Given a set of observations denoted by $x_1,...,x_n$, how to show that the mean $\bar{x}$ and the median $\tilde{x}$ are the same if the distribution is symmetrical? My book says so, but I'm having trouble seeing that from the definitions of $\bar{x}$ and $\tilde{x}$. The book doesn't provide a definition for a symmetrical distribution, but I intuit we can define it to mean that $f(\tilde{x}+c)=f(\tilde{x}-c)$ for any number $c$, where $f(x)$ is the absolute frequency of $x$. Is this definition correct? If so, how to proceed from here? Thanks.

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    $\begingroup$ The median of $n$ observations is not well-defined when $n$ is even;, say $n=2k$; it can be taken to be any number between $x_(k)$ and $x_{(k+1)}$, not necessarily the average of these two numbers. For $n=2k+1$, $x_{(k+1)}$ is the unique median for distinct observations. In any case, the mean of $n$ observations, even if the observations are drawn as independent samples from a symmetric distribution is rarely equal to the mean of the distribution or $c$, the point of symmetry of the distribution. Distinguish between sample parameters and population parameters. $\endgroup$ Feb 14, 2018 at 16:47
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    $\begingroup$ Order the observations and define the median conventionally so it is the middle value in order if $n$ is odd and the mean of the two middle values (the "comedians", according to S.M. Stigler and R. Koenker) otherwise. Then symmetry implies that each difference between the median and any value above is cancelled by a corresponding difference of opposite sign between the median and the symmetrically placed value below it. So, the sum of deviations, $x -$ median, is 0. But this is a characterisation of the mean, and hence the mean and the median are identical in this case. $\endgroup$
    – Nick Cox
    Feb 14, 2018 at 17:45
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    $\begingroup$ For any distribution with 3 or more elements, you can remove $x_1$ and $x_n$, and the median is unchanged. In a symmetric distribution, observations come in symmetrical pairs around the mean, so removing these two elements will not affect the mean either. Rinse, repeat until you have one/two centre elements left. $\endgroup$
    – cloudfeet
    Feb 14, 2018 at 17:54
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    $\begingroup$ As pointed out by @DilipSarwate, you are confused between sample mean/median, and distribution expectation/median. The result only holds for the later. $\endgroup$
    – Xi'an
    Feb 14, 2018 at 21:09
  • $\begingroup$ This question, applied to distributions rather than samples, is asked and answered in many threads here: see stats.stackexchange.com/search?q=mean+median+symmetrical. $\endgroup$
    – whuber
    Feb 14, 2018 at 23:09


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