Consider i.i.d. random variables $(X_i)$ drawn from a uniform distribution on $[0, 1].$  In the following find scaling sequences $a_n$, $b_n$ such that $a_n(M_n − b_n)$ converges in distribution to a non-trivial limit function $G$.  (a) $Y_i = X_i,$ and $M_n = \max(Y_1, . . . , Y_n);$  (b) $U_i = \frac{1}{X_i}$, and $M_n = \max(U_1, . . . , U_n)$;  For each case find first of all the probability distribution function $P(M_n ≤ u/a_n + b_n)$ as a function of $u_n = u/a_n + b_n$. After I would like to find suitable scaling sequences $a_n$, $b_n$ so that you get a non-trivial limit $G(u)$ as $n → ∞$.  I'd be very grateful and would really appreciate some help with this.
If I have an i.i.d. random variables $X_i$ from a uniform distribution on $[0,1]$.   How would I find scaling sequences $a_n$, $b_n$ such that $a_n(M_n − b_n)$ converges in distribution to a non-trivial limit function $G$ for  $ Y_i = X_i,$ and $M_n = \max(Y_1,...,Y_n) $
What circumstances can cause a question or answer to be deleted, and what does that actually mean?    How can a post be deleted? When can't I delete my own post? Can I still see my post even after it's deleted? Can I see a list of my deleted posts? How can I undelete one of my posts? What does deletion mean for a post? How do votes to delete work? What are the criteria for deletion? What else should I know about deleted posts? If I flag my question with a request to delete it, what will happen?       For more information, see the articles about  and  in the . A way of casting a base type to a derived type the slash (/) after a directory name on shell commands I have a little question here.  If I have two files, say filea and fileb, mv filea fileb would   delete fileb rename filea to fileb   Then if I have two directories, say dira and dirb, mv dira dirb would   move dira into dirb (it will become dirb/dira)   Noting that in both cases there are no notice or message, then this is pretty inconsistent to me. I think mv dira dirb should just overwrite dirb with the contents of dira (or merge the two directories under a directory named dirb).  I remember reading somewhere that a directory name with a slash (like dira/) is treated like a directory, and name with no slash (like dira) is treated like a file (to certain extents, of course). Anyway now I want to make the shell (zsh and possibly bash) respect my notation of a directory by using a slash. Is there a terminal option which enable me to enforce that?  To clarify, here is my desired behaviour:   mv dira dirb results in dirb being overwritten with the contents of dira mv dira dirb/ results in dira being moved into dirb (in dirb/dira)   Has anyone thought the same way as me? Or am I just weird? How to block and unblock https access on OS X 10.9 I am working on a test project that requires me to block https access for a specific app on my Mac (OS X 10.9).   The instruction provided are "Chinese" to me and I have no idea where to start:  (Example: Alter the iptables) Steps:   Block the port:  root@QA9:/opt/appname/appnamerouter2# iptables --list root@QA9:/opt/appname/appnamerouter2# iptables -A INPUT -p tcp --destination-port 443/8443 -j DROP root@QA9:/opt/appname/appnamerouter2# iptables --list  For Open the Port:  root@QA9:/opt/appname/appnamerouter2# iptables -D INPUT 1 Using CSS to affect div style inside iframe I'm loading an XML document in my C# application with the following:  XDocument xd1 = new XDocument(); xd1 = XDocument.Load(myfile);   but before that, I do test to make sure the file exists with:  File.Exists(myfile);   But... is there an (easy) way to test the file before the XDocument.Load() to make sure it's a valid XML file?  In other words, my user can accidentally click on a different file in the file browser and trying to load, say, a .php file causes an exception.    The only way I can think of is to load it into a StreamWriter and simple do a text search on the first few characters to make sure they say "  Thanks!  -Adeena Iterate until condition is met Why Multiplication of zero by any number is zero? If I multiply $a$ by $0$, it means $a$ is added $0$ times, but I am not able to visualize $0$ times! I see more and more servers that have no domain defined in the /etc/resolv.conf file.  Only a search entry. In what cases is this setup used ? Are there any cases where it must not be used ?  nameserver xxx.xxx.xxx.xxx nameserver yyy.yyy.yyy.yyy nameserver zzz.zzz.zzz.zzz  search domain1 domain2 domain3 ... Euclidean geometry in non-inertial frame Variational Derivation of Schrodinger Equation Wifi doesnt work on fresh ubuntu 16.04 Forgotten PostgreSQL Windows password Is there an easy way to see associativity or non-associativity from an operation's table? Most properties of a single binary operation can be easily read of from the operation's table. For example, given $$\begin{array}{c|ccccc} \cdot &amp; a &amp; b &amp; c &amp; d &amp; e\\\hline a     &amp; e &amp; d &amp; b &amp; a &amp; c\\ b     &amp; d &amp; c &amp; e &amp; b &amp; a\\ c     &amp; b &amp; e &amp; a &amp; c &amp; d\\ d     &amp; a &amp; b &amp; c &amp; d &amp; e\\ e     &amp; c &amp; a &amp; d &amp; e &amp; b \end{array}$$ it is easy to check that it is closed (no elements occur in the table which don't occur as row or column index), commutative (the table is symmetric), has a neutral element (the row and column of $d$ are copies of the index row/column), and has an inverse element for each element (there's a $d$ in each row and column). In other words, almost all important properties can immediately be seen. The only part missing is associativity.  Therefore my question: Is there a simple way to see directly from the operation's table (i.e. without doing explicitly all the calculations) if an operation is associative? A word for 'independent episodes' When we have a series of episodes where every single episode depends on and develops from the previous, we may call that a sequel, or maybe a continuum.  Now let us say that we have a collection of fables, episodes or anecdotes that have the same characters, the same milieu, and the same themes; yet, this time, the episodes are NOT interdependent. They don't depend on each other in that you can watch/read 'any' one of them be it (in terms of order) from a beginning, medial, or final season, without having to watch/read the previous to understand, appreciate, and enjoy it (the one you're watching/reading). What do we call the 'episode' in this case?   Examples: a sitcom comedy show, or a show like Dr Who. Or, in literature, the novels of Sherlock Holmes.       Is there a specific term to describe the 'independent episodes' of these works? php/mysql with multiple queries I just renewed my antivirus (Kaspersky's) yesterday. After that, I logged onto Chemistry.SE (which I frequent often) but then I was greeted with this  Which was definitely not how it looked like earlier that morning. I wasn't able to  Open the reputation and badges tab Vote on questions, answers and comments (even answers to MY questions) Comment on questions or answers. 'Favorite' a question Attach tags to a question Post questions  etc., etc. But when I logged on to the Physics and Biology SEs (which I also frequent), the page format hadn't changed, but I still faced the same problems as above. Does anyone know how to fix this? P.S- I'm not a computer/formatting whiz, so you'd have to try avoiding the jargon when helping me out, sorry.... (Also I use Chrome...in case that's important.) Also @Oded recommended checking out the browser developer tools and having a look at the CSS, but as I said, I'm no Software/formatting whiz...so I'm still lost....  On @Shadow Wizard 's advice I've taken a screen shot of the 'error' notification (I had to scroll down to get all 10 'errors' on screen, hence the need for separate screen-shots) How do I Uninstall Windows 10 or Windows 10 preview? Math contest integer triplet problem: $x + y + z + xy + yz + xz = xyz − 1$ I have two different list. In the first list are some people picker columns.  Is it possible to create a lookup column for list 2 to add the user from the first list? I know it isn't supported in Sharepoint, but i hope someone have an idea to solve my problem.