I am new to statistics. I am getting my hands dirty on

`VarianceThreshold`

. I am having a single dimensional array, containing N values. What’s the maximum and minimum values of a variance for any values present in array?I guess the minimum value will always be non-negative but I don’t know about the maximum value. I have googled it but couldn’t find a good answer.

**Answer**

I interpret this question as asking

Given a set of $N$ numbers $x_1, x_2, \ldots,x_N$, what is the minimum and maximum values that the

variance$V$, defined as

$$V = \frac 1N \sum_{k=1}^N (x_k-\bar{x})^2 ~~ \text{where}~\bar{x}=\frac 1N \sum_{k=1}^N x_k$$ can take on?

Well, the minimum value of $V$ is $0$ as Daniel Lopez’s comment points out, and it occurs if and only if all the $N$ numbers have the same value. At the other end, every finite set of real numbers has a (finite) upper bound (call it $b$) and a (finite) lower bound (call it $a$), and

$$V \leq \frac{(b-a)^2}{4} = \left(\frac{b-a}{2}\right)^2 = \left(\frac{\mathcal R}{2}\right)^2\tag{1}$$ where $\mathcal R$ is the *range* of the set of $N$ numbers. Note that it is not necessary to know the values of $b$ and $a$ *separately*; we only need the range $\mathcal R = (b-a)$ to calculate the upper bound $(1)$ on $V$.

If $N$ is an even number, there exist sets for which the bound $(1)$ holds with equality: these are sets for which $\frac N2$ of the $x_k$ have value $b$ and the other $\frac N2$ have value $a$. For odd $N$, the bound $\frac{(b-a)^2}{4}$ still applies but cannot be attained with equality if $N>1$. For odd $N>1$, the maximum value is $\frac{(b-a)^2}{4}\frac{N^2-1}{N^2}$ and is attained by a set in which $\frac{N-1}{2}$ of the $x_k$ have value $b$, $\frac{N+1}{2}$ have value $a$, or vice versa. For details, see this answer of mine on math.SE.

In another answer (and the comments on it), @Jim has argued that “A set of $N$ real numbers” tells the listener *nothing whatsoever* about the set if you don’t know what any of the values are, even the minimum or maximum, and so the only completely correct answer is that maximum possible $V$ is unbounded: *any* other answer (such as mine above or a couple of possible answers suggested by Jim) must be be festooned with caveats that the answer is based on assumptions might be unwarranted. I disagree. Even if a secretive questioner is unwilling to share any details about the set he/she is concerned about, my answer gives the questioner enough information to find the maximum possible value of $V$ for him/herself from very minimal information abut the set: just the range suffices, no need to know even what $N$ is!

EDIT: (by AHK) Corrected the maximum variance for odd $N>1$ and the corresponding choice of $x_i$‘s

**Attribution***Source : Link , Question Author : Mangu Singh Rajpurohit , Answer Author : blooraven*