What is the largest possible variance of a random variable on $[0; 1]$?  It is evident that it does not exceed $1$, but I doubt, that $1$ is actually possible. The largest variance, for which I found the example is $\frac{1}{4}$. That is the variance of a random variable $X$, such that $P(X = 1) = P(X = 0) = \frac{1}{2}$, but I doubt that  it is the largest possible one.  Why is it interesting:  Initially I wanted to find the largest possible second moment of $X - Y$, where $X$ and $Y$ are i.i.d. random variables on $[0; 1]$. Then I found: $$E(X - Y)^2 = E(X^2 - 2XY + Y^2) = EX^2 - 2EXY + EY^2 = 2(EX^2 - EXY + EXY - EXEY + Cov(X, Y)) = 2(EX^2 - {(EX)}^2) = 2VarX$$ And thats where I am now.  This question is partially inspired by:
Let $ \boldsymbol{x} $ be a vector of $n$ numbers in the range $ \left[0, c \right] $, where $ c $ is a positive real number.  What's is the maximum of the variance function of this $ n $ numbers? Maximum in the meaning what spread of the number will maximize the variance? What would be a tighter bound for other assumptions on the spread of the numbers.  The variance of the vector $ \boldsymbol{x} $ is given by:  $$ \operatorname{var} (\boldsymbol{x}) = \frac{1}{n} \sum_{i = 1}^{n} {\left( {x}_{i} - \overline{\mathbf{x}} \right )}^2 $$  Where the mean $\overline{\boldsymbol{x}}$ is given by:  $$ \overline{\boldsymbol{x}} = \frac{1}{n} \sum_{i = 1}^{n} {x}_{i} $$
How do I clear Bash's cache of paths to executables? How to set one header for each page using multicols Can you intentionally fail a skill check? This comes up to me in the context of the 3.x/PF/d20 Fly skill, even when I simply read about it: Can you intentionally fail a Fly skill check, or any other skill check for that matter, if you, or your character for that matter, decides that the effects of failing the check are what is desired in the situation?  Example:  Take a a shapeshifted Draconian cruising at high altitude, with a few friends on his back who have some means of breathing air supply, whether it be mundane or magical.  Now lets say that the air supply goes kablooie, similar to a decompression or all-packs-trip in an airliner.  For the sake of the folks on his back -- if not himself, depending on if dragon forms need supplemental air to fly at such high altitudes, the Draconian has to lose altitude, and fast.  Considering that by RAW, a Fly check failure leaves yourself falling, would saying "I initiate a rapid descent" be enough to treat the next Fly check as an intentional failure, or would you have to roll a 'reverse' Fly check, where failing the check gives you the desired outcome of a rapid descent, and succeeding leaves you stuck at altitude?  P.S. I added the rules-as-written tag because a RAW answer is highly desired here, albeit not mandatory.  I can already mostly figure it out from a simulationist standpoint, but that answer involves going in-depth into airline flying and non-normal procedures. Central limit theorem and the law of large numbers I have a very beginner's question regarding the Central Limit Theorem (CLT):  I am aware that the CLT states that a mean of i.i.d. random variables is approximately normal distributed (for $n \to \infty$, where $n$ is the index of the summands) or the standardized random variable would have a standard normal distribution.  Now the Law of Large Number states roughly speaking that the mean of i.i.d random variables converges (in probability or almost surely) to their expected value.  What I don't understand is: If, as the CLT states, the mean is approximately normally distributed, how then can it also converge to the expected value at the same time?   Convergence would imply for me that with time the probability that the mean takes a value which is not the expected value is almost zero, hence the distribution would not really be a normal but almost zero everywhere except at the expected value.  Any explanation is welcome. Pronoun question: referring to inanimate objects as 'he' or 'she' I read the following claim concerning pronouns referring to inanimate objects:      Anything that is meant to contain you, protect you or provide you with something beneficial is [often referred to as] a she; anything that is a perceived threat is a he. That's why cars, boats and some countries are she.    Is this really the case, or is it just a subjective claim? (According to the  using he/she for inanimate objects is not recommended and it should be used instead.) Is there a 'rule' for determining whether to use he, she or it based on the impression one would like to express? (I only know the 'rule' for animals: In the case where you know their gender and they are important to you, you refer to them using he/she. E.g. A dog attacked me in the street. It bit me. versus This is my dog Roger. He is 7 years old.) Have these practices changed over time?   Note 1: There is already a similar question (), but that considers only the feminine case and none of the answers offers an objective discussion of the matter.  Note 2: My native language is Czech, where the gender of the pronoun is based only on the grammatical gender of the noun. E.g. a cat (kočka) is always she, a dog (pes) is always he, a boat (loď) is always she and a car (auto) is always it. Quadratic Formula in Complex Variables What's the scope of a variable initialized in an if statement? How to quickly and safely defrost chicken? What does template  mean? Why are you not able to declare a class as static in Java? How to create a meta-package that automatically installs other packages? I'd like to be able to create a package that doesnt contain any code or programs itself, but instead installs other packages that exist in the repositories onto a computer, sort of how the ubuntu-restricted-extras package does? I am trying to retrieve list items using date range based on date column. It implicit both 'Fromdate' and 'Todate'.My CAML query is also querying on two other conditions as well. I am using the below query but it is not retrieving the data.when I remove the date range from query data is retrieved fine. But i want date range as well in my query how would i do this. Little help would work for me.   camlQuery.set_viewXml('&lt;View&gt;&lt;Query&gt;&lt;Where&gt;&lt;And&gt;&lt;And&gt;&lt;Neq&gt;&lt;FieldRef Name="ID" /&gt;&lt;Value Type="Number"&gt;0&lt;/Value&gt;&lt;/Neq&gt;&lt;Eq&gt;&lt;FieldRef Name="_ModerationStatus" /&gt;&lt;Value Type="ModStat"&gt;0&lt;/Value&gt;&lt;/Eq&gt;&lt;Geq&gt;&lt;FieldRef Name="Created" /&gt;&lt;Value Type="DateTime"&gt;' + fromDate + '&lt;/Value&gt;&lt;/Geq&gt;&lt;Leq&gt;&lt;FieldRef Name="Created" /&gt;&lt;Value Type="DateTime"&gt;' + toDate + '&lt;/Value&gt;&lt;/Leq&gt;&lt;/And&gt;&lt;/And&gt;&lt;/Where&gt;&lt;OrderBy&gt;&lt;FieldRef Name="_ModerationStatus" Ascending="True" /&gt;&lt;/OrderBy&gt;&lt;/Query&gt;&lt;/View&gt;'); Horizontal spacing issue in \listoftheorems using amsthm + thmtools Simple bijection $\mathbb N_0 \times \mathbb N_0 \to \mathbb N_0$ I'm looking for a 'simple' bijection $$ \pi \colon \mathbb N_0 \times \mathbb N_0 \to \mathbb N_0, $$  where 'simple' in this context means that it should be as easy as possible to define and it should be self-evident that said function is in fact a bijection. It's a quest for convenience - not minimal complexity in any rigorous sense.  Here are a couple of examples I came up with - none of which satisfy both requirements:   Let $\pi(m,n) = c(2^m3^{n+1})$, where $c \colon \{2^m3^{n+1} \mid m,n \in \mathbb N_0 \}\to \mathbb N_0$ is the order isomorphism (or in fact Mostowski collapse) of its domain under the natural ordering, Let $\pi(m,n) = \langle n,m \rangle$ - Gödel's pairing function, Let $\pi(m,n) = m \oplus n$ - the number resulting from '' $m,n$ (where we imagine both $m$ and $n$ as in infinite decks of cards whose $i$th element is labeled with its respective $i$th digit). ... Problem Statement  Let $f(x) = x^3 + ax^2 + bx + c$ and $g(x) = x^3 + bx^2 + cx + a$ where $a,b,c$ are integers with $c\not=0$   Suppose that the following conditions hold:   $f(1)=0$ the roots of $g(x)=0$ are the squares of the roots of $f(x)=0$      $$\text{Find the value of} \: \: a^{2013} + b^{2013} + c^{2013}$$   My attempt  Let $f(x) = (x-p)(x-q)(x-r) \: \:$ and $g(x) = (x - p^2)(x-q^2)(x-r^2)$  Expanding we get $$f(x) = x^3 - (p + q + r)x^2 + (pq + pr + qr)x - pqr$$  $$g(x) = x^3 - (p^2 + q^2 + r^2)x^2 + (p^2q^2 + p^2r^2 + r^2q^2)x - p^2q^2r^2$$  And so we have $$a = -(p+q+r) = -p^2q^2r^2$$ $$b = pq + pr + qr = p^2 + q^2 + r^2$$ $$c = -pqr = p^2q^2 + p^2r^2 + r^2q^2$$  Then we get the following relations  $$a = -c^2$$ $$b = a^2 - 2b \implies 3b = a^2 \implies b = \frac{c^4}{3}$$  From $f(1) = 0$ we have $a + b + c = -1 \implies \dfrac{c^4}{3} + c - c^2 = -1$  However, upon graphing this function, I saw that there are no integer solutions for $c$. Have I done something wrong?  Edit  For future readers, my approach can simplified alot if you replace $r$ with $1$ since $f(1) = g(1) = 0$ How to permanently set environmental variables My variables are   LD_LIBRARY_PATH=/usr/lib/oracle/11.2/client64/lib ORACLE_HOME=/usr/lib/oracle/11.2/client64   How to save these variables permanently ? How can I easily share links or text between my Android phone and my laptop? Why copy constructor not invoked here How to resolve ambiguous column names when retrieving results? How did Rita know she lost the power?