Are The two concepts really two sides of the same coin ? The latter is often referred to simply as a regression, but surely this is just an unfortunate coincidence?   The former is about predicting values of a dependent variable from a weighted combination of values of independent variables , whereas the second is Simply the observation that, with time, the average of all so far observed values of a random variable will tend to approach its true mean.
Consider the  function     $$ r(x) = \mathbb{E}(Y \mid X = x) $$   This has been called the regression function in a textbook I'm using. I'm trying to figure out the relationship between this function and the classical linear regression model.  So, I know that it is a theorem* that we may write  $$ Y = r(X) + \epsilon $$  for some random variable $\epsilon$ s.t. $\mathbb{E}(\epsilon) = 0$.  Now suppose that we have     $$ Y = \beta_0 + \beta_1 X + \epsilon $$   This is the classical 1-dimensional regression function (assuming the $\beta_0$ and $\beta_1$ minimize the residual sum of squares).  Question: Is it then a mathematical theorem that if $Y$ is defined as above, that   $$ r(X) = \mathbb{E}(Y \mid X) = (\beta_0 + \beta_1 X)? $$  And is this why the function $\mathbb{E}(Y \mid X)$ is called the "regression function"?  EDIT: The theorem that I am making use of is as follows (from All of Statistics pg. 89):     Regression models are sometimes written as      $$ Y = r(X) + \epsilon $$      where $\mathbb{E}(\epsilon) = 0$. We can always rewrite a regression model this way. To see this, define $\epsilon = Y - r(X)$ and hence $Y = Y + r(X) - r(X) = r(X) + \epsilon$. Moreover, $\mathbb{E}(\epsilon) = \mathbb{E}\mathbb{E}(\epsilon \mid X) = \mathbb{E}(\mathbb{E}(Y - r(X)) \mid X) = \mathbb{E}(\mathbb{E} ( Y \mid X) - r(X)) = \mathbb{E}(r(X) - r(X)) = 0$.
I'd like to show that, given to hermitian operators $A,G$ on a Hilbert space $\mathscr{H}$, the following identity holds: $$ e^{iG\lambda}A e^{-iG\lambda} = A + i\lambda [G,A] + \frac{\left(i\lambda\right)^2}{2!}[G,[G,A]]+\ldots+\frac{(i\lambda)^n}{n!}\underbrace{[G,[G,[G,\ldots[G}_{n\ times},A]]]\ldots]+\ldots $$ where $\lambda$ denotes a real parameter and $[\ \!,]$ indicates the commutator.  This is a proof left to the reader by Sakurai in his books on Modern Quantum Mechanics. Prove that If $m'$ is a common multiple of $s$ and $t$, then $m | m'$. Here $m$ is the LCM of $s$ and $t$.    Although the statement is intuitively clear to me I don't know how to prove. Plot $|z - i| + |z + i| = 16$ on the complex plane Two figures side by side How can I put two figures side-by-side? Not two sub-figures, but two actual figures with separate "Fig.: bla bla" captions.  A figure is supposed to spread over the entire text width, but I have two figures which are narrow and long, and I need to save the space in order to withstand the pages limit. I have  $m\times n = n + n + \dots + n$    in my text right now, but I would like to put a curly brace under the \dots to emphasize that there are m,n's involved. How would I do this? How do I with ufw deny all outgoing ports excepting the ones I need? When should I use a List vs a LinkedList How to prevent SIGPIPEs (or handle them properly) Python multiprocessing: restrict number of cores used How do you convert a matrix to a cell array of equivalent size? Find the limit of $\log\Big(\frac{(n^2+n)!}{n^2!n^{2n}}\Big)$ I have to prove that      $$\Big[\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)\Big] \to \frac 12$$   I know that $$\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)= \\ =\log\Big(\frac {n^2+1}{n^2}\Big)+ \log\Big(\frac {n^2+2}{n^2}\Big)+...+ \log\Big(\frac {n^2+n}{n^2}\Big)=\\ =\log\Big(n^2+1\Big)+ \log\Big(n^2+2\Big)+...+ \log\Big(n^2+n\Big)-2n\log n$$  Doing some manipulation we get:           $$\lim_{n \to +\infty}\log\Big(\frac{(n^2+n)!}{n^2!n^{2n}}\Big)=\frac 12$$      How can I prove this limit? $$\lim _{x \rightarrow 0} \left(\frac{ \sin x}{x}\right)^{1/x}$$  I have spent an hour on the above limit and have no work to show. I tried using L'Hopital's Rule, but just kept going around in circles. Any help would be appreciated. Thank you. When using the web interface, there is a favorite tags section where you can select a list of tags that you view most frequently.  This has become a cornerstone of my normal  on stack overflow, but it doesn't look like this is possible in the Android app.  There should be a mechanism by which you can add some favorite tags to the nav bar. Clicking these tags will take you to a list of similarly tagged questions.  Perhaps also the feed should include a section for each favorite tag.  These would be presented on their own, much like how "hot new questions" and "new unanswered questions" appear in the feed.  Perhaps something like this: How can I see the phone number dialed after I initiate a call? Difference between a Structure and a Union We know the escape velocity from the Earth is 11.2km/s. Isn't it the velocity required to escape from earth if we go normal to the surface of earth? i.e while we derive the formula for the escape velocity from earth we never consider the slanted motion of the object. So when we launch a rocket we need to use very less value of velocity compare to escape velocity to escape from earth because rocket follows a slanted path so curvature of earth has an effect?  If what I said above is right, then how we can say escape velocity is 11.2 km/s? I'm looking to book flights from) Philadelphia (PHL) to Detroit (DTW). Direct flights are quite expensive. I noticed, in expanding a search to surrounding airports, that I can book from Allentown (ABE) to Detroit for far less, and the connection is through Philadelphia.  Can I book this flight but board in Philadelphia? I'll be traveling with carry-on luggage only, so on the return, it's obviously no problem. But would I be allowed through security and on to the plane in PHL with out having flown from Allentown? If every real-valued continuous function is bounded on $X$ (metric space), then $X$ is compact. The topological space $X=\omega_1\cup \{\omega_1\}$ is not first countable but $\omega_1$ is. Here $\omega_1$ denotes the set of all countable ordinals and $\{\omega_1\}$ is the first uncountable ordinal. Cardinality of $\omega_1$ is $\aleph_1.$And also $\omega_1$ is called the limit ordinal meaning it is the limit point of the set $\omega_1.$  Here I want to prove two things. $1)$ $\omega_1$ is first countable. and $2)$ $\{\omega_1\}$ is the limit point of $\omega_1$. But since no sequence converges to $\{\omega_1\}$, using this I could say that our space $X$ is not first countable.  $1)$ The proof of this I got from  with just one question: it says we need to prove that such an $S(y)$ exists,and I'm thinking $S(y)=\cap\{u\in \omega_1:u\gt y\}.$ This is correct$?$ But here is a chance that the countable base around $y$ may be finite and not countably infinite.  $2)$ If possible let us assume that $\omega_1$ is not a limit point of the set $\omega_1.$ $U$ be a nbd of $\omega_1$ that does not intersect $\omega_1.$ Then what are the elements of $U$ like? The ones coming before $\omega_1$  cannot be uncountable because $\omega_1$  is the first. Then they must be countable. This is absurd so no such $U$ exists. Thus $\{\omega_1\}$ is a limit point.  Are my proofs correct? Thank you. There's someone much like Crazy Earl in Oasis behind a locked door. He tells me to leave until I've killed a seraph guardian or I get some sort of guardian crystal.  I just beat the DLC and did many of the sidequests and I'm still stumped as to how to get such a crystal. I assumed there would be Guardian foes just like in Borderlands 1, and they'd drop this item. After completing the story...there were no such enemies. So what do I do?