Given (non-negative) lists 'x' and 'y', if their means and their standard deviations are the same, does this imply 'x=y'? How can I (simply) prove it?

Question

Let us take a list/vector x and another one y, both of them composed of non-negative integers; e.g., x=[0,3,2] and y=[1,2,3]. And their size is equal; i.e., |x|=|y|.

Is the following statement true?

Given, non-negative x and y, where |x|=|y|: if mean of x and mean of y are the same (i.e., $\overline{x}=\overline{y}$) and their standard deviations are the same (i.e., $\sigma(x)=\sigma(y)$), then x=y.

I cannot find anything about this, but it sounds perfectly fine to me.

In other words, my question is: can there be x and y such that $x \neq y$, but $\overline{x}=\overline{y}$ and $\sigma(x)=\sigma(y)$? That is: is there a counter-example to my statement?

If not (this is what I think), how can I prove it? I guess there is simple proof, but I am not used to do this kind of reasonings.

I want the proof simple, so as I can try to encode it in a theorem prover.

Answer 1

Counterexample: $$x=[0,5,5,10], y=[1,2,8,9]$$ $$\sum (x_i^2-\overline x) = \sum (y_i^2-\overline y)=50$$

Answer 2

It is true if $\vert x\vert=\vert y\vert\le 2$, but you can find as many counterexamples as you like for sizes bigger than $2$. For example for vectors of size $3$: $$ x=(0,-\sqrt 3,\sqrt3),\quad y=(1,1,-2) $$ both have mean $0$ and variance $2$.

Answer 3

One way to get a class of counterexamples is to take a nontrivial solution to $x_1^2 + ... + x_n^2 = y_1^2 + ... + y_n^2$. You can then produce a list with by taking $[-x_1, ..., -x_n, x_1, ..., x_n]$ and similarly with $y$. The mean here is zero and the variance is proportional to $(x_1^2 + ... + x_n^2) = (y_1^2 + ... + y_n^2)$, so the two lists have the same standard deviation.

To get the requirement that the elements of the lists are non-negative, you can simply shift both of them up by the same amount. This changes the mean (by the same amount), but doesn't affect the standard deviation.

As an example, $2^2 + 3^2 + 4^2 = 0^2 + 2^2 + 5^2$, so our lists are $[-4, -3, -2, 2, 3, 4]$ and $[-5, -2, 0, 0, 2, 5]$. Shifting up by $5$, we get $[1, 2, 3, 7, 8, 9]$ and $[0, 3, 5, 5, 7, 10]$, which both have mean $5$ and standard devation $\sqrt{290}/5$