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Help me understand Bayesian prior and posterior distributions In a group of students, there are 2 out of 18 that are left-handed. Find the posterior distribution of left-handed students in the population assuming uninformative prior. Summarize the results. According to the literature 5-20% of people are left-handed. Take this information into account in your prior and calculate new posterior.    I know the beta distribution should be used here. First, with $\alpha$ and $\beta$ values as 1? The equation I found in the material for posterior is   $$\pi(r \vert Y ) \propto r^{(Y +−1)} \times (1 − r)^{(N−Y +−1)} \\ $$  $Y=2$, $N=18$  Why is that $r$ in the equation? ($r$ denoting the proportion of left-handed people). It is unknown, so how can it be in this equation? To me it seems ridiculous to calculate $r$ given $Y$ and use that $r$ in the equation giving $r$. Well, with the sample $r=2/18$ the result was $0,0019$. The $f$ should I deduce from that?  The equation giving an expected value of $R$ given known $Y$ and $N$ worked better and gave me $0,15$ which sounds about right. The equation being $E(r | X, N, α, β) = (α + X)/(α + β + N)$ with value $1$ assigned to $α$ and $β$. What values should I give $α$ and $β$ to take into account prior information?  Some tips would be much appreciated. A general lecture on prior and posterior distributions wouldn't hurt either (I have vague understanding what they are but only vague) Also bear in mind I'm not very advanced statistician (actually I'm a political scientist by my main trade) so advanced mathematics will probably fly over my head. rent a car, rent out a car, hire a car English not my native language. I'm trying to understand what would be correct in British English.  If I want to use a car for some time. I can say, "I want to rent a car". But from my knowledge there may be a  misunderstanding. I can say the same words if I want to give a car to someone to use for some time and I want to receive payment.  So what would be correct and not misunderstanding?   1) I want to use a car for some time = "I want to hire a car" 2) I want to give a car to someone to use for some time = "rent out a car"   Is my understanding correct? Python string.join(list) on object array rather than string array In Python, I can do:  &gt;&gt;&gt; list = ['a', 'b', 'c'] &gt;&gt;&gt; ', '.join(list) 'a, b, c'   Is there any easy way to do the same when I have a list of objects?  &gt;&gt;&gt; class Obj: ...     def __str__(self): ...         return 'name' ... &gt;&gt;&gt; list = [Obj(), Obj(), Obj()] &gt;&gt;&gt; ', '.join(list) Traceback (most recent call last):   File "&lt;stdin&gt;", line 1, in &lt;module&gt; TypeError: sequence item 0: expected string, instance found   Or do I have to resort to a for loop? I am an Indian citizen studying in Italy on an Italian visa that expires 19 November, 2015. I have permesso di soggiorno. I want to visit the United Kingdom for my vacations for a week during March. I know the procedures already, however I am a little bit confused about the amount that I must show as a proof of support with my bank statements. I will stay for a week with my friends who are also students in Italy and will be in London for 5 days. What would be the amount in Euros that I need to show for my visit to UK? Is this study stating red algae reduces cold virus by 92% believable? I was using this spray meant to hasten recovery from the common cold. , sponsored by the pharmaceutical company that produces the spray, stated it reduced virus by 92%, and so shortens the duration of the common cold.  Meanwhile a doctor told me that these kinds of things don't work.  Does someone have some perspective on this? Is this a believable study? Was it well designed, and has it been duplicated?   If so, why wouldn't the pharmaceutical produced a certified and expensive drug instead of a no-prescription drug? When should I use a List vs a LinkedList jQuery Event Handler created in loop Given sequence of $L-$Lipschitz functions which converges pointwise, prove uniform convergence Let $f_n:[a.b]\rightarrow \mathbb{R}$ be sequence of $L-$Lipschitz functions, that is: $$\forall x,y\in[a,b]: |f_n(x)-f_n(y)|\leq L|x-y|$$   Suppose $f_n \rightarrow f$ pointwise, prove $f_n \rightrightarrows f$   I have all the parts of the puzzle for the proof, and I'm trying to put them all together, I'm using  in my answer.  I would appreciate is you could correct my proof, and if you have an alternative proof, I would be more then happy to see it.  My proof:   Let $\epsilon&gt;0.$  $f_n$ are uniformly continuous on $[a,b]:$  $\tag{1} \exists \delta&gt;0\ \forall x,y\in[a,b]: |f_n(x)-f_n(y)|&lt;\frac{\epsilon}{3}$  $f$ is also $L-$Lipschitz:  $\tag{2} \forall x,y\in[a,b]:|f(x)-f(y)|&lt;L|x-y|=\frac{\epsilon}{3}$  Let us set a partition of $[a,b]$ such as Stephen Montgomery-Smith suggests:     Pick points $x_1,\dots,x_m \in [a,b]$ which are distance   $\frac{\epsilon}{3}$ from each other.        For each $1 \le i \le m$, find a number $N_i$ so that for all $n \ge  N_i$ we have $|f_n(x_i)-f(x_i)| \le \epsilon/3$.   Let $N = \max_{1 \le i \le m} N_i$      Now given any $x \in [a,b]$, pick $1 \le i \le m$ such that $|x-x_i| &lt;\frac{\epsilon}{3L}:  |f_n(x)-f(x)|&lt;\frac{\epsilon}{3} \tag{3}$   $$\begin{align}|f_n(y)-f(y)| &amp;=|f_n(y)-f_n(x)+f_n(x)-f(x)+f(x)-f(y)| \\ &amp;\leq |f_n(y)-f_n(x)| + |f_n(x)-f(x)|+|f(x)-f(y)| \\ &amp;&lt; \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon \\\end{align}$$ How to open terminal, split to 9 terminals and switch between them using one script? "need not to be neither atomic nor ordered" Is this double/triple negative grammatically sound?  Ordered operations require that space on disk is allocated sequentially, so spatial ordering corresponds to chronology of operations, but actual disk writes need not to be neither atomic nor ordered.   Explanation:   Words neither and nor emphasize on the fact that both conditions are (independently) allowed to remain unsatisfied, that both are optional. To be more formal, "need not be atomic and need not be ordered". But then again, it becomes a concoction of two negations joined by a conjunction. I create landing pages in order to launch PPC campaigns for my customers. However, I would like to host them all together under a sub-domain of MY website.   For example:  I want that someone who visits www.clientnumber2.com/landingpage could see the landing page and still the same URL www.clientnumber2.com/landingpage, but actually host it myself.  The same for client3, client4, and so on... I have two feature Datasets. Both are Polygons. Dataset A has very big Areas and Dataset B just small ones. Now, B still overlaps in 2 or 3 Zones from Dataset A.   I want to perform a spatial join and put the result comma seperated in just one row like this:  Zone_ID_A  |  Zone_ID_B     1       |   2,3    2       |   2,4    3       |   5    4       |   1,6   I know this is possible with ArcMap but i could not figure out a way to do this in QGIS, any ideas? I frequently need to do the following things in MySQL:   Create a non-root user Set that user's password Grant that user all privileges Make a database with the same name as the new user Allow usage only for users from the localhost.   Previously I did this using PHPmyadmin, but I would prefer doing it directly from BASH. Is there a CLI way to execute these steps? Do dual American and Mexican citizens need passports to travel between the US and Mexico? I have 2 children that were born in Mexico. They have a birth certificate abroad. So this is dual citizenship. Do they need passports?  My kids were born in Durango, Mexico. I am American. Their father is Mexican. They were both given birth certificates abroad. They are now 16 and 14. They want to go visit grandparents and aunts and uncles in Mexico.  Will they need to get a passport to enter Mexico?  Will they need it to fly? Express partial derivatives of second order (and the Laplacian) in polar coordinates $z=f(x,y)$ where $x=rcosθ$ and $y=rsinθ$  Find $ \frac{\partial z}{\partial x}$ and $ \frac{\partial^2 z}{\partial x^2}$  I'm having big troubles with using chain rule, in particularly the second derivative. Spent 2 hours on this already. What I have...  $ \frac{\partial z}{\partial x}=\frac{\partial r}{\partial x}\frac{\partial z}{\partial r}+\frac{\partial z}{\partial θ}\frac{\partial θ}{\partial x}$. So I'm guessing that $\frac{\partial z}{\partial r}$ and $\frac{\partial z}{\partial θ}$ are unkown? And I will need to use them to differentiate again. Any help is really welcome. This is simply only part of a bigger question.   Full Question: Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$ $x,y,z &gt;0$, prove   $$\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{x+y+z}{13}$$   Note: Often Stack Exchange asked to show some work before answering the question. This inequality was used as a proposal problem for National TST of an Asian country a few years back. However, upon receiving the official solution, the committee decided to drop this problem immediately. They don't believe that any students can solve this problem in 3 hour time frame.   Update 1: In this forum, somebody said  that BW is the only solution for this problem, which to the best of my knowledge is wrong. This problem is listed as "coffin problems" in my country. The official solution is very elementary and elegant.   Update 2: Although there are some solutions (or partial solution) based on numerical method, I am more interested in the approach with "pencil and papers." I think the approach by  in  may help.    Update 3: Michael has tried to apply Peter Scholze's method but not found the solution yet.  Update 4: Symbolic expanding with computer is employed and verify the inequality. However, detail solution that not involved computer has not been found. Whoever can solve this inequality using high school math knowledge will be considered as the "King of Inequality". I'm looking to schedule 16 players to play a round robin tournament with each other such that there are 4 players at each table.  I'd like for each player to play with each other player exactly once over the course of 5 rounds.  I just took a look at this post:  , and I was wondering if anyone knows if the case n=m=4 has been solved, and if so what that solution is. What is the best way to deal with very complex forms? I'm trying to give the best experience to users of a very huge system (like a CRM). It has a lot of forms and much of these have a lot of fields. My doubt is about if I should use a single column form (which raises too much vertical scrolling), or a two columns form (which needs less vertical scrolling and uses more horizontal white space) and a last one with four columns (which has few scrolling and uses almost whole white space to try to keep most of the fields above the fold).  It's important to say that our clients hate scrolling, and they always ask why we don't use the horizontal white space when we present a one column form).  So I'm little bit confused about it, because a lot of UX studies and good practices recommend to do not use/avoid multicolumn forms but the clients most of times ask multi column forms.  Our clients are very used to desktop programs and not everybody is very familiar with web based systems.  Below follows some possibilities I would like to share to illustrate an example of a complex form in the system with the three possibilities I mentioned:    In the example above I'm using the one column form (most recommended in UX studies), but with a very long vertical scroll.    In the second example I tried to reduce vertical space by using two columns.    This is the client's favorite one, 4 columns using white space and trying to keep everything above the fold.   The lengths of fields are based on the average of characters amount for each field.   I would like to know your opinion about the best approach to this case. Can we have some tools to handle link rot? There are many links posted in answers, and some in questions, that are either stale or broken.  An example is the change to  on-line documentation this year to do away with separate 'Optional' and 'Core' tasks areas in the documentation - the old links 404 now.  I've fixed a bunch of 'optionaltasks' broken links, but searching for 'coretasks' in Stack&nbsp;Overflow questions returns over 300 matches...  Are there any tools for us to review and fix these - with an emphasis on the fixing - without sifting through completely manually?  Is it time for some refactoring tools for Stack&nbsp;Overflow content, in particular, for links?  Also, I'd be interested to know if there is an analysis of links (as found in the Stack&nbsp;Overflow data dumps) out there. Answer: there is a  of a sample of links and images, but it's getting a bit out-of-date. I'm trying to  working on Linux. I have PAINSTAKINGLY installed NVidia drivers, Cuda 5.0 and incidentally I'm failing on the last step when I run make:  make  all-recursive make[1]: Entering directory `/var/progs/CudaMiner' Making all in compat make[2]: Entering directory `/var/progs/CudaMiner/compat' Making all in jansson make[3]: Entering directory `/var/progs/CudaMiner/compat/jansson' gcc -std=gnu99 -DHAVE_CONFIG_H -I. -I../..     -g -O2 -MT dump.o -MD -MP -MF .deps/dump.Tpo -c -o dump.o dump.c /bin/bash: gcc: command not found make[3]: *** [dump.o] Error 127 make[3]: Leaving directory `/var/progs/CudaMiner/compat/jansson' make[2]: *** [all-recursive] Error 1 make[2]: Leaving directory `/var/progs/CudaMiner/compat' make[1]: *** [all-recursive] Error 1 make[1]: Leaving directory `/var/progs/CudaMiner' make: *** [all] Error 2   I'm pretty sure I have gcc, in fact I have multiple versions of it. I installed version 4.6 (which is what cuda requires) but it doesn't seem to find it. I'm so close!  Thanks in advance.