Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. It only takes a minute to sign up.
Sign up to join this community
Assume you draw $N$ values from a bounded set, say $[0, 1]$ without loss of generality.
If I am right, the centered values $X-m$ will remain in range $[{1\over N}-1, 1-{1\over N}]$ (by considering the most imbalanced 0/1 cases). And the standard deviation in range $[0,{1 \over 2}]$ (even $N$) or $[0,{1\over 2}\sqrt{1-{1\over N^2}}]$ (odd $N$) (by considering the most balanced 0/1 cases).
But what is the range of the centered-reduced values ${X-m\over s}$ ? How can I address this ?
Let the data be $X_1, X_2, \ldots, X_n$ and, noting that all the $X_i$ are interchangeable, we may without loss of generality extremize $f(X_1, \ldots, X_n) = (X_1 - \bar{X})/\text{sd}(X)$ assuming $\text{sd}(X)\ne 0$ (which implies $n\gt 1$). Because $f$ is unchanged under translations and rescalings of the $X_i$, we may assume $\bar{X}=0$ and $\text{sd}(X)=1,$ whence $f(X) = X_1.$ This is a constrained linear program defined on $\mathbb{R}^n$:
Optimize
$$X_1$$
subject to
$$X_1 + X_2 + \cdots + X_n = n\bar{X} = 0$$ $$X_1^2 + X_2^2 + \cdots + X_n^2 = n(\text{sd}(X))^2 = n.$$
Introducing Lagrange multipliers $\lambda,$ $\mu,$ and $\nu$, the critical points must lie on solutions to
$$\lambda(1,0,\ldots,0)+\mu(1,1,\ldots,1)+2\nu(X_1,X_2,\ldots,X_n)=(0,0,\ldots,0).$$
It is immediate that for any solution $X_2=X_3=\cdots=X_n=-\mu/(2\nu)$. Let this common value be $Y$. Because the mean is zero,
$$X_1 = (1-n)Y.$$
Because the sum of squares of the $X_i$ is unity, we find
$$n = X_1^2 + X_2^2 + \cdots + X_n^2 = \left((1-n)Y\right)^2 + Y^2 + \cdots + Y^2 = n(n-1)Y^2,$$
yielding $Y = \pm \sqrt{1/(n-1)}$ and extreme values
$$f^{*}(X_1, X_2, \ldots, X_n) = X_1^{*} =\pm (1-n)\sqrt{1/(n-1)} = \pm\sqrt{n-1}.$$
This shows that the values of $f$ necessarily lie between $-\sqrt{n-1}$ and $\sqrt{n-1}$ and reach those extrema if and only if all but one of the $X_i$ have the same value and the remaining one has a different value. When $n\gt 2,$ $f^{*}$ can attain any value in between these extremes because $f$ is a continuous function on the connected set $\mathbb{R}^n - \{(t,t,\ldots,t)\ |\ t\in\mathbb{R}\}$. When $n=2,$ $f^{*}$ either equals $1$ or $-1$ but cannot attain any value in between.
I think I understand now. You want to know what the maximum and minimum possible values of $(X-m)/s$ can be.
My feeling (I haven't worked this out analytically) is that this will be maximized when all of the observed values are the same except for one. Suppose without loss of generality that have $N-1$ observed values of 0, and one observed value of $1$. In this case, the mean will be $1/N$, the median $m$ will be $0$, and $s=1/\sqrt{N}$. The maximum value of $$\frac{X-m}{s}$$ is therefore $\sqrt{N}$.
