Say that we have a bunch of random-values $x_1, x_2, ..., x_n$  Let $\bar{X}$ be the mean of these random-values. The variance of this sample-set is given by:  $$\sum(x_i-\bar{X})^2$$  Apparently, the above sum is smaller than if $\bar{X}$ were any other value besides the average of the $n$ observations.  I've seen a Calculus-based explanation...are there other ways to show this? Perhaps geometrically or algebraically?   Thanks.
I am trying to understand the sentence  the mean minimizes the mean squared error.   from wikipedia .  From a previous post  I can see a formal proof which according to the authors prove that the mean minimaze. But my objection on that prove is that is you substitute m with median all the reasoning still stands.  Therefore, I restricted myself to one dimension and built the following example :   MSE = 1/n[SUM(Y_hat_i - Y_i)**2]   Given the following distribution Y = [5,3,2,7,4], minimizing MSE I intended as in the following: I should find a function Y_hat such that when it is applied to the MSE formula (more specifically to each element of the Y distribution), we are sure that we obtain the minimum value. For simplicity, I could consider the Y_hat as any measure or function for the central tendency, including, mean, median, mode, etc.  Lets suppose, Y_hat = mean (=4.2 for our distribution), therefore, I did all the calculations such as:  1/5 [(4.2-5)**2 + (4.2-3)**2+ (4.2-2)**2 + (4.2-7)**2  + (4.2-4)**2 ] = 2.96   Let's suppose, Y_hat = median (=4 for our distribution), therefore, I did all the calculations such as:  1/5 [(4-5)**2 + (4-3)**2+ (4-2)**2 + (4-7)**2  + (4-4)**2 ] = 2.8   If I did all the calculation correctly, the result shows that for this case, the median is minimizing better than the mean.  I am sure, I am doing something wrong and missing important aspects of the overall reasoning.  Please, could you provide any clarification on the above topic? I would really appreciate it.  Many Thanks in advance, Best Regards, Carlo
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Thanks  This is how my "main" document looks like:  \documentclass[man,12pt,a4paper,babel,english]{apa6}  \usepackage{babel,duomasterforside} \usepackage[utf8]{inputenc} \usepackage[hidelinks]{hyperref}  % Math syntax  \usepackage{siunitx} \usepackage{upgreek} \usepackage{gensymb} \usepackage{amsmath} \usepackage{mathtools} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert}  % Figures \usepackage{graphicx} \graphicspath{{Figures/}} \usepackage{float} \usepackage{subcaption}  % Biblography and citing \usepackage[style=apa,backend=biber,sortcites=true,sorting=nyt,autocite=inline,uniquename=false]{biblatex} \addbibresource{thesisbiblo.bib}  % Frontpage \usepackage{pdfpages}   \begin{document}  % Frontpage      \includepdf[pages=1-]{titlepage.pdf}  % Abstract \input{Abstract} \newpage  % Table of contents \tableofcontents \newpage  % Document \input{Introduction.theory} \newpage \input{Methods} \newpage \input{Results} \newpage \input{Discussion} \newpage \input{Conclusion} \newpage  % References                         \printbibliography  \end{document} Limit of ${a_n}^{1/n}$ is equal to $\lim_{n\to\infty} a_{n+1}/a_n$ Today my lecturer put up on the board that:  If $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}$ exists and $a_n&gt;0$ then   $\displaystyle \limsup\limits_{n\to\infty}\left(a_n^{\frac{1}{n}}\right)=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$  however I am not sure why this is true, can somebody give me a hint or something as to how to go about proving this.  thanks for any help I have found that for only one page (so far) the spacing between the header and content is different from the spacing on all the other pages. 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