Let $\mathcal{S}$ be a set of real numbers. Let $\mu$, $m$, $\sigma$, and $r$ be mean, median, standard deviation, and range of $\mathcal{S}$ respectively.

Find $\mathcal{S}$ which maximizes $\dfrac{(\mu-m)r}{\sigma}$.

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    $\begingroup$ Is this even well defined? It seems like picking if we pick $0 <x <<<<<<y$, then take $S$ to be the set of $-x$, $y$ and arbitrarily many $0$'s we can make this quantity as large as we want. Must all the members of the sample be unique? $\endgroup$ – Joe Feb 5 '18 at 15:16
  • $\begingroup$ @Joe: If the numbers need to be unique you can perturb all your many $0$s by a tiny bit and have the same result. $\endgroup$ – Ross Millikan Feb 5 '18 at 15:31

Alright, consider the set $S = \{-y,0,x\}$ where $x,y >0$, and $x >> y$. Notice right away that the median $m = 0$. So we're left looking at the quantity:

$$\frac{ \mu r}{\sigma}$$

Now notice that for our set, we may write: \begin{equation} \sigma = \sqrt{\frac{2}{9}x^2 + \frac{4}{9}y^2 - \frac{1}{9}xy} \end{equation}

\begin{equation} \mu r = \frac{1}{3}(x^2 - y^2) \end{equation} Fixing $y$ and allowing $x$ to vary yields: $$ \sigma = O(x)$$ $$ \mu r = O(x^2)$$

Thus we have for the set $S$, $$\lim_{x \rightarrow \infty} \frac{(\mu - m)r}{\sigma} = \infty$$

Ergo the expression is unbounded and there is no maximizing set $S$.

  • $\begingroup$ Great answer - clear and smart! +1 $\endgroup$ – G Tony Jacobs Feb 5 '18 at 19:58

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