone can claim me as a dependent&quot; box on his  unchecked and claim the Recovery Rebate Credit?  This does not apply to me, but it could apply to many people this year, with the new Recovery Rebate Credit putting people in the situation where it could be advantageous to not be claimed as a dependent. Surprisingly, the  do not provide any guidance on checking this box in any way, that I can find.  I have seen it asserted many times in answers and comments here on Money.SE that because the form says, &quot;Someone CAN claim...,&quot; the box must be checked whether or not anyone actually will claim them.  But I have not yet found this explicitly stated in any official IRS website or publication.  The ideal answer here will cite an official source. Why is the sum of two random variables a convolution? What will I miss out on if I play Diablo 3 without Reaper of Souls? I own the base game of Diablo 3.  I didn't really like the "post-Inferno" content of the game, so I stopped playing shortly after I curb-stomped the final boss for the n-th time.    I understand now that a lot of things have changed in the base game.  There's also an expansion.  I don't plan to purchase the expansion, but some of these changes sound like they could make the game more fun for me.  I can guess that probably the new character and act are expansion-only content.  I know I'm probably , but if I go back and play now, what else will I be missing out on if I don't have the expansion? Who is blowing the horn to alert the Night's Watch? Statistics and causal inference? In his 1984 paper , Paul Holland raised one of the most fundamental questions in statistics:     What can a statistical model say about   causation?   This led to his motto:     NO CAUSATION WITHOUT MANIPULATION   which emphasized the importance of restrictions around experiments that consider causation.  Andrew Gelman makes :     "To find out what happens when you change something, it is necessary to change it."...There are things you learn from perturbing a system that you'll never find out from any amount of passive observation.   His ideas are summarized in .  What considerations should be made when making a causal inference from a statistical model? I have a huge text file (~1GB) and sadly the text editor I use won't read such a large file.  However, if I can just split it into two or three parts I'll be fine, so, as an exercise I wanted to write a program in python to do it.    What I think I want the program to do is to find the size of a file, divide that number into parts, and for each part, read up to that point in chunks, writing to a filename.nnn output file, then read up-to the next line-break and write that, then close the output file, etc.  Obviously the last output file just copies to the end of the input file.  Can you help me with the key filesystem related parts: filesize, reading and writing in chunks and reading to a line-break?  I'll be writing this code test-first, so there's no need to give me a complete answer, unless its a one-liner ;-) Why can a 1 element set be a member of another set but not a subset of it? I have came across this in a textbook:     $\{2\}\nsubseteq\{\{2\}\}$ but $\{2\}\in\{\{2\}\}$   I understand that $\{2\}$ is an element (member) of the other set but considering $\{2\}$ is a set itself, why is it not a subset? proof that translation of a function converges to function in $L^1$ Let $f \in L^1(\mathbb{R})$, for $a\in \mathbb{R}$ let $f_a(x)=f(x-a)$, prove that: $$\lim_{a\rightarrow 0}\|f_a -f \|_1=0$$  I know that there exists $g\in C(\mathbb{R})$ s.t $\|f-g\|_1 \leq \epsilon$, this is also true for $f_a$ and $g_a$.  Now I have the next estimation:  \begin{align*} \|f_a-f\|_1 &amp;= \int_{\mathbb{R}} |f(x)-f(x-a)| dx \\ &amp;= \int |f(x)-g(x)+g(x)-g(x-a)+g(x-a)-f(x-a)| \\ &amp;\leq \|f-g\|_1 + \|f_a-g_a\|_1 + \int |g(x)-g(x-a)| \end{align*} my question is: I can argue that $\lim_{a\to0} \int |g(x)-g(x-a)| = \lim_{a\to0} \lim_{T\to\infty} |g(x_0)-g(x_0-a)| 2T$, now I think that I can change the order of the limits here, but I am not sure why?  P.s $x_0$ is some point in $[-T,T]$, and the above is valid from the intermediate integral theorem for continuous functions, right?  Thanks. Flight reservation for Schengen visa One of the necessary documents for a Schengen visa for non EU citizens is proof of your flights, but it doesn't state the actual (e-)tickets (for which I already paid), or the reservation of flights.  So my question is, which option would be more acceptable (if I can provide any of them) to provide printed tickets to an EU member embassy or reservation of flights (reservation does not expect payment)? For $G$ group and $H$ subgroup of finite index, prove that $N \subset H$ normal subgroup of $G$ of finite index exists If a bob attached to a string is swinging in vertical circular motion, we say that the bob leaves the circular path when the tension in the string becomes zero, but even when tension becomes zero, there's always a component of the bob's weight towards the center of the circle. Why then does the bob leave the circular trajectory? I have 6 images, and I want to align them in this way  (a)image1 (c)image3 (e)image5  (b)image2 (d)image4 (f)image6  Any one know how to do that?  The problem for me is how to make (a), (b),..., (f) in vertical direction. Why is variance squared? The mean absolute deviation is:   $$\dfrac{\sum_{i=1}^{n}|x_i-\bar x|}{n}$$   The variance is: $$\dfrac{\sum_{i=1}^{n}(x_i-\bar x)^2}{n-1}$$   So the mean deviation and the variance are measuring the same thing, yet variance requires squaring the difference. Why? Squaring always gives a non-negative value, but the absolute value is also a non-negative value. Why isn't it $|x_i-\bar x|^2$, then? Squaring just enlarges, why do we need to do this?  A similar question is , but mine is a little different. Thanks. When $\min_{x \in X,y \in Y} f(x,y) = \min_{x \in X} \min_{y \in Y} f(x,y)$? Same script opened in three different ways produces three different results. Why? I have a sample script (credit goes to ):  #!/bin/bash cat &lt;(cat /etc/lsb-release)   I save this script as somename.sh in my home directory.  Now, I try to open this file in three different ways:  sh somename.sh  bash somename.sh  ./somename.sh   So, I have two questions:   Why are the outputs for the above commands different though they are running the same script?   sh produces a Syntax error bash outputs the desired result ./ gives a Permission Denied error  Why is it necessary to give the script executable permission only when running the script using ./ and not necessary in the other cases?   Thanks in advance!  EDIT: The first part might be similar to the  linked one but, I did have a second question too. How to enclosure a text block into big square brackets? everybody.  I need to enclosure a text block into big square brackets.  Something like this (I need to replicate it, more or less):   I've read that it could be possible with tikz but I don't know anything about that package.  Even, I found a code that reproduce something similar but with braces in this post:  The code is this:  \documentclass{article} \usepackage{lipsum} \usepackage[many]{tcolorbox} \usetikzlibrary{decorations.pathreplacing}  \newtcolorbox{rightbrace}{%     enhanced jigsaw,      breakable, % allow page breaks     frame hidden, % hide the default frame     overlay={%         \draw [             fill=none, % fill paper             decoration={brace,amplitude=0.5em},             decorate,             ultra thick,             gray,         ]         % right line         (frame.north east)--(frame.south east);     },     % paragraph skips obeyed within tcolorbox     parbox=false, }  \begin{document}  \begin{itemize}     \item First line           \begin{rightbrace}             \item Second line              \item Third line, which is quite long and seemingly tedious in the extreme             \item Fourth line, which isn't as long as the third            \end{rightbrace}     \item Fifth line \end{itemize}  \begin{rightbrace}     \lipsum[1] \end{rightbrace} \lipsum[2]  \end{document}   How could I put the square brackets around a block of text? (Using tikz, tcolobox or whatever package then can be useful for it).  Thanks for your help. How to apply the same style to different columns? For many of my datasets I have the same imported data duplicated numerous times, each iteration displaying the data values for a different column (in my case it's fish catch weight per gridded cell per year, so lots of iterations of the same grid, with a different pattern of values per year).  For presentation purposes, it strikes me as helpful to use the same scale for each year, so I go to the original data and divide the maximum value into (arbitrarily) 10 classes and manually input the values and labels after setting the desired colour ramp and input data field:   This is extremely tedious, not just the manual excel calculation part, but also the input interface for the values &amp; labels*, hence my intention to make use of the 'save style' option. HOWEVER: when I save the style as QML and load it for the next layer, it saves everything - including the data field (in my case, year). When I change this field, QGis automatically generates the values &amp; labels based on the data for that field, ditching all of the values &amp; labels I set.   Is there any way to avoid this happening &amp; thereby duplicate class number, values and labels for each field by simply loading a style? (somewhat meta) Am I going around this the wrong way, OR, is there a different face I could use to climb this mountain? How to pass Unicode characters as JSP/Servlet request.getParameter? Appropriate statistical analysis to use