Given the two formulas for the variance:
$V_1(x)=\frac{1}{n-1}(\sum_{j=1}^nx_j^2-n\overline x^2)$ and $V_2(x)=\frac{1}{n-1}(\sum_{j=1}^n(x_j-\overline x)^2)$,
I've got to decompose in the manner specified,
$V_1=g_1 \circ f$ and $V_2=f \circ g_2$,
where $g_1,g_2$ are differences and $f$ is the same function in both forumlas. I've tried everything but it seems to me impossible.
$f$ is just a modified square function, no? The first formula is a difference of squares and the second is a square of differences.
Explicitly, for the n dimensional vector $x$, $$f(x)=\frac{1}{n-1} \sum x_i^2$$
and $V_1$ is $f(x)-f(X)$ and $V_2$ is $f(x-X)$ where $X$ is the nvector, each of whose elements are $\bar{x}$