I want to write this in LaTeX:


However, $$\overline{S_n}^2 = \frac{1}{n}\sum\limits_{i=1}^{n}(X_i - M_n)^2$$ gives me:


whereas $$\bar{S_n}^2 = \frac{1}{n}\sum\limits_{i=1}^{n}(X_i - M_n)^2$$ gives me:


How can I fix this?

  • 1
    Put the subscript outside the \overline: \overline{S}_n^2. BTW, use \[...\] instead of $$...$$. However, my personal preference would be \bar{S}_n^2 (definitely not \bar{S_n}^2 with the subscript inside). – Ruixi Zhang Jan 22 at 13:06

Hope the below code may helps you...







enter image description here

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Using Hendrik Vogt's code in his answer to this question, you have a \widebar command which takes into account the italic angle of the glyph:

    \usepackage{mathtools, nccmath}

  \ifdim\ht0=\ht2 #3\else #2\fi
%The bar will be moved to the right by a half of \macc@kerna, which is computed by amsmath:
%If there's a superscript following the bar, then no negative kern may follow the bar;
%an additional {} makes sure that the superscript is high enough in this case:
%Use a separate algorithm for single symbols:
%Enable nesting of accents:
%If there's more than a single symbol, use the first character instead (see below):
    \if#32 \let\macc@nucleus\first@char \fi
%Determine the italic correction:
%Now \dimen@ is the italic correction of the symbol.
    \divide\dimen@ 3
%Now \@tempdima is the width of the symbol.
    \divide\@tempdima 10
%Now \dimen@ = (italic correction / 3) - (Breite / 10)
    \ifdim\dimen@>\z@ \dimen@0pt\fi
%The bar will be shortened in the case \dimen@<0 !
%Place the combined final kern (-\dimen@) if it is >0 or if a superscript follows:
      \ifdim\dimen@<\z@ \let\final@kern1\fi
      \if\final@kern1 \kern-\dimen@\fi
  \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar
  \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax}%
%The following initialises \macc@kerna and calls \mathaccent:
%If the argument consists of more than one symbol, and if the first token is
%a letter, use that letter for the computations:
    \ifcat\noexpand\first@char A\else


    \[ \widebar{S}_n^2 = \mfrac{1}{n}\sum_{i=1}^{n}(X_{i} - M_{n})^{2} \]%


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I would say



  $$\overline{S}_n^2 = \frac{1}{n}\sum\limits_{i=1}^{n}(X_i - M_n)^2$$

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